Chapter 6 Application Of Derivatives

Given:

Radius of the circle is denoted by $r$.

We are asked to find the rate of change when $r = 5$ cm.

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To Find:

The rate of change of the area of the circle per second when $r = 5$ cm.

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Solution:

The area of a circle with radius $r$ is given by the formula:

We need to find the rate of change of the area $A$ per second. This means we need to find $\frac{dA}{dt}$, where $t$ represents time in seconds.

Since the area $A$ depends on the radius $r$, and the radius $r$ changes with time $t$, we use the chain rule to find $\frac{dA}{dt}$:

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First, we find the rate of change of area with respect to radius, $\frac{dA}{dr}$. Differentiating equation (i) with respect to $r$:

Using the constant multiple rule and the power rule for differentiation ($\frac{d}{dr}(r^n) = nr^{n-1}$):

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Now, we need to find $\frac{dA}{dt}$ when $r = 5$ cm. From the chain rule, $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.

We are given $r = 5$ cm. Substituting this value into the expression:

The unit of $\frac{dA}{dt}$ is the unit of Area (cm$^2$) divided by the unit of time (s), which is cm$^2$/s.

The question asks for a numerical value for this rate. The expression we have is $10\pi$ multiplied by the rate at which the radius is changing with respect to time ($\frac{dr}{dt}$) when $r=5$ cm.

For the final answer to be $10\pi$ cm$^2$/s, the value of $\left.\frac{dr}{dt}\right|_{r=5}$ must be 1 cm/s.

Let's assume that the radius is changing at a rate of 1 cm/s at the moment when $r=5$ cm. So, $\left.\frac{dr}{dt}\right|_{r=5} = 1$ cm/s.

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Substituting this value:

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Final Answer:

The rate of change of the area of the circle per second when $r = 5$ cm is $\mathbf{10\pi}$ cm$^2$/s.

Given:

Let $x$ be the length of an edge of the cube.

Let $V$ be the volume of the cube.

Let $S$ be the surface area of the cube.

The rate of increase of the volume is given as 9 cm$^3$/s.

We need to find the rate of increase of the surface area when the length of an edge is 10 cm.

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To Find:

The rate of increase of the surface area ($\frac{dS}{dt}$) when the edge length $x = 10$ cm.

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Solution:

The volume of a cube with edge length $x$ is given by:

Differentiating equation (i) with respect to time $t$ (using the chain rule):

We are given $\frac{dV}{dt} = 9$. Substitute this value into equation (ii):

Now, we can solve for $\frac{dx}{dt}$ in terms of $x$:

Equation (iii) gives the rate at which the edge length is changing with respect to time.

The surface area of a cube with edge length $x$ is given by:

Differentiating equation (iv) with respect to time $t$ (using the chain rule):

Now, substitute the expression for $\frac{dx}{dt}$ from equation (iii) into equation (v):

Simplify the expression:

Equation (vi) gives the rate of change of the surface area in terms of the edge length $x$.

We need to find $\frac{dS}{dt}$ when $x = 10$ cm. Substitute $x = 10$ into equation (vi):

The units of $\frac{dS}{dt}$ are the units of Surface Area divided by the units of time. Surface Area is in cm$^2$ and time is in seconds, so the units are cm$^2$/s.

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Final Answer:

The rate at which the surface area is increasing when the length of an edge is 10 centimetres is $\mathbf{3.6}$ cm$^2$/s.

Given:

Let $r$ be the radius of the circular wave at time $t$ seconds.

Let $A$ be the area enclosed by the circular wave at time $t$ seconds.

The speed of the waves (rate of change of radius with respect to time) is given as 4 cm/s.

We need to find the rate of increase of the enclosed area at the instant when the radius is 10 cm.

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To Find:

The rate of increase of the enclosed area ($\frac{dA}{dt}$) when the radius $r = 10$ cm.

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Solution:

The area of a circle with radius $r$ is given by the formula:

We are looking for the rate of change of the area with respect to time, $\frac{dA}{dt}$. We can find this by differentiating equation (i) with respect to time $t$. Since $A$ is a function of $r$, and $r$ is a function of $t$, we use the chain rule:

Equation (ii) gives the rate of change of the area with respect to time in terms of the radius $r$ and the rate of change of the radius $\frac{dr}{dt}$.

We are given that $\frac{dr}{dt} = 4$ cm/s. Substitute this value into equation (ii):

Equation (iii) gives the rate of change of the area at any instant when the radius is $r$. We need to find this rate at the specific instant when $r = 10$ cm. Substitute $r = 10$ into equation (iii):

The units of $\frac{dA}{dt}$ are the units of Area (cm$^2$) divided by the units of time (s), which is cm$^2$/s.

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Final Answer:

The enclosed area is increasing at a rate of $\mathbf{80\pi}$ cm$^2$/s at the instant when the radius is 10 cm.

Given:

Let $x$ be the length of the rectangle at time $t$.

Let $y$ be the width of the rectangle at time $t$.

The rate of change of the length is given as decreasing at 3 cm/minute.

The rate of change of the width is given as increasing at 2 cm/minute.

We need to find the rates of change at the instant when $x = 10$ cm and $y = 6$ cm.

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To Find:

(a) The rate of change of the perimeter ($\frac{dP}{dt}$) when $x = 10$ cm and $y = 6$ cm.

(b) The rate of change of the area ($\frac{dA}{dt}$) when $x = 10$ cm and $y = 6$ cm.

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Solution:

(a) Rate of change of the Perimeter

The perimeter $P$ of a rectangle with length $x$ and width $y$ is given by:

To find the rate of change of the perimeter with respect to time, we differentiate equation (i) with respect to $t$:

Substitute the given values $\frac{dx}{dt} = -3$ cm/minute and $\frac{dy}{dt} = 2$ cm/minute into equation (ii):

The units for the rate of change of perimeter are units of length per unit of time, which is cm/minute.

Note that the perimeter is decreasing since the rate is negative. The values of $x$ and $y$ at the specific instant are not needed to find $\frac{dP}{dt}$, as the formula for $\frac{dP}{dt}$ only depends on $\frac{dx}{dt}$ and $\frac{dy}{dt}$.

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(b) Rate of change of the Area

The area $A$ of a rectangle with length $x$ and width $y$ is given by:

To find the rate of change of the area with respect to time, we differentiate equation (iii) with respect to $t$ using the product rule $\frac{d}{dt}(uv) = u\frac{dv}{dt} + v\frac{du}{dt}$:

Equation (iv) gives the rate of change of the area in terms of $x$, $y$, $\frac{dx}{dt}$, and $\frac{dy}{dt}$.

We need to find $\frac{dA}{dt}$ at the instant when $x = 10$ cm and $y = 6$ cm. Substitute these values, along with $\frac{dx}{dt} = -3$ cm/minute and $\frac{dy}{dt} = 2$ cm/minute, into equation (iv):

The units for the rate of change of area are units of area per unit of time, which is cm$^2$/minute.

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Final Answers:

(a) The rate of change of the perimeter of the rectangle is $\mathbf{-2}$ cm/minute (decreasing). The perimeter is decreasing at a rate of 2 cm/minute.

(b) The rate of change of the area of the rectangle is $\mathbf{2}$ cm$^2$/minute (increasing).

Given:

The total cost function $C(x)$ in Rupees, associated with the production of $x$ units, is given by:

We are asked to find the marginal cost when $x = 3$ units are produced.

Marginal cost is defined as the instantaneous rate of change of total cost with respect to the level of output $x$. This is given by the derivative of $C(x)$ with respect to $x$, denoted as $MC(x)$ or $\frac{dC}{dx}$.

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To Find:

The marginal cost, $MC(x) = \frac{dC}{dx}$, when $x = 3$ units.

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Solution:

The marginal cost is found by differentiating the total cost function $C(x)$ with respect to $x$.

Differentiating each term using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ and the constant rule $\frac{d}{dx}(c) = 0$:

Equation (ii) gives the marginal cost function for any level of output $x$.

We need to find the marginal cost when $x = 3$ units. Substitute $x = 3$ into equation (ii):

The units of marginal cost are the units of cost per unit of output, which is Rupees per unit.

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Final Answer:

The marginal cost when 3 units are produced is $\mathbf{\textsf{₹} 30.015}$.

Given:

The total revenue function $R(x)$ in Rupees, received from the sale of $x$ units of a product, is given by:

We are asked to find the marginal revenue when $x = 5$ units are sold.

Marginal revenue is defined as the rate of change of total revenue with respect to the number of items sold ($x$). This is given by the derivative of $R(x)$ with respect to $x$, denoted as $MR(x)$ or $\frac{dR}{dx}$.

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To Find:

The marginal revenue, $MR(x) = \frac{dR}{dx}$, when $x = 5$ units.

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Solution:

The marginal revenue is found by differentiating the total revenue function $R(x)$ with respect to $x$.

Differentiating each term using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ and the constant rule $\frac{d}{dx}(c) = 0$:

Equation (ii) gives the marginal revenue function for any number of units sold $x$.

We need to find the marginal revenue when $x = 5$ units. Substitute $x = 5$ into equation (ii):

The units of marginal revenue are the units of revenue per unit of item sold, which is Rupees per unit.

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Final Answer:

The marginal revenue when 5 units are sold is $\mathbf{\textsf{₹} 66}$.

Given:

Let $A$ be the area of the circle and $r$ be its radius.

We are asked to find the rate of change of the area with respect to the radius $r$ for specific values of $r$.

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To Find:

The rate of change of the area of the circle with respect to its radius $r$ when:

(a) $r = 3$ cm

(b) $r = 4$ cm

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Solution:

The formula for the area of a circle with radius $r$ is given by:

The rate of change of the area $A$ with respect to the radius $r$ is given by the derivative of $A$ with respect to $r$, denoted as $\frac{dA}{dr}$.

Differentiating equation (i) with respect to $r$:

Using the constant multiple rule and the power rule for differentiation ($\frac{d}{dr}(r^n) = nr^{n-1}$):

Equation (ii) gives the general rate of change of the area with respect to the radius for any radius $r$.

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(a) When $r = 3$ cm:

Substitute $r = 3$ into equation (ii):

The units of $\frac{dA}{dr}$ are the units of Area divided by the units of radius. Area is in cm$^2$ and radius is in cm, so the units are $\frac{\text{cm}^2}{\text{cm}} = \text{cm}$.

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(b) When $r = 4$ cm:

Substitute $r = 4$ into equation (ii):

The units are cm.

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Final Answers:

(a) When $r = 3$ cm, the rate of change of the area of the circle with respect to its radius is $\mathbf{6\pi}$ cm.

(b) When $r = 4$ cm, the rate of change of the area of the circle with respect to its radius is $\mathbf{8\pi}$ cm.

Given:

Let $x$ be the length of an edge of the cube at time $t$.

Let $V$ be the volume of the cube at time $t$.

Let $S$ be the surface area of the cube at time $t$.

The rate of increase of the volume is given as 8 cm$^3$/s.

We need to find the rate of increase of the surface area when the length of an edge is 12 cm.

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To Find:

The rate of increase of the surface area ($\frac{dS}{dt}$) when the edge length $x = 12$ cm.

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Solution:

The volume of a cube with edge length $x$ is given by:

Differentiating equation (i) with respect to time $t$ (using the chain rule):

We are given $\frac{dV}{dt} = 8$. Substitute this value into equation (ii):

Now, we can solve for $\frac{dx}{dt}$ in terms of $x$:

Equation (iii) gives the rate at which the edge length is changing with respect to time.

The surface area of a cube with edge length $x$ is given by:

Differentiating equation (iv) with respect to time $t$ (using the chain rule):

Now, substitute the expression for $\frac{dx}{dt}$ from equation (iii) into equation (v):

Simplify the expression:

Equation (vi) gives the rate of change of the surface area in terms of the edge length $x$.

We need to find $\frac{dS}{dt}$ when $x = 12$ cm. Substitute $x = 12$ into equation (vi):

Simplify the fraction:

The units of $\frac{dS}{dt}$ are the units of Surface Area (cm$^2$) divided by the units of time (s), which is cm$^2$/s.

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Final Answer:

The surface area is increasing at a rate of $\mathbf{\frac{8}{3}}$ cm$^2$/s when the length of an edge is 12 cm.

Given:

Let $r$ be the radius of the circle at time $t$ seconds.

Let $A$ be the area of the circle at time $t$ seconds.

The radius is increasing uniformly at the rate of 3 cm/s.

We need to find the rate at which the area is increasing when the radius is 10 cm.

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To Find:

The rate of increase of the area ($\frac{dA}{dt}$) when the radius $r = 10$ cm.

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Solution:

The area of a circle with radius $r$ is given by the formula:

We are looking for the rate of change of the area with respect to time, $\frac{dA}{dt}$. We can find this by differentiating equation (i) with respect to time $t$. Since $A$ is a function of $r$, and $r$ is a function of $t$, we use the chain rule:

Equation (ii) gives the rate of change of the area with respect to time in terms of the radius $r$ and the rate of change of the radius $\frac{dr}{dt}$.

We are given that $\frac{dr}{dt} = 3$ cm/s. Substitute this value into equation (ii):

Equation (iii) gives the rate of change of the area at any instant when the radius is $r$. We need to find this rate at the specific instant when $r = 10$ cm. Substitute $r = 10$ into equation (iii):

The units of $\frac{dA}{dt}$ are the units of Area (cm$^2$) divided by the units of time (s), which is cm$^2$/s.

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Final Answer:

The area of the circle is increasing at a rate of $\mathbf{60\pi}$ cm$^2$/s when the radius is 10 cm.

Given:

Let $x$ be the length of an edge of the variable cube at time $t$ seconds.

Let $V$ be the volume of the cube at time $t$ seconds.

The edge is increasing at the rate of 3 cm/s.

We need to find the rate at which the volume is increasing when the edge length is 10 cm.

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To Find:

The rate of increase of the volume ($\frac{dV}{dt}$) when the edge length $x = 10$ cm.

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Solution:

The volume of a cube with edge length $x$ is given by the formula:

We are looking for the rate of change of the volume with respect to time, $\frac{dV}{dt}$. We can find this by differentiating equation (i) with respect to time $t$. Since $V$ is a function of $x$, and $x$ is a function of $t$, we use the chain rule:

Equation (ii) gives the rate of change of the volume with respect to time in terms of the edge length $x$ and the rate of change of the edge length $\frac{dx}{dt}$.

We are given that $\frac{dx}{dt} = 3$ cm/s. Substitute this value into equation (ii):

Equation (iii) gives the rate of change of the volume at any instant when the edge length is $x$. We need to find this rate at the specific instant when $x = 10$ cm. Substitute $x = 10$ into equation (iii):

The units of $\frac{dV}{dt}$ are the units of Volume (cm$^3$) divided by the units of time (s), which is cm$^3$/s.

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Final Answer:

The volume of the cube is increasing at a rate of $\mathbf{900}$ cm$^3$/s when the edge is 10 cm long.

Given:

The speed at which the radius of the circular wave is increasing is $\frac{dr}{dt} = 5$ cm/s.

The radius of the circular wave at a specific instant is $r = 8$ cm.

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To Find:

We need to find the rate at which the enclosed area is increasing at the instant when $r=8$ cm, i.e., $\frac{dA}{dt}$ when $r=8$ cm.

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Solution:

Let $A$ be the area of the circular wave and $r$ be its radius at time $t$.

The formula for the area of a circle is given by:

$A = \pi r^2$

To find the rate at which the area is increasing, we differentiate the area $A$ with respect to time $t$ using the chain rule:

$\frac{dA}{dt} = \frac{d}{dt}(\pi r^2)$

$\frac{dA}{dt} = \pi \cdot \frac{d}{dt}(r^2)$

$\frac{dA}{dt} = \pi \cdot 2r \cdot \frac{dr}{dt}$

Now, we substitute the given values $r = 8$ cm and $\frac{dr}{dt} = 5$ cm/s into the equation for $\frac{dA}{dt}$:

$\frac{dA}{dt} = \pi \cdot 2(8 \text{ cm}) \cdot (5 \text{ cm/s})$

$\frac{dA}{dt} = \pi \cdot 16 \cdot 5 \text{ cm}^2\text{/s}$

$\frac{dA}{dt} = 80\pi \text{ cm}^2\text{/s}$

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The enclosed area is increasing at the rate of $80\pi$ cm$^2$/s at the instant when the radius is 8 cm.

Given:

The rate of increase of the radius of the circle is $\frac{dr}{dt} = 0.7$ cm/s.

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To Find:

We need to find the rate of increase of the circumference of the circle, i.e., $\frac{dC}{dt}$.

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Solution:

Let $C$ be the circumference of the circle and $r$ be its radius at time $t$.

The formula for the circumference of a circle is given by:

$C = 2\pi r$

To find the rate at which the circumference is increasing, we differentiate the circumference $C$ with respect to time $t$:

$\frac{dC}{dt} = \frac{d}{dt}(2\pi r)$

Since $2\pi$ is a constant, we can write:

$\frac{dC}{dt} = 2\pi \frac{dr}{dt}$

Now, we substitute the given value $\frac{dr}{dt} = 0.7$ cm/s into the equation for $\frac{dC}{dt}$:

$\frac{dC}{dt} = 2\pi \cdot (0.7 \text{ cm/s})$

$\frac{dC}{dt} = 1.4\pi \text{ cm/s}$

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The rate of increase of the circumference of the circle is $1.4\pi$ cm/s.

Given:

Rate of decrease of length $x$: $\frac{dx}{dt} = -5$ cm/minute (decreasing rate is negative).

Rate of increase of width $y$: $\frac{dy}{dt} = 4$ cm/minute (increasing rate is positive).

At the instant of interest: $x = 8$ cm and $y = 6$ cm.

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To Find:

(a) The rate of change of the perimeter of the rectangle, $\frac{dP}{dt}$, when $x=8$ cm and $y=6$ cm.

(b) The rate of change of the area of the rectangle, $\frac{dA}{dt}$, when $x=8$ cm and $y=6$ cm.

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Solution:

(a) Rate of change of the perimeter:

Let $P$ be the perimeter of the rectangle. The formula for the perimeter is:

$P = 2(x + y)$

To find the rate of change of the perimeter, we differentiate $P$ with respect to time $t$:

$\frac{dP}{dt} = \frac{d}{dt}(2(x + y))$

$\frac{dP}{dt} = 2\left(\frac{dx}{dt} + \frac{dy}{dt}\right)$

Substitute the given values $\frac{dx}{dt} = -5$ cm/minute and $\frac{dy}{dt} = 4$ cm/minute:

$\frac{dP}{dt} = 2(-5 \text{ cm/min} + 4 \text{ cm/min})$

$\frac{dP}{dt} = 2(-1 \text{ cm/min})$

$\frac{dP}{dt} = -2$ cm/minute

The negative sign indicates that the perimeter is decreasing.

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(b) Rate of change of the area:

Let $A$ be the area of the rectangle. The formula for the area is:

$A = xy$

To find the rate of change of the area, we differentiate $A$ with respect to time $t$ using the product rule:

$\frac{dA}{dt} = \frac{d}{dt}(xy)$

$\frac{dA}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt}$

Substitute the given values $x = 8$ cm, $y = 6$ cm, $\frac{dx}{dt} = -5$ cm/minute, and $\frac{dy}{dt} = 4$ cm/minute:

$\frac{dA}{dt} = (8 \text{ cm})(4 \text{ cm/min}) + (6 \text{ cm})(-5 \text{ cm/min})$

$\frac{dA}{dt} = 32 \text{ cm}^2\text{/min} - 30 \text{ cm}^2\text{/min}$

$\frac{dA}{dt} = 2$ cm$^2$/minute

The positive sign indicates that the area is increasing.

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At the instant when $x = 8$ cm and $y = 6$ cm:

(a) The rate of change of the perimeter is $-2$ cm/minute (decreasing at 2 cm/minute).

(b) The rate of change of the area is $2$ cm$^2$/minute (increasing at 2 cm$^2$/minute).

Given:

The balloon is spherical.

The rate at which the volume of the balloon is increasing is $\frac{dV}{dt} = 900$ cm$^3$/s.

We are interested in the instant when the radius is $r = 15$ cm.

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To Find:

We need to find the rate at which the radius of the balloon increases when $r = 15$ cm, i.e., $\frac{dr}{dt}$ when $r=15$ cm.

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Solution:

Let $V$ be the volume of the spherical balloon and $r$ be its radius at time $t$.

The formula for the volume of a sphere is given by:

$V = \frac{4}{3}\pi r^3$

To find the rate of change of the radius, we relate it to the rate of change of the volume by differentiating the volume $V$ with respect to time $t$. We use the chain rule for the derivative of $r^3$ with respect to $t$:

$\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)$

$\frac{dV}{dt} = \frac{4}{3}\pi \frac{d}{dt}(r^3)$

$\frac{dV}{dt} = \frac{4}{3}\pi (3r^2) \frac{dr}{dt}$

$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$

We are given $\frac{dV}{dt} = 900$ cm$^3$/s and we want to find $\frac{dr}{dt}$ when $r = 15$ cm. Substitute these values into the differentiated equation:

$900 \text{ cm}^3\text{/s} = 4\pi (15 \text{ cm})^2 \frac{dr}{dt}$

$900 = 4\pi (225) \frac{dr}{dt}$

$900 = 900\pi \frac{dr}{dt}$

Now, solve for $\frac{dr}{dt}$:

$\frac{dr}{dt} = \frac{900}{900\pi} \text{ cm/s}$

$\frac{dr}{dt} = \frac{1}{\pi} \text{ cm/s}$

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The rate at which the radius of the balloon increases when the radius is 15 cm is $\frac{1}{\pi}$ cm/s.

Given:

The balloon is spherical.

The radius of the balloon is variable. We are interested in the instant when the radius $r = 10$ cm.

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To Find:

We need to find the rate at which the volume of the balloon is increasing with respect to the radius, i.e., $\frac{dV}{dr}$, when $r = 10$ cm.

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Solution:

Let $V$ be the volume of the spherical balloon and $r$ be its radius.

The formula for the volume of a sphere is given by:

$V = \frac{4}{3}\pi r^3$

To find the rate at which the volume is increasing with the radius, we differentiate the volume $V$ with respect to the radius $r$:

$\frac{dV}{dr} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right)$

$\frac{dV}{dr} = \frac{4}{3}\pi \frac{d}{dr}(r^3)$

$\frac{dV}{dr} = \frac{4}{3}\pi (3r^2)$

$\frac{dV}{dr} = 4\pi r^2$

Now, we need to find this rate when the radius $r = 10$ cm. Substitute $r = 10$ into the expression for $\frac{dV}{dr}$:

$\left.\frac{dV}{dr}\right|_{r=10} = 4\pi (10 \text{ cm})^2$

$\left.\frac{dV}{dr}\right|_{r=10} = 4\pi (100 \text{ cm}^2)$

$\left.\frac{dV}{dr}\right|_{r=10} = 400\pi \text{ cm}^3\text{/cm}$

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The rate at which the volume of the balloon is increasing with the radius when the radius is 10 cm is $400\pi$ cm$^3$/cm.

Given:

Length of the ladder = 5 m.

Rate at which the bottom of the ladder is pulled away from the wall, $\frac{dx}{dt} = 2$ cm/s.

The instant when the foot of the ladder is 4 m away from the wall, i.e., $x = 4$ m.

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To Find:

The rate at which the height of the ladder on the wall is decreasing when $x = 4$ m, i.e., $\frac{dy}{dt}$ at that instant.

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Solution:

Let $x$ be the distance of the foot of the ladder from the wall and $y$ be the height of the top of the ladder on the wall at time $t$.

The ladder, the wall, and the ground form a right-angled triangle. By the Pythagorean theorem, we have:

Differentiate equation (i) with respect to time $t$:

$\frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(25)$

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

Divide by 2:

We are given that $\frac{dx}{dt} = 2$ cm/s. We need to use consistent units. Let's convert everything to meters and seconds. So, $\frac{dx}{dt} = 2$ cm/s $= 0.02$ m/s.

We are interested in the instant when $x = 4$ m.

First, find the value of $y$ when $x = 4$ m using equation (i):

$4^2 + y^2 = 5^2$

$16 + y^2 = 25$

$y^2 = 25 - 16$

$y^2 = 9$

$y = \sqrt{9} = 3$ m (Since height must be positive)

Now, substitute $x = 4$ m, $y = 3$ m, and $\frac{dx}{dt} = 0.02$ m/s into equation (ii):

$(4 \text{ m})(0.02 \text{ m/s}) + (3 \text{ m})\frac{dy}{dt} = 0$

$0.08 \text{ m}^2\text{/s} + 3 \text{ m}\frac{dy}{dt} = 0$

$3 \text{ m}\frac{dy}{dt} = -0.08 \text{ m}^2\text{/s}$

$\frac{dy}{dt} = \frac{-0.08}{3} \text{ m/s}$

The negative sign indicates that the height $y$ is decreasing. To express the rate of decrease in cm/s, we multiply by 100:

$\frac{dy}{dt} = -\frac{0.08}{3} \times 100 \text{ cm/s}$

$\frac{dy}{dt} = -\frac{8}{3} \text{ cm/s}$

The rate of decrease of the height is the absolute value of $\frac{dy}{dt}$.

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The height on the wall is decreasing at the rate of $\frac{8}{3}$ cm/s when the foot of the ladder is 4 m away from the wall.

Given:

The equation of the curve along which the particle moves is $6y = x^3 + 2$.

The rate of change of the y-coordinate with respect to time $t$ is 8 times the rate of change of the x-coordinate with respect to time $t$. This can be expressed as $\frac{dy}{dt} = 8\frac{dx}{dt}$.

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To Find:

The coordinates of the points $(x, y)$ on the curve where the given condition $\frac{dy}{dt} = 8\frac{dx}{dt}$ holds true.

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Solution:

The equation of the curve is:

We are given the condition relating the rates of change of $y$ and $x$ with respect to time $t$:

To relate the equation of the curve to the rates of change, we differentiate equation (i) with respect to time $t$ using the chain rule:

$\frac{d}{dt}(6y) = \frac{d}{dt}(x^3 + 2)$

$6\frac{dy}{dt} = 3x^2\frac{dx}{dt} + \frac{d}{dt}(2)$

$6\frac{dy}{dt} = 3x^2\frac{dx}{dt} + 0$

Now, substitute the expression for $\frac{dy}{dt}$ from equation (ii) into equation (iii):

$6\left(8\frac{dx}{dt}\right) = 3x^2\frac{dx}{dt}$

$48\frac{dx}{dt} = 3x^2\frac{dx}{dt}$

Rearrange the equation to find the values of $x$ that satisfy this condition:

$48\frac{dx}{dt} - 3x^2\frac{dx}{dt} = 0$

Factor out $\frac{dx}{dt}$:

$\frac{dx}{dt}(48 - 3x^2) = 0$

This equation is satisfied if either $\frac{dx}{dt} = 0$ or $48 - 3x^2 = 0$.

If $\frac{dx}{dt} = 0$, then from equation (ii), $\frac{dy}{dt} = 8 \times 0 = 0$. This corresponds to an instant where the particle is not moving, and the condition $0=8 \times 0$ is satisfied. However, we are typically interested in points where there is a non-zero rate of change related by the factor 8. So, we consider the second case.

Consider the case where $48 - 3x^2 = 0$ (assuming $\frac{dx}{dt} \neq 0$):

$3x^2 = 48$

$x^2 = \frac{48}{3}$

$x^2 = 16$

Taking the square root of both sides:

$x = \pm \sqrt{16}$

$x = \pm 4$

Now we need to find the corresponding $y$-coordinates for these $x$ values using the original curve equation (i), $6y = x^3 + 2$.

Case 1: When $x = 4$

$6y = (4)^3 + 2$

$6y = 64 + 2$

$6y = 66$

$y = \frac{66}{6}$

$y = 11$

So, one point is $(4, 11)$.

Case 2: When $x = -4$

$6y = (-4)^3 + 2$

$6y = -64 + 2$

$6y = -62$

$y = \frac{-62}{6}$

$y = -\frac{31}{3}$

So, the other point is $(-4, -\frac{31}{3})$.

---

The points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate are $(4, 11)$ and $(-4, -\frac{31}{3})$.

Given:

The air bubble is spherical.

The rate at which the radius of the air bubble is increasing is $\frac{dr}{dt} = \frac{1}{2}$ cm/s.

We are interested in the instant when the radius is $r = 1$ cm.

---

To Find:

We need to find the rate at which the volume of the bubble is increasing when $r = 1$ cm, i.e., $\frac{dV}{dt}$ when $r=1$ cm.

---

Solution:

Let $V$ be the volume of the spherical air bubble and $r$ be its radius at time $t$.

The formula for the volume of a sphere is given by:

$V = \frac{4}{3}\pi r^3$

To find the rate at which the volume is increasing, we differentiate the volume $V$ with respect to time $t$ using the chain rule:

$\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)$

$\frac{dV}{dt} = \frac{4}{3}\pi \frac{d}{dt}(r^3)$

$\frac{dV}{dt} = \frac{4}{3}\pi (3r^2) \frac{dr}{dt}$

$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$

We are given $\frac{dr}{dt} = \frac{1}{2}$ cm/s and we want to find $\frac{dV}{dt}$ when $r = 1$ cm. Substitute these values into the differentiated equation:

$\frac{dV}{dt} = 4\pi (1 \text{ cm})^2 \left(\frac{1}{2} \text{ cm/s}\right)$

$\frac{dV}{dt} = 4\pi (1 \text{ cm}^2) \left(\frac{1}{2} \text{ cm/s}\right)$

$\frac{dV}{dt} = 4\pi \cdot \frac{1}{2} \text{ cm}^3\text{/s}$

$\frac{dV}{dt} = 2\pi \text{ cm}^3\text{/s}$

---

The rate at which the volume of the bubble is increasing when the radius is 1 cm is $2\pi$ cm$^3$/s.

Given:

The balloon is spherical.

The variable diameter of the balloon is given by $D = \frac{3}{2}(2x + 1)$.

---

To Find:

The rate of change of its volume with respect to $x$, i.e., $\frac{dV}{dx}$.

---

Solution:

The diameter of the spherical balloon is $D = \frac{3}{2}(2x + 1)$.

The radius $r$ of the balloon is half of its diameter:

$r = \frac{D}{2} = \frac{1}{2} \left(\frac{3}{2}(2x + 1)\right) = \frac{3}{4}(2x + 1)$

The volume $V$ of a sphere with radius $r$ is given by the formula:

$V = \frac{4}{3}\pi r^3$

Substitute the expression for $r$ in terms of $x$ into the volume formula:

$V = \frac{4}{3}\pi \left(\frac{3}{4}(2x + 1)\right)^3$

$V = \frac{4}{3}\pi \left(\frac{3^3}{4^3}(2x + 1)^3\right)$

$V = \frac{4}{3}\pi \left(\frac{27}{64}(2x + 1)^3\right)$

$V = \left(\frac{4}{3} \cdot \frac{27}{64}\right) \pi (2x + 1)^3$

Simplify the fraction $\frac{4}{3} \cdot \frac{27}{64}$:

$\frac{4}{3} \cdot \frac{27}{64} = \frac{\cancel{4}^{1}}{\cancel{3}^{1}} \cdot \frac{\cancel{27}^{9}}{\cancel{64}^{16}} = \frac{1 \cdot 9}{1 \cdot 16} = \frac{9}{16}$

So, the volume in terms of $x$ is:

$V = \frac{9\pi}{16}(2x + 1)^3$

To find the rate of change of the volume with respect to $x$, we differentiate $V$ with respect to $x$:

$\frac{dV}{dx} = \frac{d}{dx}\left(\frac{9\pi}{16}(2x + 1)^3\right)$

$\frac{dV}{dx} = \frac{9\pi}{16} \frac{d}{dx}((2x + 1)^3)$

Using the chain rule, $\frac{d}{dx}((2x + 1)^3) = 3(2x + 1)^{3-1} \cdot \frac{d}{dx}(2x + 1)$.

The derivative of $(2x + 1)$ with respect to $x$ is $\frac{d}{dx}(2x + 1) = 2$.

So, $\frac{d}{dx}((2x + 1)^3) = 3(2x + 1)^2 \cdot 2 = 6(2x + 1)^2$.

Substitute this back into the expression for $\frac{dV}{dx}$:

$\frac{dV}{dx} = \frac{9\pi}{16} \cdot 6(2x + 1)^2$

$\frac{dV}{dx} = \frac{54\pi}{16}(2x + 1)^2$

Simplify the fraction $\frac{54}{16}$ by dividing the numerator and denominator by their greatest common divisor, 2:

$\frac{54}{16} = \frac{27}{8}$

Therefore, the rate of change of the volume with respect to $x$ is:

$\frac{dV}{dx} = \frac{27\pi}{8}(2x + 1)^2$

---

The rate of change of the volume of the balloon with respect to $x$ is $\frac{27\pi}{8}(2x + 1)^2$.

Given:

The rate at which the volume of sand is increasing is $\frac{dV}{dt} = 12$ cm$^3$/s.

The height ($h$) of the cone is always one-sixth of the radius ($r$) of the base, which means $h = \frac{1}{6}r$. This can be rewritten as $r = 6h$.

We are interested in the instant when the height is $h = 4$ cm.

---

To Find:

The rate at which the height of the sand cone is increasing when $h = 4$ cm, i.e., $\frac{dh}{dt}$ when $h=4$ cm.

---

Solution:

Let $V$ be the volume of the sand cone, $r$ be the radius of its base, and $h$ be its height at time $t$.

The formula for the volume of a cone is given by:

$V = \frac{1}{3}\pi r^2 h$

We are given the relationship between $h$ and $r$:

From equation (i), we can express the radius in terms of the height:

$r = 6h$

Substitute this expression for $r$ into the volume formula to express $V$ in terms of $h$ only:

$V = \frac{1}{3}\pi (6h)^2 h$

$V = \frac{1}{3}\pi (36h^2) h$

To find the rate at which the height is increasing, we differentiate the volume $V$ with respect to time $t$. We use the chain rule for the derivative of $h^3$ with respect to $t$:

$\frac{dV}{dt} = \frac{d}{dt}(12\pi h^3)$

$\frac{dV}{dt} = 12\pi \frac{d}{dt}(h^3)$

$\frac{dV}{dt} = 12\pi (3h^2) \frac{dh}{dt}$

We are given that sand is pouring at the rate of 12 cm$^3$/s, so $\frac{dV}{dt} = 12$. We want to find $\frac{dh}{dt}$ when $h = 4$ cm.

Substitute these values into equation (iii):

$12 = 36\pi (4)^2 \frac{dh}{dt}$

$12 = 36\pi (16) \frac{dh}{dt}$

$12 = 576\pi \frac{dh}{dt}$

Now, solve for $\frac{dh}{dt}$:

$\frac{dh}{dt} = \frac{12}{576\pi}$

Simplify the fraction by dividing the numerator and denominator by 12:

$\frac{dh}{dt} = \frac{\cancel{12}^{1}}{\cancel{576}_{48}\pi}$

$\frac{dh}{dt} = \frac{1}{48\pi}$ cm/s

---

The height of the sand cone is increasing at the rate of $\frac{1}{48\pi}$ cm/s when the height is 4 cm.

Given:

The total cost function is $C(x) = 0.007x^3 - 0.003x^2 + 15x + 4000$, where $x$ is the number of units produced.

We need to find the marginal cost when $x = 17$ units.

---

To Find:

The marginal cost, which is the rate of change of total cost with respect to the number of units produced, when $x=17$.

---

Solution:

The marginal cost (MC) is given by the derivative of the total cost function with respect to $x$, i.e., $MC(x) = C'(x)$.

Differentiate the cost function $C(x)$ with respect to $x$:

$C'(x) = \frac{d}{dx}(0.007x^3 - 0.003x^2 + 15x + 4000)$

$C'(x) = \frac{d}{dx}(0.007x^3) - \frac{d}{dx}(0.003x^2) + \frac{d}{dx}(15x) + \frac{d}{dx}(4000)$

$C'(x) = 0.007(3x^{3-1}) - 0.003(2x^{2-1}) + 15(1) + 0$

$C'(x) = 0.021x^2 - 0.006x + 15$

Now, we need to find the marginal cost when $x = 17$. Substitute $x = 17$ into the expression for $C'(x)$:

$C'(17) = 0.021(17)^2 - 0.006(17) + 15$

$C'(17) = 0.021(289) - 0.006(17) + 15$

$C'(17) = 6.069 - 0.102 + 15$

$C'(17) = 5.967 + 15$

$C'(17) = 20.967$

The marginal cost is in Rupees per unit.

---

The marginal cost when 17 units are produced is $\textsf{₹} 20.967$.

Given:

The total revenue function is $R(x) = 13x^2 + 26x + 15$, where $x$ is the number of units sold.

We need to find the marginal revenue when $x = 7$ units.

---

To Find:

The marginal revenue when $x = 7$.

---

Solution:

The marginal revenue (MR) is defined as the rate of change of total revenue with respect to the number of units sold. It is given by the derivative of the total revenue function with respect to $x$, i.e., $MR(x) = R'(x)$.

Differentiate the total revenue function $R(x)$ with respect to $x$:

$R'(x) = \frac{d}{dx}(13x^2 + 26x + 15)$

$R'(x) = \frac{d}{dx}(13x^2) + \frac{d}{dx}(26x) + \frac{d}{dx}(15)$

$R'(x) = 13(2x) + 26(1) + 0$

$R'(x) = 26x + 26$

Now, we need to find the marginal revenue when $x = 7$. Substitute $x = 7$ into the expression for $R'(x)$:

$R'(7) = 26(7) + 26$

$R'(7) = 182 + 26$

$R'(7) = 208$

The marginal revenue is in Rupees.

---

The marginal revenue when $x = 7$ is $\textsf{₹} 208$.

Given:

The radius of the circle is $r$.

We need to find the rate of change of the area at $r = 6$ cm.

---

To Find:

The rate of change of the area of a circle with respect to its radius, $\frac{dA}{dr}$, when $r=6$ cm.

---

Solution:

Let $A$ be the area of the circle and $r$ be its radius.

The formula for the area of a circle is given by:

$A = \pi r^2$

To find the rate of change of the area with respect to the radius, we differentiate $A$ with respect to $r$:

$\frac{dA}{dr} = \frac{d}{dr}(\pi r^2)$

$\frac{dA}{dr} = \pi \frac{d}{dr}(r^2)$

$\frac{dA}{dr} = \pi (2r)$

$\frac{dA}{dr} = 2\pi r$

Now, we need to find this rate when the radius $r = 6$ cm. Substitute $r = 6$ into the expression for $\frac{dA}{dr}$:

$\left.\frac{dA}{dr}\right|_{r=6} = 2\pi (6 \text{ cm})$

$\left.\frac{dA}{dr}\right|_{r=6} = 12\pi$ cm$^2$/cm

---

The rate of change of the area of the circle with respect to its radius at $r = 6$ cm is $12\pi$.

This matches option (B).

The correct answer is (B) 12$\pi$.

Given:

The total revenue function is $R(x) = 3x^2 + 36x + 5$, where $x$ is the number of units sold.

We need to find the marginal revenue when $x = 15$ units.

---

To Find:

The marginal revenue when $x = 15$.

---

Solution:

The marginal revenue (MR) is defined as the rate of change of total revenue with respect to the number of units sold. It is given by the derivative of the total revenue function with respect to $x$, i.e., $MR(x) = R'(x)$.

Differentiate the total revenue function $R(x)$ with respect to $x$:

$R'(x) = \frac{d}{dx}(3x^2 + 36x + 5)$

$R'(x) = \frac{d}{dx}(3x^2) + \frac{d}{dx}(36x) + \frac{d}{dx}(5)$

$R'(x) = 3(2x^{2-1}) + 36(1) + 0$

$R'(x) = 6x + 36$

Now, we need to find the marginal revenue when $x = 15$. Substitute $x = 15$ into the expression for $R'(x)$:

$R'(15) = 6(15) + 36$

$R'(15) = 90 + 36$

$R'(15) = 126$

The marginal revenue is in Rupees.

---

The marginal revenue when $x = 15$ is $\textsf{₹} 126$.

This matches option (D).

The correct answer is (D) 126.

Given:

The function is $f(x) = 7x - 3$.

---

To Prove:

The function $f(x)$ is strictly increasing on the set of real numbers $\mathbb{R}$.

---

Solution:

A function $f(x)$ is considered strictly increasing on an interval if its derivative $f'(x)$ is positive for all values of $x$ in that interval.

We find the first derivative of the given function $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(7x - 3)$

Using the rules of differentiation:

$f'(x) = \frac{d}{dx}(7x) - \frac{d}{dx}(3)$

$f'(x) = 7 \cdot \frac{d}{dx}(x) - 0$

$f'(x) = 7 \cdot 1$

$f'(x) = 7$

We observe that the derivative $f'(x) = 7$ is a constant value.

Since $7 > 0$, the derivative $f'(x)$ is positive for all real values of $x$ (i.e., for all $x \in \mathbb{R}$).

Therefore, according to the condition for a strictly increasing function, $f(x) = 7x - 3$ is strictly increasing on $\mathbb{R}$.

---

Hence, the function $f(x) = 7x - 3$ is strictly increasing on $\mathbb{R}$.

Given:

The function is $f(x) = x^3 - 3x^2 + 4x$, where $x \in \mathbb{R}$.

---

To Prove:

The function $f(x)$ is strictly increasing on the set of real numbers $\mathbb{R}$.

---

Solution:

To determine if the function is strictly increasing, we need to find its derivative $f'(x)$ and check its sign on the interval $\mathbb{R}$.

Differentiate the function $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 4x)$

$f'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(3x^2) + \frac{d}{dx}(4x)$

$f'(x) = 3x^{3-1} - 3(2x^{2-1}) + 4(1)$

$f'(x) = 3x^2 - 6x + 4$

Now, we need to show that $f'(x) > 0$ for all $x \in \mathbb{R}$.

The expression for $f'(x)$ is a quadratic in $x$. We can analyze its sign by completing the square or by examining its discriminant and the coefficient of $x^2$.

Let's complete the square for $f'(x) = 3x^2 - 6x + 4$:

$f'(x) = 3(x^2 - 2x) + 4$

Complete the square for the terms inside the parenthesis $(x^2 - 2x)$. Add and subtract $(\frac{-2}{2})^2 = (-1)^2 = 1$ inside the parenthesis:

$f'(x) = 3(x^2 - 2x + 1 - 1) + 4$

$f'(x) = 3((x - 1)^2 - 1) + 4$

$f'(x) = 3(x - 1)^2 - 3 + 4$

$f'(x) = 3(x - 1)^2 + 1$

For any real number $x$, the term $(x - 1)^2$ is always non-negative, i.e., $(x - 1)^2 \ge 0$.

Multiplying by 3 (which is positive) does not change the inequality: $3(x - 1)^2 \ge 0$.

Adding 1 to both sides of the inequality:

$3(x - 1)^2 + 1 \ge 0 + 1$

$3(x - 1)^2 + 1 \ge 1$

So, $f'(x) = 3(x - 1)^2 + 1$. Since $3(x - 1)^2 + 1 \ge 1$, it follows that $f'(x) > 0$ for all $x \in \mathbb{R}$.

Alternatively, consider the discriminant of the quadratic $f'(x) = 3x^2 - 6x + 4$. The discriminant $\Delta$ is given by $\Delta = b^2 - 4ac$, where $a=3$, $b=-6$, and $c=4$.

$\Delta = (-6)^2 - 4(3)(4)$

$\Delta = 36 - 48$

$\Delta = -12$

Since the discriminant $\Delta = -12 < 0$ and the coefficient of $x^2$ is $a = 3 > 0$, the quadratic $f'(x) = 3x^2 - 6x + 4$ is always positive for all real values of $x$.

Both methods show that $f'(x) > 0$ for all $x \in \mathbb{R}$.

A function is strictly increasing on an interval if its derivative is positive on that interval.

Since $f'(x) > 0$ for all $x \in \mathbb{R}$, the function $f(x)$ is strictly increasing on $\mathbb{R}$.

---

Thus, the function $f(x) = x^3 - 3x^2 + 4x$ is strictly increasing on $\mathbb{R}$.

Given:

The function is $f(x) = \cos x$.

---

To Prove:

The function $f(x)$ is decreasing in $(0, \pi)$, increasing in $(\pi, 2\pi)$, and neither increasing nor decreasing in $(0, 2\pi)$.

---

Solution:

To determine the intervals where the function is increasing or decreasing, we need to find its derivative $f'(x)$ and analyze its sign.

The derivative of $f(x) = \cos x$ with respect to $x$ is:

$f'(x) = \frac{d}{dx}(\cos x)$

We will now examine the sign of $f'(x) = -\sin x$ on the given intervals.

(a) Interval $(0, \pi)$:

In the interval $(0, \pi)$, the value of $\sin x$ is positive.

Therefore, from equation (i), the sign of $f'(x)$ is:

Since $f'(x) < 0$ for all $x$ in $(0, \pi)$, the function $f(x) = \cos x$ is strictly decreasing on $(0, \pi)$. A strictly decreasing function is also a decreasing function.

Thus, $f(x) = \cos x$ is decreasing in $(0, \pi)$.

(b) Interval $(\pi, 2\pi)$:

In the interval $(\pi, 2\pi)$, the value of $\sin x$ is negative.

Therefore, from equation (i), the sign of $f'(x)$ is:

Since $f'(x) > 0$ for all $x$ in $(\pi, 2\pi)$, the function $f(x) = \cos x$ is strictly increasing on $(\pi, 2\pi)$. A strictly increasing function is also an increasing function.

Thus, $f(x) = \cos x$ is increasing in $(\pi, 2\pi)$.

(c) Interval $(0, 2\pi)$:

The interval $(0, 2\pi)$ includes both the interval $(0, \pi)$ and the interval $(\pi, 2\pi)$.

From part (a), $f(x)$ is decreasing in $(0, \pi)$.

From part (b), $f(x)$ is increasing in $(\pi, 2\pi)$.

Since the function is decreasing over a part of the interval $(0, 2\pi)$ and increasing over another part of the same interval, it is neither increasing throughout the entire interval $(0, 2\pi)$ nor decreasing throughout the entire interval $(0, 2\pi)$.

For instance, consider $x_1 = \frac{\pi}{2}$ and $x_2 = \frac{3\pi}{2}$ in $(0, 2\pi)$, where $x_1 < x_2$.

$f(x_1) = f(\frac{\pi}{2}) = \cos(\frac{\pi}{2}) = 0$

$f(x_2) = f(\frac{3\pi}{2}) = \cos(\frac{3\pi}{2}) = 0$

Here $f(x_1) = f(x_2)$, which violates the condition for strict monotonicity.

Consider $x_3 = \frac{\pi}{4}$ and $x_4 = \frac{\pi}{2}$ in $(0, \pi)$, where $x_3 < x_4$.

$f(x_3) = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$

$f(x_4) = \cos(\frac{\pi}{2}) = 0$

Here $f(x_3) > f(x_4)$, which shows decreasing behavior.

Consider $x_5 = \pi$ and $x_6 = \frac{3\pi}{2}$ in $(\pi, 2\pi)$, where $x_5 < x_6$.

$f(x_5) = \cos(\pi) = -1$

$f(x_6) = \cos(\frac{3\pi}{2}) = 0$

Here $f(x_5) < f(x_6)$, which shows increasing behavior.

Because the function exhibits both decreasing and increasing behavior within the interval $(0, 2\pi)$, it is neither increasing nor decreasing on the entire interval.

---

Therefore, the function $f(x) = \cos x$ is decreasing in $(0, \pi)$, increasing in $(\pi, 2\pi)$, and neither increasing nor decreasing in $(0, 2\pi)$.

Given:

The function is $f(x) = x^2 - 4x + 6$.

---

To Find:

The intervals in which the function $f(x)$ is:

(a) increasing

(b) decreasing

---

Solution:

To find the intervals where the function is increasing or decreasing, we need to find its derivative $f'(x)$ and determine its sign.

Differentiate the function $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(x^2 - 4x + 6)$

$f'(x) = \frac{d}{dx}(x^2) - \frac{d}{dx}(4x) + \frac{d}{dx}(6)$

$f'(x) = 2x - 4 + 0$

To find the critical points where the function might change its increasing/decreasing nature, we set the derivative equal to zero:

$f'(x) = 0$

$2x - 4 = 0$

$2x = 4$

$x = \frac{4}{2}$

$x = 2$

This critical point $x=2$ divides the real number line into two disjoint intervals: $(-\infty, 2)$ and $(2, \infty)$. We will test the sign of $f'(x)$ in each interval.

(a) Intervals for increasing:

A function is increasing if $f'(x) > 0$.

From equation (i), $f'(x) = 2x - 4$. We need $2x - 4 > 0$ for the function to be increasing.

$2x > 4$

$x > \frac{4}{2}$

$x > 2$

This condition is satisfied for $x$ in the interval $(2, \infty)$.

Thus, the function $f(x)$ is strictly increasing on $(2, \infty)$. For an increasing function (non-strict), we consider the closed interval $[2, \infty)$.

(b) Intervals for decreasing:

A function is decreasing if $f'(x) < 0$.

From equation (i), $f'(x) = 2x - 4$. We need $2x - 4 < 0$ for the function to be decreasing.

$2x < 4$

$x < \frac{4}{2}$

$x < 2$

This condition is satisfied for $x$ in the interval $(-\infty, 2)$.

Thus, the function $f(x)$ is strictly decreasing on $(-\infty, 2)$. For a decreasing function (non-strict), we consider the closed interval $(-\infty, 2]$.

---

Based on the analysis of the sign of the derivative $f'(x)$:

(a) The function $f(x) = x^2 - 4x + 6$ is strictly increasing on the interval $(2, \infty)$. It is increasing on $[2, \infty)$.

(b) The function $f(x) = x^2 - 4x + 6$ is strictly decreasing on the interval $(-\infty, 2)$. It is decreasing on $(-\infty, 2]$.

Given:

The function is $f(x) = 4x^3 - 6x^2 - 72x + 30$.

---

To Find:

The intervals in which the function $f(x)$ is:

(a) increasing

(b) decreasing

---

Solution:

To find the intervals where the function is increasing or decreasing, we first find its derivative $f'(x)$ and determine its sign.

Differentiate the function $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(4x^3 - 6x^2 - 72x + 30)$

$f'(x) = \frac{d}{dx}(4x^3) - \frac{d}{dx}(6x^2) - \frac{d}{dx}(72x) + \frac{d}{dx}(30)$

$f'(x) = 4(3x^{3-1}) - 6(2x^{2-1}) - 72(1) + 0$

$f'(x) = 12x^2 - 12x - 72$

To find the critical points where the function might change its increasing/decreasing nature, we set the derivative equal to zero:

$f'(x) = 0$

$12x^2 - 12x - 72 = 0$

Divide the entire equation by 12:

$\frac{12x^2}{12} - \frac{12x}{12} - \frac{72}{12} = 0$

$x^2 - x - 6 = 0$

Factor the quadratic expression:

$(x - 3)(x + 2) = 0$

The critical points are the values of $x$ that make the derivative zero:

$x - 3 = 0 \implies x = 3$

$x + 2 = 0 \implies x = -2$

These critical points, $x = -2$ and $x = 3$, divide the real number line into three disjoint intervals: $(-\infty, -2)$, $(-2, 3)$, and $(3, \infty)$. We now determine the sign of $f'(x)$ in each of these intervals.

We can write $f'(x) = 12(x^2 - x - 6) = 12(x - 3)(x + 2)$.

Interval 1: $(-\infty, -2)$

Choose a test value, say $x = -3$.

$f'(-3) = 12((-3) - 3)((-3) + 2) = 12(-6)(-1) = 12(6) = 72$

Since $f'(x) > 0$ in this interval, the function is strictly increasing on $(-\infty, -2)$.

Interval 2: $(-2, 3)$

Choose a test value, say $x = 0$.

$f'(0) = 12((0) - 3)((0) + 2) = 12(-3)(2) = 12(-6) = -72$

Since $f'(x) < 0$ in this interval, the function is strictly decreasing on $(-2, 3)$.

Interval 3: $(3, \infty)$

Choose a test value, say $x = 4$.

$f'(4) = 12((4) - 3)((4) + 2) = 12(1)(6) = 12(6) = 72$

Since $f'(x) > 0$ in this interval, the function is strictly increasing on $(3, \infty)$.

For non-strict increasing/decreasing intervals, we include the critical points.

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Based on the sign analysis of $f'(x)$:

(a) Increasing intervals: The function $f(x)$ is increasing on the intervals where $f'(x) \ge 0$. This occurs on $(-\infty, -2]$ and $[3, \infty)$. We can write this as $(-\infty, -2] \cup [3, \infty)$.

(b) Decreasing intervals: The function $f(x)$ is decreasing on the intervals where $f'(x) \le 0$. This occurs on $[-2, 3]$.

Given:

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To Find:

Intervals where $f(x)$ is (a) increasing and (b) decreasing.

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Solution:

To find the intervals where the function is increasing or decreasing, we need to find the first derivative $f'(x)$ and analyze its sign.

The given function is $f(x) = \sin 3x$.

Differentiate $f(x)$ with respect to $x$. We use the chain rule.

$f'(x) = \frac{d}{dx}(\sin 3x)$

Let $u = 3x$. Then $\frac{du}{dx} = \frac{d}{dx}(3x) = 3$.

$\frac{d}{dx}(\sin u) = \cos u \cdot \frac{du}{dx}$

$f'(x) = \cos(3x) \cdot 3$

$f'(x) = 3 \cos 3x$

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Now, we find the critical points by setting $f'(x) = 0$.

$3 \cos 3x = 0$

$\cos 3x = 0$

The general solution for $\cos \theta = 0$ is $\theta = (2n + 1)\frac{\pi}{2}$ or $\theta = \frac{\pi}{2} + n\pi$, where $n$ is an integer.

So, $3x = \frac{\pi}{2} + n\pi$

$x = \frac{\pi}{6} + \frac{n\pi}{3}$

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We need to find the integer values of $n$ for which $x$ lies in the given interval $\left[ 0, \frac{\pi}{2} \right]$.

$0 \leq \frac{\pi}{6} + \frac{n\pi}{3} \leq \frac{\pi}{2}$

Divide by $\pi$ (since $\pi > 0$, the inequality signs do not change):

$0 \leq \frac{1}{6} + \frac{n}{3} \leq \frac{1}{2}$

Subtract $\frac{1}{6}$ from all parts:

$0 - \frac{1}{6} \leq \frac{n}{3} \leq \frac{1}{2} - \frac{1}{6}$

$-\frac{1}{6} \leq \frac{n}{3} \leq \frac{3}{6} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$

Multiply by 3 (since $3 > 0$, the inequality signs do not change):

$-\frac{1}{6} \times 3 \leq n \leq \frac{1}{3} \times 3$

$-\frac{3}{6} \leq n \leq 1$

$-\frac{1}{2} \leq n \leq 1$

Since $n$ must be an integer, the possible values for $n$ are $0$ and $1$.

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For $n=0$: $x = \frac{\pi}{6} + \frac{0\pi}{3} = \frac{\pi}{6}$. This is in the interval $\left[ 0, \frac{\pi}{2} \right]$.

For $n=1$: $x = \frac{\pi}{6} + \frac{1\pi}{3} = \frac{\pi}{6} + \frac{2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}$. This is in the interval $\left[ 0, \frac{\pi}{2} \right]$.

The critical points in the interval $\left[ 0, \frac{\pi}{2} \right]$ are $x = \frac{\pi}{6}$ and $x = \frac{\pi}{2}$. These points divide the interval into subintervals.

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The relevant subintervals within $\left[ 0, \frac{\pi}{2} \right]$ are $\left[ 0, \frac{\pi}{6} \right]$ and $\left[ \frac{\pi}{6}, \frac{\pi}{2} \right]$.

We now check the sign of $f'(x) = 3 \cos 3x$ in the open intervals $\left( 0, \frac{\pi}{6} \right)$ and $\left( \frac{\pi}{6}, \frac{\pi}{2} \right)$.

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Consider the interval $\left( 0, \frac{\pi}{6} \right)$. Choose a test value, say $x = \frac{\pi}{12}$.

$f'\left(\frac{\pi}{12}\right) = 3 \cos \left( 3 \cdot \frac{\pi}{12} \right) = 3 \cos \left( \frac{\pi}{4} \right) = 3 \cdot \frac{\sqrt{2}}{2}$.

Since $3 \cdot \frac{\sqrt{2}}{2} > 0$, $f'(x) > 0$ for $x \in \left( 0, \frac{\pi}{6} \right)$.

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Consider the interval $\left( \frac{\pi}{6}, \frac{\pi}{2} \right)$. Choose a test value, say $x = \frac{\pi}{4}$.

$f'\left(\frac{\pi}{4}\right) = 3 \cos \left( 3 \cdot \frac{\pi}{4} \right) = 3 \cos \left( \frac{3\pi}{4} \right) = 3 \cdot \left( -\frac{\sqrt{2}}{2} \right)$.

Since $3 \cdot \left( -\frac{\sqrt{2}}{2} \right) < 0$, $f'(x) < 0$ for $x \in \left( \frac{\pi}{6}, \frac{\pi}{2} \right)$.

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(a) The function $f(x)$ is increasing in the interval where $f'(x) \geq 0$. Based on our analysis, $f'(x) > 0$ on $\left(0, \frac{\pi}{6}\right)$, and $f'(x)=0$ at $x=0$ and $x=\frac{\pi}{6}$. Thus, $f(x)$ is increasing on $\left[ 0, \frac{\pi}{6} \right]$.

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(b) The function $f(x)$ is decreasing in the interval where $f'(x) \leq 0$. Based on our analysis, $f'(x) < 0$ on $\left(\frac{\pi}{6}, \frac{\pi}{2}\right)$, and $f'(x)=0$ at $x=\frac{\pi}{6}$ and $x=\frac{\pi}{2}$. Thus, $f(x)$ is decreasing on $\left[ \frac{\pi}{6}, \frac{\pi}{2} \right]$.

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Final Answer:

(a) The function is increasing in the interval $\boxed{\left[ 0, \frac{\pi}{6} \right]}$.

(b) The function is decreasing in the interval $\boxed{\left[ \frac{\pi}{6}, \frac{\pi}{2} \right]}$.

Given:

The function is $f(x) = \sin x + \cos x$.

The domain for $x$ is the closed interval $[0, 2\pi]$.

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To Find:

The intervals within $[0, 2\pi]$ where the function $f(x)$ is increasing or decreasing.

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Solution:

To find the intervals where the function is increasing or decreasing, we need to find its derivative $f'(x)$ and determine its sign within the given domain $[0, 2\pi]$.

Differentiate the function $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(\sin x + \cos x)$

$f'(x) = \frac{d}{dx}(\sin x) + \frac{d}{dx}(\cos x)$

To find the critical points within the interval $[0, 2\pi]$ where the function might change its increasing/decreasing nature, we set the derivative equal to zero:

$f'(x) = 0$

$\cos x - \sin x = 0$

$\cos x = \sin x$

Dividing by $\cos x$ (assuming $\cos x \neq 0$ at the critical points), we get:

$\frac{\sin x}{\cos x} = 1$

$\tan x = 1$

We need to find the values of $x$ in the interval $[0, 2\pi]$ where $\tan x = 1$. The general solution for $\tan x = 1$ is $x = n\pi + \frac{\pi}{4}$, where $n$ is an integer.

For $n=0$, $x = 0\pi + \frac{\pi}{4} = \frac{\pi}{4}$. This is in $[0, 2\pi]$.

For $n=1$, $x = 1\pi + \frac{\pi}{4} = \frac{4\pi + \pi}{4} = \frac{5\pi}{4}$. This is in $[0, 2\pi]$.

For $n=2$, $x = 2\pi + \frac{\pi}{4} = \frac{9\pi}{4}$. This is outside $[0, 2\pi]$.

The critical points within the interval $[0, 2\pi]$ are $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$.

These critical points divide the interval $[0, 2\pi]$ into three disjoint open sub-intervals: $\left(0, \frac{\pi}{4}\right)$, $\left(\frac{\pi}{4}, \frac{5\pi}{4}\right)$, and $\left(\frac{5\pi}{4}, 2\pi\right)$. We analyze the sign of $f'(x) = \cos x - \sin x$ in each of these intervals.

Interval 1: $\left(0, \frac{\pi}{4}\right)$

In this interval, $x$ is in the first quadrant, and $0 < x < \frac{\pi}{4}$. In the first quadrant, both $\cos x$ and $\sin x$ are positive. For $0 < x < \frac{\pi}{4}$, we know that $\cos x > \sin x$.

So, $f'(x) = \cos x - \sin x > 0$ in $\left(0, \frac{\pi}{4}\right)$.

Thus, $f(x)$ is strictly increasing on $\left(0, \frac{\pi}{4}\right)$.

Interval 2: $\left(\frac{\pi}{4}, \frac{5\pi}{4}\right)$

In this interval, $x$ ranges from the first quadrant (after $\frac{\pi}{4}$) through the second, third, and into the fourth quadrant (before $\frac{5\pi}{4}$). For $\frac{\pi}{4} < x < \frac{5\pi}{4}$, we know that $\cos x < \sin x$ (e.g., at $x=\frac{\pi}{2}$, $\cos x=0, \sin x=1$; at $x=\pi$, $\cos x=-1, \sin x=0$). So, $f'(x) = \cos x - \sin x < 0$ in $\left(\frac{\pi}{4}, \frac{5\pi}{4}\right)$.

Thus, $f(x)$ is strictly decreasing on $\left(\frac{\pi}{4}, \frac{5\pi}{4}\right)$.

Interval 3: $\left(\frac{5\pi}{4}, 2\pi\right)$

In this interval, $x$ is in the fourth quadrant, and $\frac{5\pi}{4} < x < 2\pi$. In the fourth quadrant, $\cos x$ is positive and $\sin x$ is negative. Also, for $\frac{5\pi}{4} < x < 2\pi$, we know that $\cos x > \sin x$.

So, $f'(x) = \cos x - \sin x > 0$ in $\left(\frac{5\pi}{4}, 2\pi\right)$.

Thus, $f(x)$ is strictly increasing on $\left(\frac{5\pi}{4}, 2\pi\right)$.

Considering the closed interval $[0, 2\pi]$ and the critical points where $f'(x)=0$, we can state the intervals for non-strict monotonicity:

The function $f(x)$ is increasing on the intervals where $f'(x) \ge 0$. This occurs on $\left[0, \frac{\pi}{4}\right]$ and $\left[\frac{5\pi}{4}, 2\pi\right]$.

The function $f(x)$ is decreasing on the interval where $f'(x) \le 0$. This occurs on $\left[\frac{\pi}{4}, \frac{5\pi}{4}\right]$.

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Based on the sign analysis of $f'(x)$:

The function $f(x) = \sin x + \cos x$ is increasing on $\left[0, \frac{\pi}{4}\right] \cup \left[\frac{5\pi}{4}, 2\pi\right]$.

The function $f(x) = \sin x + \cos x$ is decreasing on $\left[\frac{\pi}{4}, \frac{5\pi}{4}\right]$.

Given:

The function is $f(x) = 3x + 17$.

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To Prove:

The function $f(x)$ is strictly increasing on the set of real numbers $\mathbb{R}$.

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Solution:

A function $f(x)$ is strictly increasing on an interval if its derivative $f'(x)$ is positive for all values of $x$ in that interval.

We find the first derivative of the given function $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(3x + 17)$

Using the rules of differentiation:

$f'(x) = \frac{d}{dx}(3x) + \frac{d}{dx}(17)$

$f'(x) = 3 \cdot \frac{d}{dx}(x) + 0$

$f'(x) = 3 \cdot 1$

$f'(x) = 3$

We observe that the derivative $f'(x) = 3$ is a constant value.

Since $3 > 0$, the derivative $f'(x)$ is positive for all real values of $x$ (i.e., for all $x \in \mathbb{R}$).

Therefore, according to the condition for a strictly increasing function, $f(x) = 3x + 17$ is strictly increasing on $\mathbb{R}$.

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Hence, the function $f(x) = 3x + 17$ is strictly increasing on $\mathbb{R}$.

Given the function $f(x) = e^{2x}$.

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To show that the function $f(x)$ is strictly increasing on the set of real numbers $R$, we can examine the sign of its derivative $f'(x)$.

A function $f$ is strictly increasing on an interval if its derivative $f'(x)$ is strictly positive for all $x$ in that interval.

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Let's find the derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(e^{2x})$

Using the chain rule, we differentiate the outer function ($e^u$) and multiply by the derivative of the inner function ($u = 2x$):

$\frac{d}{dx}(e^{2x}) = e^{2x} \cdot \frac{d}{dx}(2x)$

$f'(x) = e^{2x} \cdot 2$

$f'(x) = 2e^{2x}$

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Now, let's analyze the sign of $f'(x)$ for all real numbers $x \in R$.

The exponential function $e^u$ is always positive for any real value of $u$. This is a fundamental property of the exponential function.

Therefore, $e^{2x} > 0$ for all $x \in R$.

Since $e^{2x}$ is positive and we are multiplying it by the positive constant 2, the product $2e^{2x}$ must also be positive.

So, $f'(x) = 2e^{2x} > 0$ for all $x \in R$.

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Since the derivative $f'(x)$ is strictly positive for all values of $x$ in the domain $R$, the function $f(x) = e^{2x}$ is strictly increasing on $R$.

Given the function $f(x) = \sin x$.

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To determine whether the function is increasing or decreasing in a given interval, we examine the sign of its derivative, $f'(x)$, in that interval.

The derivative of $f(x) = \sin x$ with respect to $x$ is:

$f'(x) = \frac{d}{dx}(\sin x) = \cos x$

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(a) Increasing in $\left( 0, \frac{\pi}{2} \right)$

Consider the interval $\left( 0, \frac{\pi}{2} \right)$.

In this interval, the values of $x$ are in the first quadrant.

The cosine function is positive in the first quadrant.

So, for all $x \in \left( 0, \frac{\pi}{2} \right)$, we have $\cos x > 0$.

Therefore, $f'(x) = \cos x > 0$ for all $x \in \left( 0, \frac{\pi}{2} \right)$.

Since the derivative is strictly positive in this interval, the function $f(x) = \sin x$ is strictly increasing in $\left( 0, \frac{\pi}{2} \right)$.

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(b) Decreasing in $\left( \frac{\pi}{2}, \pi \right)$

Consider the interval $\left( \frac{\pi}{2}, \pi \right)$.

In this interval, the values of $x$ are in the second quadrant.

The cosine function is negative in the second quadrant.

So, for all $x \in \left( \frac{\pi}{2}, \pi \right)$, we have $\cos x < 0$.

Therefore, $f'(x) = \cos x < 0$ for all $x \in \left( \frac{\pi}{2}, \pi \right)$.

Since the derivative is strictly negative in this interval, the function $f(x) = \sin x$ is strictly decreasing in $\left( \frac{\pi}{2}, \pi \right)$.

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(c) Neither increasing nor decreasing in $(0, \pi)$

Consider the interval $(0, \pi)$. This interval is the union of $(0, \frac{\pi}{2})$ and $(\frac{\pi}{2}, \pi)$, excluding the endpoint $\frac{\pi}{2}$.

From parts (a) and (b), we found that the function $f(x) = \sin x$ is strictly increasing in $(0, \frac{\pi}{2})$ and strictly decreasing in $(\frac{\pi}{2}, \pi)$.

Since the function is increasing in one part of the interval $(0, \pi)$ and decreasing in another part of the same interval, it is neither increasing nor decreasing over the entire interval $(0, \pi)$.

For a function to be strictly increasing or strictly decreasing over an interval, its derivative must maintain the same strict sign (either strictly positive or strictly negative) throughout that interval.

Given the function $f(x) = 2x^2 - 3x$.

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To find the intervals where the function is increasing or decreasing, we need to find the derivative of the function and determine its sign.

Calculate the derivative $f'(x)$:

$f'(x) = \frac{d}{dx}(2x^2 - 3x)$

$f'(x) = 2 \frac{d}{dx}(x^2) - 3 \frac{d}{dx}(x)$

$f'(x) = 2(2x) - 3(1)$

$f'(x) = 4x - 3$

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To find the critical points, we set $f'(x) = 0$:

$4x - 3 = 0$

$4x = 3$

$x = \frac{3}{4}$

The critical point $x = \frac{3}{4}$ divides the real number line into two intervals: $\left(-\infty, \frac{3}{4}\right)$ and $\left(\frac{3}{4}, \infty\right)$.

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Now, we analyze the sign of $f'(x) = 4x - 3$ in each interval.

(a) Increasing intervals:

A function is strictly increasing in an interval if $f'(x) > 0$ for all $x$ in that interval.

We need to find $x$ such that $4x - 3 > 0$:

$4x > 3$

$x > \frac{3}{4}$

So, $f'(x) > 0$ for $x \in \left(\frac{3}{4}, \infty\right)$.

Therefore, the function $f(x)$ is increasing in the interval $\left[\frac{3}{4}, \infty\right)$. (Note: For increasing/decreasing intervals, we include endpoints if the derivative is zero there and continuous, but for *strictly* increasing/decreasing, we use open intervals). Based on standard convention when asked for increasing, we use closed interval where the derivative is 0, if the function is continuous there.

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(b) Decreasing intervals:

A function is strictly decreasing in an interval if $f'(x) < 0$ for all $x$ in that interval.

We need to find $x$ such that $4x - 3 < 0$:

$4x < 3$

$x < \frac{3}{4}$

So, $f'(x) < 0$ for $x \in \left(-\infty, \frac{3}{4}\right)$.

Therefore, the function $f(x)$ is decreasing in the interval $\left(-\infty, \frac{3}{4}\right]$.

(Note: Similar to the increasing case, we include the endpoint $\frac{3}{4}$ for the decreasing interval).

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Summary of intervals:

The function $f(x) = 2x^2 - 3x$ is:

Increasing on $\left[\frac{3}{4}, \infty\right)$

Decreasing on $\left(-\infty, \frac{3}{4}\right]$

Given the function $f(x) = 2x^3 – 3x^2 – 36x + 7$.

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To find the intervals where the function is increasing or decreasing, we need to find the first derivative of the function, $f'(x)$.

Calculate the derivative $f'(x)$:

$f'(x) = \frac{d}{dx}(2x^3 – 3x^2 – 36x + 7)$

$f'(x) = 2 \frac{d}{dx}(x^3) - 3 \frac{d}{dx}(x^2) - 36 \frac{d}{dx}(x) + \frac{d}{dx}(7)$

$f'(x) = 2(3x^2) - 3(2x) - 36(1) + 0$

$f'(x) = 6x^2 - 6x - 36$

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To find the critical points, we set the derivative $f'(x)$ equal to zero and solve for $x$.

$6x^2 - 6x - 36 = 0$

Divide the entire equation by 6:

$\frac{6x^2}{6} - \frac{6x}{6} - \frac{36}{6} = \frac{0}{6}$

$x^2 - x - 6 = 0$

Now, we factor the quadratic expression. We look for two numbers that multiply to -6 and add up to -1. These numbers are -3 and 2.

$(x - 3)(x + 2) = 0$

Setting each factor equal to zero gives the critical points:

$x - 3 = 0 \Rightarrow x = 3$

$x + 2 = 0 \Rightarrow x = -2$

The critical points are $x = -2$ and $x = 3$. These points divide the real number line $(-\infty, \infty)$ into three open intervals:

$(-\infty, -2)$, $(-2, 3)$, and $(3, \infty)$.

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We now analyze the sign of $f'(x) = 6(x - 3)(x + 2)$ in each of these intervals to determine where the function is increasing ($f'(x) > 0$) and decreasing ($f'(x) < 0$).

Consider the intervals:

Interval 1: $(-\infty, -2)$

Choose a test value in this interval, for example, $x = -3$.

$f'(-3) = 6(-3 - 3)(-3 + 2) = 6(-6)(-1) = 36$

Since $f'(-3) = 36 > 0$, the derivative is positive in this interval.

Interval 2: $(-2, 3)$

Choose a test value in this interval, for example, $x = 0$.

$f'(0) = 6(0 - 3)(0 + 2) = 6(-3)(2) = -36$

Since $f'(0) = -36 < 0$, the derivative is negative in this interval.

Interval 3: $(3, \infty)$

Choose a test value in this interval, for example, $x = 4$.

$f'(4) = 6(4 - 3)(4 + 2) = 6(1)(6) = 36$

Since $f'(4) = 36 > 0$, the derivative is positive in this interval.

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Based on the sign analysis of $f'(x)$:

(a) Increasing intervals:

The function $f(x)$ is increasing where $f'(x) \ge 0$. This occurs in the intervals where $f'(x)$ is positive, including the critical points where $f'(x) = 0$.

$f'(x) > 0$ on $(-\infty, -2)$ and $(3, \infty)$.

$f'(x) = 0$ at $x = -2$ and $x = 3$.

Therefore, the function is increasing on the closed intervals including the critical points:

The function $f(x)$ is increasing on $(-\infty, -2] \cup [3, \infty)$.

(b) Decreasing intervals:

The function $f(x)$ is decreasing where $f'(x) \le 0$. This occurs in the interval where $f'(x)$ is negative, including the critical points where $f'(x) = 0$.

$f'(x) < 0$ on $(-2, 3)$.

$f'(x) = 0$ at $x = -2$ and $x = 3$.

Therefore, the function is decreasing on the closed interval including the critical points:

The function $f(x)$ is decreasing on $[-2, 3]$.

To find the intervals where a function $f(x)$ is strictly increasing or strictly decreasing, we need to find its first derivative $f'(x)$ and analyze its sign.

$f(x)$ is strictly increasing if $f'(x) > 0$.

$f(x)$ is strictly decreasing if $f'(x) < 0$.

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(a) $f(x) = x^2 + 2x – 5$

Find the derivative:

$f'(x) = \frac{d}{dx}(x^2 + 2x - 5) = 2x + 2$

Find critical points by setting $f'(x) = 0$:

$2x + 2 = 0$

$2x = -2$

$x = -1$

The critical point divides the real line into intervals $(-\infty, -1)$ and $(-1, \infty)$.

Analyze the sign of $f'(x) = 2x + 2$:

In $(-\infty, -1)$, take $x = -2$. $f'(-2) = 2(-2) + 2 = -4 + 2 = -2 < 0$.

In $(-1, \infty)$, take $x = 0$. $f'(0) = 2(0) + 2 = 2 > 0$.

Thus, $f'(x) < 0$ on $(-\infty, -1)$ and $f'(x) > 0$ on $(-1, \infty)$.

The function is strictly decreasing on $(-\infty, -1)$.

The function is strictly increasing on $(-1, \infty)$.

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(b) $f(x) = 10 – 6x – 2x^2$

Find the derivative:

$f'(x) = \frac{d}{dx}(10 - 6x - 2x^2) = -6 - 4x$

Find critical points by setting $f'(x) = 0$:

$-6 - 4x = 0$

$-4x = 6$

$x = -\frac{6}{4} = -\frac{3}{2}$

The critical point divides the real line into intervals $(-\infty, -\frac{3}{2})$ and $(-\frac{3}{2}, \infty)$.

Analyze the sign of $f'(x) = -6 - 4x$:

In $(-\infty, -\frac{3}{2})$, take $x = -2$. $f'(-2) = -6 - 4(-2) = -6 + 8 = 2 > 0$.

In $(-\frac{3}{2}, \infty)$, take $x = 0$. $f'(0) = -6 - 4(0) = -6 < 0$.

Thus, $f'(x) > 0$ on $(-\infty, -\frac{3}{2})$ and $f'(x) < 0$ on $(-\frac{3}{2}, \infty)$.

The function is strictly increasing on $(-\infty, -\frac{3}{2})$.

The function is strictly decreasing on $(-\frac{3}{2}, \infty)$.

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(c) $f(x) = –2x^3 – 9x^2 – 12x + 1$

Find the derivative:

$f'(x) = \frac{d}{dx}(-2x^3 - 9x^2 - 12x + 1) = -6x^2 - 18x - 12$

Find critical points by setting $f'(x) = 0$:

$-6x^2 - 18x - 12 = 0$

Divide by -6:

$x^2 + 3x + 2 = 0$

Factor the quadratic:

$(x + 1)(x + 2) = 0$

Critical points are $x = -1$ and $x = -2$. Ordered, they are $x = -2, x = -1$.

These points divide the real line into intervals $(-\infty, -2)$, $(-2, -1)$, and $(-1, \infty)$.

Analyze the sign of $f'(x) = -6(x + 1)(x + 2)$:

In $(-\infty, -2)$, take $x = -3$. $f'(-3) = -6(-3 + 1)(-3 + 2) = -6(-2)(-1) = -12 < 0$.

In $(-2, -1)$, take $x = -1.5$. $f'(-1.5) = -6(-1.5 + 1)(-1.5 + 2) = -6(-0.5)(0.5) = 1.5 > 0$.

In $(-1, \infty)$, take $x = 0$. $f'(0) = -6(0 + 1)(0 + 2) = -6(1)(2) = -12 < 0$.

Thus, $f'(x) < 0$ on $(-\infty, -2)$ and $(-1, \infty)$, and $f'(x) > 0$ on $(-2, -1)$.

The function is strictly decreasing on $(-\infty, -2) \cup (-1, \infty)$.

The function is strictly increasing on $(-2, -1)$.

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(d) $f(x) = 6 – 9x – x^2$

Find the derivative:

$f'(x) = \frac{d}{dx}(6 - 9x - x^2) = -9 - 2x$

Find critical points by setting $f'(x) = 0$:

$-9 - 2x = 0$

$-2x = 9$

$x = -\frac{9}{2}$

The critical point divides the real line into intervals $(-\infty, -\frac{9}{2})$ and $(-\frac{9}{2}, \infty)$.

Analyze the sign of $f'(x) = -9 - 2x$:

In $(-\infty, -\frac{9}{2})$, take $x = -5$. $f'(-5) = -9 - 2(-5) = -9 + 10 = 1 > 0$.

In $(-\frac{9}{2}, \infty)$, take $x = 0$. $f'(0) = -9 - 2(0) = -9 < 0$.

Thus, $f'(x) > 0$ on $(-\infty, -\frac{9}{2})$ and $f'(x) < 0$ on $(-\frac{9}{2}, \infty)$.

The function is strictly increasing on $(-\infty, -\frac{9}{2})$.

The function is strictly decreasing on $(-\frac{9}{2}, \infty)$.

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(e) $f(x) = (x + 1)^3 (x – 3)^3$

We can rewrite the function as $f(x) = ((x+1)(x-3))^3 = (x^2 - 2x - 3)^3$.

Find the derivative using the chain rule:

$f'(x) = 3(x^2 - 2x - 3)^2 \cdot \frac{d}{dx}(x^2 - 2x - 3)$

$f'(x) = 3(x^2 - 2x - 3)^2 \cdot (2x - 2)$

$f'(x) = 3(x - 3)^2 (x + 1)^2 \cdot 2(x - 1)$

$f'(x) = 6(x - 3)^2 (x + 1)^2 (x - 1)$

Find critical points by setting $f'(x) = 0$:

$6(x - 3)^2 (x + 1)^2 (x - 1) = 0$

This gives $x - 3 = 0$ or $x + 1 = 0$ or $x - 1 = 0$.

Critical points are $x = 3$, $x = -1$, and $x = 1$. Ordered, they are $x = -1, x = 1, x = 3$.

These points divide the real line into intervals $(-\infty, -1)$, $(-1, 1)$, $(1, 3)$, and $(3, \infty)$.

Analyze the sign of $f'(x) = 6(x - 3)^2 (x + 1)^2 (x - 1)$. Note that $(x - 3)^2 \ge 0$ and $(x + 1)^2 \ge 0$. The sign of $f'(x)$ depends mainly on the sign of $(x - 1)$.

In $(-\infty, -1)$, take $x = -2$. $x - 1 = -3 < 0$. Since the other factors squared are positive, $f'(-2) < 0$.

In $(-1, 1)$, take $x = 0$. $x - 1 = -1 < 0$. Since the other factors squared are positive, $f'(0) < 0$.

In $(1, 3)$, take $x = 2$. $x - 1 = 1 > 0$. Since the other factors squared are positive, $f'(2) > 0$.

In $(3, \infty)$, take $x = 4$. $x - 1 = 3 > 0$. Since the other factors squared are positive, $f'(4) > 0$.

Thus, $f'(x) < 0$ on $(-\infty, -1)$ and $(-1, 1)$, and $f'(x) > 0$ on $(1, 3)$ and $(3, \infty)$.

The function is strictly decreasing on $(-\infty, -1) \cup (-1, 1)$ which can be written as $(-\infty, 1)$.

The function is strictly increasing on $(1, 3) \cup (3, \infty)$ which can be written as $(1, \infty)$.

Given the function $y = f(x) = \log (1 + x) - \frac{2x}{2 + x}$ for $x > -1$.

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To show that the function is increasing throughout its domain, we need to find its derivative $f'(x)$ and show that $f'(x) \ge 0$ for all $x$ in the domain $(-\infty, -1)$. Wait, the domain is $x > -1$. The interval is $(-1, \infty)$.

Let's calculate the derivative $f'(x)$ with respect to $x$.

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The derivative of $\log(1 + x)$ using the chain rule is $\frac{1}{1+x} \cdot \frac{d}{dx}(1+x) = \frac{1}{1+x} \cdot 1 = \frac{1}{1+x}$.

The derivative of $\frac{2x}{2 + x}$ can be found using the quotient rule $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$.

Here, $u = 2x$ and $v = 2 + x$. So, $u' = 2$ and $v' = 1$.

$\frac{d}{dx}\left(\frac{2x}{2 + x}\right) = \frac{2(2 + x) - 2x(1)}{(2 + x)^2} = \frac{4 + 2x - 2x}{(2 + x)^2} = \frac{4}{(2 + x)^2}$.

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Now, we find the derivative of the given function $f(x)$:

$f'(x) = \frac{d}{dx}\left(\log (1 + x) - \frac{2x}{2 + x}\right)$

$f'(x) = \frac{1}{1 + x} - \frac{4}{(2 + x)^2}$

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To determine the sign of $f'(x)$, we combine the terms by finding a common denominator:

$f'(x) = \frac{(2 + x)^2 - 4(1 + x)}{(1 + x)(2 + x)^2}$

Expand the numerator:

Numerator $= (2^2 + 2 \cdot 2 \cdot x + x^2) - (4 \cdot 1 + 4 \cdot x)$

Numerator $= (4 + 4x + x^2) - (4 + 4x)$

Numerator $= 4 + 4x + x^2 - 4 - 4x$

Numerator $= x^2$

So, the derivative is:

$f'(x) = \frac{x^2}{(1 + x)(2 + x)^2}$

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Now, we analyze the sign of $f'(x)$ for $x > -1$ (the domain of the function).

For $x > -1$, we have:

1. The numerator $x^2$: For any real number $x$, $x^2 \ge 0$. $x^2 = 0$ only when $x = 0$.

2. The term $(1 + x)$: Since $x > -1$, $1 + x > 0$. This term is always positive in the domain.

3. The term $(2 + x)^2$: Since $x > -1$, $2 + x > 1$. Thus, $(2 + x)^2 > 1^2 = 1 > 0$. This term is always positive in the domain.

So, for all $x > -1$, the numerator is non-negative ($x^2 \ge 0$) and the denominator is positive ($(1 + x)(2 + x)^2 > 0$).

Therefore, $f'(x) = \frac{x^2}{(1 + x)(2 + x)^2} \ge 0$ for all $x > -1$.

The derivative $f'(x)$ is zero only at $x = 0$, which is a single point in the domain.

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Since $f'(x) \ge 0$ for all $x > -1$, the function $y = \log (1 + x) - \frac{2x}{2 + x}$ is an increasing function throughout its domain $(-\infty, -1)$. Oh, domain is $(-1, \infty)$, not $(-\infty, -1)$. The domain is $x > -1$, which is the interval $(-1, \infty)$.

Since $f'(x) \ge 0$ for all $x > -1$, the function $y = \log (1 + x) - \frac{2x}{2 + x}$ is an increasing function throughout its domain $(-1, \infty)$.

Given the function $y = f(x) = [x(x – 2)]^2$.

We can rewrite the function as $f(x) = (x^2 - 2x)^2$.

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To find the values of $x$ for which the function is increasing, we need to find the derivative $f'(x)$ and determine where $f'(x) \ge 0$.

Calculate the derivative $f'(x)$ using the chain rule:

$f'(x) = \frac{d}{dx}(x^2 - 2x)^2$

$f'(x) = 2(x^2 - 2x) \cdot \frac{d}{dx}(x^2 - 2x)$

$f'(x) = 2(x^2 - 2x) \cdot (2x - 2)$

$f'(x) = 2x(x - 2) \cdot 2(x - 1)$

$f'(x) = 4x(x - 1)(x - 2)$

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To find the critical points, we set the derivative $f'(x)$ equal to zero:

$4x(x - 1)(x - 2) = 0$

This equation is satisfied if any of the factors are zero:

$x = 0$ or $x - 1 = 0$ or $x - 2 = 0$

The critical points are $x = 0$, $x = 1$, and $x = 2$.

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These critical points divide the real number line into four open intervals:

$(-\infty, 0)$, $(0, 1)$, $(1, 2)$, and $(2, \infty)$.

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We now analyze the sign of $f'(x) = 4x(x - 1)(x - 2)$ in each of these intervals.

Interval $(-\infty, 0)$: Choose a test value, e.g., $x = -1$.

$f'(-1) = 4(-1)(-1 - 1)(-1 - 2) = 4(-1)(-2)(-3) = -24$. The sign is negative.

Interval $(0, 1)$: Choose a test value, e.g., $x = 0.5$.

$f'(0.5) = 4(0.5)(0.5 - 1)(0.5 - 2) = 4(0.5)(-0.5)(-1.5) = (2)(-0.5)(-1.5) = (-1)(-1.5) = 1.5$. The sign is positive.

Interval $(1, 2)$: Choose a test value, e.g., $x = 1.5$.

$f'(1.5) = 4(1.5)(1.5 - 1)(1.5 - 2) = 4(1.5)(0.5)(-0.5) = (6)(0.5)(-0.5) = (3)(-0.5) = -1.5$. The sign is negative.

Interval $(2, \infty)$: Choose a test value, e.g., $x = 3$.

$f'(3) = 4(3)(3 - 1)(3 - 2) = 4(3)(2)(1) = 24$. The sign is positive.

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The function $f(x)$ is increasing where $f'(x) \ge 0$.

Based on the sign analysis, $f'(x)$ is positive in the intervals $(0, 1)$ and $(2, \infty)$.

$f'(x)$ is zero at the critical points $x=0, x=1, x=2$.

Including the critical points where the derivative is zero, the function is increasing on the closed intervals.

The function $y = [x(x – 2)]^2$ is increasing for $x \in [0, 1] \cup [2, \infty)$.

Given the function $y = f(\theta) = \frac{4 \sin \theta}{(2 + \cos\theta)} - \theta$ for $\theta \in \left[ 0,\frac{\pi}{2} \right]$.

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To prove that the function is increasing in the given interval, we need to find its derivative with respect to $\theta$, denoted as $\frac{dy}{d\theta}$, and show that $\frac{dy}{d\theta} \ge 0$ for all $\theta \in \left[ 0,\frac{\pi}{2} \right]$.

Let's find the derivative of each term:

For the first term, $\frac{4 \sin \theta}{2 + \cos\theta}$, we use the quotient rule $\frac{d}{d\theta}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$, where $u = 4 \sin \theta$ and $v = 2 + \cos\theta$.

$u' = \frac{d}{d\theta}(4 \sin \theta) = 4 \cos \theta$

$v' = \frac{d}{d\theta}(2 + \cos\theta) = -\sin \theta$

So, the derivative of the first term is:

$\frac{d}{d\theta}\left(\frac{4 \sin \theta}{2 + \cos\theta}\right) = \frac{(4 \cos \theta)(2 + \cos\theta) - (4 \sin \theta)(-\sin \theta)}{(2 + \cos\theta)^2}$

$= \frac{8 \cos \theta + 4 \cos^2 \theta + 4 \sin^2 \theta}{(2 + \cos\theta)^2}$

Using the identity $\cos^2 \theta + \sin^2 \theta = 1$:

$= \frac{8 \cos \theta + 4(\cos^2 \theta + \sin^2 \theta)}{(2 + \cos\theta)^2} = \frac{8 \cos \theta + 4}{(2 + \cos\theta)^2}$

The derivative of the second term, $-\theta$, is $\frac{d}{d\theta}(-\theta) = -1$.

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Now, combine the derivatives to find $\frac{dy}{d\theta}$:

$\frac{dy}{d\theta} = \frac{8 \cos \theta + 4}{(2 + \cos\theta)^2} - 1$

Combine the terms by finding a common denominator:

$\frac{dy}{d\theta} = \frac{8 \cos \theta + 4 - (2 + \cos\theta)^2}{(2 + \cos\theta)^2}$

Expand the numerator:

Numerator $= 8 \cos \theta + 4 - (4 + 4 \cos \theta + \cos^2 \theta)$

Numerator $= 8 \cos \theta + 4 - 4 - 4 \cos \theta - \cos^2 \theta$

Numerator $= 4 \cos \theta - \cos^2 \theta$

Numerator $= \cos \theta (4 - \cos \theta)$

So, the derivative is:

$\frac{dy}{d\theta} = \frac{\cos \theta (4 - \cos \theta)}{(2 + \cos\theta)^2}$

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Now, let's analyze the sign of $\frac{dy}{d\theta}$ in the interval $\left[ 0,\frac{\pi}{2} \right]$.

Consider $\theta \in \left[ 0,\frac{\pi}{2} \right]$.

1. Denominator: $(2 + \cos\theta)^2$

In the interval $\left[ 0,\frac{\pi}{2} \right]$, $0 \le \cos \theta \le 1$.

So, $2 + \cos\theta$ is between $2+0=2$ and $2+1=3$. $2 \le 2 + \cos\theta \le 3$.

Thus, $(2 + \cos\theta)^2$ is between $2^2=4$ and $3^2=9$.

Therefore, $(2 + \cos\theta)^2 > 0$ for all $\theta \in \left[ 0,\frac{\pi}{2} \right]$.

2. Numerator: $\cos \theta (4 - \cos \theta)$

In the interval $\left[ 0,\frac{\pi}{2} \right]$:

- $\cos \theta$: $\cos \theta \ge 0$. $\cos \theta > 0$ for $\theta \in \left[ 0,\frac{\pi}{2} \right)$ and $\cos \theta = 0$ at $\theta = \frac{\pi}{2}$.

- $4 - \cos \theta$: Since $0 \le \cos \theta \le 1$, $4 - \cos \theta$ is between $4-1=3$ and $4-0=4$. $3 \le 4 - \cos \theta \le 4$.

Therefore, $4 - \cos \theta > 0$ for all $\theta \in \left[ 0,\frac{\pi}{2} \right]$.

Combining the terms in the numerator, we have the product of a non-negative term ($\cos \theta$) and a positive term ($4 - \cos \theta$).

So, $\cos \theta (4 - \cos \theta) \ge 0$ for all $\theta \in \left[ 0,\frac{\pi}{2} \right]$.

The numerator is zero only when $\cos \theta = 0$, which happens at $\theta = \frac{\pi}{2}$ in the given interval.

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Since the numerator is non-negative ($\ge 0$) and the denominator is positive ($> 0$) for all $\theta \in \left[ 0,\frac{\pi}{2} \right]$, the derivative $\frac{dy}{d\theta}$ is non-negative in this interval.

$\frac{dy}{d\theta} = \frac{\cos \theta (4 - \cos \theta)}{(2 + \cos\theta)^2} \ge 0$ for all $\theta \in \left[ 0,\frac{\pi}{2} \right]$.

The derivative is zero only at $\theta = \frac{\pi}{2}$.

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Since $\frac{dy}{d\theta} \ge 0$ for all $\theta \in \left[ 0,\frac{\pi}{2} \right]$, the function $y = \frac{4 \sin \theta}{(2 + \cos\theta)} - \theta$ is an increasing function in the interval $\left[ 0,\frac{\pi}{2} \right]$.

Let the logarithmic function be $f(x) = \log_b x$, where $b > 1$ is the base of the logarithm. The domain of the logarithmic function is $(0, \infty)$.

(Commonly, the logarithmic function refers to the natural logarithm, $f(x) = \ln x$, which has base $e \approx 2.718 > 1$. We will use $f(x) = \ln x$ for this proof.)

Given the function $f(x) = \ln x$.

The domain of $f(x)$ is $(0, \infty)$, which means $x > 0$.

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To prove that the function is increasing on $(0, \infty)$, we need to find its derivative $f'(x)$ and show that $f'(x) > 0$ for all $x$ in the domain $(0, \infty)$.

Calculate the derivative of $f(x) = \ln x$ with respect to $x$:

$f'(x) = \frac{d}{dx}(\ln x)$

$f'(x) = \frac{1}{x}$

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Now, we analyze the sign of the derivative $f'(x) = \frac{1}{x}$ on the domain $(0, \infty)$.

The domain is defined by $x > 0$.

For any value of $x$ such that $x > 0$, the reciprocal $\frac{1}{x}$ is also a positive number.

Therefore, for all $x \in (0, \infty)$, we have $f'(x) = \frac{1}{x} > 0$.

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Since the derivative $f'(x)$ is strictly positive for all $x$ in the interval $(0, \infty)$, the function $f(x) = \ln x$ is strictly increasing on $(0, \infty)$.

A strictly increasing function is also considered an increasing function.

Thus, the logarithmic function is increasing on $(0, \infty)$.

Given the function $f(x) = x^2 – x + 1$.

We are considering the interval $(-1, 1)$, which means $-1 < x < 1$.

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To determine if the function is strictly increasing or decreasing on the interval, we find its first derivative, $f'(x)$.

Calculate the derivative $f'(x)$:

$f'(x) = \frac{d}{dx}(x^2 - x + 1)$

$f'(x) = 2x - 1$

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To find the critical points, we set $f'(x) = 0$:

$2x - 1 = 0$

$2x = 1$

$x = \frac{1}{2}$

The critical point $x = \frac{1}{2}$ lies within the given interval $(-1, 1)$, since $-1 < \frac{1}{2} < 1$.

This critical point divides the interval $(-1, 1)$ into two sub-intervals: $(-1, \frac{1}{2})$ and $(\frac{1}{2}, 1)$.

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Now, we analyze the sign of $f'(x) = 2x - 1$ in each of these sub-intervals.

Sub-interval 1: $(-1, \frac{1}{2})$

Choose a test value in this interval, for example, $x = 0$.

$f'(0) = 2(0) - 1 = -1$

Since $f'(0) = -1 < 0$, the derivative is negative in this sub-interval.

Thus, $f(x)$ is strictly decreasing on $(-1, \frac{1}{2})$.

Sub-interval 2: $(\frac{1}{2}, 1)$

Choose a test value in this interval, for example, $x = 0.75 = \frac{3}{4}$.

$f'(\frac{3}{4}) = 2(\frac{3}{4}) - 1 = \frac{6}{4} - 1 = \frac{3}{2} - 1 = \frac{3 - 2}{2} = \frac{1}{2}$

Since $f'(\frac{3}{4}) = \frac{1}{2} > 0$, the derivative is positive in this sub-interval.

Thus, $f(x)$ is strictly increasing on $(\frac{1}{2}, 1)$.

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Since the function $f(x)$ is strictly decreasing on the interval $(-1, \frac{1}{2})$ and strictly increasing on the interval $(\frac{1}{2}, 1)$, it changes its behavior within the interval $(-1, 1)$.

Therefore, the function $f(x) = x^2 - x + 1$ is neither strictly increasing nor strictly decreasing on the entire interval $(-1, 1)$.

To determine if a function $f(x)$ is decreasing on an interval, we find its derivative $f'(x)$ and check if $f'(x) \le 0$ on that interval. For strict decreasing, we check if $f'(x) < 0$ (except possibly at isolated points).

The given interval is $\left( 0, \frac{\pi}{2} \right)$.

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(A) $f(x) = \cos x$

The derivative is $f'(x) = \frac{d}{dx}(\cos x) = -\sin x$.

For $x \in \left( 0, \frac{\pi}{2} \right)$, the sine function $\sin x$ is positive.

Therefore, $f'(x) = -\sin x < 0$ for all $x \in \left( 0, \frac{\pi}{2} \right)$.

Since the derivative is strictly negative, the function $f(x) = \cos x$ is strictly decreasing on $\left( 0, \frac{\pi}{2} \right)$.

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(B) $f(x) = \cos 2x$

The derivative is $f'(x) = \frac{d}{dx}(\cos 2x) = -\sin(2x) \cdot \frac{d}{dx}(2x) = -2 \sin(2x)$.

For $x \in \left( 0, \frac{\pi}{2} \right)$, the argument $2x$ is in the interval $\left( 2 \cdot 0, 2 \cdot \frac{\pi}{2} \right) = (0, \pi)$.

In the interval $(0, \pi)$, the sine function $\sin(2x)$ is positive.

Therefore, $f'(x) = -2 \sin(2x) < 0$ for all $x \in \left( 0, \frac{\pi}{2} \right)$.

Since the derivative is strictly negative, the function $f(x) = \cos 2x$ is strictly decreasing on $\left( 0, \frac{\pi}{2} \right)$.

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(C) $f(x) = \cos 3x$

The derivative is $f'(x) = \frac{d}{dx}(\cos 3x) = -\sin(3x) \cdot \frac{d}{dx}(3x) = -3 \sin(3x)$.

For $x \in \left( 0, \frac{\pi}{2} \right)$, the argument $3x$ is in the interval $\left( 3 \cdot 0, 3 \cdot \frac{\pi}{2} \right) = \left( 0, \frac{3\pi}{2} \right)$.

In the interval $\left( 0, \frac{3\pi}{2} \right)$, $\sin(3x)$ is positive for $3x \in (0, \pi)$ (i.e., $x \in (0, \frac{\pi}{3})$) and negative for $3x \in (\pi, \frac{3\pi}{2})$ (i.e., $x \in (\frac{\pi}{3}, \frac{\pi}{2})$).

For $x \in \left( 0, \frac{\pi}{3} \right)$, $\sin(3x) > 0$, so $f'(x) = -3 \sin(3x) < 0$. The function is strictly decreasing.

For $x \in \left( \frac{\pi}{3}, \frac{\pi}{2} \right)$, $\sin(3x) < 0$, so $f'(x) = -3 \sin(3x) > 0$. The function is strictly increasing.

Since the function is decreasing on a part of the interval and increasing on another part, it is neither increasing nor decreasing on the entire interval $\left( 0, \frac{\pi}{2} \right)$.

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(D) $f(x) = \tan x$

The derivative is $f'(x) = \frac{d}{dx}(\tan x) = \sec^2 x = \frac{1}{\cos^2 x}$.

For $x \in \left( 0, \frac{\pi}{2} \right)$, the cosine function $\cos x$ is positive, so $\cos^2 x$ is positive.

Therefore, $f'(x) = \frac{1}{\cos^2 x} > 0$ for all $x \in \left( 0, \frac{\pi}{2} \right)$.

Since the derivative is strictly positive, the function $f(x) = \tan x$ is strictly increasing on $\left( 0, \frac{\pi}{2} \right)$.

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Based on the analysis, the functions that are decreasing on $\left( 0, \frac{\pi}{2} \right)$ are those for which the derivative is $\le 0$ on the interval.

Options (A) and (B) have derivatives that are strictly negative throughout the interval $\left( 0, \frac{\pi}{2} \right)$, making them strictly decreasing (and thus decreasing) on this interval.

The correct options are (A) cos x and (B) cos 2x.

Given the function $f(x) = x^{100} + \sin x – 1$.

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To find the intervals where the function is decreasing, we need to find the derivative $f'(x)$ and determine where $f'(x) \le 0$.

Calculate the derivative $f'(x)$:

$f'(x) = \frac{d}{dx}(x^{100} + \sin x - 1)$

$f'(x) = 100x^{99} + \cos x$

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Now, we analyze the sign of $f'(x)$ in the given intervals.

(A) Interval $(0, 1)$

For $x \in (0, 1)$:

The term $x^{99}$ is positive because $x > 0$. So, $100x^{99} > 0$.

Since $0 < x < 1$, and $1$ radian is approximately $57.3^\circ$, the interval $(0, 1)$ is contained within the first quadrant $\left(0, \frac{\pi}{2}\right)$, where $\cos x > 0$.

Therefore, for $x \in (0, 1)$, $100x^{99} > 0$ and $\cos x > 0$.

$f'(x) = 100x^{99} + \cos x > 0$ for all $x \in (0, 1)$.

The function is strictly increasing on $(0, 1)$.

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(B) Interval $\left( \frac{\pi}{2},\pi \right)$

For $x \in \left( \frac{\pi}{2},\pi \right)$:

The term $x^{99}$ is positive because $x > \frac{\pi}{2} \approx 1.57 > 0$. So, $100x^{99} > 0$.

For $x \in \left( \frac{\pi}{2},\pi \right)$, $x$ is in the second quadrant, where $\cos x < 0$. The value of $\cos x$ is between -1 (exclusive at $\pi$) and 0 (exclusive at $\frac{\pi}{2}$).

So, $f'(x) = 100x^{99} + \cos x$. Here, $100x^{99}$ is positive and $\cos x$ is negative.

Let's consider the magnitude of the terms. For $x \in \left( \frac{\pi}{2},\pi \right)$, $x > \frac{\pi}{2} \approx 1.57$. Thus $x^{99} > (\frac{\pi}{2})^{99}$. $(\frac{\pi}{2})^{99}$ is a very large positive number.

The value of $|\cos x|$ is at most 1 for $x \in \left( \frac{\pi}{2},\pi \right)$.

So, $100x^{99}$ is a very large positive number, while $\cos x$ is a negative number between -1 and 0.

$f'(x) = 100x^{99} + \cos x > 100(\frac{\pi}{2})^{99} + (-1)$.

Since $100(\frac{\pi}{2})^{99}$ is much greater than 1, the sum $100(\frac{\pi}{2})^{99} - 1$ is positive.

Therefore, $f'(x) > 0$ for all $x \in \left( \frac{\pi}{2},\pi \right)$.

The function is strictly increasing on $\left( \frac{\pi}{2},\pi \right)$.

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(C) Interval $\left( 0,\frac{\pi}{2} \right)$

For $x \in \left( 0,\frac{\pi}{2} \right)$:

The term $x^{99}$ is positive because $x > 0$. So, $100x^{99} > 0$.

For $x \in \left( 0,\frac{\pi}{2} \right)$, $x$ is in the first quadrant, where $\cos x > 0$.

Therefore, for $x \in \left( 0,\frac{\pi}{2} \right)$, $100x^{99} > 0$ and $\cos x > 0$.

$f'(x) = 100x^{99} + \cos x > 0$ for all $x \in \left( 0,\frac{\pi}{2} \right)$.

The function is strictly increasing on $\left( 0,\frac{\pi}{2} \right)$.

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In all the given intervals (A), (B), and (C), the derivative $f'(x) = 100x^{99} + \cos x$ is strictly positive. This means the function is strictly increasing on these intervals.

Since the function is strictly increasing on these intervals, it is not decreasing on any of them.

Thus, the function is decreasing on none of the given intervals.

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The correct option is (D) None of these.

Given the function $f(x) = x^2 + ax + 1$.

We are given the interval $(1, 2)$.

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For a function to be increasing on an interval, its derivative must be greater than or equal to zero on that interval.

Find the derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(x^2 + ax + 1)$

$f'(x) = 2x + a$

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For $f(x)$ to be increasing on the interval $(1, 2)$, we require $f'(x) \ge 0$ for all $x \in (1, 2)$.

$2x + a \ge 0$ for all $x \in (1, 2)$

$a \ge -2x$ for all $x \in (1, 2)$

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The inequality $a \ge -2x$ must hold for every value of $x$ in the interval $(1, 2)$.

Consider the expression $-2x$ for $x \in (1, 2)$.

As $x$ increases in the interval $(1, 2)$, the value of $-2x$ decreases.

For $x = 1$, $-2x = -2(1) = -2$.

For $x$ approaching 2 from the left, $-2x$ approaches $-2(2) = -4$.

The range of values for $-2x$ on the interval $(1, 2)$ is $(-4, -2)$.

So, we need $a$ to be greater than or equal to every value in the interval $(-4, -2)$.

The largest value that $-2x$ approaches in the interval is $-2$ (as $x$ approaches 1).

Therefore, $a$ must be greater than or equal to the supremum (least upper bound) of the values of $-2x$ on the interval $(1, 2)$.

The supremum of $(-4, -2)$ is $-2$.

So, we need $a \ge -2$.

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Alternatively, consider the function $g(x) = -2x$. On the interval $(1, 2)$, $g(x)$ is a decreasing function.

The values of $-2x$ are in the interval $(-4, -2)$.

For $a \ge -2x$ to hold for all $x \in (1, 2)$, $a$ must be greater than or equal to the maximum value of $-2x$ over the interval $(1, 2)$. The maximum value of $-2x$ on $(1, 2)$ is approached as $x$ approaches 1, which is $-2$.

So, $a \ge -2$.

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The function $f(x) = x^2 + ax + 1$ is increasing on $(1, 2)$ if $f'(x) = 2x + a \ge 0$ for all $x \in (1, 2)$.

This inequality $2x + a \ge 0$ is equivalent to $a \ge -2x$.

For this to hold for all $x \in (1, 2)$, $a$ must be greater than or equal to the maximum value of $-2x$ on the interval $(1, 2)$.

The function $g(x) = -2x$ is decreasing on $(1, 2)$. Its values range from $-2(1) = -2$ (exclusive, as $x$ approaches 1) down to $-2(2) = -4$ (exclusive, as $x$ approaches 2).

The values of $-2x$ are in the interval $(-4, -2)$.

We need $a \ge \text{sup}(-4, -2)$.

$\text{sup}(-4, -2) = -2$.

Thus, $a \ge -2$.

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The values of $a$ for which the function $f(x) = x^2 + ax + 1$ is increasing on (1, 2) are $a \ge -2$.

Given the function $f(x) = x + \frac{1}{x}$.

The domain of the function is all real numbers except 0, i.e., $x \in (-\infty, 0) \cup (0, \infty)$.

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We are given an interval I that is disjoint from $[-1, 1]$.

Being disjoint from $[-1, 1]$ means the interval I must lie entirely within the set $(-\infty, -1) \cup (1, \infty)$.

Thus, I is a sub-interval of either $(-\infty, -1)$ or $(1, \infty)$.

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To prove that the function $f(x)$ is increasing on I, we find its derivative $f'(x)$ and check its sign on the intervals $(-\infty, -1)$ and $(1, \infty)$.

Calculate the derivative $f'(x)$:

$f'(x) = \frac{d}{dx}\left(x + \frac{1}{x}\right)$

$f'(x) = \frac{d}{dx}(x + x^{-1})$

$f'(x) = 1 - x^{-2}$

$f'(x) = 1 - \frac{1}{x^2}$

$f'(x) = \frac{x^2 - 1}{x^2}$

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Now, let's analyze the sign of $f'(x) = \frac{x^2 - 1}{x^2}$ for $x$ in the set $(-\infty, -1) \cup (1, \infty)$.

The denominator, $x^2$, is always positive for any $x \neq 0$. Since the interval I is disjoint from $[-1, 1]$, $x$ cannot be 0.

The sign of $f'(x)$ is determined solely by the sign of the numerator, $x^2 - 1$.

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Consider the case where I is a sub-interval of $(-\infty, -1)$.

For any $x \in (-\infty, -1)$, we have $x < -1$. Squaring both sides of an inequality with negative numbers reverses the inequality sign, so $x^2 > (-1)^2 = 1$.

Therefore, $x^2 - 1 > 0$ for all $x \in (-\infty, -1)$.

Since the numerator ($x^2 - 1$) is positive and the denominator ($x^2$) is positive, $f'(x) = \frac{x^2 - 1}{x^2} > 0$ for all $x \in (-\infty, -1)$.

If I $\subseteq (-\infty, -1)$, then $f'(x) > 0$ for all $x \in I$.

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Consider the case where I is a sub-interval of $(1, \infty)$.

For any $x \in (1, \infty)$, we have $x > 1$. Squaring both sides of an inequality with positive numbers maintains the inequality sign, so $x^2 > 1^2 = 1$.

Therefore, $x^2 - 1 > 0$ for all $x \in (1, \infty)$.

Since the numerator ($x^2 - 1$) is positive and the denominator ($x^2$) is positive, $f'(x) = \frac{x^2 - 1}{x^2} > 0$ for all $x \in (1, \infty)$.

If I $\subseteq (1, \infty)$, then $f'(x) > 0$ for all $x \in I$.

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In both possible cases for the interval I (being a sub-interval of $(-\infty, -1)$ or $(1, \infty)$), the derivative $f'(x)$ is strictly positive for all $x \in I$.

Since $f'(x) > 0$ for all $x \in I$, the function $f(x) = x + \frac{1}{x}$ is strictly increasing on I.

A strictly increasing function is also considered an increasing function.

Thus, the function $f(x) = x + \frac{1}{x}$ is increasing on any interval I disjoint from [-1, 1].

Given the function $f(x) = \log (\sin x)$.

The domain of $\log u$ is $u > 0$. So, for $f(x)$ to be defined, we must have $\sin x > 0$.

In the interval $(0, \pi)$, $\sin x > 0$, so the function is defined on $(0, \pi)$. We are asked to prove its behavior on sub-intervals of this domain.

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To determine whether the function is increasing or decreasing on an interval, we find its derivative $f'(x)$ and check its sign in that interval.

Calculate the derivative of $f(x)$ with respect to $x$ using the chain rule:

$f'(x) = \frac{d}{dx}(\log (\sin x))$

Let $u = \sin x$. Then $f(x) = \log u$.

$f'(x) = \frac{d}{du}(\log u) \cdot \frac{du}{dx}$

$f'(x) = \frac{1}{u} \cdot \frac{d}{dx}(\sin x)$

$f'(x) = \frac{1}{\sin x} \cdot \cos x$

$f'(x) = \frac{\cos x}{\sin x} = \cot x$

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Now, we analyze the sign of $f'(x) = \cot x$ in the given intervals.

Interval $\left( 0,\frac{\pi}{2} \right)$

For $x \in \left( 0,\frac{\pi}{2} \right)$, $x$ is in the first quadrant.

In the first quadrant, both $\cos x$ and $\sin x$ are positive.

$\cos x > 0$ for $x \in \left( 0,\frac{\pi}{2} \right)$.

$\sin x > 0$ for $x \in \left( 0,\frac{\pi}{2} \right)$.

Therefore, $f'(x) = \cot x = \frac{\cos x}{\sin x} > 0$ for all $x \in \left( 0,\frac{\pi}{2} \right)$.

Since the derivative is strictly positive on this interval, the function $f(x) = \log (\sin x)$ is increasing on $\left( 0,\frac{\pi}{2} \right)$.

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Interval $\left( \frac{\pi}{2},\pi \right)$

For $x \in \left( \frac{\pi}{2},\pi \right)$, $x$ is in the second quadrant.

In the second quadrant, $\cos x$ is negative and $\sin x$ is positive.

$\cos x < 0$ for $x \in \left( \frac{\pi}{2},\pi \right)$.

$\sin x > 0$ for $x \in \left( \frac{\pi}{2},\pi \right)$.

Therefore, $f'(x) = \cot x = \frac{\cos x}{\sin x} < 0$ for all $x \in \left( \frac{\pi}{2},\pi \right)$.

Since the derivative is strictly negative on this interval, the function $f(x) = \log (\sin x)$ is decreasing on $\left( \frac{\pi}{2},\pi \right)$.

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We have shown that $f'(x) > 0$ on $\left( 0,\frac{\pi}{2} \right)$ and $f'(x) < 0$ on $\left( \frac{\pi}{2},\pi \right)$.

Hence, the function $f(x) = \log (\sin x)$ is increasing on $\left( 0,\frac{\pi}{2} \right)$ and decreasing on $\left( \frac{\pi}{2},\pi \right)$.

Given the function $f(x) = \log |\cos x|$. The base of the logarithm is assumed to be $e$ (natural logarithm).

The domain of $f(x)$ requires $|\cos x| > 0$, which means $\cos x \neq 0$. This is true for the given intervals $\left( 0,\frac{\pi}{2} \right)$ and $\left( \frac{3\pi}{2}, 2 \pi \right)$.

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To determine if the function is increasing or decreasing on an interval, we find its derivative $f'(x)$ and check its sign in that interval.

Calculate the derivative of $f(x)$ with respect to $x$ using the chain rule. The derivative of $\log |u|$ with respect to $u$ is $\frac{1}{u}$.

Let $u = \cos x$. Then $\frac{du}{dx} = -\sin x$.

$f'(x) = \frac{d}{dx}(\log |\cos x|)$

$f'(x) = \frac{1}{\cos x} \cdot \frac{d}{dx}(\cos x)$

$f'(x) = \frac{1}{\cos x} \cdot (-\sin x)$

$f'(x) = -\frac{\sin x}{\cos x}$

$f'(x) = -\tan x$

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Now, we analyze the sign of $f'(x) = -\tan x$ in the given intervals.

Interval $\left( 0,\frac{\pi}{2} \right)$

For $x \in \left( 0,\frac{\pi}{2} \right)$, $x$ is in the first quadrant.

In the first quadrant, $\tan x$ is positive ($\tan x > 0$).

Therefore, $f'(x) = -\tan x$ is negative for all $x \in \left( 0,\frac{\pi}{2} \right)$.

Since $f'(x) < 0$ for all $x \in \left( 0,\frac{\pi}{2} \right)$, the function $f(x) = \log |\cos x|$ is strictly decreasing on $\left( 0,\frac{\pi}{2} \right)$. A strictly decreasing function is also a decreasing function.

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Interval $\left( \frac{3\pi}{2}, 2 \pi \right)$

For $x \in \left( \frac{3\pi}{2}, 2 \pi \right)$, $x$ is in the fourth quadrant.

In the fourth quadrant, $\cos x$ is positive and $\sin x$ is negative.

$\tan x = \frac{\sin x}{\cos x}$ is negative in the fourth quadrant ($\tan x < 0$).

Therefore, $f'(x) = -\tan x$. Since $\tan x < 0$, $-\tan x$ is positive.

$f'(x) = -\tan x > 0$ for all $x \in \left( \frac{3\pi}{2}, 2 \pi \right)$.

Since $f'(x) > 0$ for all $x \in \left( \frac{3\pi}{2}, 2 \pi \right)$, the function $f(x) = \log |\cos x|$ is strictly increasing on $\left( \frac{3\pi}{2}, 2 \pi \right)$. A strictly increasing function is also an increasing function.

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We have shown that $f'(x) = -\tan x < 0$ on $\left( 0,\frac{\pi}{2} \right)$ and $f'(x) = -\tan x > 0$ on $\left( \frac{3\pi}{2}, 2 \pi \right)$.

Hence, the function $f(x) = \log |\cos x|$ is decreasing on $\left( 0,\frac{\pi}{2} \right)$ and increasing on $\left( \frac{3\pi}{2}, 2 \pi \right)$.

Given the function $f(x) = x^3 – 3x^2 + 3x – 100$.

We need to prove that this function is increasing in $R$, which is the set of all real numbers $(-\infty, \infty)$.

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To prove that the function is increasing on $R$, we find its derivative $f'(x)$ and show that $f'(x) \ge 0$ for all $x \in R$.

Calculate the derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 3x - 100)$

$f'(x) = \frac{d}{dx}(x^3) - 3\frac{d}{dx}(x^2) + 3\frac{d}{dx}(x) - \frac{d}{dx}(100)$

$f'(x) = 3x^2 - 3(2x) + 3(1) - 0$

$f'(x) = 3x^2 - 6x + 3$

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Now, we analyze the sign of $f'(x) = 3x^2 - 6x + 3$ for all $x \in R$.

We can factor the expression for $f'(x)$:

$f'(x) = 3(x^2 - 2x + 1)$

The expression inside the parentheses is a perfect square trinomial:

$x^2 - 2x + 1 = (x - 1)^2$

So, the derivative is:

$f'(x) = 3(x - 1)^2$

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Now, consider the sign of $f'(x) = 3(x - 1)^2$ for all real numbers $x$.

For any real number $x$, the term $(x - 1)$ is a real number. The square of any real number is always non-negative.

Therefore, $(x - 1)^2 \ge 0$ for all $x \in R$.

Since we are multiplying $(x - 1)^2$ by the positive constant 3, the product is also non-negative.

$f'(x) = 3(x - 1)^2 \ge 0$ for all $x \in R$.

The derivative $f'(x)$ is equal to zero only when $(x - 1)^2 = 0$, which means $x - 1 = 0$, so $x = 1$. This is an isolated point where the derivative is zero.

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Since $f'(x) \ge 0$ for all $x \in R$, and $f'(x) = 0$ only at an isolated point $x=1$, the function $f(x)$ is strictly increasing on $R$.

A strictly increasing function is also considered an increasing function.

Thus, the function $f(x) = x^3 – 3x^2 + 3x – 100$ is increasing in $R$.

Given the function $y = f(x) = x^2 e^{-x}$.

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To find the interval where the function is increasing, we need to find the derivative $f'(x)$ and determine where $f'(x) \ge 0$.

Calculate the derivative $f'(x)$ using the product rule.

Let $u = x^2$ and $v = e^{-x}$. Then $\frac{du}{dx} = 2x$ and $\frac{dv}{dx} = -e^{-x}$.

$f'(x) = \frac{d}{dx}(x^2 e^{-x})$

$f'(x) = \frac{du}{dx} \cdot v + u \cdot \frac{dv}{dx}$

$f'(x) = (2x)(e^{-x}) + (x^2)(-e^{-x})$

$f'(x) = 2xe^{-x} - x^2e^{-x}$

Factor out the common term $xe^{-x}$:

$f'(x) = xe^{-x}(2 - x)$

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To find the critical points, we set $f'(x) = 0$:

$xe^{-x}(2 - x) = 0$

Since the exponential term $e^{-x}$ is always positive for any real $x$, the equation holds if and only if $x(2 - x) = 0$.

This implies $x = 0$ or $2 - x = 0$, which gives $x = 0$ or $x = 2$.

The critical points are $x = 0$ and $x = 2$. These points divide the real number line into three intervals: $(-\infty, 0)$, $(0, 2)$, and $(2, \infty)$.

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Now, we analyze the sign of $f'(x) = xe^{-x}(2 - x)$ in each interval. Note that $e^{-x} > 0$ for all real $x$. The sign of $f'(x)$ is determined by the sign of $x(2 - x)$.

Interval $(-\infty, 0)$: Choose a test value, e.g., $x = -1$.

$x = -1$ (negative)

$2 - x = 2 - (-1) = 3$ (positive)

$x(2 - x) = (-)(+) = (-)$. Thus, $f'(x) < 0$ in this interval.

Interval $(0, 2)$: Choose a test value, e.g., $x = 1$.

$x = 1$ (positive)

$2 - x = 2 - 1 = 1$ (positive)

$x(2 - x) = (+)(+) = (+)$. Thus, $f'(x) > 0$ in this interval.

Interval $(2, \infty)$: Choose a test value, e.g., $x = 3$.

$x = 3$ (positive)

$2 - x = 2 - 3 = -1$ (negative)

$x(2 - x) = (+)(-) = (-)$. Thus, $f'(x) < 0$ in this interval.

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The function $y = f(x)$ is increasing where $f'(x) \ge 0$.

Based on the sign analysis, $f'(x) > 0$ on the interval $(0, 2)$.

$f'(x) = 0$ at $x = 0$ and $x = 2$.

The function is strictly increasing on $(0, 2)$ and increasing on $[0, 2]$.

Among the given options, $(0, 2)$ is the interval where $f'(x)$ is strictly positive.

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The interval in which $y = x^2 e^{-x}$ is increasing is (0, 2).

This corresponds to option (D).

Given the function $f(x) = x^2$ for $x \in R$.

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To find the maximum and minimum values, we first find the critical points by calculating the derivative $f'(x)$ and setting it to zero.

$f'(x) = \frac{d}{dx}(x^2) = 2x$

Set $f'(x) = 0$ to find critical points:

$2x = 0$

$x = 0$

So, $x = 0$ is the only critical point.

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Now, we analyze the behavior of the function around the critical point $x = 0$. We can use the first derivative test.

For $x < 0$, $f'(x) = 2x < 0$. This means $f(x)$ is decreasing for $x < 0$.

For $x > 0$, $f'(x) = 2x > 0$. This means $f(x)$ is increasing for $x > 0$.

Since the function changes from decreasing to increasing at $x = 0$, this point corresponds to a local minimum.

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Evaluate the function at the local minimum:

$f(0) = (0)^2 = 0$

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Now, let's check for global maximum and minimum values.

Since $x^2 \ge 0$ for all $x \in R$, the smallest possible value of $f(x) = x^2$ is 0, which occurs at $x = 0$. Thus, the minimum value of the function is 0.

As $x \to \infty$, $f(x) = x^2 \to \infty$.

As $x \to -\infty$, $f(x) = x^2 \to \infty$.

Since the function values increase without bound as $x$ moves away from 0 in either direction, there is no largest value that the function attains.

Therefore, the function has no maximum value.

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Conclusion:

The minimum value of the function $f(x) = x^2$ is $f(0) = 0$.

The function has no maximum value.

Given the function $f(x) = |x|$ for $x \in R$, the set of all real numbers.

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The definition of the absolute value function is:

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By the definition of absolute value, we know that $|x| \ge 0$ for every real number $x$.

This means that the function $f(x) = |x|$ is always greater than or equal to 0.

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The smallest possible value of $|x|$ is 0.

This minimum value occurs when $x = 0$, since $f(0) = |0| = 0$.

For any other value of $x$, $|x| > 0$.

Thus, the minimum value of the function $f(x) = |x|$ is 0, and it is attained at $x = 0$.

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Now, let's consider if the function has a maximum value.

As $x$ takes larger and larger positive values, $f(x) = |x| = x$ also takes larger and larger positive values (e.g., $f(10) = 10, f(100) = 100$).

As $x$ takes larger and larger negative values (in magnitude), $f(x) = |x| = -x$ takes larger and larger positive values (e.g., $f(-10) = |-10| = 10, f(-100) = |-100| = 100$).

The value of $f(x) = |x|$ can become arbitrarily large. For any large number $M$, we can always find an $x$ (e.g., $x=M$) such that $f(x) = M$.

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Since the function values can increase without bound, there is no single largest value that the function attains.

Therefore, the function $f(x) = |x|$ has no maximum value.

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Conclusion:

The minimum value of the function $f(x) = |x|$ is 0.

The function has no maximum value.

Given the function $f(x) = x$ on the interval $(0, 1)$.

The domain for this function is the open interval $(0, 1)$, which means $0 < x < 1$.

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The function $f(x) = x$ is a simple identity function. For any value of $x$ in the domain, the function value $f(x)$ is equal to $x$.

Since the domain is the interval $(0, 1)$, the values of the function $f(x)$ are also within the interval $(0, 1)$.

That is, for all $x \in (0, 1)$, we have $0 < f(x) < 1$.

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Consider the lower bound of the interval $(0, 1)$. The values of $x$ are strictly greater than 0. As $x$ approaches 0 (from the right), $f(x) = x$ approaches 0.

However, the function $f(x) = x$ never actually takes the value 0 within the open interval $(0, 1)$, because $x$ is strictly greater than 0.

Since the function never attains its greatest lower bound (infimum) on the interval, there is no minimum value.

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Consider the upper bound of the interval $(0, 1)$. The values of $x$ are strictly less than 1. As $x$ approaches 1 (from the left), $f(x) = x$ approaches 1.

However, the function $f(x) = x$ never actually takes the value 1 within the open interval $(0, 1)$, because $x$ is strictly less than 1.

Since the function never attains its least upper bound (supremum) on the interval, there is no maximum value.

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Conclusion:

For the function $f(x) = x$ on the open interval $(0, 1)$, there is no maximum value and no minimum value.

Given the function $f(x) = x^3 – 3x + 3$.

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To find the points of local maxima and local minima, we first find the derivative of the function and the critical points.

Calculate the derivative $f'(x)$:

$f'(x) = \frac{d}{dx}(x^3 - 3x + 3)$

$f'(x) = 3x^2 - 3$

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To find the critical points, we set $f'(x) = 0$ and solve for $x$:

$3x^2 - 3 = 0$

$3(x^2 - 1) = 0$}

$x^2 - 1 = 0$}

$(x - 1)(x + 1) = 0$

This gives the critical points $x = 1$ and $x = -1$.

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We can use the first derivative test to determine if these critical points are local maxima or minima.

The critical points $x = -1$ and $x = 1$ divide the real number line into three intervals: $(-\infty, -1)$, $(-1, 1)$, and $(1, \infty)$.

We analyze the sign of $f'(x) = 3x^2 - 3 = 3(x^2 - 1)$ in each interval.

Interval $(-\infty, -1)$: Choose a test value, e.g., $x = -2$.

$f'(-2) = 3((-2)^2 - 1) = 3(4 - 1) = 3(3) = 9$.

Since $f'(-2) > 0$, $f'(x) > 0$ on $(-\infty, -1)$. The function is increasing in this interval.

Interval $(-1, 1)$: Choose a test value, e.g., $x = 0$.

$f'(0) = 3((0)^2 - 1) = 3(0 - 1) = 3(-1) = -3$.

Since $f'(0) < 0$, $f'(x) < 0$ on $(-1, 1)$. The function is decreasing in this interval.

Interval $(1, \infty)$: Choose a test value, e.g., $x = 2$.

$f'(2) = 3((2)^2 - 1) = 3(4 - 1) = 3(3) = 9$.

Since $f'(2) > 0$, $f'(x) > 0$ on $(1, \infty)$. The function is increasing in this interval.

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Applying the first derivative test:

At $x = -1$, the derivative $f'(x)$ changes sign from positive (on the left of -1) to negative (on the right of -1). Therefore, $x = -1$ is a point of local maximum.

The local maximum value is $f(-1) = (-1)^3 - 3(-1) + 3 = -1 + 3 + 3 = 5$.

At $x = 1$, the derivative $f'(x)$ changes sign from negative (on the left of 1) to positive (on the right of 1). Therefore, $x = 1$ is a point of local minimum.

The local minimum value is $f(1) = (1)^3 - 3(1) + 3 = 1 - 3 + 3 = 1$.

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Alternate Method (Using Second Derivative Test)

Find the second derivative $f''(x)$:

$f'(x) = 3x^2 - 3$

$f''(x) = \frac{d}{dx}(3x^2 - 3) = 6x$

Evaluate $f''(x)$ at the critical points $x = -1$ and $x = 1$.

At $x = -1$: $f''(-1) = 6(-1) = -6$.

Since $f''(-1) < 0$, $x = -1$ is a point of local maximum.

The local maximum value is $f(-1) = 5$.

At $x = 1$: $f''(1) = 6(1) = 6$.

Since $f''(1) > 0$, $x = 1$ is a point of local minimum.

The local minimum value is $f(1) = 1$.

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Conclusion:

The function has a local maximum at $x = -1$ with a value of 5.

The function has a local minimum at $x = 1$ with a value of 1.

Given the function $f(x) = 2x^3 – 6x^2 + 6x + 5$.

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To find the points of local maxima and local minima, we first find the derivative of the function and the critical points.

Calculate the derivative $f'(x)$:

$f'(x) = \frac{d}{dx}(2x^3 - 6x^2 + 6x + 5)$

$f'(x) = 2\frac{d}{dx}(x^3) - 6\frac{d}{dx}(x^2) + 6\frac{d}{dx}(x) + \frac{d}{dx}(5)$

$f'(x) = 2(3x^2) - 6(2x) + 6(1) + 0$

$f'(x) = 6x^2 - 12x + 6$

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To find the critical points, we set $f'(x) = 0$ and solve for $x$:

$6x^2 - 12x + 6 = 0$

Factor out the common factor of 6:

$6(x^2 - 2x + 1) = 0$

The expression inside the parentheses is a perfect square trinomial:

$x^2 - 2x + 1 = (x - 1)^2$

So, the equation becomes:

$6(x - 1)^2 = 0$

This equation is satisfied when $(x - 1)^2 = 0$, which means $x - 1 = 0$.

The only critical point is $x = 1$.

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We use the first derivative test to classify the critical point $x = 1$. We analyze the sign of $f'(x) = 6(x - 1)^2$ around $x = 1$.

For $x < 1$, let's choose $x = 0$.

$f'(0) = 6(0 - 1)^2 = 6(-1)^2 = 6(1) = 6$.

Since $f'(0) > 0$, $f'(x) > 0$ for $x < 1$. The function is increasing on $(-\infty, 1)$.

For $x > 1$, let's choose $x = 2$.

$f'(2) = 6(2 - 1)^2 = 6(1)^2 = 6(1) = 6$.

Since $f'(2) > 0$, $f'(x) > 0$ for $x > 1$. The function is increasing on $(1, \infty)$.

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Since the sign of $f'(x)$ does not change as $x$ passes through the critical point $x = 1$ (it remains positive on both sides), the point $x = 1$ is neither a point of local maximum nor a point of local minimum.

It is a point of inflection with a horizontal tangent.

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Alternatively, using the Second Derivative Test:

The second derivative is $f''(x) = \frac{d}{dx}(6x^2 - 12x + 6) = 12x - 12$.

Evaluate $f''(x)$ at the critical point $x = 1$:

$f''(1) = 12(1) - 12 = 12 - 12 = 0$.

Since $f''(1) = 0$, the second derivative test is inconclusive. We must rely on the first derivative test, which we did above.

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Conclusion:

The function $f(x) = 2x^3 – 6x^2 + 6x + 5$ has no points of local maxima or local minima.

Given the function $f(x) = 3 + |x|$ for $x \in R$, the set of all real numbers.

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The absolute value function $|x|$ is defined as:

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A key property of the absolute value function is that $|x| \ge 0$ for all real numbers $x$.

The minimum value of $|x|$ is 0, which occurs when $x = 0$.

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Now, consider the function $f(x) = 3 + |x|$.

Since $|x| \ge 0$, we have $3 + |x| \ge 3 + 0$.

So, $f(x) \ge 3$ for all $x \in R$.

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The smallest value that $f(x)$ can take is 3.

This value is attained when $|x|$ is at its minimum, which is at $x = 0$.

$f(0) = 3 + |0| = 3 + 0 = 3$.

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For any $x$ in a neighborhood around 0 (e.g., $(- \delta, \delta)$ for some small $\delta > 0$), $x \neq 0$ implies $|x| > 0$. Thus, $f(x) = 3 + |x| > 3$.

Since $f(0) = 3$ and $f(x) > 3$ for $x$ close to 0 but not equal to 0, $x=0$ is a point of local minimum.

The value of the function at this point is the local minimum value.

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The local minimum value of the function $f(x) = 3 + |x|$ is $f(0) = 3$.

Note that this is also the global minimum value of the function.

Given the function $f(x) = 3x^4 + 4x^3 – 12x^2 + 12$.

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To find the local maximum and local minimum values, we first find the derivative of the function and the critical points.

Calculate the first derivative $f'(x)$:

$f'(x) = \frac{d}{dx}(3x^4 + 4x^3 - 12x^2 + 12)$

$f'(x) = 12x^3 + 12x^2 - 24x$

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To find the critical points, we set $f'(x) = 0$ and solve for $x$:

$12x^3 + 12x^2 - 24x = 0$

Factor out the common term $12x$:

$12x(x^2 + x - 2) = 0$

Factor the quadratic expression $x^2 + x - 2$: We look for two numbers that multiply to -2 and add to 1. These are 2 and -1.

$x^2 + x - 2 = (x + 2)(x - 1)$

So, the factored equation is:

$12x(x + 2)(x - 1) = 0$

Setting each factor to zero gives the critical points:

$12x = 0 \Rightarrow x = 0$

$x + 2 = 0 \Rightarrow x = -2$

$x - 1 = 0 \Rightarrow x = 1$

The critical points are $x = -2, x = 0, x = 1$.

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We can use the first derivative test to determine the nature of these critical points. The critical points divide the real line into intervals: $(-\infty, -2)$, $(-2, 0)$, $(0, 1)$, and $(1, \infty)$. We examine the sign of $f'(x) = 12x(x + 2)(x - 1)$ in each interval.

Interval $(-\infty, -2)$: Choose a test value, e.g., $x = -3$.

$f'(-3) = 12(-3)(-3 + 2)(-3 - 1) = (-)(-) (-) (-) = (-)$. $f'(-3) = -144 < 0$.

$f'(x) < 0$ on $(-\infty, -2)$. $f(x)$ is decreasing.

Interval $(-2, 0)$: Choose a test value, e.g., $x = -1$.

$f'(-1) = 12(-1)(-1 + 2)(-1 - 1) = (-)(+) (-) = (+)$. $f'(-1) = 24 > 0$.

$f'(x) > 0$ on $(-2, 0)$. $f(x)$ is increasing.

Since $f'(x)$ changes from negative to positive at $x = -2$, $x = -2$ is a point of local minimum.

Interval $(0, 1)$: Choose a test value, e.g., $x = 0.5$.

$f'(0.5) = 12(0.5)(0.5 + 2)(0.5 - 1) = (+)(+) (-) = (-)$. $f'(0.5) = -7.5 < 0$.

$f'(x) < 0$ on $(0, 1)$. $f(x)$ is decreasing.

Since $f'(x)$ changes from positive to negative at $x = 0$, $x = 0$ is a point of local maximum.

Interval $(1, \infty)$: Choose a test value, e.g., $x = 2$.

$f'(2) = 12(2)(2 + 2)(2 - 1) = (+)(+) (+) = (+)$. $f'(2) = 96 > 0$.

$f'(x) > 0$ on $(1, \infty)$. $f(x)$ is increasing.

Since $f'(x)$ changes from negative to positive at $x = 1$, $x = 1$ is a point of local minimum.

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Now, we calculate the function values at these local extreme points.

Local minimum value at $x = -2$:

$f(-2) = 3(-2)^4 + 4(-2)^3 - 12(-2)^2 + 12$

$f(-2) = 3(16) + 4(-8) - 12(4) + 12$

$f(-2) = 48 - 32 - 48 + 12$

$f(-2) = -20$

Local maximum value at $x = 0$:

$f(0) = 3(0)^4 + 4(0)^3 - 12(0)^2 + 12$

$f(0) = 0 + 0 - 0 + 12$

$f(0) = 12$

Local minimum value at $x = 1$:

$f(1) = 3(1)^4 + 4(1)^3 - 12(1)^2 + 12$

$f(1) = 3(1) + 4(1) - 12(1) + 12$

$f(1) = 3 + 4 - 12 + 12$

$f(1) = 7$

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Alternate Method (Using Second Derivative Test)

Calculate the second derivative $f''(x)$:

$f'(x) = 12x^3 + 12x^2 - 24x$

$f''(x) = \frac{d}{dx}(12x^3 + 12x^2 - 24x)$

$f''(x) = 36x^2 + 24x - 24$

Evaluate $f''(x)$ at each critical point:

At $x = -2$:

$f''(-2) = 36(-2)^2 + 24(-2) - 24 = 36(4) - 48 - 24 = 144 - 72 = 72$

Since $f''(-2) = 72 > 0$, $x = -2$ is a point of local minimum. The local minimum value is $f(-2) = -20$.

At $x = 0$:

$f''(0) = 36(0)^2 + 24(0) - 24 = 0 + 0 - 24 = -24$

Since $f''(0) = -24 < 0$, $x = 0$ is a point of local maximum. The local maximum value is $f(0) = 12$.

At $x = 1$:

$f''(1) = 36(1)^2 + 24(1) - 24 = 36 + 24 - 24 = 36$

Since $f''(1) = 36 > 0$, $x = 1$ is a point of local minimum. The local minimum value is $f(1) = 7$.

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Conclusion:

The function has local minimum values at $x = -2$ and $x = 1$. The local minimum value at $x = -2$ is -20, and the local minimum value at $x = 1$ is 7.

The function has a local maximum value at $x = 0$. The local maximum value is 12.

Given the function $f(x) = 2x^3 – 6x^2 + 6x + 5$.

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To find the points of local maxima and local minima, we first find the derivative of the function and the critical points.

Calculate the first derivative $f'(x)$:

$f'(x) = \frac{d}{dx}(2x^3 - 6x^2 + 6x + 5)$

$f'(x) = 2\frac{d}{dx}(x^3) - 6\frac{d}{dx}(x^2) + 6\frac{d}{dx}(x) + \frac{d}{dx}(5)$

$f'(x) = 2(3x^2) - 6(2x) + 6(1) + 0$

$f'(x) = 6x^2 - 12x + 6$

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To find the critical points, we set $f'(x) = 0$ and solve for $x$:

$6x^2 - 12x + 6 = 0$

Factor out the common factor of 6:

$6(x^2 - 2x + 1) = 0$

The expression inside the parentheses is a perfect square trinomial:

$x^2 - 2x + 1 = (x - 1)^2$

So, the equation becomes:

$6(x - 1)^2 = 0$

This equation is satisfied when $(x - 1)^2 = 0$, which means $x - 1 = 0$.

The only critical point is $x = 1$.

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We use the first derivative test to classify the critical point $x = 1$. We analyze the sign of $f'(x) = 6(x - 1)^2$ around $x = 1$.

For $x < 1$, $(x - 1)$ is negative, but $(x - 1)^2$ is positive. Thus, $f'(x) = 6(x - 1)^2 > 0$.

For $x > 1$, $(x - 1)$ is positive, and $(x - 1)^2$ is positive. Thus, $f'(x) = 6(x - 1)^2 > 0$.

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Since the sign of $f'(x)$ does not change as $x$ passes through the critical point $x = 1$ (it remains positive on both sides), the point $x = 1$ is neither a point of local maximum nor a point of local minimum.

It is a point of inflection with a horizontal tangent.

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Alternatively, using the Second Derivative Test:

Calculate the second derivative $f''(x)$:

$f'(x) = 6x^2 - 12x + 6$

$f''(x) = \frac{d}{dx}(6x^2 - 12x + 6) = 12x - 12$

Evaluate $f''(x)$ at the critical point $x = 1$:

$f''(1) = 12(1) - 12 = 12 - 12 = 0$.

Since $f''(1) = 0$, the second derivative test is inconclusive. We must rely on the first derivative test, which indicated no local extremum.

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Conclusion:

The function $f(x) = 2x^3 – 6x^2 + 6x + 5$ has no points of local maxima or local minima.

Let the two positive numbers be $x$ and $y$.

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Given the condition that their sum is 15:

Since the numbers are positive, we have $x > 0$ and $y > 0$.

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We want to minimize the sum of their squares. Let $S$ be the sum of their squares:

$S = x^2 + y^2$

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We need to express $S$ as a function of a single variable. From the sum condition, we can express $y$ in terms of $x$:

$y = 15 - x$

Since $y > 0$, we have $15 - x > 0$, which implies $x < 15$.

Also, since $x$ is a positive number, $x > 0$.

Thus, the variable $x$ is in the interval $(0, 15)$.

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Substitute $y = 15 - x$ into the expression for $S$:

$S(x) = x^2 + (15 - x)^2$

$S(x) = x^2 + (15^2 - 2 \cdot 15 \cdot x + x^2)$

$S(x) = x^2 + (225 - 30x + x^2)$

$S(x) = 2x^2 - 30x + 225$

We want to find the minimum value of $S(x)$ for $x \in (0, 15)$.

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To find the minimum value, we find the derivative of $S(x)$ with respect to $x$ and find the critical points.

$S'(x) = \frac{d}{dx}(2x^2 - 30x + 225)$

$S'(x) = 4x - 30$

Set $S'(x) = 0$ to find the critical point:

$4x - 30 = 0$

$4x = 30$

$x = \frac{30}{4} = \frac{15}{2}$

The critical point is $x = \frac{15}{2} = 7.5$. This point is within the domain $(0, 15)$.

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We use the second derivative test to determine if this critical point corresponds to a minimum.

$S''(x) = \frac{d}{dx}(4x - 30) = 4$

Evaluate $S''(x)$ at the critical point $x = \frac{15}{2}$:

$S''(\frac{15}{2}) = 4$

Since $S''(\frac{15}{2}) = 4 > 0$, the critical point $x = \frac{15}{2}$ corresponds to a local minimum.

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Since the domain is an open interval $(0, 15)$ and there is only one critical point within this interval, this local minimum is also the absolute minimum on the interval.

The value of the first number is $x = \frac{15}{2}$.

Now, find the value of the second number $y$ using $y = 15 - x$:

$y = 15 - \frac{15}{2} = \frac{30}{2} - \frac{15}{2} = \frac{15}{2}$

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The two numbers are $\frac{15}{2}$ and $\frac{15}{2}$.

Both numbers are positive and their sum is $\frac{15}{2} + \frac{15}{2} = \frac{30}{2} = 15$.

The sum of their squares is minimum when the numbers are $\frac{15}{2}$ and $\frac{15}{2}$.

The minimum sum of squares is $(\frac{15}{2})^2 + (\frac{15}{2})^2 = \frac{225}{4} + \frac{225}{4} = \frac{450}{4} = \frac{225}{2}$.

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The two positive numbers whose sum is 15 and the sum of whose squares is minimum are $\frac{15}{2}$ and $\frac{15}{2}$.

Let the given point be $P = (0, c)$, where $\frac{1}{2} \le c \le 5$.

Let $Q$ be any point on the parabola $y = x^2$. The coordinates of $Q$ can be represented as $(x, x^2)$ for some real number $x$.

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The distance between the points $P(0, c)$ and $Q(x, x^2)$ is given by the distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$d = \sqrt{(x - 0)^2 + (x^2 - c)^2}$

$d = \sqrt{x^2 + (x^2 - c)^2}$

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To minimize the distance $d$, it is equivalent to minimize the square of the distance, $d^2$. Let $D(x) = d^2$.

$D(x) = x^2 + (x^2 - c)^2$

We want to find the minimum value of $D(x)$ for $x \in R$.

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To find the minimum value, we find the critical points by calculating the derivative $D'(x)$ and setting it to zero.

$D(x) = x^2 + (x^4 - 2cx^2 + c^2)$

$D(x) = x^4 + x^2 - 2cx^2 + c^2$

$D(x) = x^4 + (1 - 2c)x^2 + c^2$

Calculate the derivative $D'(x)$ with respect to $x$:

$D'(x) = \frac{d}{dx}(x^4 + (1 - 2c)x^2 + c^2)$

$D'(x) = 4x^3 + 2(1 - 2c)x + 0$

$D'(x) = 4x^3 + (2 - 4c)x$

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Set $D'(x) = 0$ to find the critical points:

$4x^3 + (2 - 4c)x = 0$

Factor out $2x$:

$2x (2x^2 + 1 - 2c) = 0$

This equation is satisfied if $2x = 0$ or $2x^2 + 1 - 2c = 0$.

Case 1: $2x = 0 \Rightarrow x = 0$.

Case 2: $2x^2 + 1 - 2c = 0 \Rightarrow 2x^2 = 2c - 1 \Rightarrow x^2 = c - \frac{1}{2}$.

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The existence of real solutions for $x$ in Case 2 depends on the value of $c$. We are given $\frac{1}{2} \le c \le 5$.

If $c = \frac{1}{2}$, then $x^2 = \frac{1}{2} - \frac{1}{2} = 0$, which gives $x = 0$. In this case, the only critical point is $x = 0$.

If $c > \frac{1}{2}$, then $c - \frac{1}{2} > 0$, so $x^2 = c - \frac{1}{2}$ has two real solutions: $x = \pm\sqrt{c - \frac{1}{2}}$. In this case, there are three critical points: $x = -\sqrt{c - \frac{1}{2}}$, $x = 0$, and $x = \sqrt{c - \frac{1}{2}}$.

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Now, we evaluate the square of the distance $D(x)$ at these critical points to find the minimum value.

$D(x) = x^4 + (1 - 2c)x^2 + c^2$.

At $x = 0$:

$D(0) = (0)^4 + (1 - 2c)(0)^2 + c^2 = c^2$

At $x = \pm\sqrt{c - \frac{1}{2}}$ (valid for $c \ge \frac{1}{2}$): Let $x^2 = c - \frac{1}{2}$.

$D\left(\pm\sqrt{c - \frac{1}{2}}\right) = \left(c - \frac{1}{2}\right)^2 + (1 - 2c)\left(c - \frac{1}{2}\right) + c^2$

$D\left(\pm\sqrt{c - \frac{1}{2}}\right) = \left(c - \frac{1}{2}\right)^2 - 2\left(c - \frac{1}{2}\right)\left(c - \frac{1}{2}\right) + c^2$

$D\left(\pm\sqrt{c - \frac{1}{2}}\right) = \left(c - \frac{1}{2}\right)^2 - 2\left(c - \frac{1}{2}\right)^2 + c^2$

$D\left(\pm\sqrt{c - \frac{1}{2}}\right) = -\left(c - \frac{1}{2}\right)^2 + c^2$

$D\left(\pm\sqrt{c - \frac{1}{2}}\right) = -\left(c^2 - c + \frac{1}{4}\right) + c^2$

$D\left(\pm\sqrt{c - \frac{1}{2}}\right) = -c^2 + c - \frac{1}{4} + c^2$

$D\left(\pm\sqrt{c - \frac{1}{2}}\right) = c - \frac{1}{4}$

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We need to find the minimum value of $D(x)$ for $x \in R$ when $\frac{1}{2} \le c \le 5$.

If $c = \frac{1}{2}$, the only critical point is $x=0$. $D(0) = (\frac{1}{2})^2 = \frac{1}{4}$. Using the second critical point value formula: $c - \frac{1}{4} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$. Both give the same value, which is the minimum when $c = \frac{1}{2}$.

If $c > \frac{1}{2}$, we compare $D(0) = c^2$ and $D(\pm\sqrt{c - \frac{1}{2}}) = c - \frac{1}{4}$.

Consider the difference $D(0) - D(\pm\sqrt{c - \frac{1}{2}}) = c^2 - (c - \frac{1}{4}) = c^2 - c + \frac{1}{4} = \left(c - \frac{1}{2}\right)^2$.

Since $c > \frac{1}{2}$, $c - \frac{1}{2} > 0$, so $\left(c - \frac{1}{2}\right)^2 > 0$.

Thus, $c^2 > c - \frac{1}{4}$ when $c > \frac{1}{2}$.

This means $D(0) > D(\pm\sqrt{c - \frac{1}{2}})$ when $c > \frac{1}{2}$.

So, for $\frac{1}{2} \le c \le 5$, the minimum value of the square of the distance is $c - \frac{1}{4}$, which occurs at $x = \pm\sqrt{c - \frac{1}{2}}$ (and at $x=0$ when $c=\frac{1}{2}$).

The minimum square of the distance is $c - \frac{1}{4}$.

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The shortest distance is the square root of the minimum square of the distance.

Shortest distance $= \sqrt{c - \frac{1}{4}}$

Since $\frac{1}{2} \le c \le 5$, $c - \frac{1}{4} \ge \frac{1}{2} - \frac{1}{4} = \frac{1}{4} > 0$, so the value under the square root is non-negative.

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The shortest distance of the point $(0, c)$ from the parabola $y = x^2$, where $\frac{1}{2} \le c \le 5$, is $\sqrt{c - \frac{1}{4}}$.

Let the point A be at the origin (0, 0). Since AB = 20 m and the poles are vertical, we can place the points on a coordinate plane.

Let the line segment AB lie on the x-axis.

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AP is a vertical pole at A with height 16 m. So, point P is at $(0, 16)$.

BQ is a vertical pole at B with height 22 m. So, point Q is at $(20, 22)$.

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Let R be a point on the line segment AB. Since R is on AB and AB is on the x-axis, the coordinates of R are $(x, 0)$ for some value of $x$.

Since R is on the segment AB, the x-coordinate $x$ must be between 0 and 20, inclusive.

So, $0 \le x \le 20$.

The distance of R from point A is the x-coordinate of R, which is $x$. We need to find the value of $x$ that minimizes the given expression.

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We want to minimize the sum of the squares of the distances $RP^2 + RQ^2$.

The coordinates are $R(x, 0)$, $P(0, 16)$, and $Q(20, 22)$.

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Calculate $RP^2$ using the distance formula squared:

$RP^2 = (x - 0)^2 + (0 - 16)^2 = x^2 + (-16)^2 = x^2 + 256$

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Calculate $RQ^2$ using the distance formula squared:

$RQ^2 = (x - 20)^2 + (0 - 22)^2 = (x - 20)^2 + (-22)^2 = (x - 20)^2 + 484$

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Let $S(x)$ be the sum of the squares of the distances:

$S(x) = RP^2 + RQ^2$

$S(x) = (x^2 + 256) + ((x - 20)^2 + 484)$

$S(x) = x^2 + 256 + (x^2 - 40x + 400) + 484$

$S(x) = x^2 + 256 + x^2 - 40x + 400 + 484$

$S(x) = 2x^2 - 40x + 1140$

We need to find the minimum value of $S(x)$ on the closed interval $[0, 20]$.

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To find the minimum, we calculate the derivative $S'(x)$ and find the critical points.

$S'(x) = \frac{d}{dx}(2x^2 - 40x + 1140)$

$S'(x) = 2(2x) - 40(1) + 0$

$S'(x) = 4x - 40$

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Set $S'(x) = 0$ to find the critical point:

The critical point is $x = 10$. This point lies within the interval $[0, 20]$.

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We can use the second derivative test to confirm if this critical point is a minimum.

$S''(x) = \frac{d}{dx}(4x - 40) = 4$

Evaluate $S''(x)$ at the critical point $x = 10$:

$S''(10) = 4$

Since $S''(10) = 4 > 0$, the critical point $x = 10$ corresponds to a local minimum.

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To find the absolute minimum on the closed interval $[0, 20]$, we evaluate the function $S(x)$ at the critical point within the interval and at the endpoints of the interval.

Critical point $x = 10$:

$S(10) = 2(10)^2 - 40(10) + 1140 = 2(100) - 400 + 1140 = 200 - 400 + 1140 = 940$

Endpoint $x = 0$:

$S(0) = 2(0)^2 - 40(0) + 1140 = 0 - 0 + 1140 = 1140$

Endpoint $x = 20$:

$S(20) = 2(20)^2 - 40(20) + 1140 = 2(400) - 800 + 1140 = 800 - 800 + 1140 = 1140$

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Comparing the values $S(10) = 940$, $S(0) = 1140$, and $S(20) = 1140$, the minimum value of $S(x)$ on the interval $[0, 20]$ is 940, which occurs at $x = 10$.

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The value of $x$ that minimizes $RP^2 + RQ^2$ is $x = 10$. The variable $x$ represents the distance of the point R from point A.

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Therefore, the distance of point R on AB from the point A such that $RP^2 + RQ^2$ is minimum is 10 m.

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The final answer is $\boxed{10 \text{ m}}$.

Let the trapezium be denoted by ABCD, where AB is parallel to DC. Let the lengths of the three equal sides (other than the base) be AD = BC = CD = 10 cm.

Let AB be the base of the trapezium.

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Draw perpendiculars from D and C to AB, meeting AB at points E and F, respectively.

Since AD = BC = 10 cm and DE = CF (height), $\triangle ADE$ is congruent to $\triangle BCF$ (RHS congruence rule).

The quadrilateral DCFE is a rectangle, so EF = DC = 10 cm.

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Let AE = $x$ cm. Then FB = $x$ cm (from congruent triangles).

The length of the base AB is AB = AE + EF + FB = $x + 10 + x = (10 + 2x)$ cm.

Since lengths must be non-negative, $x \ge 0$.

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Let $h$ cm be the height of the trapezium (DE or CF).

In the right-angled triangle ADE, by the Pythagorean theorem:

$10^2 = x^2 + h^2$

$100 = x^2 + h^2$

$h^2 = 100 - x^2$

For the height to be real and positive (for a valid trapezium with non-zero height), $100 - x^2 > 0$, which means $x^2 < 100$, so $-10 < x < 10$.

Combined with $x \ge 0$, the variable $x$ lies in the interval $[0, 10)$.

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The area of the trapezium, $A$, is given by:

$A = \frac{1}{2}$ (Sum of parallel sides) $\times$ (Height)

$A = \frac{1}{2}(CD + AB)h$

$A(x) = \frac{1}{2}(10 + (10 + 2x))\sqrt{100 - x^2}$

$A(x) = \frac{1}{2}(20 + 2x)\sqrt{100 - x^2}$

$A(x) = (10 + x)\sqrt{100 - x^2}$

We want to maximize the area $A(x)$ for $x \in [0, 10)$.

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To find the maximum value, we find the derivative of $A(x)$ with respect to $x$ and find the critical points.

Using the product rule, $A(x) = (10 + x)(100 - x^2)^{\frac{1}{2}}$:

$A'(x) = \frac{d}{dx}(10 + x) \cdot \sqrt{100 - x^2} + (10 + x) \cdot \frac{d}{dx}(\sqrt{100 - x^2})$

$A'(x) = 1 \cdot \sqrt{100 - x^2} + (10 + x) \cdot \frac{1}{2\sqrt{100 - x^2}} \cdot \frac{d}{dx}(100 - x^2)$

$A'(x) = \sqrt{100 - x^2} + (10 + x) \cdot \frac{1}{2\sqrt{100 - x^2}} \cdot (-2x)$

$A'(x) = \sqrt{100 - x^2} - \frac{2x(10 + x)}{2\sqrt{100 - x^2}}$

$A'(x) = \sqrt{100 - x^2} - \frac{x(10 + x)}{\sqrt{100 - x^2}}$

Combine the terms by finding a common denominator:

$A'(x) = \frac{\sqrt{100 - x^2} \cdot \sqrt{100 - x^2} - x(10 + x)}{\sqrt{100 - x^2}}$

$A'(x) = \frac{(100 - x^2) - (10x + x^2)}{\sqrt{100 - x^2}}$

$A'(x) = \frac{100 - x^2 - 10x - x^2}{\sqrt{100 - x^2}}$

$A'(x) = \frac{100 - 10x - 2x^2}{\sqrt{100 - x^2}}$

Factor the numerator:

$A'(x) = \frac{-2(x^2 + 5x - 50)}{\sqrt{100 - x^2}}$

Factor the quadratic expression $x^2 + 5x - 50$: We look for two numbers that multiply to -50 and add up to 5. These are 10 and -5.

$x^2 + 5x - 50 = (x + 10)(x - 5)$.

So, $A'(x) = \frac{-2(x + 10)(x - 5)}{\sqrt{100 - x^2}}$.

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Set $A'(x) = 0$ to find critical points:

The numerator must be zero: $-2(x + 10)(x - 5) = 0$.

This gives $x + 10 = 0$ or $x - 5 = 0$.

So, $x = -10$ or $x = 5$.

The point $x = -10$ is not in the domain $[0, 10)$.

The only critical point in the interval $(0, 10)$ is $x = 5$.

The derivative $A'(x)$ is undefined at $x = 10$ (where the denominator is zero), which is an endpoint of the domain $[0, 10)$.

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To find the maximum area, we evaluate the area function $A(x)$ at the critical point within the interval and at the endpoint(s) of the closed interval or consider the limit at the boundary of the open interval.

The domain is $[0, 10)$. We evaluate at $x=0$ and $x=5$. We also consider the limit as $x$ approaches the boundary $x=10$ from the left.

At the endpoint $x = 0$:

$A(0) = (10 + 0)\sqrt{100 - 0^2} = 10\sqrt{100} = 10 \times 10 = 100$ cm$^2$.

In this case, $x=0$, so the base AB = $10 + 2(0) = 10$. All four sides are 10 cm. This is a square with side 10 cm.

At the critical point $x = 5$:

$A(5) = (10 + 5)\sqrt{100 - 5^2} = 15\sqrt{100 - 25} = 15\sqrt{75}$

$A(5) = 15\sqrt{25 \cdot 3} = 15 \cdot 5\sqrt{3} = 75\sqrt{3}$ cm$^2$.

In this case, $x=5$. The non-parallel sides are 10 cm, the shorter base is 10 cm, and the longer base is $10 + 2(5) = 20$ cm. The height is $h = \sqrt{100-5^2} = \sqrt{75} = 5\sqrt{3}$ cm.

As $x$ approaches 10 from the left ($x \to 10^-$):

$\lim\limits_{x \to 10^-} A(x) = \lim\limits_{x \to 10^-} (10 + x)\sqrt{100 - x^2} = (10 + 10)\sqrt{100 - 10^2} = 20\sqrt{0} = 0$.

In this limiting case, the trapezium collapses into a line segment.

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Comparing the values:

$A(0) = 100$

$A(5) = 75\sqrt{3}$

We know that $\sqrt{3} \approx 1.732$.

$75\sqrt{3} \approx 75 \times 1.732 = 129.9$

Since $129.9 > 100$ and the limit as $x \to 10^-$ is 0, the maximum area occurs at the critical point $x = 5$.

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The maximum area of the trapezium is $75\sqrt{3}$ cm$^2$.

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The final answer is $\boxed{75\sqrt{3} \text{ cm}^2}$.

Given:

A right circular cone with a given radius R and a given height H.

A right circular cylinder inscribed in the cone.

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To Prove:

The radius of the cylinder with the greatest curved surface area is half the radius of the cone (i.e., $r = \frac{R}{2}$, where r is the cylinder's radius and R is the cone's radius).

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Solution:

Let R be the radius of the base of the cone and H be its height. These are constants.

Let r be the radius of the base of the inscribed cylinder and h be its height. These are variables.

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The curved surface area of the cylinder, denoted by A, is given by the formula:

We want to maximize this area A.

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To relate the variables r and h to the constants R and H, consider a cross-section of the cone and the inscribed cylinder through the axis of the cone. This cross-section will show a triangle (representing the cone) and a rectangle inscribed within it (representing the cylinder).

Let the vertex of the cone be at the origin (0, 0) and the axis of the cone be along the y-axis pointing upwards. The base of the cone is parallel to the x-axis at $y=H$. The points on the base are $(\pm R, H)$. The line forming the slant height in the first quadrant connects (0, 0) and (R, H). The equation of this line is $y = \frac{H}{R}x$.

If the base of the cylinder lies on the base of the cone, its top circular edge will be at a height $h$ from the base of the cone, or at a height $H-h$ from the vertex. The radius of the cylinder at this height is r.

A point on the top edge of the cylinder's cross-section in the first quadrant is $(r, H-h)$. This point must lie on the slant height line of the cone.

Substitute $(r, H-h)$ into the equation of the line $y = \frac{H}{R}x$:

Divide by H (since $H > 0$):

Rearrange to express h in terms of r, R, and H:

The cylinder is inscribed, so its radius r must be positive and cannot exceed the cone's radius R. Also, its height h must be positive. Since $h = H(1 - \frac{r}{R})$ and $H > 0$, $h > 0$ implies $1 - \frac{r}{R} > 0$, which means $\frac{r}{R} < 1$, so $r < R$.

Thus, the possible values for the radius r are in the interval $(0, R)$.

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Substitute the expression for h from equation (ii) into the area formula (i):

$A(r) = 2 \pi r \left[ H\left(1 - \frac{r}{R}\right) \right]$

$A(r) = 2 \pi H r \left(1 - \frac{r}{R}\right)$

$A(r) = 2 \pi H \left(r - \frac{r^2}{R}\right)$

We want to maximize this function $A(r)$ for $r \in (0, R)$.

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To find the maximum value, we find the derivative of $A(r)$ with respect to r and find the critical points.

$A'(r) = \frac{d}{dr} \left[ 2 \pi H \left(r - \frac{r^2}{R}\right) \right]$

$A'(r) = 2 \pi H \left( \frac{d}{dr}(r) - \frac{1}{R}\frac{d}{dr}(r^2) \right)$

$A'(r) = 2 \pi H \left( 1 - \frac{1}{R}(2r) \right)$

$A'(r) = 2 \pi H \left( 1 - \frac{2r}{R} \right)$

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Set $A'(r) = 0$ to find critical points:

Since $2 \pi H \neq 0$ (R, H are from a given cone, implies $H > 0$), we can divide by $2 \pi H$:

This critical point $r = \frac{R}{2}$ is in the domain $(0, R)$ since R is the cone's radius ($R>0$).

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To confirm that this critical point corresponds to a maximum, we use the second derivative test.

$A'(r) = 2 \pi H \left( 1 - \frac{2r}{R} \right)$

$A''(r) = \frac{d}{dr} \left[ 2 \pi H \left( 1 - \frac{2r}{R} \right) \right]$

$A''(r) = 2 \pi H \left( \frac{d}{dr}(1) - \frac{2}{R}\frac{d}{dr}(r) \right)$

$A''(r) = 2 \pi H \left( 0 - \frac{2}{R}(1) \right)$

$A''(r) = -\frac{4 \pi H}{R}$

Evaluate $A''(r)$ at the critical point $r = \frac{R}{2}$:

$A''\left(\frac{R}{2}\right) = -\frac{4 \pi H}{R}$

Since R and H are positive constants (from a given cone), $A''\left(\frac{R}{2}\right) = -\frac{4 \pi H}{R}$ is a negative constant.

According to the second derivative test, $r = \frac{R}{2}$ corresponds to a local maximum.

Since this is the only critical point in the open interval $(0, R)$ and the second derivative is always negative on this interval, this local maximum is also the absolute maximum on $(0, R)$.

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The curved surface area of the inscribed cylinder is greatest when its radius is $r = \frac{R}{2}$.

This proves that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone.

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Hence, proven.

Given the function $f(x) = 2x^3 – 15x^2 + 36x + 1$ on the closed interval $[1, 5]$.

Since $f(x)$ is a polynomial, it is continuous and differentiable everywhere, including the interval $[1, 5]$.

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To find the absolute maximum and minimum values on a closed interval, we need to evaluate the function at the critical points that lie within the interval and at the endpoints of the interval.

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Step 1: Find the critical points.

Calculate the derivative $f'(x)$:

$f'(x) = \frac{d}{dx}(2x^3 - 15x^2 + 36x + 1)$

$f'(x) = 6x^2 - 30x + 36$

Set $f'(x) = 0$ to find the critical points:

$6x^2 - 30x + 36 = 0$

Divide by 6:

$x^2 - 5x + 6 = 0$

Factor the quadratic equation. We look for two numbers that multiply to 6 and add up to -5. These numbers are -2 and -3.

$(x - 2)(x - 3) = 0$

This gives the critical points $x = 2$ and $x = 3$.

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Step 2: Check which critical points lie within the given interval [1, 5].

The interval is $[1, 5]$.

For $x = 2$: $1 \le 2 \le 5$. So, $x = 2$ is in the interval.

For $x = 3$: $1 \le 3 \le 5$. So, $x = 3$ is in the interval.

Both critical points are within the interval [1, 5].

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Step 3: Evaluate the function $f(x)$ at the critical points within the interval and at the endpoints of the interval.

The endpoints of the interval are $x = 1$ and $x = 5$.

The critical points within the interval are $x = 2$ and $x = 3$.

Evaluate $f(x)$ at $x = 1$ (left endpoint):

$f(1) = 2(1)^3 - 15(1)^2 + 36(1) + 1 = 2 - 15 + 36 + 1 = 24$

Evaluate $f(x)$ at $x = 2$ (critical point):

$f(2) = 2(2)^3 - 15(2)^2 + 36(2) + 1 = 2(8) - 15(4) + 72 + 1 = 16 - 60 + 72 + 1 = 29$

Evaluate $f(x)$ at $x = 3$ (critical point):

$f(3) = 2(3)^3 - 15(3)^2 + 36(3) + 1 = 2(27) - 15(9) + 108 + 1 = 54 - 135 + 108 + 1 = 28$

Evaluate $f(x)$ at $x = 5$ (right endpoint):

$f(5) = 2(5)^3 - 15(5)^2 + 36(5) + 1 = 2(125) - 15(25) + 180 + 1 = 250 - 375 + 180 + 1 = 56$

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Step 4: Identify the absolute maximum and minimum values from the evaluated function values.

The values are $f(1) = 24$, $f(2) = 29$, $f(3) = 28$, and $f(5) = 56$.

The largest of these values is 56.

The smallest of these values is 24.

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Conclusion:

The absolute maximum value of the function $f(x) = 2x^3 – 15x^2 + 36x + 1$ on the interval [1, 5] is 56, which occurs at $x = 5$.

The absolute minimum value of the function on the interval [1, 5] is 24, which occurs at $x = 1$.

Given the function $f(x) = 12x^{\frac{4}{3}} - 6x^{\frac{1}{3}}$ on the closed interval $[-1, 1]$.

Since $f(x)$ involves fractional exponents with odd denominators (3), it is defined and continuous for all real numbers, including the interval $[-1, 1]$.

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To find the absolute maximum and minimum values on a closed interval, we need to evaluate the function at the critical points within the interval and at the endpoints of the interval.

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Step 1: Find the critical points.

Calculate the derivative $f'(x)$:

$f'(x) = \frac{d}{dx}(12x^{\frac{4}{3}} - 6x^{\frac{1}{3}})$

$f'(x) = 12 \cdot \frac{4}{3} x^{\frac{4}{3} - 1} - 6 \cdot \frac{1}{3} x^{\frac{1}{3} - 1}$

$f'(x) = 16 x^{\frac{1}{3}} - 2 x^{-\frac{2}{3}}$

$f'(x) = 16 x^{\frac{1}{3}} - \frac{2}{x^{\frac{2}{3}}}$

To find the critical points, we set $f'(x) = 0$ or find where $f'(x)$ is undefined.

Set $f'(x) = 0$:

$16 x^{\frac{1}{3}} - \frac{2}{x^{\frac{2}{3}}} = 0$

$16 x^{\frac{1}{3}} = \frac{2}{x^{\frac{2}{3}}}$

Multiply both sides by $x^{\frac{2}{3}}$ (assuming $x \neq 0$):

$16 x^{\frac{1}{3}} \cdot x^{\frac{2}{3}} = 2$

$16 x^{\frac{1}{3} + \frac{2}{3}} = 2$

$16 x^1 = 2$

$16x = 2$

$x = \frac{2}{16} = \frac{1}{8}$

This is one critical point.

Now, check where $f'(x)$ is undefined.

$f'(x) = 16 x^{\frac{1}{3}} - \frac{2}{x^{\frac{2}{3}}} = 16 \sqrt[3]{x} - \frac{2}{\sqrt[3]{x^2}}$

The term $\frac{2}{\sqrt[3]{x^2}}$ is undefined when the denominator is zero, i.e., when $x^2 = 0$, which means $x = 0$.

So, $x = 0$ is also a critical point because the derivative is undefined there.

The critical points are $x = \frac{1}{8}$ and $x = 0$.

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Step 2: Check which critical points lie within the given interval [-1, 1].

The interval is $[-1, 1]$.

For $x = \frac{1}{8}$: $-1 \le \frac{1}{8} \le 1$. So, $x = \frac{1}{8}$ is in the interval.

For $x = 0$: $-1 \le 0 \le 1$. So, $x = 0$ is in the interval.

Both critical points are within the interval [-1, 1].

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Step 3: Evaluate the function $f(x)$ at the critical points within the interval and at the endpoints of the interval.

The endpoints of the interval are $x = -1$ and $x = 1$.

The critical points within the interval are $x = 0$ and $x = \frac{1}{8}$.

Evaluate $f(x)$ at $x = -1$ (left endpoint):

$f(-1) = 12(-1)^{\frac{4}{3}} - 6(-1)^{\frac{1}{3}}$

$(-1)^{\frac{4}{3}} = (\sqrt[3]{-1})^4 = (-1)^4 = 1$

$(-1)^{\frac{1}{3}} = \sqrt[3]{-1} = -1$

$f(-1) = 12(1) - 6(-1) = 12 + 6 = 18$

Evaluate $f(x)$ at $x = 0$ (critical point):

$f(0) = 12(0)^{\frac{4}{3}} - 6(0)^{\frac{1}{3}} = 12(0) - 6(0) = 0 - 0 = 0$

Evaluate $f(x)$ at $x = \frac{1}{8}$ (critical point):

$f\left(\frac{1}{8}\right) = 12\left(\frac{1}{8}\right)^{\frac{4}{3}} - 6\left(\frac{1}{8}\right)^{\frac{1}{3}}$

$\left(\frac{1}{8}\right)^{\frac{1}{3}} = \sqrt[3]{\frac{1}{8}} = \frac{1}{2}$

$\left(\frac{1}{8}\right)^{\frac{4}{3}} = \left(\left(\frac{1}{8}\right)^{\frac{1}{3}}\right)^4 = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$

$f\left(\frac{1}{8}\right) = 12\left(\frac{1}{16}\right) - 6\left(\frac{1}{2}\right) = \frac{12}{16} - \frac{6}{2} = \frac{3}{4} - 3 = \frac{3 - 12}{4} = -\frac{9}{4} = -2.25$

Evaluate $f(x)$ at $x = 1$ (right endpoint):

$f(1) = 12(1)^{\frac{4}{3}} - 6(1)^{\frac{1}{3}} = 12(1) - 6(1) = 12 - 6 = 6$

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Step 4: Identify the absolute maximum and minimum values from the evaluated function values.

The values are $f(-1) = 18$, $f(0) = 0$, $f(\frac{1}{8}) = -\frac{9}{4}$, and $f(1) = 6$.

Comparing these values:

$18 > 6 > 0 > -\frac{9}{4}$

The largest value is 18.

The smallest value is $-\frac{9}{4}$.

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Conclusion:

The absolute maximum value of the function $f(x) = 12x^{\frac{4}{3}} - 6x^{\frac{1}{3}}$ on the interval [-1, 1] is 18, which occurs at $x = -1$.

The absolute minimum value of the function on the interval [-1, 1] is $-\frac{9}{4}$, which occurs at $x = \frac{1}{8}$.

Given:

Position of the soldier: $S = (3, 7)$.

Curve along which the helicopter is flying: $y = x^2 + 7$.

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To Find:

The nearest distance between the soldier and the helicopter.

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Solution:

Let $P(x, y)$ be any point on the curve $y = x^2 + 7$. Since $y = x^2 + 7$, the coordinates of a point on the curve can be written as $P(x, x^2 + 7)$ for some real number $x$.

The position of the soldier is $S(3, 7)$.

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The distance between the soldier at $S(3, 7)$ and the helicopter at $P(x, x^2 + 7)$ is given by the distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$d = \sqrt{(x - 3)^2 + ((x^2 + 7) - 7)^2}$

$d = \sqrt{(x - 3)^2 + (x^2)^2}$

$d = \sqrt{(x - 3)^2 + x^4}$

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To minimize the distance $d$, it is equivalent to minimize the square of the distance, $d^2$, as the square root function is increasing. Let $D(x) = d^2$.

$D(x) = (x - 3)^2 + x^4$

$D(x) = (x^2 - 6x + 9) + x^4$

$D(x) = x^4 + x^2 - 6x + 9$

We want to find the minimum value of $D(x)$ for $x \in R$.

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To find the minimum value, we find the derivative of $D(x)$ with respect to $x$ and find the critical points by setting the derivative to zero.

$D'(x) = \frac{d}{dx}(x^4 + x^2 - 6x + 9)$

$D'(x) = 4x^3 + 2x - 6$

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Set $D'(x) = 0$ to find the critical points:

Divide the equation by 2:

We can try integer values that are divisors of the constant term (-3) to find a root. Let's test $x=1$:

$2(1)^3 + (1) - 3 = 2(1) + 1 - 3 = 2 + 1 - 3 = 0$.

Since substituting $x=1$ into the equation results in 0, $x=1$ is a root of the equation. This means $(x - 1)$ is a factor of the polynomial $2x^3 + x - 3$.

We can divide the polynomial $2x^3 + x - 3$ by $(x - 1)$ to find the other factor(s). The result of the division is $2x^2 + 2x + 3$.

So, the equation $2x^3 + x - 3 = 0$ can be written as $(x - 1)(2x^2 + 2x + 3) = 0$.

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The critical points are found by setting each factor to zero.

Case 1: $x - 1 = 0 \implies x = 1$.

Case 2: $2x^2 + 2x + 3 = 0$. We check the discriminant of this quadratic equation: $\Delta = b^2 - 4ac$. Here $a=2, b=2, c=3$.

$\Delta = (2)^2 - 4(2)(3) = 4 - 24 = -20$.

Since the discriminant is negative ($\Delta < 0$) and the leading coefficient (2) is positive, the quadratic expression $2x^2 + 2x + 3$ is always positive for all real values of $x$. It has no real roots.

Thus, the only real critical point is $x = 1$.

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We can use the first derivative test to confirm that $x = 1$ corresponds to a minimum for $D(x)$.

$D'(x) = 2(x - 1)(2x^2 + 2x + 3)$.

Since $2x^2 + 2x + 3 > 0$ for all real $x$, the sign of $D'(x)$ depends only on the sign of $(x - 1)$.

For $x < 1$, $(x - 1) < 0$, so $D'(x) = 2 \cdot (\text{negative}) \cdot (\text{positive}) < 0$. This means $D(x)$ is decreasing for $x < 1$.

For $x > 1$, $(x - 1) > 0$, so $D'(x) = 2 \cdot (\text{positive}) \cdot (\text{positive}) > 0$. This means $D(x)$ is increasing for $x > 1$.

Since $D'(x)$ changes from negative to positive at $x = 1$, $x = 1$ corresponds to a local minimum of $D(x)$.

Since there is only one critical point where the derivative is zero and the function changes from decreasing to increasing, this local minimum is also the absolute minimum of $D(x)$ on the set of all real numbers R.

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The minimum value of the square of the distance $D(x)$ occurs at $x = 1$.

Substitute $x = 1$ into the expression for $D(x)$:

Minimum $D = D(1) = (1)^4 + (1)^2 - 6(1) + 9 = 1 + 1 - 6 + 9 = 2 - 6 + 9 = -4 + 9 = 5$.

The minimum value of the square of the distance is 5.

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The nearest distance is the square root of the minimum square of the distance.

Nearest distance $= \sqrt{\text{Minimum } D} = \sqrt{5}$.

The units are meters, as the coordinates are in meters (implied by the context of "soldier at (3, 7)" and "helicopter flying along y=x^2+7" in a problem likely set up in a coordinate system where units are consistent).

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The nearest distance between the soldier and the helicopter is $\sqrt{5}$ meters.

The point on the curve nearest to the soldier is when $x=1$, so the point is $(1, 1^2 + 7) = (1, 1 + 7) = (1, 8)$. The distance between the soldier at $(3, 7)$ and this point $(1, 8)$ is $\sqrt{(1-3)^2 + (8-7)^2} = \sqrt{(-2)^2 + 1^2} = \sqrt{4+1} = \sqrt{5}$.

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The final answer is $\boxed{\sqrt{5}}$.

To find the maximum and minimum values of a function, we analyze its behavior using calculus (derivatives) or by examining the properties of the function itself.

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(i) $f(x) = (2x – 1)^2 + 3$

The domain of this function is the set of all real numbers, $R$.

We know that the square of any real number is always non-negative.

Adding 3 to both sides of the inequality:

$f(x) \ge 3$ for all $x \in R$.

The minimum value of $f(x)$ is 3. This minimum value is attained when $(2x - 1)^2 = 0$, which means $2x - 1 = 0$, so $x = \frac{1}{2}$.

Thus, the minimum value is $f(\frac{1}{2}) = (2(\frac{1}{2}) - 1)^2 + 3 = (1 - 1)^2 + 3 = 0^2 + 3 = 3$.

As $x$ tends to positive or negative infinity, $(2x - 1)^2$ tends to infinity, and hence $f(x) = (2x - 1)^2 + 3$ also tends to infinity.

The function values increase without bound.

Therefore, there is no maximum value for this function.

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(ii) $f(x) = 9x^2 + 12x + 2$

The domain of this function is the set of all real numbers, $R$.

This is a quadratic function in the form $ax^2 + bx + c$ with $a = 9$, $b = 12$, $c = 2$. Since $a = 9 > 0$, the parabola opens upwards, which means it has a minimum value but no maximum value.

The minimum value occurs at the vertex of the parabola. The x-coordinate of the vertex is given by $x = -\frac{b}{2a}$.

$x = -\frac{12}{2 \times 9} = -\frac{12}{18} = -\frac{2}{3}$

The minimum value is $f(-\frac{2}{3})$.

$f(-\frac{2}{3}) = 9\left(-\frac{2}{3}\right)^2 + 12\left(-\frac{2}{3}\right) + 2$

$f(-\frac{2}{3}) = 9\left(\frac{4}{9}\right) - \frac{24}{3} + 2$

$f(-\frac{2}{3}) = 4 - 8 + 2 = -2$

Thus, the minimum value is -2.

As $|x|$ tends to infinity, $9x^2$ tends to infinity, and $f(x) = 9x^2 + 12x + 2$ also tends to infinity.

The function values increase without bound.

Therefore, there is no maximum value for this function.

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(iii) $f(x) = – (x – 1)^2 + 10$

The domain of this function is the set of all real numbers, $R$.

We know that the square of any real number is always non-negative.

Multiplying by -1 reverses the inequality sign:

Adding 10 to both sides:

$f(x) \le 10$ for all $x \in R$.

The maximum value of $f(x)$ is 10. This maximum value is attained when $-(x - 1)^2 = 0$, which means $(x - 1)^2 = 0$, so $x = 1$.

Thus, the maximum value is $f(1) = -(1 - 1)^2 + 10 = -0^2 + 10 = 10$.

As $|x|$ tends to infinity, $(x - 1)^2$ tends to infinity, and $-(x - 1)^2$ tends to negative infinity. Hence, $f(x) = -(x - 1)^2 + 10$ also tends to negative infinity.

The function values decrease without bound.

Therefore, there is no minimum value for this function.

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(iv) $g(x) = x^3 + 1$

The domain of this function is the set of all real numbers, $R$.

Let's examine the behavior of the function as $x$ approaches infinity and negative infinity.

As $x \to \infty$, $g(x) = x^3 + 1 \to \infty$.

As $x \to -\infty$, $g(x) = x^3 + 1 \to -\infty$.

Since the function values can be arbitrarily large (positive infinity) and arbitrarily small (negative infinity), the function has neither a maximum value nor a minimum value on its domain R.

Alternatively, consider the derivative:

$g'(x) = \frac{d}{dx}(x^3 + 1) = 3x^2$

For all $x \in R$, $x^2 \ge 0$, so $g'(x) = 3x^2 \ge 0$.

The derivative is non-negative everywhere. The derivative is zero only at $x = 0$, which is an isolated point.

Since $g'(x) \ge 0$ for all $x \in R$ and $g'(x) = 0$ only at $x=0$, the function $g(x)$ is strictly increasing on $R$.

A strictly increasing function on an unbounded domain like R has no maximum and no minimum value.

Therefore, there is no maximum value and no minimum value for this function.

To find the maximum and minimum values of a function, we analyze its behavior over its domain.

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(i) $f(x) = |x + 2| – 1$

The domain is $R$. We know that the absolute value of any real number is non-negative, i.e., $|u| \ge 0$ for all $u \in R$.

So, $|x + 2| \ge 0$ for all $x \in R$.

Adding -1 to both sides, we get $|x + 2| - 1 \ge 0 - 1$, which means $f(x) \ge -1$.

The minimum value of $f(x)$ is -1. This minimum is attained when $|x + 2| = 0$, which occurs when $x + 2 = 0$, so $x = -2$.

As $x \to \infty$ or $x \to -\infty$, $|x + 2| \to \infty$, so $f(x) = |x + 2| - 1 \to \infty$.

Thus, the function increases without bound.

Minimum value: -1 (at $x = -2$)

Maximum value: None

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(ii) $g(x) = – |x + 1| + 3$

The domain is $R$. We know that $|x + 1| \ge 0$ for all $x \in R$.

Multiplying by -1 reverses the inequality: $-|x + 1| \le 0$ for all $x \in R$.

Adding 3 to both sides, we get $-|x + 1| + 3 \le 0 + 3$, which means $g(x) \le 3$.

The maximum value of $g(x)$ is 3. This maximum is attained when $-|x + 1| = 0$, which occurs when $|x + 1| = 0$, so $x + 1 = 0$, which means $x = -1$.

As $x \to \infty$ or $x \to -\infty$, $|x + 1| \to \infty$, so $-|x + 1| \to -\infty$, and $g(x) = -|x + 1| + 3 \to -\infty$.

Thus, the function decreases without bound.

Maximum value: 3 (at $x = -1$)

Minimum value: None

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(iii) $h(x) = \sin(2x) + 5$

The domain is $R$. We know that the range of the sine function is $[-1, 1]$.

So, $-1 \le \sin(2x) \le 1$ for all $x \in R$.

Adding 5 to all parts of the inequality:

$-1 + 5 \le \sin(2x) + 5 \le 1 + 5$

$4 \le h(x) \le 6$

The function $h(x)$ is bounded between 4 and 6, inclusive.

The minimum value is 4, which is attained when $\sin(2x) = -1$.

The maximum value is 6, which is attained when $\sin(2x) = 1$.

Maximum value: 6

Minimum value: 4

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(iv) $f(x) = |\sin 4x + 3|$

The domain is $R$. We know that the range of the sine function is $[-1, 1]$.

So, $-1 \le \sin(4x) \le 1$ for all $x \in R$.

Adding 3 to all parts of the inequality:

$-1 + 3 \le \sin(4x) + 3 \le 1 + 3$

$2 \le \sin(4x) + 3 \le 4$

Now, consider the absolute value. Since the expression $\sin(4x) + 3$ is always between 2 and 4 (which are positive), the absolute value does not change the inequality.

$|2| \le |\sin(4x) + 3| \le |4|$

$2 \le f(x) \le 4$

The function $f(x)$ is bounded between 2 and 4, inclusive.

The minimum value is 2, which is attained when $\sin(4x) + 3 = 2$, i.e., $\sin(4x) = -1$.

The maximum value is 4, which is attained when $\sin(4x) + 3 = 4$, i.e., $\sin(4x) = 1$.

Maximum value: 4

Minimum value: 2

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(v) $h(x) = x + 1$, x ∈ (– 1, 1)

The domain is the open interval $(-1, 1)$, which means $-1 < x < 1$.

The function is $h(x) = x + 1$. This is a linear function, which is strictly increasing.

For the given domain, we can find the range of the function values:

Starting with $-1 < x < 1$, add 1 to all parts:

$-1 + 1 < x + 1 < 1 + 1$

$0 < h(x) < 2$

The range of the function on the interval $(-1, 1)$ is the open interval $(0, 2)$.

The values of the function get arbitrarily close to 0 as $x$ approaches -1, but they never reach 0.

The values of the function get arbitrarily close to 2 as $x$ approaches 1, but they never reach 2.

On an open interval, a continuous function does not necessarily attain its boundary values.

Maximum value: None

Minimum value: None

To find local maxima and minima, we use the first derivative test or the second derivative test. First, we find the critical points by setting the first derivative equal to zero or finding where it's undefined.

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(i) $f(x) = x^2$

Domain: $R$.

First derivative: $f'(x) = \frac{d}{dx}(x^2) = 2x$.

Critical points: Set $f'(x) = 0 \Rightarrow 2x = 0 \Rightarrow x = 0$.

Sign of $f'(x)$: For $x < 0$, $f'(x) = 2x < 0$ (decreasing). For $x > 0$, $f'(x) = 2x > 0$ (increasing).

Since $f'(x)$ changes sign from negative to positive at $x = 0$, $x = 0$ is a point of local minimum.

Local minimum value: $f(0) = (0)^2 = 0$.

There is no local maximum.

Local minimum: at $x = 0$, value is 0.

Local maximum: None.

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(ii) $g(x) = x^3 – 3x$

Domain: $R$.

First derivative: $g'(x) = \frac{d}{dx}(x^3 - 3x) = 3x^2 - 3$.

Critical points: Set $g'(x) = 0 \Rightarrow 3x^2 - 3 = 0 \Rightarrow 3(x^2 - 1) = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$.

Critical points are $x = -1$ and $x = 1$.

Sign of $g'(x) = 3(x - 1)(x + 1)$:

• Interval $(-\infty, -1)$: Choose $x = -2$, $g'(-2) = 3(-3)(-1) = 9 > 0$ (increasing).
• Interval $(-1, 1)$: Choose $x = 0$, $g'(0) = 3(-1)(1) = -3 < 0$ (decreasing).
• Interval $(1, \infty)$: Choose $x = 2$, $g'(2) = 3(1)(3) = 9 > 0$ (increasing).

At $x = -1$, $g'(x)$ changes from positive to negative, so $x = -1$ is a point of local maximum.

Local maximum value: $g(-1) = (-1)^3 - 3(-1) = -1 + 3 = 2$.

At $x = 1$, $g'(x)$ changes from negative to positive, so $x = 1$ is a point of local minimum.

Local minimum value: $g(1) = (1)^3 - 3(1) = 1 - 3 = -2$.

Local maximum: at $x = -1$, value is 2.

Local minimum: at $x = 1$, value is -2.

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(iii) $h(x) = \sin x + \cos x$, $0 < x < \frac{\pi}{2}$

Domain: $(0, \frac{\pi}{2})$.

First derivative: $h'(x) = \frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x$.

Critical points: Set $h'(x) = 0 \Rightarrow \cos x - \sin x = 0 \Rightarrow \cos x = \sin x$.

For $0 < x < \frac{\pi}{2}$, we can divide by $\cos x$ (since $\cos x > 0$ in this interval) to get $\frac{\sin x}{\cos x} = 1 \Rightarrow \tan x = 1$.

In the interval $(0, \frac{\pi}{2})$, the only solution is $x = \frac{\pi}{4}$.

Sign of $h'(x) = \cos x - \sin x$ in $(0, \frac{\pi}{2})$:

• Interval $(0, \frac{\pi}{4})$: For $x < \frac{\pi}{4}$, $\cos x > \sin x$, so $\cos x - \sin x > 0$. $h'(x) > 0$ (increasing).
• Interval $(\frac{\pi}{4}, \frac{\pi}{2})$: For $x > \frac{\pi}{4}$, $\cos x < \sin x$, so $\cos x - \sin x < 0$. $h'(x) < 0$ (decreasing).

At $x = \frac{\pi}{4}$, $h'(x)$ changes from positive to negative, so $x = \frac{\pi}{4}$ is a point of local maximum.

Local maximum value: $h(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

There is no local minimum in the open interval $(0, \frac{\pi}{2})$.

Local maximum: at $x = \frac{\pi}{4}$, value is $\sqrt{2}$.

Local minimum: None.

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(iv) $f(x) = \sin x – \cos x$, $0 < x < 2\pi$

Domain: $(0, 2\pi)$.

First derivative: $f'(x) = \frac{d}{dx}(\sin x - \cos x) = \cos x - (-\sin x) = \cos x + \sin x$.

Critical points: Set $f'(x) = 0 \Rightarrow \cos x + \sin x = 0 \Rightarrow \sin x = -\cos x$.

Divide by $\cos x$ (assuming $\cos x \neq 0$) to get $\tan x = -1$.

In the interval $(0, 2\pi)$, the solutions for $\tan x = -1$ are $x = \frac{3\pi}{4}$ (in Quadrant II) and $x = \frac{7\pi}{4}$ (in Quadrant IV).

Also, $\cos x = 0$ at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$. At these points, $\sin x = \pm 1$, so $\cos x + \sin x \neq 0$. Thus, there are no critical points where $f'(x)$ is undefined.

Critical points are $x = \frac{3\pi}{4}$ and $x = \frac{7\pi}{4}$.

We can use the second derivative test.

Second derivative: $f''(x) = \frac{d}{dx}(\cos x + \sin x) = -\sin x + \cos x$.

At $x = \frac{3\pi}{4}$:

$f''(\frac{3\pi}{4}) = -\sin(\frac{3\pi}{4}) + \cos(\frac{3\pi}{4}) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2}$.

Since $f''(\frac{3\pi}{4}) < 0$, $x = \frac{3\pi}{4}$ is a point of local maximum.

Local maximum value: $f(\frac{3\pi}{4}) = \sin(\frac{3\pi}{4}) - \cos(\frac{3\pi}{4}) = \frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}}) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$.

At $x = \frac{7\pi}{4}$:

$f''(\frac{7\pi}{4}) = -\sin(\frac{7\pi}{4}) + \cos(\frac{7\pi}{4}) = -(-\frac{1}{\sqrt{2}}) + \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$.

Wait, there's a mistake in the second derivative value. $-\sin(\frac{7\pi}{4}) = -(-\frac{1}{\sqrt{2}}) = \frac{1}{\sqrt{2}}$. $\cos(\frac{7\pi}{4}) = \frac{1}{\sqrt{2}}$. So $f''(\frac{7\pi}{4}) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$. Something is wrong. $\cos x + \sin x = 0$. If $\cos x = \frac{1}{\sqrt{2}}$, $\sin x = -\frac{1}{\sqrt{2}}$ (for $x = \frac{7\pi}{4}$). $\cos x - \sin x = \frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}}) = \sqrt{2}$. $f''(x) = \cos x - \sin x$. At $x=\frac{7\pi}{4}$, $f''(\frac{7\pi}{4}) = \cos(\frac{7\pi}{4}) - \sin(\frac{7\pi}{4}) = \frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}}) = \frac{2}{\sqrt{2}} = \sqrt{2}$.

The second derivative is $f''(x) = -\sin x + \cos x$.

At $x = \frac{3\pi}{4}$: $f''(\frac{3\pi}{4}) = -\sin(\frac{3\pi}{4}) + \cos(\frac{3\pi}{4}) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2} < 0$. Local Maximum.

At $x = \frac{7\pi}{4}$: $f''(\frac{7\pi}{4}) = -\sin(\frac{7\pi}{4}) + \cos(\frac{7\pi}{4}) = -(-\frac{1}{\sqrt{2}}) + \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} > 0$. Local Minimum.

Local minimum value at $x = \frac{7\pi}{4}$: $f(\frac{7\pi}{4}) = \sin(\frac{7\pi}{4}) - \cos(\frac{7\pi}{4}) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2}$.

Local maximum: at $x = \frac{3\pi}{4}$, value is $\sqrt{2}$.

Local minimum: at $x = \frac{7\pi}{4}$, value is $-\sqrt{2}$.

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(v) $f(x) = x^3 – 6x^2 + 9x + 15$

Domain: $R$.

First derivative: $f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 9x + 15) = 3x^2 - 12x + 9$.

Critical points: Set $f'(x) = 0 \Rightarrow 3x^2 - 12x + 9 = 0 \Rightarrow 3(x^2 - 4x + 3) = 0$.

Factor the quadratic: $x^2 - 4x + 3 = (x - 1)(x - 3)$.

So, $3(x - 1)(x - 3) = 0$. Critical points are $x = 1$ and $x = 3$.

Sign of $f'(x) = 3(x - 1)(x - 3)$:

• Interval $(-\infty, 1)$: Choose $x = 0$, $f'(0) = 3(-1)(-3) = 9 > 0$ (increasing).
• Interval $(1, 3)$: Choose $x = 2$, $f'(2) = 3(1)(-1) = -3 < 0$ (decreasing).
• Interval $(3, \infty)$: Choose $x = 4$, $f'(4) = 3(3)(1) = 9 > 0$ (increasing).

At $x = 1$, $f'(x)$ changes from positive to negative, so $x = 1$ is a point of local maximum.

Local maximum value: $f(1) = (1)^3 - 6(1)^2 + 9(1) + 15 = 1 - 6 + 9 + 15 = 19$.

At $x = 3$, $f'(x)$ changes from negative to positive, so $x = 3$ is a point of local minimum.

Local minimum value: $f(3) = (3)^3 - 6(3)^2 + 9(3) + 15 = 27 - 6(9) + 27 + 15 = 27 - 54 + 27 + 15 = 15$.

Local maximum: at $x = 1$, value is 19.

Local minimum: at $x = 3$, value is 15.

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(vi) $g(x) = \frac{x}{2} + \frac{2}{x}$, $x > 0$

Domain: $(0, \infty)$.

First derivative: $g'(x) = \frac{d}{dx}(\frac{1}{2}x + 2x^{-1}) = \frac{1}{2} - 2x^{-2} = \frac{1}{2} - \frac{2}{x^2} = \frac{x^2 - 4}{2x^2}$.

Critical points: Set $g'(x) = 0 \Rightarrow \frac{x^2 - 4}{2x^2} = 0 \Rightarrow x^2 - 4 = 0 \Rightarrow (x - 2)(x + 2) = 0$.

This gives $x = 2$ or $x = -2$.

Also, $g'(x)$ is undefined at $x = 0$.

Considering the domain $x > 0$, the only critical point is $x = 2$. The boundary $x=0$ is excluded from the domain.

Sign of $g'(x) = \frac{(x - 2)(x + 2)}{2x^2}$ for $x > 0$:

• Interval $(0, 2)$: Choose $x = 1$, $g'(1) = \frac{(1 - 2)(1 + 2)}{2(1)^2} = \frac{(-1)(3)}{2} = -\frac{3}{2} < 0$ (decreasing).
• Interval $(2, \infty)$: Choose $x = 3$, $g'(3) = \frac{(3 - 2)(3 + 2)}{2(3)^2} = \frac{(1)(5)}{18} = \frac{5}{18} > 0$ (increasing).

At $x = 2$, $g'(x)$ changes from negative to positive, so $x = 2$ is a point of local minimum.

Local minimum value: $g(2) = \frac{2}{2} + \frac{2}{2} = 1 + 1 = 2$.

There is no local maximum on the domain $(0, \infty)$.

Local minimum: at $x = 2$, value is 2.

Local maximum: None.

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(vii) $g(x) = \frac{1}{x^{2}+2}$

Domain: $R$. The denominator $x^2 + 2$ is never zero ($x^2 \ge 0 \Rightarrow x^2 + 2 \ge 2$).

First derivative: $g'(x) = \frac{d}{dx}((x^2 + 2)^{-1}) = -1(x^2 + 2)^{-2} \cdot \frac{d}{dx}(x^2 + 2) = -(x^2 + 2)^{-2} \cdot 2x = -\frac{2x}{(x^2 + 2)^2}$.

Critical points: Set $g'(x) = 0 \Rightarrow -\frac{2x}{(x^2 + 2)^2} = 0 \Rightarrow -2x = 0 \Rightarrow x = 0$.

$g'(x)$ is defined for all real $x$. The only critical point is $x = 0$.

Sign of $g'(x) = -\frac{2x}{(x^2 + 2)^2}$. The denominator $(x^2 + 2)^2$ is always positive for all real $x$. The sign depends on the numerator $-2x$.

• Interval $(-\infty, 0)$: Choose $x = -1$, $g'(-1) = -\frac{2(-1)}{((-1)^2 + 2)^2} = \frac{2}{(1 + 2)^2} = \frac{2}{9} > 0$ (increasing).
• Interval $(0, \infty)$: Choose $x = 1$, $g'(1) = -\frac{2(1)}{(1^2 + 2)^2} = -\frac{2}{(1 + 2)^2} = -\frac{2}{9} < 0$ (decreasing).

At $x = 0$, $g'(x)$ changes from positive to negative, so $x = 0$ is a point of local maximum.

Local maximum value: $g(0) = \frac{1}{(0)^2 + 2} = \frac{1}{2}$.

There is no local minimum. Note that as $|x| \to \infty$, $g(x) = \frac{1}{x^2 + 2} \to 0$. The function approaches 0 but never reaches it.

Local maximum: at $x = 0$, value is $\frac{1}{2}$.

Local minimum: None.

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(viii) $f(x) = x\sqrt{1-x}$, $0 < x < 1$

Domain: $(0, 1)$. Note that $\sqrt{1-x}$ requires $1-x \ge 0 \Rightarrow x \le 1$. The domain given is $0 < x < 1$, which means $x \in (0, 1)$. The function is defined and differentiable in this open interval.

First derivative: $f'(x) = \frac{d}{dx}(x(1-x)^{1/2})$. Use the product rule.

$f'(x) = \frac{d}{dx}(x) \cdot \sqrt{1-x} + x \cdot \frac{d}{dx}((1-x)^{1/2})$

$f'(x) = 1 \cdot \sqrt{1-x} + x \cdot \frac{1}{2}(1-x)^{-1/2} \cdot (-1)$

$f'(x) = \sqrt{1-x} - \frac{x}{2\sqrt{1-x}}$

$f'(x) = \frac{\sqrt{1-x} \cdot 2\sqrt{1-x} - x}{2\sqrt{1-x}}$

$f'(x) = \frac{2(1-x) - x}{2\sqrt{1-x}}$

$f'(x) = \frac{2 - 2x - x}{2\sqrt{1-x}} = \frac{2 - 3x}{2\sqrt{1-x}}$

Critical points: Set $f'(x) = 0 \Rightarrow \frac{2 - 3x}{2\sqrt{1-x}} = 0$.

The numerator must be zero: $2 - 3x = 0 \Rightarrow 3x = 2 \Rightarrow x = \frac{2}{3}$.

The critical point is $x = \frac{2}{3}$. This point is in the domain $(0, 1)$, since $0 < \frac{2}{3} < 1$.

$f'(x)$ is undefined when the denominator is zero: $2\sqrt{1-x} = 0 \Rightarrow \sqrt{1-x} = 0 \Rightarrow 1-x = 0 \Rightarrow x = 1$. However, $x = 1$ is not in the open domain $(0, 1)$.

Sign of $f'(x) = \frac{2 - 3x}{2\sqrt{1-x}}$ in $(0, 1)$. The denominator $2\sqrt{1-x}$ is positive for $x \in (0, 1)$. The sign depends on the numerator $2 - 3x$.

• Interval $(0, \frac{2}{3})$: Choose $x = 0.5 = \frac{1}{2}$. $2 - 3(\frac{1}{2}) = 2 - \frac{3}{2} = \frac{1}{2} > 0$. $f'(x) > 0$ (increasing).
• Interval $(\frac{2}{3}, 1)$: Choose $x = 0.75 = \frac{3}{4}$. $2 - 3(\frac{3}{4}) = 2 - \frac{9}{4} = \frac{8 - 9}{4} = -\frac{1}{4} < 0$. $f'(x) < 0$ (decreasing).

At $x = \frac{2}{3}$, $f'(x)$ changes from positive to negative, so $x = \frac{2}{3}$ is a point of local maximum.

Local maximum value: $f(\frac{2}{3}) = \frac{2}{3}\sqrt{1 - \frac{2}{3}} = \frac{2}{3}\sqrt{\frac{1}{3}} = \frac{2}{3} \cdot \frac{1}{\sqrt{3}} = \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{9}$.

There is no local minimum in the open interval $(0, 1)$.

Local maximum: at $x = \frac{2}{3}$, value is $\frac{2\sqrt{3}}{9}$.

Local minimum: None.

A continuous and differentiable function has local maxima or minima only at its critical points (where the first derivative is zero or undefined).

If a function is strictly increasing or strictly decreasing over its entire domain, it does not have any local extrema (and thus no global extrema unless the domain is a closed interval).

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(i) $f(x) = e^x$

Domain: $R = (-\infty, \infty)$.

First derivative: $f'(x) = \frac{d}{dx}(e^x) = e^x$.

For all real numbers $x$, the exponential function $e^x$ is strictly positive ($e^x > 0$).

Since $f'(x) = e^x > 0$ for all $x \in R$, the function $f(x) = e^x$ is strictly increasing on its entire domain $R$.

A strictly increasing function on an unbounded domain has no local maximum and no local minimum.

As $x \to \infty$, $f(x) = e^x \to \infty$. There is no maximum value.

As $x \to -\infty$, $f(x) = e^x \to 0$. The function approaches 0 but never reaches it, so there is no minimum value.

Therefore, the function $f(x) = e^x$ does not have maxima or minima.

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(ii) $g(x) = \log x$

Domain: $(0, \infty)$. Note that $\log x$ typically refers to the natural logarithm $\ln x$. If it is $\log_b x$ with $b>1$, the result is similar.

Let's assume $g(x) = \ln x$.

First derivative: $g'(x) = \frac{d}{dx}(\ln x) = \frac{1}{x}$.

The domain is $x > 0$. For all $x$ in the domain $(0, \infty)$, $\frac{1}{x}$ is strictly positive ($\frac{1}{x} > 0$).

Since $g'(x) = \frac{1}{x} > 0$ for all $x \in (0, \infty)$, the function $g(x) = \ln x$ is strictly increasing on its entire domain $(0, \infty)$.

A strictly increasing function on an unbounded domain has no local maximum and no local minimum.

As $x \to \infty$, $g(x) = \ln x \to \infty$. There is no maximum value.

As $x \to 0^+$ (from the right, since domain is $x>0$), $g(x) = \ln x \to -\infty$. There is no minimum value.

Therefore, the function $g(x) = \log x$ does not have maxima or minima.

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(iii) $h(x) = x^3 + x^2 + x + 1$

Domain: $R = (-\infty, \infty)$.

First derivative: $h'(x) = \frac{d}{dx}(x^3 + x^2 + x + 1) = 3x^2 + 2x + 1$.

To find critical points, we set $h'(x) = 0$: $3x^2 + 2x + 1 = 0$.

This is a quadratic equation. We check the discriminant to find the nature of the roots.

Discriminant $\Delta = b^2 - 4ac = (2)^2 - 4(3)(1) = 4 - 12 = -8$.

Since the discriminant is negative ($\Delta < 0$) and the coefficient of $x^2$ (which is 3) is positive, the quadratic expression $3x^2 + 2x + 1$ is always positive for all real values of $x$.

So, $h'(x) = 3x^2 + 2x + 1 > 0$ for all $x \in R$.

Since $h'(x) > 0$ for all $x \in R$, the function $h(x) = x^3 + x^2 + x + 1$ is strictly increasing on its entire domain $R$.

A strictly increasing function on an unbounded domain has no local maximum and no local minimum.

As $x \to \infty$, $h(x) \to \infty$. There is no maximum value.

As $x \to -\infty$, $h(x) \to -\infty$. There is no minimum value.

Therefore, the function $h(x) = x^3 + x^2 + x + 1$ does not have maxima or minima.

To find the absolute maximum and minimum values of a continuous function on a closed interval, we evaluate the function at the critical points within the interval and at the endpoints of the interval. The largest value is the absolute maximum, and the smallest value is the absolute minimum.

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(i) $f(x) = x^3$, x ∈ [– 2, 2]

Interval: $[-2, 2]$.

First derivative: $f'(x) = \frac{d}{dx}(x^3) = 3x^2$.

Critical points: Set $f'(x) = 0 \Rightarrow 3x^2 = 0 \Rightarrow x = 0$.

The critical point $x = 0$ is within the interval $[-2, 2]$.

Endpoints: $x = -2$ and $x = 2$.

Evaluate $f(x)$ at the critical point and endpoints:

$f(0) = (0)^3 = 0$

$f(-2) = (-2)^3 = -8$

$f(2) = (2)^3 = 8$

Comparing the values: -8, 0, 8.

Absolute maximum value is 8.

Absolute minimum value is -8.

Absolute maximum value: 8

Absolute minimum value: -8

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(ii) $f (x) = \sin x + \cos x$, x ∈ [0, π]

Interval: $[0, \pi]$.

First derivative: $f'(x) = \frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x$.

Critical points: Set $f'(x) = 0 \Rightarrow \cos x - \sin x = 0 \Rightarrow \cos x = \sin x$.

Divide by $\cos x$ (assuming $\cos x \neq 0$) to get $\tan x = 1$.

In the interval $[0, \pi]$, the only solution is $x = \frac{\pi}{4}$. (Note: $\cos x = 0$ at $x = \frac{\pi}{2}$, where $\sin x = 1$, so $\cos x - \sin x = -1 \neq 0$. Thus, $x=\frac{\pi}{2}$ is not a critical point from this equation).

The critical point $x = \frac{\pi}{4}$ is within the interval $[0, \pi]$.

Endpoints: $x = 0$ and $x = \pi$.

Evaluate $f(x)$ at the critical point and endpoints:

$f(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$

$f(0) = \sin(0) + \cos(0) = 0 + 1 = 1$

$f(\pi) = \sin(\pi) + \cos(\pi) = 0 + (-1) = -1$

Comparing the values: $\sqrt{2} \approx 1.414$, 1, -1.

Absolute maximum value is $\sqrt{2}$.

Absolute minimum value is -1.

Absolute maximum value: $\sqrt{2}$

Absolute minimum value: -1

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(iii) f(x) = 4x - $\frac{1}{2}$ x² , x ∈ $\left[ -2,\frac{9}{2} \right]$

Interval: $\left[ -2,\frac{9}{2} \right]$. Note $\frac{9}{2} = 4.5$. The interval is $[-2, 4.5]$.

First derivative: $f'(x) = \frac{d}{dx}(4x - \frac{1}{2}x^2) = 4 - \frac{1}{2}(2x) = 4 - x$.

Critical points: Set $f'(x) = 0 \Rightarrow 4 - x = 0 \Rightarrow x = 4$.

The critical point $x = 4$ is within the interval $[-2, 4.5]$.

Endpoints: $x = -2$ and $x = \frac{9}{2} = 4.5$.

Evaluate $f(x)$ at the critical point and endpoints:

$f(4) = 4(4) - \frac{1}{2}(4)^2 = 16 - \frac{1}{2}(16) = 16 - 8 = 8$

$f(-2) = 4(-2) - \frac{1}{2}(-2)^2 = -8 - \frac{1}{2}(4) = -8 - 2 = -10$

$f(\frac{9}{2}) = 4(\frac{9}{2}) - \frac{1}{2}(\frac{9}{2})^2 = \frac{36}{2} - \frac{1}{2}(\frac{81}{4}) = 18 - \frac{81}{8} = \frac{144 - 81}{8} = \frac{63}{8} = 7.875$

Comparing the values: 8, -10, $\frac{63}{8} = 7.875$.

Absolute maximum value is 8.

Absolute minimum value is -10.

Absolute maximum value: 8

Absolute minimum value: -10

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(iv) f(x) = (x - 1)² + 3, x ∈ [-3, 1]

Interval: $[-3, 1]$.

First derivative: $f'(x) = \frac{d}{dx}((x - 1)^2 + 3) = 2(x - 1) \cdot \frac{d}{dx}(x - 1) + 0 = 2(x - 1)$.

Critical points: Set $f'(x) = 0 \Rightarrow 2(x - 1) = 0 \Rightarrow x - 1 = 0 \Rightarrow x = 1$.

The critical point $x = 1$ is one of the endpoints of the interval $[-3, 1]$.

Endpoints: $x = -3$ and $x = 1$.

Evaluate $f(x)$ at the critical point/endpoint and the other endpoint:

$f(1) = (1 - 1)^2 + 3 = 0^2 + 3 = 3$

$f(-3) = (-3 - 1)^2 + 3 = (-4)^2 + 3 = 16 + 3 = 19$

Comparing the values: 3, 19.

Absolute maximum value is 19.

Absolute minimum value is 3.

Absolute maximum value: 19

Absolute minimum value: 3

Given the profit function $p(x) = 41 – 72x – 18x^2$.

This is a quadratic function of the form $p(x) = ax^2 + bx + c$, where $a = -18$, $b = -72$, and $c = 41$.

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Since the coefficient of $x^2$ is negative ($a = -18 < 0$), the graph of the function is a parabola that opens downwards. A parabola opening downwards has a maximum value at its vertex.

The x-coordinate of the vertex is given by the formula $x = -\frac{b}{2a}$.

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Calculate the x-coordinate of the vertex:

$x = -\frac{-72}{2(-18)}$

$x = -\frac{-72}{-36}$

$x = -2$

The maximum profit occurs at $x = -2$. While $x$ often represents a quantity (like the number of units produced), which is typically non-negative, the question asks for the maximum profit of the given function. If $x$ were restricted (e.g., $x \ge 0$), we would need to check the vertex and boundary/critical points within the feasible domain. However, based solely on the function form and the request for the maximum profit (implying it exists), we find the vertex of the parabola.

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Substitute this value of $x$ into the profit function $p(x)$ to find the maximum profit:

$p(-2) = 41 - 72(-2) - 18(-2)^2$

$p(-2) = 41 + 144 - 18(4)$

$p(-2) = 41 + 144 - 72$

$p(-2) = 185 - 72$

$p(-2) = 113$

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Alternatively, using calculus:

Find the first derivative $p'(x)$:

$p'(x) = \frac{d}{dx}(41 - 72x - 18x^2) = -72 - 36x$

Set $p'(x) = 0$ to find critical points:

$-72 - 36x = 0$

$-36x = 72$

$x = \frac{72}{-36} = -2$

The critical point is $x = -2$.

Find the second derivative $p''(x)$:

$p''(x) = \frac{d}{dx}(-72 - 36x) = -36$

Evaluate $p''(x)$ at the critical point $x = -2$:

$p''(-2) = -36$

Since $p''(-2) < 0$, the critical point $x = -2$ corresponds to a local maximum. Since it's a quadratic function, this local maximum is also the absolute maximum.

The maximum profit is $p(-2) = 113$.

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The maximum profit that a company can make, according to the given profit function, is 113.

Let the function be $f(x) = 3x^4 – 8x^3 + 12x^2 – 48x + 25$.

We need to find the absolute maximum and minimum values of this function on the closed interval $[0, 3]$.

Since $f(x)$ is a polynomial, it is continuous on $[0, 3]$ and differentiable on $(0, 3)$.

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Step 1: Find the critical points in the interval (0, 3).

Calculate the first derivative $f'(x)$:

$f'(x) = \frac{d}{dx}(3x^4 - 8x^3 + 12x^2 - 48x + 25)$

$f'(x) = 12x^3 - 24x^2 + 24x - 48$

Set $f'(x) = 0$ to find the critical points:

$12x^3 - 24x^2 + 24x - 48 = 0$

Factor out the common factor 12:

$12(x^3 - 2x^2 + 2x - 4) = 0$

Divide by 12:

$x^3 - 2x^2 + 2x - 4 = 0$

Factor by grouping:

$(x^3 - 2x^2) + (2x - 4) = 0$

$x^2(x - 2) + 2(x - 2) = 0$

$(x - 2)(x^2 + 2) = 0$

Set each factor to zero to find the critical points:

Case 1: $x - 2 = 0 \Rightarrow x = 2$.

Case 2: $x^2 + 2 = 0 \Rightarrow x^2 = -2$. This equation has no real solutions.

The only real critical point is $x = 2$.

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Step 2: Check if the critical point lies within the given interval [0, 3].

The interval is $[0, 3]$.

For $x = 2$: $0 \le 2 \le 3$. So, $x = 2$ is in the interval.

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Step 3: Evaluate the function $f(x)$ at the critical point within the interval and at the endpoints of the interval.

The endpoints of the interval are $x = 0$ and $x = 3$.

The critical point within the interval is $x = 2$.

Evaluate $f(x)$ at $x = 0$ (left endpoint):

$f(0) = 3(0)^4 - 8(0)^3 + 12(0)^2 - 48(0) + 25 = 0 - 0 + 0 - 0 + 25 = 25$

Evaluate $f(x)$ at $x = 2$ (critical point):

$f(2) = 3(2)^4 - 8(2)^3 + 12(2)^2 - 48(2) + 25$

$f(2) = 3(16) - 8(8) + 12(4) - 96 + 25$

$f(2) = 48 - 64 + 48 - 96 + 25$

$f(2) = (48 + 48 + 25) - (64 + 96)$

$f(2) = 121 - 160$

$f(2) = -39$

Evaluate $f(x)$ at $x = 3$ (right endpoint):

$f(3) = 3(3)^4 - 8(3)^3 + 12(3)^2 - 48(3) + 25$

$f(3) = 3(81) - 8(27) + 12(9) - 144 + 25$

$f(3) = 243 - 216 + 108 - 144 + 25$

$f(3) = (243 + 108 + 25) - (216 + 144)$

$f(3) = 376 - 360$

$f(3) = 16$

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Step 4: Identify the absolute maximum and minimum values from the evaluated function values.

The values are $f(0) = 25$, $f(2) = -39$, and $f(3) = 16$.

Comparing these values: $25 > 16 > -39$.

The largest value is 25.

The smallest value is -39.

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Conclusion:

The absolute maximum value of the function on the interval [0, 3] is 25.

The absolute minimum value of the function on the interval [0, 3] is -39.

Given the function $f(x) = \sin(2x)$ on the interval $[0, 2\pi]$.

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The sine function, $\sin u$, has a maximum value of 1.

So, the maximum value of $f(x) = \sin(2x)$ is also 1.

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We need to find the values of $x$ in the interval $[0, 2\pi]$ such that $\sin(2x) = 1$.

The general solution for $\sin u = 1$ is given by $u = \frac{\pi}{2} + 2n\pi$, where $n$ is an integer.

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In our case, the argument is $2x$. So, we have:

Divide equation (i) by 2 to solve for $x$:

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Now, we find the integer values of $n$ for which the value of $x$ from equation (ii) lies within the given interval $[0, 2\pi]$.

$0 \le \frac{\pi}{4} + n\pi \le 2\pi$

Subtract $\frac{\pi}{4}$ from all parts of the inequality:

$0 - \frac{\pi}{4} \le n\pi \le 2\pi - \frac{\pi}{4}$

$-\frac{\pi}{4} \le n\pi \le \frac{8\pi}{4} - \frac{\pi}{4}$

$-\frac{\pi}{4} \le n\pi \le \frac{7\pi}{4}$

Divide all parts by $\pi$ (which is positive):

$-\frac{1}{4} \le n \le \frac{7}{4}$

Since $n$ must be an integer, the possible values for $n$ are $0$ and $1$.

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Substitute these values of $n$ back into the equation for $x$ (ii):

For $n = 0$: $x = \frac{\pi}{4} + 0\pi = \frac{\pi}{4}$.

For $n = 1$: $x = \frac{\pi}{4} + 1\pi = \frac{\pi}{4} + \frac{4\pi}{4} = \frac{5\pi}{4}$.

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Both points, $\frac{\pi}{4}$ and $\frac{5\pi}{4}$, are within the interval $[0, 2\pi]$.

At these points, $f(\frac{\pi}{4}) = \sin(2 \cdot \frac{\pi}{4}) = \sin(\frac{\pi}{2}) = 1$ and $f(\frac{5\pi}{4}) = \sin(2 \cdot \frac{5\pi}{4}) = \sin(\frac{5\pi}{2}) = \sin(2\pi + \frac{\pi}{2}) = \sin(\frac{\pi}{2}) = 1$.

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The function $f(x) = \sin(2x)$ attains its maximum value in the interval $[0, 2\pi]$ at the points $\frac{\pi}{4}$ and $\frac{5\pi}{4}$.

Let the function be $f(x) = \sin x + \cos x$. The domain of the function is the set of all real numbers, $R$.

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Method 1: Using Calculus

Find the first derivative $f'(x)$:

$f'(x) = \frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x$

Set $f'(x) = 0$ to find critical points:

$\cos x - \sin x = 0$

$\cos x = \sin x$

Divide by $\cos x$ (assuming $\cos x \neq 0$) to get $\tan x = 1$.

The general solution for $\tan x = 1$ is $x = \frac{\pi}{4} + n\pi$, where $n$ is an integer.

Find the second derivative $f''(x)$:

$f''(x) = \frac{d}{dx}(\cos x - \sin x) = -\sin x - \cos x = -(\sin x + \cos x)$

Evaluate $f''(x)$ at the critical points $x = \frac{\pi}{4} + n\pi$.

When $n$ is an even integer, let $n = 2k$. Then $x = \frac{\pi}{4} + 2k\pi$.

$\sin(\frac{\pi}{4} + 2k\pi) = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$

$\cos(\frac{\pi}{4} + 2k\pi) = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$

$f''(\frac{\pi}{4} + 2k\pi) = -(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = -\frac{2}{\sqrt{2}} = -\sqrt{2}$. Since $f''(x) < 0$, these points are local maxima.

When $n$ is an odd integer, let $n = 2k + 1$. Then $x = \frac{\pi}{4} + (2k + 1)\pi = \frac{\pi}{4} + \pi + 2k\pi = \frac{5\pi}{4} + 2k\pi$.

$\sin(\frac{5\pi}{4} + 2k\pi) = \sin(\frac{5\pi}{4}) = -\frac{1}{\sqrt{2}}$

$\cos(\frac{5\pi}{4} + 2k\pi) = \cos(\frac{5\pi}{4}) = -\frac{1}{\sqrt{2}}$

$f''(\frac{5\pi}{4} + 2k\pi) = -(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = -(-\frac{2}{\sqrt{2}}) = \sqrt{2}$. Since $f''(x) > 0$, these points are local minima.

The local maximum values occur at $x = \frac{\pi}{4} + 2k\pi$. The value is $f(\frac{\pi}{4} + 2k\pi) = \sin(\frac{\pi}{4} + 2k\pi) + \cos(\frac{\pi}{4} + 2k\pi) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$.

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Method 2: Using Trigonometric Identity

We can express $f(x) = \sin x + \cos x$ in the form $R \sin(x + \alpha)$ or $R \cos(x + \alpha)$.

Let $\sin x + \cos x = R \sin(x + \alpha) = R (\sin x \cos \alpha + \cos x \sin \alpha)$.

Comparing coefficients of $\sin x$ and $\cos x$:

Square and add (i) and (ii): $R^2 \cos^2 \alpha + R^2 \sin^2 \alpha = 1^2 + 1^2$

$R^2(\cos^2 \alpha + \sin^2 \alpha) = 1 + 1$

$R^2(1) = 2$

$R = \sqrt{2}$ (since R is amplitude, $R > 0$)

Divide (ii) by (i): $\frac{R \sin \alpha}{R \cos \alpha} = \frac{1}{1} \Rightarrow \tan \alpha = 1$. We can choose $\alpha = \frac{\pi}{4}$.

So, $f(x) = \sqrt{2} \sin(x + \frac{\pi}{4})$.

The range of the sine function is $[-1, 1]$.

So, $-1 \le \sin(x + \frac{\pi}{4}) \le 1$ for all $x \in R$.

Multiply by $\sqrt{2}$:

$-\sqrt{2} \le \sqrt{2} \sin(x + \frac{\pi}{4}) \le \sqrt{2}$

$-\sqrt{2} \le f(x) \le \sqrt{2}$

The maximum value of $f(x)$ is $\sqrt{2}$.

The minimum value of $f(x)$ is $-\sqrt{2}$.

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The maximum value of the function $\sin x + \cos x$ is $\sqrt{2}$.

Let the function be $f(x) = 2x^3 – 24x + 107$. We need to find the absolute maximum value of this function on two different closed intervals: $[1, 3]$ and $[-3, -1]$.

Since $f(x)$ is a polynomial, it is continuous and differentiable everywhere.

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First, we find the critical points of the function.

Calculate the first derivative $f'(x)$:

$f'(x) = \frac{d}{dx}(2x^3 - 24x + 107)$

$f'(x) = 6x^2 - 24$

Set $f'(x) = 0$ to find the critical points:

$6x^2 - 24 = 0$

$6(x^2 - 4) = 0$

$x^2 - 4 = 0$

$(x - 2)(x + 2) = 0$

The critical points are $x = 2$ and $x = -2$.

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Maximum value in the interval [1, 3]:

The interval is $[1, 3]$. We need to check the critical points within this interval and the endpoints.

The critical points are $x = 2$ and $x = -2$.

Only $x = 2$ is within the interval $[1, 3]$.

The endpoints are $x = 1$ and $x = 3$.

Evaluate the function at $x = 1, 2, 3$:

$f(1) = 2(1)^3 - 24(1) + 107 = 2 - 24 + 107 = 85$

$f(2) = 2(2)^3 - 24(2) + 107 = 2(8) - 48 + 107 = 16 - 48 + 107 = -32 + 107 = 75$

$f(3) = 2(3)^3 - 24(3) + 107 = 2(27) - 72 + 107 = 54 - 72 + 107 = -18 + 107 = 89$

Comparing the values: 85, 75, 89. The maximum value is 89.

The maximum value of the function $f(x)$ in the interval [1, 3] is 89.

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Maximum value in the interval [–3, –1]:

The interval is $[-3, -1]$. We need to check the critical points within this interval and the endpoints.

The critical points are $x = 2$ and $x = -2$.

Only $x = -2$ is within the interval $[-3, -1]$.

The endpoints are $x = -3$ and $x = -1$.

Evaluate the function at $x = -3, -2, -1$:

$f(-3) = 2(-3)^3 - 24(-3) + 107 = 2(-27) + 72 + 107 = -54 + 72 + 107 = 18 + 107 = 125$

$f(-2) = 2(-2)^3 - 24(-2) + 107 = 2(-8) + 48 + 107 = -16 + 48 + 107 = 32 + 107 = 139$

$f(-1) = 2(-1)^3 - 24(-1) + 107 = 2(-1) + 24 + 107 = -2 + 24 + 107 = 22 + 107 = 129$

Comparing the values: 125, 139, 129. The maximum value is 139.

The maximum value of the function $f(x)$ in the interval [-3, -1] is 139.

Given the function $f(x) = x^4 – 62x^2 + ax + 9$ on the interval $[0, 2]$.

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We are given that the function attains its maximum value at $x = 1$ in the interval $[0, 2]$.

The interval $[0, 2]$ is a closed interval.

The point $x = 1$ is an interior point of the interval $[0, 2]$ since $0 < 1 < 2$.

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According to the theory of extrema, if a differentiable function $f(x)$ attains its maximum or minimum value at an interior point $c$ of an interval, then that point must be a critical point. A critical point is where the derivative $f'(x)$ is either zero or undefined.

Since $f(x)$ is a polynomial, its derivative $f'(x)$ exists for all real numbers $x$, and hence is defined at $x=1$.

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Therefore, if $f(x)$ attains its maximum at $x = 1$, then $f'(1)$ must be equal to 0.

First, let's find the derivative of $f(x)$:

$f'(x) = \frac{d}{dx}(x^4 - 62x^2 + ax + 9)$

$f'(x) = 4x^{4-1} - 62 \cdot 2x^{2-1} + a \cdot 1 + 0$

$f'(x) = 4x^3 - 124x + a$

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Now, we use the condition that $f'(1) = 0$:

Substitute $x = 1$ into the expression for $f'(x)$:

$f'(1) = 4(1)^3 - 124(1) + a$

$f'(1) = 4(1) - 124 + a$

$f'(1) = 4 - 124 + a$

$f'(1) = -120 + a$

Setting the derivative at $x=1$ to zero:

$a = 120$

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Thus, the value of $a$ for which the function $f(x) = x^4 – 62x^2 + ax + 9$ attains its maximum value at $x = 1$ on the interval $[0, 2]$ is 120.

Let the function be $f(x) = x + \sin(2x)$. The given interval is the closed interval $[0, 2\pi]$.

Since $f(x)$ is a sum of polynomial ($x$) and trigonometric ($\sin(2x)$) functions, it is continuous on the closed interval $[0, 2\pi]$ and differentiable on the open interval $(0, 2\pi)$.

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To find the absolute maximum and minimum values, we evaluate the function at the critical points within the open interval $(0, 2\pi)$ and at the endpoints of the interval $[0, 2\pi]$.

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Step 1: Find the critical points in the interval $(0, 2\pi)$.

Calculate the first derivative $f'(x)$:

$f'(x) = \frac{d}{dx}(x + \sin(2x))$

$f'(x) = 1 + \cos(2x) \cdot \frac{d}{dx}(2x)$

$f'(x) = 1 + 2\cos(2x)$

Set $f'(x) = 0$ to find the critical points:

$1 + 2\cos(2x) = 0$

$2\cos(2x) = -1$

$\cos(2x) = -\frac{1}{2}$

We need to find the values of $2x$ such that $\cos(2x) = -\frac{1}{2}$. In the interval $[0, 4\pi]$ (corresponding to $x \in [0, 2\pi]$), the general solutions for $\cos u = -\frac{1}{2}$ are $u = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}$.

Setting $2x$ equal to these values:

$2x = \frac{2\pi}{3} \Rightarrow x = \frac{\pi}{3}$

$2x = \frac{4\pi}{3} \Rightarrow x = \frac{2\pi}{3}$

$2x = \frac{8\pi}{3} \Rightarrow x = \frac{4\pi}{3}$

$2x = \frac{10\pi}{3} \Rightarrow x = \frac{5\pi}{3}$

The critical points in the interval $(0, 2\pi)$ are $x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$.

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Step 2: Evaluate the function $f(x)$ at the critical points within the interval and at the endpoints of the interval.

The endpoints are $x = 0$ and $x = 2\pi$.

The critical points in $[0, 2\pi]$ are $x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$ and the endpoints $0, 2\pi$.

Evaluate $f(x)$ at $x = 0$ (left endpoint):

$f(0) = 0 + \sin(2 \cdot 0) = 0 + \sin(0) = 0 + 0 = 0$

Evaluate $f(x)$ at $x = \frac{\pi}{3}$ (critical point):

$f(\frac{\pi}{3}) = \frac{\pi}{3} + \sin(2 \cdot \frac{\pi}{3}) = \frac{\pi}{3} + \sin(\frac{2\pi}{3})$

$f(\frac{\pi}{3}) = \frac{\pi}{3} + \frac{\sqrt{3}}{2}$

Evaluate $f(x)$ at $x = \frac{2\pi}{3}$ (critical point):

$f(\frac{2\pi}{3}) = \frac{2\pi}{3} + \sin(2 \cdot \frac{2\pi}{3}) = \frac{2\pi}{3} + \sin(\frac{4\pi}{3})$

$f(\frac{2\pi}{3}) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}$

Evaluate $f(x)$ at $x = \frac{4\pi}{3}$ (critical point):

$f(\frac{4\pi}{3}) = \frac{4\pi}{3} + \sin(2 \cdot \frac{4\pi}{3}) = \frac{4\pi}{3} + \sin(\frac{8\pi}{3})$

$f(\frac{4\pi}{3}) = \frac{4\pi}{3} + \sin(2\pi + \frac{2\pi}{3}) = \frac{4\pi}{3} + \sin(\frac{2\pi}{3})$

$f(\frac{4\pi}{3}) = \frac{4\pi}{3} + \frac{\sqrt{3}}{2}$

Evaluate $f(x)$ at $x = \frac{5\pi}{3}$ (critical point):

$f(\frac{5\pi}{3}) = \frac{5\pi}{3} + \sin(2 \cdot \frac{5\pi}{3}) = \frac{5\pi}{3} + \sin(\frac{10\pi}{3})$

$f(\frac{5\pi}{3}) = \frac{5\pi}{3} + \sin(3\pi + \frac{\pi}{3}) = \frac{5\pi}{3} - \sin(\frac{\pi}{3})$

$f(\frac{5\pi}{3}) = \frac{5\pi}{3} - \frac{\sqrt{3}}{2}$

Evaluate $f(x)$ at $x = 2\pi$ (right endpoint):

$f(2\pi) = 2\pi + \sin(2 \cdot 2\pi) = 2\pi + \sin(4\pi) = 2\pi + 0 = 2\pi$

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Step 3: Identify the absolute maximum and minimum values from the evaluated function values.

The values are:

$f(0) = 0$

$f(\frac{\pi}{3}) = \frac{\pi}{3} + \frac{\sqrt{3}}{2}$

$f(\frac{2\pi}{3}) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}$

$f(\frac{4\pi}{3}) = \frac{4\pi}{3} + \frac{\sqrt{3}}{2}$

$f(\frac{5\pi}{3}) = \frac{5\pi}{3} - \frac{\sqrt{3}}{2}$

$f(2\pi) = 2\pi$

Comparing these values:

The minimum value is 0, which occurs at $x = 0$. All other values are positive:

$\frac{\pi}{3} + \frac{\sqrt{3}}{2} > 0$

$\frac{2\pi}{3} - \frac{\sqrt{3}}{2}$. Since $\frac{2\pi}{3} \approx 2.094$ and $\frac{\sqrt{3}}{2} \approx 0.866$, this value is positive.

$\frac{4\pi}{3} + \frac{\sqrt{3}}{2} > 0$

$\frac{5\pi}{3} - \frac{\sqrt{3}}{2}$. Since $\frac{5\pi}{3} \approx 5.236$ and $\frac{\sqrt{3}}{2} \approx 0.866$, this value is positive.

$2\pi \approx 6.28 > 0$

So, the absolute minimum value is 0.

The maximum value is $2\pi$, which occurs at $x = 2\pi$. Compare $2\pi$ with the other values:

$\frac{\pi}{3} + \frac{\sqrt{3}}{2} \approx 1.913 < 2\pi \approx 6.28$

$\frac{2\pi}{3} - \frac{\sqrt{3}}{2} \approx 1.228 < 2\pi \approx 6.28$

$\frac{4\pi}{3} + \frac{\sqrt{3}}{2} \approx 5.055 < 2\pi \approx 6.28$

$\frac{5\pi}{3} - \frac{\sqrt{3}}{2} \approx 4.370 < 2\pi \approx 6.28$

So, the absolute maximum value is $2\pi$.

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Conclusion:

The absolute maximum value of the function $f(x) = x + \sin(2x)$ on the interval $[0, 2\pi]$ is $2\pi$.

The absolute minimum value of the function on the interval $[0, 2\pi]$ is 0.

Given:

Two numbers whose sum is 24.

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To Find:

The two numbers such that their product is maximized.

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Solution:

Let the two numbers be $x$ and $y$.

According to the problem statement, their sum is 24:

$x + y = 24$

We want to maximize their product $P$, where:

$P = x \times y$

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From the sum equation, we can express one variable in terms of the other. Let's express $y$ in terms of $x$:

$y = 24 - x$

Now, substitute this expression for $y$ into the product equation:

$P(x) = x(24 - x)$

$P(x) = 24x - x^2$

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The product $P(x)$ is a quadratic function of $x$. We can rewrite it as $P(x) = -x^2 + 24x$. This is in the form $ax^2 + bx + c$, with $a = -1$, $b = 24$, and $c = 0$.

Since the coefficient of the $x^2$ term ($a = -1$) is negative, the parabola represented by this function opens downwards. The maximum value of a downward-opening parabola occurs at its vertex.

The x-coordinate of the vertex of a quadratic function $ax^2 + bx + c$ is given by the formula $x = \frac{-b}{2a}$.

Using this formula for $P(x) = -x^2 + 24x$ (where $a = -1$ and $b = 24$), the value of $x$ that maximizes the product is:

$x = \frac{-(24)}{2(-1)}$

$x = \frac{-24}{-2}$

$x = 12$

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Now that we have found the value of $x$ that maximizes the product, we can find the corresponding value of the second number, $y$, using the sum equation $y = 24 - x$:

$y = 24 - 12$

$y = 12$

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So, the two numbers are 12 and 12.

Their sum is $12 + 12 = 24$, which satisfies the given condition.

Their product is $12 \times 12 = 144$. This is the largest possible product for two numbers summing to 24.

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Alternate Solution using AM-GM Inequality:

For any two non-negative numbers, the Arithmetic Mean (AM) is always greater than or equal to the Geometric Mean (GM). The equality holds if and only if the two numbers are equal.

Let the two non-negative numbers be $x$ and $y$.

The AM is $\frac{x+y}{2}$.

The GM is $\sqrt{xy}$.

The AM-GM inequality states:

$\frac{x+y}{2} \geq \sqrt{xy}$

Given that the sum of the two numbers is 24, $x+y = 24$. Substitute this value into the inequality:

$\frac{24}{2} \geq \sqrt{xy}$

$12 \geq \sqrt{xy}$

To find the maximum value of the product $xy$, we square both sides of the inequality:

$12^2 \geq (\sqrt{xy})^2$

$144 \geq xy$

This inequality shows that the product $xy$ is always less than or equal to 144. The maximum value the product can attain is 144.

The equality (when the product is maximized) holds if and only if the two numbers are equal, i.e., $x = y$.

Since the sum is $x + y = 24$ and $x = y$, we have $x + x = 24$, which simplifies to $2x = 24$.

Solving for $x$:

$x = \frac{24}{2}$

$x = 12$

Since $x=y$, the value of $y$ is also 12.

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The two numbers whose sum is 24 and whose product is as large as possible are 12 and 12.

Given:

Two positive numbers $x$ and $y$ such that their sum is 60.

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To Find:

The values of $x$ and $y$ such that the expression $xy^3$ is maximized.

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Solution:

Let the two positive numbers be $x$ and $y$.

According to the problem, their sum is 60:

We are asked to maximize the product $P = xy^3$.

From equation (i), we can express $x$ in terms of $y$:

$x = 60 - y$

Since $x$ and $y$ must be positive numbers, we have $x > 0$ and $y > 0$.

The condition $x > 0$ implies $60 - y > 0$, which means $y < 60$.

So, the variable $y$ must satisfy $0 < y < 60$.

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Now, substitute the expression for $x$ into the product we want to maximize:

$P(y) = (60 - y)y^3$

Expanding the expression:

To find the maximum value of $P(y)$ on the interval $(0, 60)$, we need to find the critical points by taking the derivative of $P(y)$ with respect to $y$ and setting it equal to zero.

Using the power rule for differentiation, we find the first derivative:

$P'(y) = \frac{d}{dy}(60y^3 - y^4) = 3 \times 60y^{3-1} - 4 \times y^{4-1}$

Now, set the first derivative equal to zero to find the critical points:

$180y^2 - 4y^3 = 0$

Factor out the common term $4y^2$:

From equation (iv), the possible values for $y$ are when $4y^2 = 0$ or $45 - y = 0$.

$y = 0$ or $y = 45$

Since we are looking for positive numbers, $y > 0$, so $y=0$ is not a valid solution in the domain $(0, 60)$.

The critical point within the domain is $y = 45$.

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To confirm that $y=45$ corresponds to a maximum, we can use the First Derivative Test. We examine the sign of $P'(y)$ around $y=45$ for $y \in (0, 60)$.

For $0 < y < 45$, choose a test value, say $y=10$. $P'(10) = 4(10^2)(45 - 10) = 4(100)(35) > 0$. Thus, $P(y)$ is increasing on $(0, 45)$.

For $45 < y < 60$, choose a test value, say $y=50$. $P'(50) = 4(50^2)(45 - 50) = 4(2500)(-5) < 0$. Thus, $P(y)$ is decreasing on $(45, 60)$.

Since $P'(y)$ changes from positive to negative at $y=45$, there is a local maximum at $y=45$. Because this is the only critical point in the domain $(0, 60)$, it corresponds to the absolute maximum on this interval.

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Now that we have found the value of $y$ that maximizes the expression, we find the corresponding value of $x$ using equation (i):

$x = 60 - y$

$x = 60 - 45$

$x = 15$

Both numbers $x=15$ and $y=45$ are positive, and their sum is $15 + 45 = 60$, satisfying the given condition.

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The two positive numbers are 15 and 45.

Given:

Two positive numbers $x$ and $y$ such that their sum is 35.

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To Find:

The values of $x$ and $y$ such that the expression $x^2y^5$ is maximized.

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Solution:

Let the two positive numbers be $x$ and $y$.

According to the problem, their sum is 35:

We are asked to maximize the product $P = x^2y^5$.

From equation (i), we can express $x$ in terms of $y$:

$x = 35 - y$

Since $x$ and $y$ must be positive numbers, we have $x > 0$ and $y > 0$.

The condition $x > 0$ implies $35 - y > 0$, which means $y < 35$.

So, the variable $y$ must satisfy $0 < y < 35$.

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Now, substitute the expression for $x$ into the product we want to maximize:

$P(y) = (35 - y)^2 y^5$

To find the maximum value of $P(y)$ on the interval $(0, 35)$, we need to find the critical points by taking the derivative of $P(y)$ with respect to $y$ and setting it equal to zero.

Using the product rule for differentiation, $(uv)' = u'v + uv'$, where $u = (35 - y)^2$ and $v = y^5$:

First, find the derivatives of $u$ and $v$ with respect to $y$:

$u' = \frac{d}{dy}[(35 - y)^2] = 2(35 - y) \cdot \frac{d}{dy}(35 - y) = 2(35 - y)(-1) = -2(35 - y)$

$v' = \frac{d}{dy}[y^5] = 5y^{5-1} = 5y^4$

Now, apply the product rule to find $P'(y)$:

$P'(y) = u'v + uv'$

$P'(y) = [-2(35 - y)](y^5) + [(35 - y)^2](5y^4)$

$P'(y) = -2y^5(35 - y) + 5y^4(35 - y)^2$

Factor out the common terms $y^4$ and $(35 - y)$:

$P'(y) = y^4(35 - y)[-2y + 5(35 - y)]$

$P'(y) = y^4(35 - y)[-2y + 175 - 5y]$

Now, set the first derivative equal to zero to find the critical points:

$y^4(35 - y)(175 - 7y) = 0$

This equation is satisfied if any of the factors are zero:

$y^4 = 0 \Rightarrow y = 0$

$35 - y = 0 \Rightarrow y = 35$

$175 - 7y = 0 \Rightarrow 7y = 175 \Rightarrow y = \frac{175}{7} = 25$

We are looking for positive numbers, so $y > 0$. Also, the domain is $0 < y < 35$. The critical point within this domain is $y = 25$. The endpoints $y=0$ and $y=35$ result in the product being zero, which is clearly not the maximum for positive numbers.

---

To confirm that $y=25$ corresponds to a maximum, we can use the First Derivative Test. Let's rewrite $P'(y)$ as $P'(y) = 7y^4(35 - y)(25 - y)$. We examine the sign of $P'(y)$ around $y=25$ within the interval $(0, 35)$.

For $0 < y < 25$, choose a test value, say $y=10$. $P'(10) = 7(10)^4(35 - 10)(25 - 10) = 7(10000)(25)(15)$. This is positive ($P'(y) > 0$). So, $P(y)$ is increasing on $(0, 25)$.

For $25 < y < 35$, choose a test value, say $y=30$. $P'(30) = 7(30)^4(35 - 30)(25 - 30) = 7(810000)(5)(-5)$. This is negative ($P'(y) < 0$). So, $P(y)$ is decreasing on $(25, 35)$.

Since $P'(y)$ changes from positive to negative at $y=25$, there is a local maximum at $y=25$. As this is the only critical point in the domain $(0, 35)$, it corresponds to the absolute maximum on this interval.

---

Now that we have found the value of $y$ that maximizes the expression, we find the corresponding value of $x$ using equation (i):

$x = 35 - y$

$x = 35 - 25$

$x = 10$

Both numbers $x=10$ and $y=25$ are positive, and their sum is $10 + 25 = 35$, satisfying the given condition.

---

The two positive numbers such that their sum is 35 and the product $x^2y^5$ is a maximum are 10 and 25.

Given:

Two positive numbers whose sum is 16.

---

To Find:

The two positive numbers such that the sum of their cubes is minimum.

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Solution:

Let the two positive numbers be $x$ and $y$.

According to the problem statement, their sum is 16:

We want to minimize the sum of their cubes, let's call this sum $S$:

$S = x^3 + y^3$

---

From equation (i), we can express one variable in terms of the other. Let's express $x$ in terms of $y$:

$x = 16 - y$

Since $x$ and $y$ must be positive numbers, we have $x > 0$ and $y > 0$.

The condition $x > 0$ implies $16 - y > 0$, which means $y < 16$.

So, the variable $y$ must satisfy $0 < y < 16$.

---

Now, substitute the expression for $x$ into the expression for $S$:

$S(y) = (16 - y)^3 + y^3$

To find the minimum value of $S(y)$ on the interval $(0, 16)$, we need to find the critical points by taking the derivative of $S(y)$ with respect to $y$ and setting it equal to zero.

Find the first derivative $S'(y)$:

$S'(y) = \frac{d}{dy}[(16 - y)^3 + y^3]$

Using the chain rule and the power rule:

$S'(y) = 3(16 - y)^{3-1} \cdot \frac{d}{dy}(16 - y) + 3y^{3-1}$

$S'(y) = 3(16 - y)^2 \cdot (-1) + 3y^2$

Now, set the first derivative equal to zero to find the critical points:

$-3(16 - y)^2 + 3y^2 = 0$

Divide both sides by 3:

$-(16 - y)^2 + y^2 = 0$

Rearrange the terms:

$y^2 - (16 - y)^2 = 0$

Factor using the difference of squares formula $a^2 - b^2 = (a-b)(a+b)$:

$[y - (16 - y)][y + (16 - y)] = 0$

$[y - 16 + y][y + 16 - y] = 0$

$(2y - 16)(16) = 0$

Since $16 \neq 0$, we must have:

$2y - 16 = 0$

$2y = 16$

$y = 8$

This critical point $y=8$ lies within our domain $(0, 16)$.

---

To determine if $y=8$ corresponds to a minimum, we can use the Second Derivative Test.

Find the second derivative $S''(y)$ by differentiating $S'(y) = -3(16 - y)^2 + 3y^2$:

$S''(y) = \frac{d}{dy}[-3(16 - y)^2 + 3y^2]$

$S''(y) = -3 \cdot \frac{d}{dy}[(16 - y)^2] + \frac{d}{dy}[3y^2]$

$S''(y) = -3 \cdot [2(16 - y) \cdot (-1)] + 6y$

$S''(y) = -3 \cdot [-2(16 - y)] + 6y$

$S''(y) = 6(16 - y) + 6y$

$S''(y) = 96 - 6y + 6y$

$S''(y) = 96$

Now, evaluate the second derivative at the critical point $y=8$:

$S''(8) = 96$

Since $S''(8) = 96 > 0$, the function $S(y)$ has a local minimum at $y=8$. Because this is the only critical point in the domain $(0, 16)$, it represents the absolute minimum on this interval.

---

Now that we have found the value of $y$ that minimizes the sum of cubes, we find the corresponding value of $x$ using equation (i):

$x = 16 - y$

$x = 16 - 8$

$x = 8$

Both numbers $x=8$ and $y=8$ are positive, and their sum is $8 + 8 = 16$, satisfying the given condition.

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The two positive numbers whose sum is 16 and the sum of whose cubes is minimum are 8 and 8.

Given:

A square piece of tin with a side length of 18 cm.

---

To Find:

The side length of the square to be cut off from each corner to maximize the volume of the box formed.

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Solution:

Let the side length of the square cut off from each corner be $x$ cm.

When a square of side $x$ is cut from each of the four corners, the remaining shape, when the flaps are folded up, forms an open-top box.

The dimensions of the base of the box will be the original side length minus the length cut off from both ends along that side.

Length of the base = $18 - x - x = 18 - 2x$ cm.

Width of the base = $18 - x - x = 18 - 2x$ cm.

The height of the box will be the side length of the square that was cut off from the corners.

Height of the box = $x$ cm.

For the dimensions of the box to be physically possible and positive, we must have:

$x > 0$

and

$18 - 2x > 0$

$18 > 2x$

$9 > x$

So, the valid range for $x$ is $0 < x < 9$.

---

The volume $V$ of the box is given by the product of its length, width, and height:

$V = \text{Length} \times \text{Width} \times \text{Height}$

Simplifying the expression for the volume:

$V(x) = (18 - 2x)^2 x$

$V(x) = (18^2 - 2(18)(2x) + (2x)^2) x$

$V(x) = (324 - 72x + 4x^2) x$

$V(x) = 324x - 72x^2 + 4x^3$

To find the value of $x$ that maximizes the volume $V(x)$, we need to find the critical points by taking the first derivative of $V(x)$ with respect to $x$ and setting it equal to zero.

$V'(x) = \frac{d}{dx}(324x - 72x^2 + 4x^3)$

$V'(x) = 324 - 144x + 12x^2$

Set the first derivative equal to zero:

Divide equation (ii) by 12 to simplify it:

$\frac{12x^2}{12} - \frac{144x}{12} + \frac{324}{12} = 0$

$x^2 - 12x + 27 = 0$

We can solve this quadratic equation by factoring. We need two numbers that multiply to 27 and add up to -12. These numbers are -3 and -9.

$(x - 3)(x - 9) = 0$

This gives two possible values for $x$:

$x - 3 = 0 \Rightarrow x = 3$

$x - 9 = 0 \Rightarrow x = 9$

We must consider the domain of $x$, which is $0 < x < 9$. The value $x=9$ is on the boundary of the domain and results in a box with base dimensions $18 - 2(9) = 0$, leading to a volume of 0. This is clearly not the maximum volume.

The only critical point within the valid domain $(0, 9)$ is $x = 3$.

---

To confirm that $x=3$ corresponds to a maximum volume, we can use the Second Derivative Test.

Find the second derivative $V''(x)$ by differentiating the first derivative $V'(x) = 324 - 144x + 12x^2$:

$V''(x) = \frac{d}{dx}(324 - 144x + 12x^2)$

$V''(x) = -144 + 24x$

Now, evaluate the second derivative at the critical point $x = 3$:

$V''(3) = -144 + 24(3)$

$V''(3) = -144 + 72$

$V''(3) = -72$

Since $V''(3) = -72 < 0$, the function $V(x)$ has a local maximum at $x = 3$. As this is the only critical point within the domain $(0, 9)$, it represents the absolute maximum volume.

---

The side of the square to be cut off to maximize the volume of the box is 3 cm.

The dimensions of the box will be $18 - 2(3) = 12$ cm for length and width, and the height will be 3 cm. The maximum volume is $12 \times 12 \times 3 = 432$ cm$^3$.

Given:

Dimensions of the rectangular tin sheet: Length = 45 cm, Width = 24 cm.

---

To Find:

The side length of the square to be cut off from each corner to maximize the volume of the resulting open-top box.

---

Solution:

Let the side length of the square cut off from each corner be $x$ cm.

When a square of side $x$ is cut from each of the four corners, the remaining shape, when the flaps are folded up, forms an open-top box.

The dimensions of the base of the box will be the original length minus the length cut off from both ends (twice $x$) and the original width minus the length cut off from both ends (twice $x$). The height of the box will be the side length of the square that was cut off.

Length of the base = $45 - x - x = 45 - 2x$ cm.

Width of the base = $24 - x - x = 24 - 2x$ cm.

Height of the box = $x$ cm.

For the dimensions of the box to be physically possible and positive, the side length $x$ must satisfy:

$x > 0$

$45 - 2x > 0 \implies 2x < 45 \implies x < \frac{45}{2} = 22.5$

$24 - 2x > 0 \implies 2x < 24 \implies x < \frac{24}{2} = 12$

Combining these conditions, the valid range for $x$ is $0 < x < 12$.

---

The volume $V$ of the box is given by the product of its length, width, and height:

$V = \text{Length} \times \text{Width} \times \text{Height}$

Expand the expression for the volume:

$V(x) = (45 \times 24 - 45 \times 2x - 2x \times 24 + (-2x)(-2x))x$

$V(x) = (1080 - 90x - 48x + 4x^2)x$

$V(x) = (4x^2 - 138x + 1080)x$

$V(x) = 4x^3 - 138x^2 + 1080x$

---

To find the value of $x$ that maximizes the volume $V(x)$ on the interval $(0, 12)$, we need to find the critical points by taking the first derivative of $V(x)$ with respect to $x$ and setting it equal to zero.

$V'(x) = \frac{d}{dx}(4x^3 - 138x^2 + 1080x)$

$V'(x) = 3 \cdot 4x^{3-1} - 2 \cdot 138x^{2-1} + 1 \cdot 1080x^{1-1}$

Set the first derivative equal to zero to find the critical points:

$12x^2 - 276x + 1080 = 0$

Divide the equation by the greatest common divisor of the coefficients (12) to simplify:

$\frac{12x^2}{12} - \frac{276x}{12} + \frac{1080}{12} = \frac{0}{12}$

$x^2 - 23x + 90 = 0$

Solve this quadratic equation by factoring. We need two numbers whose product is 90 and whose sum is -23. These numbers are -5 and -18.

$(x - 5)(x - 18) = 0$

This equation gives two possible values for $x$:

$x - 5 = 0 \implies x = 5$

$x - 18 = 0 \implies x = 18$

We must consider the valid domain for $x$, which is $0 < x < 12$. The value $x = 18$ falls outside this domain ($18 \not< 12$), so it is not a physically possible or relevant solution for this problem.

The only critical point within the valid domain $(0, 12)$ is $x = 5$.

---

To confirm that $x=5$ corresponds to a maximum volume, we can use the Second Derivative Test.

Find the second derivative $V''(x)$ by differentiating the first derivative $V'(x) = 12x^2 - 276x + 1080$:

$V''(x) = \frac{d}{dx}(12x^2 - 276x + 1080)$

$V''(x) = 2 \cdot 12x^{2-1} - 1 \cdot 276x^{1-1} + 0$

$V''(x) = 24x - 276$

Now, evaluate the second derivative at the critical point $x = 5$:

$V''(5) = 24(5) - 276$

$V''(5) = 120 - 276$

$V''(5) = -156$

Since $V''(5) = -156$ is less than 0, the function $V(x)$ has a local maximum at $x = 5$. As this is the only critical point within the valid domain $(0, 12)$, it corresponds to the absolute maximum volume on this interval.

---

The side of the square that should be cut off from each corner to maximize the volume of the box is 5 cm.

The dimensions of the box with maximum volume will be: Length = $45 - 2(5) = 35$ cm, Width = $24 - 2(5) = 14$ cm, Height = 5 cm.

The maximum volume is $35 \times 14 \times 5 = 2450$ cm$^3$.

Given:

A fixed circle with a given radius. Let the radius of the circle be $R$.

Rectangles inscribed in the circle.

---

To Prove:

The rectangle inscribed in the circle that has the maximum area is a square.

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Solution:

Consider a circle with its center at the origin $(0,0)$ and a fixed radius $R$. The equation of the circle is $x^2 + y^2 = R^2$.

Let a rectangle be inscribed in this circle. The vertices of the rectangle lie on the circle.

Let the sides of the rectangle be $2a$ and $2b$. Since the vertices lie on the circle, the vertices can be represented as $(a, b)$, $(-a, b)$, $(-a, -b)$, and $(a, -b)$.

Since the point $(a, b)$ lies on the circle $x^2 + y^2 = R^2$, we have:

$a^2 + b^2 = R^2$

The lengths of the sides of the rectangle are $L = 2a$ and $W = 2b$. Note that $a > 0$ and $b > 0$ for a non-degenerate rectangle.

The relation between the sides and the radius of the circle is also evident from the fact that the diagonal of the inscribed rectangle is a diameter of the circle (length $2R$). Using the Pythagorean theorem on the rectangle's sides $2a$ and $2b$ and its diagonal $2R$: $(2a)^2 + (2b)^2 = (2R)^2$, which simplifies to $4a^2 + 4b^2 = 4R^2$, or $a^2 + b^2 = R^2$. This confirms our setup.

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The area of the rectangle is given by $A = \text{Length} \times \text{Width}$.

$A = (2a)(2b) = 4ab$

We want to maximize the area $A = 4ab$, subject to the constraint $a^2 + b^2 = R^2$. Since $a$ and $b$ are positive side lengths, we have $a > 0$ and $b > 0$. From the constraint, $a^2 < R^2$ and $b^2 < R^2$, so $0 < a < R$ and $0 < b < R$.

---

From the constraint $a^2 + b^2 = R^2$, we can express $b^2$ as $b^2 = R^2 - a^2$. Since $b > 0$, $b = \sqrt{R^2 - a^2}$.

Substitute this into the area formula:

$A(a) = 4a\sqrt{R^2 - a^2}$

The domain for $a$ is $0 < a < R$. To maximize $A(a)$, we can maximize $A(a)^2$ because the area is non-negative. Let $S(a) = A(a)^2$.

$S(a) = (4a\sqrt{R^2 - a^2})^2 = 16a^2(R^2 - a^2) = 16R^2 a^2 - 16a^4$

To simplify differentiation, let $u = a^2$. Since $0 < a < R$, we have $0 < a^2 < R^2$, so the domain for $u$ is $0 < u < R^2$. The function to maximize becomes:

$S(u) = 16R^2 u - 16u^2$

To find the maximum, we take the derivative of $S(u)$ with respect to $u$ and set it to zero:

$S'(u) = \frac{d}{du}(16R^2 u - 16u^2) = 16R^2 - 32u$

Set $S'(u) = 0$:

$16R^2 - 32u = 0$

$32u = 16R^2$

$u = \frac{16R^2}{32} = \frac{R^2}{2}$

This critical point $u = \frac{R^2}{2}$ is within the domain $0 < u < R^2$.

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To verify that this critical point corresponds to a maximum, we use the Second Derivative Test.

Find the second derivative of $S(u)$:

$S''(u) = \frac{d}{du}(16R^2 - 32u) = -32$

Since $S''(u) = -32 < 0$ for all values of $u$, the function $S(u)$ has a local maximum at $u = \frac{R^2}{2}$. Since this is the only critical point in the domain, it corresponds to the absolute maximum on the interval $(0, R^2)$.

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Now, we translate back from $u$ to $a$. We have $u = a^2$, so $a^2 = \frac{R^2}{2}$. Since $a > 0$, we take the positive square root:

$a = \sqrt{\frac{R^2}{2}} = \frac{R}{\sqrt{2}} = \frac{R\sqrt{2}}{2}$

Now find the corresponding value of $b$ using the constraint $a^2 + b^2 = R^2$:

$(\frac{R}{\sqrt{2}})^2 + b^2 = R^2$

$\frac{R^2}{2} + b^2 = R^2$

$b^2 = R^2 - \frac{R^2}{2} = \frac{R^2}{2}$

Since $b > 0$, we have:

$b = \sqrt{\frac{R^2}{2}} = \frac{R}{\sqrt{2}} = \frac{R\sqrt{2}}{2}$

---

We found that the area of the rectangle is maximized when $a = \frac{R\sqrt{2}}{2}$ and $b = \frac{R\sqrt{2}}{2}$.

The sides of the rectangle are $L = 2a$ and $W = 2b$.

$L = 2 \left(\frac{R\sqrt{2}}{2}\right) = R\sqrt{2}$

$W = 2 \left(\frac{R\sqrt{2}}{2}\right) = R\sqrt{2}$

Since the length and width of the rectangle are equal ($L = W = R\sqrt{2}$), the rectangle is a square.

Thus, the square inscribed in a fixed circle has the maximum area among all inscribed rectangles.

---

The maximum area is $A = LW = (R\sqrt{2})(R\sqrt{2}) = 2R^2$.

Given:

A right circular cylinder with a given fixed surface area $S$.

---

To Prove:

The right circular cylinder with the maximum volume is such that its height is equal to the diameter of its base.

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Solution:

Let $r$ be the radius of the base and $h$ be the height of the right circular cylinder.

The surface area of the cylinder (assuming total surface area, as is standard unless specified otherwise) is given by:

The volume of the cylinder is given by:

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From the constraint equation (i), we express $h$ in terms of $r$ and $S$:

$2\pi rh = S - 2\pi r^2$

Substitute the expression for $h$ from (iii) into the volume equation (ii) to get the volume as a function of $r$:

$V(r) = \pi r^2 \left( \frac{S}{2\pi r} - r \right)$

$V(r) = \pi r^2 \left( \frac{S}{2\pi r} \right) - \pi r^2 (r)$

$V(r) = \frac{S}{2} r - \pi r^3$

---

For the dimensions to be valid, the radius $r$ and height $h$ must be positive. $r > 0$. From (iii), $h > 0$ implies $\frac{S}{2\pi r} - r > 0$, which means $\frac{S}{2\pi r} > r$, or $S > 2\pi r^2$. Since $S$ is a positive surface area and $r > 0$, this means $r^2 < \frac{S}{2\pi}$, so $r < \sqrt{\frac{S}{2\pi}}$.

The domain for $r$ is $(0, \sqrt{\frac{S}{2\pi}})$.

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To find the maximum volume, we find the critical points by taking the derivative of $V(r)$ with respect to $r$ and setting it to zero:

$V'(r) = \frac{d}{dr} \left( \frac{S}{2} r - \pi r^3 \right)$

Set $V'(r) = 0$:

$\frac{S}{2} - 3\pi r^2 = 0$

$3\pi r^2 = \frac{S}{2}$

$r^2 = \frac{S}{6\pi}$

Since $r > 0$, we take the positive square root:

This critical point lies within the domain $(0, \sqrt{\frac{S}{2\pi}})$ because $\frac{S}{6\pi} < \frac{S}{2\pi}$.

---

To confirm that this critical point corresponds to a maximum, we use the Second Derivative Test. Find the second derivative of $V(r)$:

$V''(r) = \frac{d}{dr} \left( \frac{S}{2} - 3\pi r^2 \right)$

$V''(r) = -6\pi r$

Evaluate the second derivative at the critical point $r = \sqrt{\frac{S}{6\pi}}$:

$V''\left(\sqrt{\frac{S}{6\pi}}\right) = -6\pi \sqrt{\frac{S}{6\pi}}$

Since $r > 0$ and $\pi > 0$, $V''\left(\sqrt{\frac{S}{6\pi}}\right) < 0$.

According to the Second Derivative Test, a negative second derivative at the critical point indicates a local maximum. Since this is the only critical point in the domain, it corresponds to the absolute maximum volume.

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Now, we find the height $h$ corresponding to this radius $r$ using the constraint equation $S = 2\pi r^2 + 2\pi rh$. At the maximum volume, we found $S = 6\pi r^2$ (from $3\pi r^2 = S/2$). Substitute this value of $S$ back into the constraint equation:

Subtract $2\pi r^2$ from both sides of equation (v):

$6\pi r^2 - 2\pi r^2 = 2\pi rh$

$4\pi r^2 = 2\pi rh$

Since $r > 0$, we can divide both sides by $2\pi r$:

$\frac{\cancel{4\pi r^2}^{2r}}{\cancel{2\pi r}_{1}} = \frac{\cancel{2\pi rh}^{h}}{\cancel{2\pi r}_{1}}$

$2r = h$

---

The diameter of the base is $d = 2r$. We have shown that the height $h = 2r$.

Therefore, the height of the cylinder is equal to the diameter of the base when its volume is maximum for a given surface area.

This completes the proof.

Given:

Volume of the closed cylindrical can, $V = 100$ cubic centimetres.

---

To Find:

The dimensions (radius and height) of the can which has the minimum surface area.

---

Solution:

Let $r$ be the radius of the base and $h$ be the height of the closed right circular cylinder.

The volume of the cylinder is given by $V = \pi r^2 h$.

We are given $V = 100 \text{ cm}^3$.

From equation (1), we can express the height $h$ in terms of the radius $r$:

The surface area $S$ of a closed cylinder is given by the formula:

$S = \text{Area of top circle} + \text{Area of bottom circle} + \text{Lateral surface area}$

$S = \pi r^2 + \pi r^2 + 2\pi r h$

Substitute the expression for $h$ from equation (2) into equation (3):

$S(r) = 2\pi r^2 + 2\pi r \left(\frac{100}{\pi r^2}\right)$

$S(r) = 2\pi r^2 + \frac{200}{r}$

To find the minimum surface area, we need to find the critical points by differentiating $S(r)$ with respect to $r$ and setting the derivative equal to zero.

$\frac{dS}{dr} = \frac{d}{dr}(2\pi r^2 + 200r^{-1})$

$\frac{dS}{dr} = 4\pi r - 200r^{-2}$

Set $\frac{dS}{dr} = 0$ to find the critical point(s):

$4\pi r - \frac{200}{r^2} = 0$

$4\pi r = \frac{200}{r^2}$

$4\pi r^3 = 200$

$r^3 = \frac{200}{4\pi}$

$r^3 = \frac{50}{\pi}$

Now, we use the second derivative test to confirm if this value of $r$ corresponds to a minimum surface area.

Differentiate equation (4) with respect to $r$:

$\frac{d^2S}{dr^2} = \frac{d}{dr}(4\pi r - 200r^{-2})$

$\frac{d^2S}{dr^2} = 4\pi - 200(-2)r^{-3}$

Substitute the value of $r^3 = \frac{50}{\pi}$ (from equation 5) into equation (6):

$\frac{d^2S}{dr^2} = 4\pi + \frac{400}{(50/\pi)} = 4\pi + \frac{400\pi}{50}$

$\frac{d^2S}{dr^2} = 4\pi + 8\pi = 12\pi$

Since $12\pi > 0$, the surface area $S$ is minimum when $r = \left(\frac{50}{\pi}\right)^{1/3}$.

Now, we find the corresponding height $h$ using equation (2):

$h = \frac{100}{\pi r^2}$

Substitute the value of $r$ from equation (5) into the expression for $h$:

$h = \frac{100}{\pi \left(\left(\frac{50}{\pi}\right)^{1/3}\right)^2} = \frac{100}{\pi \left(\frac{50}{\pi}\right)^{2/3}}$

$h = \frac{100}{\pi^{1} \cdot 50^{2/3} \cdot \pi^{-2/3}} = \frac{100}{\pi^{1 - 2/3} 50^{2/3}} = \frac{100}{\pi^{1/3} 50^{2/3}}$

Alternatively, from equation (2), $h = \frac{100}{\pi r^2}$. Multiply numerator and denominator by $r$:

$h = \frac{100r}{\pi r^3}$

Substitute $r^3 = \frac{50}{\pi}$ from equation (5) into this expression:

$h = \frac{100r}{\pi (50/\pi)} = \frac{100r}{50}$

$h = 2r$

So, the minimum surface area occurs when the height of the cylinder is equal to its diameter.

---

Conclusion:

The dimensions of the can which has the minimum surface area for a given volume of $100 \text{ cm}^3$ are:

Radius: $r = \left(\frac{50}{\pi}\right)^{1/3}$ cm

Height: $h = 2 \left(\frac{50}{\pi}\right)^{1/3}$ cm

Given:

Total length of the wire = 28 m.

The wire is cut into two pieces. One piece is used to form a square, and the other is used to form a circle.

---

To Find:

The lengths of the two pieces such that the combined area of the square and the circle is minimum.

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Solution:

Let the length of the first piece of wire be $x$ metres. This piece is used to form the square.

The perimeter of the square is equal to the length of the wire piece, so Perimeter = $x$ m.

If $s$ is the side length of the square, then $4s = x$, which means $s = \frac{x}{4}$.

The area of the square is $A_s = s^2 = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16}$.

The length of the second piece of wire is the total length minus the length of the first piece, which is $(28-x)$ metres. This piece is used to form the circle.

The circumference of the circle is equal to the length of the wire piece, so Circumference = $(28-x)$ m.

If $r$ is the radius of the circle, then the circumference is $2\pi r$. So, $2\pi r = 28-x$.

This gives the radius $r = \frac{28-x}{2\pi}$.

The area of the circle is $A_c = \pi r^2 = \pi \left(\frac{28-x}{2\pi}\right)^2 = \pi \frac{(28-x)^2}{4\pi^2} = \frac{(28-x)^2}{4\pi}$.

The combined area of the square and the circle is the sum of their areas:

The variable $x$ represents the length of a piece of wire, so it must be non-negative and cannot exceed the total length of the wire. Thus, the domain of $A(x)$ is $0 \le x \le 28$.

To find the minimum combined area, we need to find the minimum value of the function $A(x)$ on the closed interval $[0, 28]$. We find the critical points by differentiating $A(x)$ with respect to $x$ and setting the derivative equal to zero.

$\frac{dA}{dx} = \frac{d}{dx}\left(\frac{x^2}{16} + \frac{(28-x)^2}{4\pi}\right)$

Using the chain rule for the second term, $\frac{d}{dx}(28-x)^2 = 2(28-x) \cdot (-1) = -2(28-x)$.

$\frac{dA}{dx} = \frac{2x}{16} + \frac{1}{4\pi} (-2(28-x))$

Set $\frac{dA}{dx} = 0$ to find the critical points:

$\frac{x}{8} - \frac{28-x}{2\pi} = 0$

$\frac{x}{8} = \frac{28-x}{2\pi}$

$2\pi x = 8(28-x)$

$2\pi x = 224 - 8x$

$2\pi x + 8x = 224$

$x(2\pi + 8) = 224$

$2x(\pi + 4) = 224$

$x(\pi + 4) = 112$

This value $x = \frac{112}{\pi + 4}$ is approximately $\frac{112}{3.14159 + 4} = \frac{112}{7.14159} \approx 15.68$, which lies within the interval $[0, 28]$.

Now, we apply the second derivative test to determine if this critical point corresponds to a minimum.

Differentiate equation (2) with respect to $x$:

$\frac{d^2A}{dx^2} = \frac{d}{dx}\left(\frac{x}{8} - \left(\frac{28}{2\pi} - \frac{x}{2\pi}\right)\right)$

$\frac{d^2A}{dx^2} = \frac{d}{dx}\left(\frac{x}{8} - \frac{14}{\pi} + \frac{x}{2\pi}\right)$

Since $\pi > 0$, both $\frac{1}{8}$ and $\frac{1}{2\pi}$ are positive constants. Therefore, $\frac{d^2A}{dx^2} = \frac{1}{8} + \frac{1}{2\pi} > 0$ for all values of $x$.

The second derivative is positive at the critical point, which confirms that $x = \frac{112}{\pi + 4}$ corresponds to a local minimum.

To find the absolute minimum on the closed interval $[0, 28]$, we compare the value of $A(x)$ at the critical point and at the endpoints $x=0$ and $x=28$.

At the critical point $x = \frac{112}{\pi + 4}$:

The length of the first piece is $x = \frac{112}{\pi + 4}$ m.

The length of the second piece is $28-x = 28 - \frac{112}{\pi + 4} = \frac{28(\pi+4) - 112}{\pi+4} = \frac{28\pi + 112 - 112}{\pi+4} = \frac{28\pi}{\pi+4}$ m.

At the endpoint $x=0$ (entire wire forms the circle):

$A(0) = \frac{0^2}{16} + \frac{(28-0)^2}{4\pi} = 0 + \frac{28^2}{4\pi} = \frac{784}{4\pi} = \frac{196}{\pi}$

At the endpoint $x=28$ (entire wire forms the square):

$A(28) = \frac{28^2}{16} + \frac{(28-28)^2}{4\pi} = \frac{784}{16} + 0 = 49$

At the critical point $x = \frac{112}{\pi + 4}$:

$A\left(\frac{112}{\pi + 4}\right) = \frac{1}{16}\left(\frac{112}{\pi + 4}\right)^2 + \frac{1}{4\pi}\left(28 - \frac{112}{\pi + 4}\right)^2$

$A\left(\frac{112}{\pi + 4}\right) = \frac{1}{16}\left(\frac{112}{\pi + 4}\right)^2 + \frac{1}{4\pi}\left(\frac{28\pi}{\pi + 4}\right)^2$

$A\left(\frac{112}{\pi + 4}\right) = \frac{112^2}{16(\pi + 4)^2} + \frac{(28\pi)^2}{4\pi(\pi + 4)^2}$

$A\left(\frac{112}{\pi + 4}\right) = \frac{12544}{16(\pi + 4)^2} + \frac{784\pi^2}{4\pi(\pi + 4)^2}$

$A\left(\frac{112}{\pi + 4}\right) = \frac{784}{(\pi + 4)^2} + \frac{196\pi}{(\pi + 4)^2}$

$A\left(\frac{112}{\pi + 4}\right) = \frac{784 + 196\pi}{(\pi + 4)^2} = \frac{196(4 + \pi)}{(\pi + 4)^2} = \frac{196}{\pi + 4}$

Now we compare the areas: $\frac{196}{\pi}$, $49$, and $\frac{196}{\pi + 4}$.

Since $\pi \approx 3.14$, $\pi + 4 \approx 7.14$.

$A(0) = \frac{196}{\pi} \approx \frac{196}{3.14} \approx 62.42$

$A(28) = 49$

$A\left(\frac{112}{\pi + 4}\right) = \frac{196}{\pi + 4} \approx \frac{196}{7.14} \approx 27.45$

Clearly, the minimum combined area is $\frac{196}{\pi + 4}$, which occurs when the length of the piece for the square is $x = \frac{112}{\pi + 4}$ m.

---

Conclusion:

To minimize the combined area of the square and the circle, the wire should be cut into two pieces with the following lengths:

Length of the piece used for the square: $\frac{112}{\pi + 4}$ metres.

Length of the piece used for the circle: $\frac{28\pi}{\pi + 4}$ metres.

Given:

A sphere of radius $R$.

---

To Prove:

The volume of the largest cone that can be inscribed in the sphere is $\frac{8}{27}$ of the volume of the sphere.

---

Solution:

Let the radius of the sphere be $R$ and its center be O.

Consider a cone inscribed in the sphere. Let the radius of the base of the cone be $r$ and the height of the cone be $h$. Let the vertex of the cone be A and the center of its base be B.

We can place the center of the sphere O at the origin $(0,0,0)$. Let the vertex A be at $(0, R)$. The base of the cone will be a circle parallel to the xy-plane, with its center B at $(0, y)$ for some $y$. The height of the cone is $h = R - y$. The radius of the base is $r$.

A point C on the circumference of the cone's base has coordinates $(r, y)$ in a 2D cross-section through the axis of the cone (e.g., in the yz-plane, C could be at $(0, y, r)$ or $(0, y, -r)$). This point C also lies on the sphere.

The equation of the sphere is $x^2 + y^2 + z^2 = R^2$. In the yz-plane cross-section, any point on the sphere satisfies $y^2 + z^2 = R^2$. The point C on the base circumference, with coordinates $(y, r)$ in this 2D cross-section (where $z$ coordinate is $r$ in 3D), must satisfy this equation:

From this, we can express $r^2$ in terms of $y$ and $R$:

The height of the cone is $h = R - y$, which means $y = R - h$. Substitute this expression for $y$ into the equation for $r^2$:

The volume of the cone, $V$, is given by the formula:

Substitute the expression for $r^2$ from equation (i) into equation (ii):

The height $h$ of the inscribed cone can range from slightly greater than 0 (a flat disk) to $2R$ (a degenerate cone with base radius 0, vertex at one pole and base at the other). Thus, $0 < h \le 2R$. To find the maximum volume, we differentiate $V(h)$ with respect to $h$ and set the derivative to zero:

Set $\frac{dV}{dh} = 0$:

Since $\frac{\pi}{3} \neq 0$, we have:

This gives two solutions: $h = 0$ or $4R - 3h = 0$. $h = 0$ corresponds to zero volume, which is a minimum. The other solution is:

This value of $h$ is between $0$ and $2R$ ($4/3 < 2$), so it is a valid height for an inscribed cone. To confirm this is a maximum, we find the second derivative:

Evaluate the second derivative at $h = \frac{4R}{3}$:

Since $R > 0$, the second derivative is negative, confirming that $h = \frac{4R}{3}$ gives a maximum volume.

Now substitute $h = \frac{4R}{3}$ back into the volume formula $V(h) = \frac{\pi}{3} (2h^2R - h^3)$ to find the maximum volume:

Find a common denominator (27) for the terms inside the parenthesis:

The volume of the sphere is given by $V_{sphere} = \frac{4}{3} \pi R^3$.

We need to show that $V_{max} = \frac{8}{27} V_{sphere}$. Let's calculate $\frac{8}{27} V_{sphere}$:

Comparing the maximum volume of the cone with $\frac{8}{27}$ of the volume of the sphere:

Thus, we have shown that $V_{max} = \frac{8}{27} V_{sphere}$.

Hence Proved.

Given:

A right circular cone with a given constant volume $V$.

---

To Prove:

The altitude $h$ of the cone of least curved surface area is $\sqrt{2}$ times the radius of its base $r$. That is, $h = \sqrt{2}r$.

---

Solution:

Let $r$ be the radius of the base of the cone, $h$ be its altitude (height), $l$ be its slant height, $V$ be its volume, and $S$ be its curved surface area.

The volume of a cone is given by:

$V = \frac{1}{3}\pi r^2 h$

Since the volume $V$ is given and is constant, we can express $h$ in terms of $r$ and $V$:

The curved surface area of a cone is given by:

$S = \pi r l$

The slant height $l$ is related to the radius $r$ and height $h$ by the Pythagorean theorem:

$l = \sqrt{r^2 + h^2}$

So, the curved surface area can be written as:

$S = \pi r \sqrt{r^2 + h^2}$

To minimize $S$, we can equivalently minimize $S^2$, as $S$ is always a positive value. Let $f(r) = S^2$.

Substitute the expression for $h$ from equation (i) into the expression for $f(r)$:

To find the value of $r$ that minimizes $f(r)$, we differentiate $f(r)$ with respect to $r$ and set the derivative equal to zero:

Setting $\frac{df}{dr} = 0$ for critical points:

Now we relate this condition back to the relationship between $h$ and $r$. From equation (i), $h = \frac{3V}{\pi r^2}$, we can express $V$ in terms of $h$ and $r$: $V = \frac{\pi r^2 h}{3}$. Squaring this gives $V^2 = \frac{\pi^2 r^4 h^2}{9}$.

Substitute this expression for $V^2$ into equation (iii):

Since $r$ is a radius, $r \neq 0$. We can divide both sides of the equation by $r^4$:

Taking the positive square root of both sides (since $h > 0$ and $r > 0$):

To confirm that this critical point corresponds to a minimum curved surface area, we examine the second derivative of $f(r)$.

Since $r > 0$ and $V > 0$, $12\pi^2 r^2 > 0$ and $54V^2 r^{-4} > 0$. Therefore, $\frac{d^2f}{dr^2} > 0$ for all $r > 0$. This confirms that the critical point found corresponds to a minimum value of $f(r)$ (and thus a minimum value of $S$).

Hence, the right circular cone of least curved surface area for a given volume has an altitude equal to $\sqrt{2}$ times the radius of its base.

Hence Proved.

Given:

A right circular cone with a given constant slant height $l$.

---

To Prove:

The semi-vertical angle $\theta$ of the cone of maximum volume is $\tan^{-1} \sqrt{2}$.

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Solution:

Let $r$ be the radius of the base of the cone, $h$ be its altitude (height), $l$ be its slant height, and $\theta$ be the semi-vertical angle.

From the geometry of the cone, we have the following relationships between $r$, $h$, $l$, and $\theta$:

The volume of the cone, $V$, is given by:

Substitute the expressions for $r$ from (i) and $h$ from (ii) into the volume formula (iii):

Since $l$ is a given constant, the volume $V$ is a function of the semi-vertical angle $\theta$. To find the maximum volume, we need to differentiate $V(\theta)$ with respect to $\theta$ and set the derivative to zero.

Let $f(\theta) = \sin^2 \theta \cos \theta$. We maximize $f(\theta)$ for $0 \le \theta \le \frac{\pi}{2}$.

Differentiating $f(\theta)$ with respect to $\theta$ using the product rule:

Set the derivative to zero to find the critical points:

Factor out $\sin \theta$:

This gives two possibilities:

Case 1: $\sin \theta = 0$

If $\sin \theta = 0$, then $\theta = 0$ (since $0 \le \theta \le \frac{\pi}{2}$). In this case, $r = l \sin 0 = 0$, which means the cone has zero base radius and zero volume. This corresponds to a minimum volume.

Case 2: $2 \cos^2 \theta - \sin^2 \theta = 0$

Rearrange the equation:

Divide both sides by $\cos^2 \theta$ (assuming $\cos \theta \neq 0$. If $\cos \theta = 0$, then $\theta = \frac{\pi}{2}$, which means $h = l \cos \frac{\pi}{2} = 0$, giving zero volume - another minimum):

Taking the square root of both sides:

Since $\theta$ is the semi-vertical angle of a cone, $0 < \theta < \frac{\pi}{2}$, so $\tan \theta$ must be positive.

This value of $\theta$ gives the maximum volume. The semi-vertical angle is:

(We can confirm this is a maximum using the second derivative test, which shows the second derivative is negative at this value of $\theta$).

Thus, the semi-vertical angle of the cone of the maximum volume and of given slant height is $\tan^{-1} \sqrt{2}$.

Hence Proved.

Given:

A right circular cone with a given constant total surface area $S$.

---

To Prove:

The semi-vertical angle $\theta$ of the cone of maximum volume is $\sin^{-1} \left(\frac{1}{3}\right)$.

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Solution:

Let $r$ be the radius of the base, $h$ be the height (altitude), $l$ be the slant height, $V$ be the volume, and $\theta$ be the semi-vertical angle of the cone.

The total surface area $S$ of the cone is given by:

$S = \text{Area of base} + \text{Curved surface area}$

Since $S$ is constant, we can express $l$ in terms of $S$ and $r$ from equation (i):

The volume of the cone is given by:

The height $h$ is related to the radius $r$ and slant height $l$ by the Pythagorean theorem: $h^2 = l^2 - r^2$. Since $h = \sqrt{l^2 - r^2}$, we substitute this into the volume formula (iii):

Substitute the expression for $l$ from equation (ii) into this volume equation:

To maximize $V(r)$, we can maximize $V(r)^2$, as $V(r)$ must be positive. Let $f(r) = V(r)^2$.

Differentiate $f(r)$ with respect to $r$ to find critical points:

Set the derivative to zero:

Factor out $2Sr$:

Since $r > 0$ (for a cone to exist) and $S > 0$ (given surface area), we must have:

This condition gives the radius for maximum volume (checked by the second derivative test, which yields $\frac{d^2f}{dr^2} = \frac{1}{9}(2S^2 - 24S\pi r^2)$, substituting $S=4\pi r^2$ gives $\frac{1}{9}(2(4\pi r^2)^2 - 24(4\pi r^2)\pi r^2) = \frac{1}{9}(32\pi^2 r^4 - 96\pi^2 r^4) = -\frac{64}{9}\pi^2 r^4 < 0$, confirming a maximum).

Now we find the semi-vertical angle $\theta$. The semi-vertical angle $\theta$ is related to $r$ and $l$ by:

Substitute the condition $S = 4\pi r^2$ (from equation iv) into the surface area formula $S = \pi r^2 + \pi r l$ (from equation i):

Subtract $\pi r^2$ from both sides:

Divide both sides by $\pi r$ (since $\pi \neq 0$ and $r \neq 0$):

Now substitute this relationship between $r$ and $l$ into the expression for $\sin \theta$:

Therefore, the semi-vertical angle is:

Thus, the semi-vertical angle of the right circular cone of given surface area and maximum volume is $\sin^{-1} \left( \frac{1}{3} \right)$.

Hence Proved.

The correct answer is: (A) (2$\sqrt{2}$ , 4)

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Solution:

The given curve is $x^2 = 2y$. Any point P on this curve can be represented by the coordinates $(x, y)$. Since $x^2 = 2y$, we have $y = \frac{x^2}{2}$. Thus, a point P on the curve can be written as $\left(x, \frac{x^2}{2}\right)$.

We want to find the point P on the curve that is nearest to the point $Q(0, 5)$.

Let $D$ be the distance between the point $P\left(x, \frac{x^2}{2}\right)$ and the point $Q(0, 5)$. The distance formula is $D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

---

To minimize the distance $D$, we can minimize the square of the distance, $D^2$. Let $f(x) = D^2$.

Expand the term $\left(\frac{x^2}{2} - 5\right)^2$:

Substitute this back into the expression for $f(x)$:

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To find the minimum value of $f(x)$, we find the critical points by taking the derivative with respect to $x$ and setting it equal to zero.

---

Set $f'(x) = 0$:

Factor out $x$:

This gives the critical points $x = 0$ or $x^2 - 8 = 0 \implies x^2 = 8 \implies x = \pm \sqrt{8} = \pm 2\sqrt{2}$.

---

Now we evaluate the function $f(x) = \frac{x^4}{4} - 4x^2 + 25$ at these critical points to find the minimum squared distance.

For $x = 0$:

For $x = 2\sqrt{2}$ (where $x^2 = 8$):

For $x = -2\sqrt{2}$ (where $x^2 = 8$):

The minimum value of $D^2$ is 9, which occurs when $x = 2\sqrt{2}$ or $x = -2\sqrt{2}$.

---

Now we find the corresponding y-coordinates for these x-values using the equation of the curve $y = \frac{x^2}{2}$.

If $x = 0$, $y = \frac{0^2}{2} = 0$. The point is $(0, 0)$. $D^2 = 25$.

If $x = 2\sqrt{2}$, $y = \frac{(2\sqrt{2})^2}{2} = \frac{8}{2} = 4$. The point is $(2\sqrt{2}, 4)$. $D^2 = 9$.

If $x = -2\sqrt{2}$, $y = \frac{(-2\sqrt{2})^2}{2} = \frac{8}{2} = 4$. The point is $(-2\sqrt{2}, 4)$. $D^2 = 9$.

The points on the curve nearest to $(0, 5)$ are $(2\sqrt{2}, 4)$ and $(-2\sqrt{2}, 4)$, as these yield the minimum squared distance of 9.

---

Looking at the options provided:

(A) $(2\sqrt{2}, 4)$ - This is one of the points with the minimum distance.

(B) $(2\sqrt{2}, 0)$ - Check if on the curve: $(2\sqrt{2})^2 = 8$, $2(0) = 0$. $8 \neq 0$. Not on the curve.

(C) $(0, 0)$ - This point is on the curve ($0^2 = 2(0)$), but the squared distance is 25, which is not the minimum.

(D) $(2, 2)$ - Check if on the curve: $2^2 = 4$, $2(2) = 4$. $4 = 4$. This point is on the curve. The squared distance is $(2-0)^2 + (2-5)^2 = 2^2 + (-3)^2 = 4 + 9 = 13$. This is not the minimum.

Therefore, the point on the curve $x^2 = 2y$ nearest to $(0, 5)$ from the given options is $(2\sqrt{2}, 4)$.

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The final answer is (A) (2$\sqrt{2}$ , 4).

The correct answer is: (D) $\frac{1}{3}$

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Solution:

Let the given function be $y = f(x)$.

We want to find the minimum value of $y$ for all real values of $x$. We can rewrite the equation to express $x$ in terms of $y$.

Rearrange the terms to form a quadratic equation in $x$:

For $x$ to be a real number, the discriminant of this quadratic equation in $x$ must be greater than or equal to zero. The discriminant $D$ is given by $D = b^2 - 4ac$, where $a = (y-1)$, $b = (y+1)$, and $c = (y-1)$.

For real values of $x$, we require $D \ge 0$:

Multiply by -1 and reverse the inequality sign:

To find the values of $y$ that satisfy this inequality, we find the roots of the quadratic equation $3y^2 - 10y + 3 = 0$.

The roots are $y = \frac{1}{3}$ and $y = 3$.

Since the quadratic expression $3y^2 - 10y + 3$ has a positive leading coefficient (3), the parabola opens upwards. The inequality $3y^2 - 10y + 3 \le 0$ is satisfied when $y$ is between or equal to the roots.

This inequality gives the range of the function $y = \frac{1-x+x^{2}}{1+x+x^{2}}$ for all real values of $x$.

The minimum value of the function is the smallest value in this range, which is $\frac{1}{3}$.

Note: The denominator $1+x+x^2$ is always positive for all real $x$ because its discriminant is $1^2 - 4(1)(1) = -3 < 0$ and the coefficient of $x^2$ is positive (1). So the function is well-defined for all real $x$. The case $y=1$ in the quadratic $(y-1)x^2 + (y+1)x + (y-1) = 0$ leads to $2x=0$, $x=0$, and $f(0)=1$, which is within the range $[1/3, 3]$.

---

The minimum value of the function is $\frac{1}{3}$.

---

The final answer is (D) $\frac{1}{3}$.

The correct answer is: (C) 1

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Solution:

Let the given function be $f(x) = \left[ x(x-1)+1 \right]^{\frac{1}{3}}$.

The function is defined on the interval $0 \le x \le 1$.

Let's simplify the expression inside the cube root:

So, $f(x) = [g(x)]^{\frac{1}{3}} = [x^2 - x + 1]^{\frac{1}{3}}$.

To find the maximum value of $f(x)$ on the interval $[0, 1]$, we can find the maximum value of $g(x) = x^2 - x + 1$ on the same interval, since the cube root function $y = z^{1/3}$ is a strictly increasing function.

The function $g(x) = x^2 - x + 1$ is a quadratic function. Its graph is a parabola opening upwards because the coefficient of $x^2$ is positive (1).

The vertex of the parabola $Ax^2 + Bx + C$ is located at $x = -\frac{B}{2A}$. For $g(x) = x^2 - x + 1$, the vertex is at $x = -\frac{-1}{2(1)} = \frac{1}{2}$.

The interval we are considering is $[0, 1]$. The vertex $x = \frac{1}{2}$ lies within this interval.

For a parabola opening upwards, the minimum value on a closed interval occurs at the vertex if the vertex is within the interval. The maximum value occurs at one or both of the endpoints of the interval.

Let's evaluate $g(x)$ at the endpoints of the interval $[0, 1]$ and at the vertex $x = \frac{1}{2}$:

At $x = 0$ (endpoint):

At $x = 1$ (endpoint):

At $x = \frac{1}{2}$ (vertex):

Comparing the values: $g(0)=1$, $g(1)=1$, $g(1/2)=3/4$.

The minimum value of $g(x)$ on $[0, 1]$ is $\frac{3}{4}$ at $x = \frac{1}{2}$.

The maximum value of $g(x)$ on $[0, 1]$ is $1$ at $x = 0$ and $x = 1$.

---

Since $f(x) = [g(x)]^{1/3}$, the maximum value of $f(x)$ occurs when $g(x)$ is at its maximum.

The maximum value of $g(x)$ is 1.

So, the maximum value of $f(x)$ is $(1)^{1/3} = 1$.

---

Comparing with the given options:

(A) $\left( \frac{1}{3} \right)^{\frac{1}{3}}$

(B) $\frac{1}{2}$

(C) 1

(D) 0

The maximum value is 1, which corresponds to option (C).

---

The final answer is (C) 1.

Given:

The distance $x$, in metres, covered by the car in $t$ seconds is given by the function:

The car starts from point P at $t=0$ and stops at point Q.

---

To Find:

The time taken by the car to reach Q.

The distance between P and Q.

---

Solution:

The car stops when its velocity becomes zero.

The velocity $v(t)$ is the rate of change of distance with respect to time, i.e., $v(t) = \frac{dx}{dt}$.

First, let's expand the expression for $x(t)$:

Now, differentiate $x(t)$ with respect to $t$ to find the velocity $v(t)$:

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The car stops at point Q, which means its velocity at that point is zero.

Set $v(t) = 0$ to find the time when the car stops:

Factor the equation:

This equation gives two possible values for $t$: $t = 0$ or $4 - t = 0 \implies t = 4$.

At $t = 0$ seconds, the car starts from point P. This is the initial time.

At $t = 4$ seconds, the velocity is zero again, which means the car has stopped at point Q.

Therefore, the time taken by the car to reach Q is 4 seconds.

---

Now, we need to find the distance between P and Q. The distance is given by the function $x(t)$ evaluated at the time the car stops, minus the distance at the starting point.

The distance at the starting point P ($t=0$) is:

The distance covered at the time the car stops ($t=4$ seconds) is:

The distance between P and Q is $x(4) - x(0)$.

---

The time taken by the car to reach Q is 4 seconds.

The distance between P and Q is $\frac{32}{3}$ metres.

Given:

Shape of the tank: Inverted right circular cone.

Axis: Vertical, vertex lowermost.

Semi-vertical angle: $\theta = \tan^{-1}(0.5) = \tan^{-1}\left(\frac{1}{2}\right)$.

Rate of pouring water: $\frac{dV}{dt} = 5 \, \text{m}^3/\text{h}$, where $V$ is the volume of water in the cone and $t$ is time in hours.

Instant at which the rate of rising level is required: when depth of water $h = 4 \, \text{m}$.

---

To Find:

The rate at which the level of the water is rising at the instant when $h = 4 \, \text{m}$, i.e., $\frac{dh}{dt}$ at $h = 4 \, \text{m}$.

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Solution:

Let $r$ be the radius of the water surface and $h$ be the depth of water in the tank at time $t$. The semi-vertical angle is $\theta$.

From the geometry of the cone, we have:

We are given $\theta = \tan^{-1}(0.5) = \tan^{-1}\left(\frac{1}{2}\right)$. Therefore, $\tan \theta = \frac{1}{2}$.

This gives us a relationship between $r$ and $h$:

The volume of water in the cone at depth $h$ is given by the formula for the volume of a cone:

Substitute the expression for $r$ from equation (i) into the volume formula (ii) so that $V$ is expressed only in terms of $h$:

Now, we need to find the rate at which the water level is rising, which is $\frac{dh}{dt}$. We differentiate the volume equation with respect to time $t$:

We are given $\frac{dV}{dt} = 5 \, \text{m}^3/\text{h}$ and we want to find $\frac{dh}{dt}$ when $h = 4 \, \text{m}$. Substitute these values into equation (iii):

Now, solve for $\frac{dh}{dt}$:

The units for $\frac{dh}{dt}$ will be metres per hour.

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The rate at which the level of the water is rising at the instant when the depth of water in the tank is 4 m is $\frac{5}{4\pi}$ m/h.

Given:

Height of the lamp post = 6 metres.

Height of the man = 2 metres.

Speed of the man walking away from the lamp post = $\frac{dx}{dt} = 5 \, \text{km/h}$, where $x$ is the distance of the man from the base of the lamp post.

---

To Find:

The rate at which the length of his shadow increases, i.e., $\frac{ds}{dt}$, where $s$ is the length of the shadow.

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Solution:

Let the lamp post be denoted by AB and the man by CD. Let the length of the shadow be DE, where E is the tip of the shadow on the ground.

Let AB = 6 m and CD = 2 m.

Let $x$ be the distance of the man from the lamp post base, so BD = $x$.

Let $s$ be the length of the man's shadow, so DE = $s$.

The points A, C, and E are collinear (the light ray from the top of the lamp post passes through the top of the man's head to the tip of the shadow).

Consider the triangles $\triangle$ABE and $\triangle$CDE. Both are right-angled triangles (at B and D respectively, assuming the ground is horizontal and the lamp post and man are vertical).

Also, $\angle$AEB is common to both triangles (or $\angle$AEB = $\angle$CED).

Therefore, by AA similarity criterion, $\triangle$ABE is similar to $\triangle$CDE.

For similar triangles, the ratio of corresponding sides is equal.

Now, we establish a relationship between $x$ and $s$:

We are given the rate at which the man walks away from the lamp post, which is $\frac{dx}{dt} = 5 \, \text{km/h}$. We need to find the rate at which the length of his shadow increases, $\frac{ds}{dt}$.

Differentiate equation (i) with respect to time $t$:

Substitute the given value of $\frac{dx}{dt}$:

Solve for $\frac{ds}{dt}$:

The rate is in km/h, consistent with the given speed of the man.

---

The rate at which the length of the man's shadow increases is 2.5 km/h.

Given:

The function $f(x) = \frac{3}{10} x^4 - \frac{4}{5} x^3 - 3x^2 + \frac{36}{5} x + 11$.

---

To Find:

Intervals in which $f(x)$ is (a) increasing and (b) decreasing.

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Solution:

To find the intervals where the function is increasing or decreasing, we first find the derivative of the function, $f'(x)$.

---

To find the critical points, we set $f'(x) = 0$ and solve for $x$.

Multiply the entire equation by 5 to eliminate the fractions:

Divide the equation by 6:

We can test integer roots that are divisors of the constant term (6), such as $\pm 1, \pm 2, \pm 3, \pm 6$.

For $x=1$: $(1)^3 - 2(1)^2 - 5(1) + 6 = 1 - 2 - 5 + 6 = 0$. So, $x=1$ is a root.

For $x=-2$: $(-2)^3 - 2(-2)^2 - 5(-2) + 6 = -8 - 2(4) + 10 + 6 = -8 - 8 + 10 + 6 = 0$. So, $x=-2$ is a root.

For $x=3$: $(3)^3 - 2(3)^2 - 5(3) + 6 = 27 - 2(9) - 15 + 6 = 27 - 18 - 15 + 6 = 9 - 15 + 6 = 0$. So, $x=3$ is a root.

The critical points are $x = -2, 1, 3$. These points divide the real number line into four open intervals: $(-\infty, -2)$, $(-2, 1)$, $(1, 3)$, and $(3, \infty)$.

---

Now we examine the sign of $f'(x)$ in each interval. We can use the factored form of $f'(x)$: $f'(x) = \frac{6}{5}(x^3 - 2x^2 - 5x + 6) = \frac{6}{5}(x+2)(x-1)(x-3)$.

Interval	Test Value ($x$)	Sign of $(x+2)$	Sign of $(x-1)$	Sign of $(x-3)$	Sign of $f'(x) = \frac{6}{5}(x+2)(x-1)(x-3)$	Nature of $f(x)$
$(-\infty, -2)$	$-3$	$-$	$-$	$-$	$\frac{6}{5}(-)(-) (-) = -$	Decreasing
$(-2, 1)$	$0$	$+$	$-$	$-$	$\frac{6}{5}(+)(-) (-) = +$	Increasing
$(1, 3)$	$2$	$+$	$+$	$-$	$\frac{6}{5}(+)(+) (-) = -$	Decreasing
$(3, \infty)$	$4$	$+$	$+$	$+$	$\frac{6}{5}(+)(+) (+) = +$	Increasing

Based on the sign analysis of $f'(x)$:

(a) The function $f(x)$ is increasing when $f'(x) > 0$. This occurs in the intervals $(-2, 1)$ and $(3, \infty)$. Including the critical points where $f'(x) = 0$, the function is increasing on the closed intervals $[-2, 1]$ and $[3, \infty)$.

(b) The function $f(x)$ is decreasing when $f'(x) < 0$. This occurs in the intervals $(-\infty, -2)$ and $(1, 3)$. Including the critical points where $f'(x) = 0$, the function is decreasing on the closed intervals $(-\infty, -2]$ and $[1, 3]$.

---

The function $f(x)$ is:

(a) increasing on $[-2, 1] \cup [3, \infty)$.

(b) decreasing on $(-\infty, -2] \cup [1, 3]$.

Given:

The function $f(x) = \tan^{-1} (\sin x + \cos x)$.

The interval is $(0, \frac{\pi}{4})$.

---

To Show:

The function $f(x)$ is always an increasing function in the interval $\left( 0 , \frac{\pi}{4} \right)$.

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Solution:

To show that the function $f(x)$ is increasing in the interval $\left( 0 , \frac{\pi}{4} \right)$, we need to find its derivative $f'(x)$ and show that $f'(x) > 0$ for all $x \in \left( 0 , \frac{\pi}{4} \right)$.

The function is given by $f(x) = \tan^{-1} (\sin x + \cos x)$.

Using the chain rule, the derivative of $\tan^{-1}(u)$ is $\frac{1}{1+u^2} \frac{du}{dx}$. Here, let $u = \sin x + \cos x$.

First, find the derivative of $u$ with respect to $x$:

Now, find the derivative of $f(x)$:

Simplify the term $(\sin x + \cos x)^2$ in the denominator:

Substitute this back into the expression for $f'(x)$:

---

Now, we need to determine the sign of $f'(x)$ for $x \in \left( 0 , \frac{\pi}{4} \right)$.

Consider the denominator, $2 + \sin(2x)$. For $x \in \left( 0 , \frac{\pi}{4} \right)$, the range of $2x$ is $\left( 0 , \frac{\pi}{2} \right)$.

In the interval $\left( 0 , \frac{\pi}{2} \right)$, the sine function $\sin(2x)$ is positive and ranges from 0 to 1 (exclusive of endpoints). So, $0 < \sin(2x) < 1$.

Therefore, the denominator $2 + \sin(2x)$ is always greater than $2 + 0 = 2$ and less than $2 + 1 = 3$. Thus, the denominator is always positive in the interval $\left( 0 , \frac{\pi}{4} \right)$.

Now, consider the numerator, $\cos x - \sin x$. For $x \in \left( 0 , \frac{\pi}{4} \right)$, the value of $\cos x$ is greater than the value of $\sin x$. As $x$ increases from 0 to $\frac{\pi}{4}$, $\cos x$ decreases from 1 to $\frac{\sqrt{2}}{2}$, while $\sin x$ increases from 0 to $\frac{\sqrt{2}}{2}$. For any $x$ in the open interval $\left( 0 , \frac{\pi}{4} \right)$, $\cos x > \sin x$.

Since the numerator $(\cos x - \sin x)$ is positive and the denominator $(2 + \sin(2x))$ is positive for all $x \in \left( 0 , \frac{\pi}{4} \right)$, their quotient, $f'(x)$, must be positive in this interval.

Since $f'(x) > 0$ for all $x$ in the interval $\left( 0 , \frac{\pi}{4} \right)$, the function $f(x)$ is strictly increasing in this interval.

Hence Showed.

Given:

Radius of the circular disc = $r$ cm.

Initial radius = 3 cm (This value is for context, not directly used in the rate calculation at a specific instant).

Rate at which the radius increases = $\frac{dr}{dt} = 0.05 \, \text{cm/s}$.

Instant at which the rate of area increase is required: when radius $r = 3.2 \, \text{cm}$.

---

To Find:

The rate at which the area of the disc is increasing when the radius is 3.2 cm, i.e., $\frac{dA}{dt}$ at $r = 3.2 \, \text{cm}$.

---

Solution:

The area $A$ of a circular disc with radius $r$ is given by the formula:

We need to find the rate of change of the area with respect to time, $\frac{dA}{dt}$. We differentiate equation (i) with respect to time $t$ using the chain rule:

We are given $\frac{dr}{dt} = 0.05 \, \text{cm/s}$ and we want to find $\frac{dA}{dt}$ at the instant when $r = 3.2 \, \text{cm}$. Substitute these values into equation (ii):

Calculate the product:

The units for the rate of area increase are cm$^2$/s, consistent with the units of radius (cm) and time (s).

---

The rate at which the area of the circular disc is increasing when the radius is 3.2 cm is $0.32 \pi \, \text{cm}^2/\text{s}$.

Given:

A rectangular sheet of aluminium with dimensions 3 metres by 8 metres.

Equal squares are removed from each corner.

The sides are folded up to construct an open-topped box.

---

To Find:

The volume of the largest such box.

---

Solution:

Let the side length of the equal squares removed from each corner be $x$ metres.

When squares of side $x$ are removed from the corners, the dimensions of the base of the open box will be:

Length = $8 - 2x$ metres

Width = $3 - 2x$ metres

The height of the box will be the side length of the folded squares, which is $x$ metres.

For the dimensions to be positive, we must have:

So, the possible range for $x$ is $0 < x < 1.5$.

The volume $V$ of the open-topped box is given by:

Expand the expression for $V(x)$:

We want to find the maximum value of $V(x)$ for $x \in (0, 1.5)$. To do this, we find the derivative of $V(x)$ with respect to $x$ and set it equal to zero.

Set $\frac{dV}{dx} = 0$:

Divide the equation by 4:

We can solve this quadratic equation for $x$ using factoring or the quadratic formula. Let's factor it:

This gives two possible values for $x$: $3x - 2 = 0 \implies x = \frac{2}{3}$ or $x - 3 = 0 \implies x = 3$.

We need to consider the values of $x$ within the interval $(0, 1.5)$.

$x = \frac{2}{3} = 0.666...$ which is in the interval $(0, 1.5)$.

$x = 3$ which is not in the interval $(0, 1.5)$.

So, the critical point within the valid range is $x = \frac{2}{3}$.

---

To determine if $x = \frac{2}{3}$ corresponds to a maximum volume, we can use the second derivative test.

Evaluate the second derivative at the critical point $x = \frac{2}{3}$:

Since the second derivative is negative at $x = \frac{2}{3}$, this value of $x$ corresponds to a local maximum of the volume function $V(x)$.

---

Now, we find the maximum volume by substituting $x = \frac{2}{3}$ into the volume formula $V(x) = (8 - 2x)(3 - 2x)(x)$.

Calculate the terms in the parentheses:

Substitute these values back:

The maximum volume is $\frac{200}{27}$ cubic metres.

---

The volume of the largest open-topped box that can be constructed is $\frac{200}{27}$ m$^3$.

Given:

Number of items sold = $x$.

Selling price per item = $\textsf{₹} \left( 5 - \frac{x}{100} \right)$.

Cost price of $x$ items = $\textsf{₹} \left( \frac{x}{5} + 500 \right)$.

---

To Find:

The number of items ($x$) the manufacturer should sell to earn maximum profit.

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Solution:

The total revenue ($R$) from selling $x$ items is given by the number of items multiplied by the selling price per item:

The cost price ($C$) of $x$ items is given as:

The profit ($P$) is given by the total revenue minus the total cost:

Substitute the expressions for $R(x)$ and $C(x)$ from equations (i) and (ii):

Combine the terms with $x$:

We want to find the value of $x$ that maximizes the profit $P(x)$. To do this, we find the derivative of $P(x)$ with respect to $x$ and set it equal to zero.

Set $\frac{dP}{dx} = 0$:

Solve for $x$:

The critical point is $x = 240$. Since $x$ represents the number of items, it must be a non-negative integer. In optimization problems involving continuous functions for discrete quantities, we usually find the critical points as if $x$ were continuous and then consider the nearest integer values if the critical point is not an integer. Here, the critical point is an integer, 240.

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To confirm that $x = 240$ corresponds to a maximum profit, we use the second derivative test.

The second derivative is $\frac{d^2P}{dx^2} = -\frac{1}{50}$, which is a negative constant. Since the second derivative is always negative, the profit function $P(x)$ is concave down, and the critical point $x = 240$ corresponds to a global maximum.

The number of items must also satisfy the condition that the selling price per item is non-negative: $5 - \frac{x}{100} \ge 0 \implies 5 \ge \frac{x}{100} \implies 500 \ge x$. The number of items must also be non-negative, $x \ge 0$. So the valid range for $x$ is $0 \le x \le 500$. Our critical point $x=240$ lies within this range.

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The manufacturer should sell 240 items to earn the maximum profit.

Given: The function is $f(x) = \frac{\log x}{x}$.

To Show: The function $f(x)$ has a maximum at $x = e$.

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Solution:

To find the maximum of the function, we first find its derivative $f'(x)$.

The function is $f(x) = \frac{\log x}{x}$. The domain of the function is $x > 0$.

Using the quotient rule for differentiation, $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$, where $u = \log x$ and $v = x$.

Here, $u' = \frac{d}{dx}(\log x) = \frac{1}{x}$ and $v' = \frac{d}{dx}(x) = 1$.

The first derivative is:

$f'(x) = \frac{\frac{1}{x} \cdot x - (\log x) \cdot 1}{x^2}$

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To find the critical points, we set the first derivative $f'(x)$ equal to zero and solve for $x$.

Since the domain requires $x > 0$, $x^2$ is never zero. Thus, we can multiply both sides by $x^2$:

Converting the logarithmic equation to exponential form, we get:

So, the only critical point in the domain of $f(x)$ is $x = e$.

---

Now we use the second derivative test to determine if this critical point corresponds to a maximum or minimum.

We need to find the second derivative, $f''(x)$, from the first derivative $f'(x) = \frac{1 - \log x}{x^2}$.

Using the quotient rule again, with $u = 1 - \log x$ and $v = x^2$.

Here, $u' = \frac{d}{dx}(1 - \log x) = -\frac{1}{x}$ and $v' = \frac{d}{dx}(x^2) = 2x$.

The second derivative is:

$f''(x) = \frac{(-\frac{1}{x}) \cdot x^2 - (1 - \log x) \cdot 2x}{(x^2)^2}$

$f''(x) = \frac{-x - 2x(1 - \log x)}{x^4}$

$f''(x) = \frac{-x - 2x + 2x \log x}{x^4}$

$f''(x) = \frac{-3x + 2x \log x}{x^4}$

Factor out $x$ from the numerator:

$f''(x) = \frac{x(-3 + 2 \log x)}{x^4}$

---

Finally, we evaluate the second derivative $f''(x)$ at the critical point $x = e$ (from equation (ii)).

$f''(e) = \frac{-3 + 2 \log e}{e^3}$

Since $\log e = 1$, we substitute this value:

$f''(e) = \frac{-3 + 2(1)}{e^3}$

$f''(e) = \frac{-3 + 2}{e^3}$

$f''(e) = \frac{-1}{e^3}$

Since $e$ is the base of the natural logarithm and is approximately $2.718$, $e > 0$, which implies $e^3 > 0$.

Therefore, $f''(e) = \frac{-1}{e^3}$ is a negative value, i.e., $f''(e) < 0$.

---

Conclusion:

According to the second derivative test, if $f'(c) = 0$ at a critical point $c$, and $f''(c) < 0$, then the function $f(x)$ has a local maximum at $x = c$.

In this case, we found that $f'(e) = 0$ and $f''(e) = \frac{-1}{e^3} < 0$.

Thus, the function $f(x) = \frac{\log x}{x}$ has a local maximum at $x = e$.

Since $x=e$ is the only critical point for $x>0$, this local maximum is also the absolute maximum in the domain.

Hence, the function has a maximum at $x = e$.

Given:

Let the isosceles triangle be $\triangle ABC$, where $AB = AC = x$ and the base $BC = b$ (fixed). The equal sides $x$ are decreasing at the rate of 3 cm/sec.

The base $b$ is fixed, so $\frac{db}{dt} = 0$.

---

To Find:

The rate at which the area of the triangle is decreasing when the two equal sides are equal to the base, i.e., when $x = b$. We need to find $\frac{dA}{dt}$ when $x=b$, where $A$ is the area of the triangle.

---

Solution:

Let $A$ be the area of the isosceles triangle. We can find the area using the formula involving the base and height. Let $h$ be the altitude from $A$ to $BC$, meeting $BC$ at $D$. In $\triangle ABD$, $AD = h$, $AB = x$, and $BD = \frac{b}{2}$ (since the altitude from the vertex angle bisects the base in an isosceles triangle).

By the Pythagorean theorem in $\triangle ABD$:

$h^2 + \left(\frac{b}{2}\right)^2 = x^2$

$h^2 = x^2 - \frac{b^2}{4}$

The area of the triangle is $A = \frac{1}{2} \times \text{base} \times \text{height}$.

$A = \frac{1}{2} \times b \times h$

Substitute the expression for $h$:

We need to find $\frac{dA}{dt}$. Differentiate equation (i) with respect to $t$ using the chain rule and remembering that $b$ is a constant.

$\frac{dA}{dt} = \frac{d}{dt} \left( \frac{1}{2} b \sqrt{x^2 - \frac{b^2}{4}} \right)$

$\frac{dA}{dt} = \frac{1}{2} b \frac{d}{dt} \left( (x^2 - \frac{b^2}{4})^{1/2} \right)$

Using the chain rule, $\frac{d}{dt}(u^{1/2}) = \frac{1}{2} u^{-1/2} \frac{du}{dt}$, where $u = x^2 - \frac{b^2}{4}$.

$\frac{du}{dt} = \frac{d}{dt} (x^2 - \frac{b^2}{4}) = 2x \frac{dx}{dt} - 0 = 2x \frac{dx}{dt}$ (since $b$ is constant, $\frac{db}{dt}=0$)

So,

$\frac{dA}{dt} = \frac{1}{2} b \cdot \frac{1}{2} \left(x^2 - \frac{b^2}{4}\right)^{-1/2} \cdot \left(2x \frac{dx}{dt}\right)$

$\frac{dA}{dt} = \frac{1}{4} b \cdot \frac{1}{\sqrt{x^2 - \frac{b^2}{4}}} \cdot \left(2x \frac{dx}{dt}\right)$

$\frac{dA}{dt} = \frac{bx}{2 \sqrt{x^2 - \frac{b^2}{4}}} \frac{dx}{dt}$

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We are given that $\frac{dx}{dt} = -3$ cm/sec. We need to find $\frac{dA}{dt}$ when $x = b$.

Substitute $x = b$ and $\frac{dx}{dt} = -3$ into equation (ii):

$\left. \frac{dA}{dt} \right|_{x=b} = \frac{b(b)}{\sqrt{4(b)^2 - b^2}} (-3)$

$\left. \frac{dA}{dt} \right|_{x=b} = \frac{b^2}{\sqrt{4b^2 - b^2}} (-3)$

$\left. \frac{dA}{dt} \right|_{x=b} = \frac{b^2}{\sqrt{3b^2}} (-3)$

$\left. \frac{dA}{dt} \right|_{x=b} = \frac{b^2}{\sqrt{3}b} (-3)$

For a valid triangle with equal sides $x$ and base $b$, we must have $2x > b$. At the moment when $x=b$, this condition becomes $2b > b$, which is true for $b > 0$. Also, for the square root $\sqrt{x^2 - b^2/4}$ to be real, we need $x^2 > b^2/4$, which means $x > b/2$. At $x=b$, this is $b > b/2$, true for $b>0$. Since $b$ is a fixed base length, $b>0$.

Therefore, we can simplify the expression:

$\left. \frac{dA}{dt} \right|_{x=b} = \frac{b}{\sqrt{3}} (-3)$

Rationalize the denominator:

$\left. \frac{dA}{dt} \right|_{x=b} = -\frac{3b}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$\left. \frac{dA}{dt} \right|_{x=b} = -\frac{3b\sqrt{3}}{3}$

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The rate of change of area is $-b\sqrt{3}$ cm$^2$/sec. The negative sign indicates that the area is decreasing.

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Answer:

The area of the triangle is decreasing at the rate of $\mathbf{b\sqrt{3}}$ cm$^2$/sec when the two equal sides are equal to the base.

Given: The function is $f(x) = \frac{4 \sin x − 2x − x \cos x}{2 + \cos x}$.

To Find: The intervals where $f(x)$ is (i) increasing and (ii) decreasing.

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Solution:

To find the intervals of increasing or decreasing, we need to find the first derivative of the function, $f'(x)$. We will use the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$, where $u = 4 \sin x - 2x - x \cos x$ and $v = 2 + \cos x$.

First, find the derivatives of $u$ and $v$ with respect to $x$:

$u' = \frac{d}{dx}(4 \sin x - 2x - x \cos x)$

$u' = 4 \cos x - 2 - (\frac{d}{dx}(x)\cos x + x\frac{d}{dx}(\cos x))$

$u' = 4 \cos x - 2 - (1 \cdot \cos x + x(-\sin x))$

$u' = 4 \cos x - 2 - \cos x + x \sin x$

$v' = \frac{d}{dx}(2 + \cos x)$

Now, apply the quotient rule:

$f'(x) = \frac{(3 \cos x - 2 + x \sin x)(2 + \cos x) - (4 \sin x - 2x - x \cos x)(-\sin x)}{(2 + \cos x)^2}$

Expand the numerator:

Numerator $= (6 \cos x + 3 \cos^2 x - 4 - 2 \cos x + 2x \sin x + x \sin x \cos x) - (-4 \sin^2 x + 2x \sin x + x \sin x \cos x)$

Numerator $= 6 \cos x + 3 \cos^2 x - 4 - 2 \cos x + 2x \sin x + x \sin x \cos x + 4 \sin^2 x - 2x \sin x - x \sin x \cos x$

Combine like terms:

Numerator $= (6 \cos x - 2 \cos x) + (3 \cos^2 x + 4 \sin^2 x) - 4 + (2x \sin x - 2x \sin x) + (x \sin x \cos x - x \sin x \cos x)$

Numerator $= 4 \cos x + 3 \cos^2 x + 4 \sin^2 x - 4$

Using the identity $\sin^2 x = 1 - \cos^2 x$:

Numerator $= 4 \cos x + 3 \cos^2 x + 4(1 - \cos^2 x) - 4$

Numerator $= 4 \cos x + 3 \cos^2 x + 4 - 4 \cos^2 x - 4$

Numerator $= 4 \cos x - \cos^2 x$

Factor the numerator:

Numerator $= \cos x (4 - \cos x)$

So, the first derivative is:

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Now, we analyze the sign of $f'(x)$ to find the intervals of increasing and decreasing.

The denominator $(2 + \cos x)^2$ is always non-negative. Since $\cos x$ is in the range $[-1, 1]$, $2 + \cos x$ is in the range $[1, 3]$. Thus, $2 + \cos x \neq 0$, and $(2 + \cos x)^2 > 0$ for all $x$ where $\cos x$ is defined.

The term $(4 - \cos x)$ is also always positive, since $\cos x \leq 1$, so $4 - \cos x \geq 4 - 1 = 3 > 0$.

Therefore, the sign of $f'(x)$ is determined solely by the sign of $\cos x$.

(i) Increasing Intervals:

The function $f(x)$ is increasing when $f'(x) > 0$.

Since $(4 - \cos x) > 0$ and $(2 + \cos x)^2 > 0$, this inequality holds when $\cos x > 0$.

The general solution for $\cos x > 0$ is given by the intervals $(\frac{-\pi}{2} + 2n\pi, \frac{\pi}{2} + 2n\pi)$, where $n$ is an integer ($n \in \mathbb{Z}$).

So, $f(x)$ is increasing on the intervals $\left(2n\pi - \frac{\pi}{2}, 2n\pi + \frac{\pi}{2}\right)$, $n \in \mathbb{Z}$.

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(ii) Decreasing Intervals:

The function $f(x)$ is decreasing when $f'(x) < 0$.

Since $(4 - \cos x) > 0$ and $(2 + \cos x)^2 > 0$, this inequality holds when $\cos x < 0$.

The general solution for $\cos x < 0$ is given by the intervals $(\frac{\pi}{2} + 2n\pi, \frac{3\pi}{2} + 2n\pi)$, where $n$ is an integer ($n \in \mathbb{Z}$).

So, $f(x)$ is decreasing on the intervals $\left(2n\pi + \frac{\pi}{2}, 2n\pi + \frac{3\pi}{2}\right)$, $n \in \mathbb{Z}$.

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Answer:

(i) The function $f(x)$ is increasing on the intervals $\left(2n\pi - \frac{\pi}{2}, 2n\pi + \frac{\pi}{2}\right)$, where $n \in \mathbb{Z}$.

(ii) The function $f(x)$ is decreasing on the intervals $\left(2n\pi + \frac{\pi}{2}, 2n\pi + \frac{3\pi}{2}\right)$, where $n \in \mathbb{Z}$.

Given: The function is $f(x) = x^3 + \frac{1}{x^3}$, with domain $x \neq 0$.

To Find: The intervals where $f(x)$ is (i) increasing and (ii) decreasing.

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Solution:

To determine the intervals of increasing or decreasing, we first need to find the derivative of the function, $f'(x)$.

The function can be written as $f(x) = x^3 + x^{-3}$.

Differentiating with respect to $x$, we get:

$f'(x) = \frac{d}{dx}(x^3 + x^{-3})$

$f'(x) = 3x^{3-1} + (-3)x^{-3-1}$

$f'(x) = 3x^2 - 3x^{-4}$

To analyze the sign of $f'(x)$, we can rewrite it with a common denominator:

$f'(x) = \frac{3x^2 \cdot x^4}{x^4} - \frac{3}{x^4}$

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The function is increasing when $f'(x) > 0$ and decreasing when $f'(x) < 0$.

From equation (i), $f'(x) = \frac{3(x^6 - 1)}{x^4}$.

For $x \neq 0$, the term $x^4$ is always positive ($x^4 > 0$). The constant factor $3$ is also positive.

Therefore, the sign of $f'(x)$ depends entirely on the sign of the term $(x^6 - 1)$.

We need to find when $x^6 - 1 > 0$ and when $x^6 - 1 < 0$.

$x^6 - 1 > 0 \implies x^6 > 1$. This inequality holds when $x > 1$ or $x < -1$.

$x^6 - 1 < 0 \implies x^6 < 1$. This inequality holds when $-1 < x < 1$.

Considering the domain $x \neq 0$, the intervals based on the sign of $x^6 - 1$ are:

1. $(-\infty, -1)$: In this interval, $x^6 - 1 > 0$. Since $x^4 > 0$, $f'(x) = \frac{3(positive)}{positive} > 0$.

2. $(-1, 0)$: In this interval, $-1 < x < 0$. So, $x^6 < 1$, which means $x^6 - 1 < 0$. Since $x^4 > 0$, $f'(x) = \frac{3(negative)}{positive} < 0$.

3. $(0, 1)$: In this interval, $0 < x < 1$. So, $x^6 < 1$, which means $x^6 - 1 < 0$. Since $x^4 > 0$, $f'(x) = \frac{3(negative)}{positive} < 0$.

4. $(1, \infty)$: In this interval, $x > 1$. So, $x^6 > 1$, which means $x^6 - 1 > 0$. Since $x^4 > 0$, $f'(x) = \frac{3(positive)}{positive} > 0$.

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(i) Increasing Intervals:

The function $f(x)$ is increasing when $f'(x) > 0$. This occurs in the intervals where $x^6 - 1 > 0$.

These intervals are $(-\infty, -1)$ and $(1, \infty)$.

So, $f(x)$ is increasing on $\mathbf{(-\infty, -1) \cup (1, \infty)}$.

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(ii) Decreasing Intervals:

The function $f(x)$ is decreasing when $f'(x) < 0$. This occurs in the intervals where $x^6 - 1 < 0$, considering the domain $x \neq 0$.

These intervals are $(-1, 0)$ and $(0, 1)$.

So, $f(x)$ is decreasing on $\mathbf{(-1, 0) \cup (0, 1)}$.

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Answer:

(i) The function $f(x)$ is increasing on the intervals $\mathbf{(-\infty, -1) \cup (1, \infty)}$.

(ii) The function $f(x)$ is decreasing on the intervals $\mathbf{(-1, 0) \cup (0, 1)}$.

Given: An ellipse with equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. An isosceles triangle is inscribed in the ellipse with one vertex at an end of the major axis.

To Find: The maximum area of such an isosceles triangle.

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Solution:

Assume the major axis is along the x-axis, so $a > b$. The ends of the major axis are $(\pm a, 0)$. Let the vertex of the isosceles triangle be $A = (a, 0)$.

Let the other two vertices of the isosceles triangle be $P$ and $Q$ on the ellipse. For the triangle $APQ$ to be isosceles with $AP = AQ$, and $A$ on the x-axis, the points $P$ and $Q$ must be symmetric with respect to the x-axis. Let $P = (x_0, y_0)$ and $Q = (x_0, -y_0)$, where $y_0 > 0$.

Since $P(x_0, y_0)$ lies on the ellipse, it satisfies the equation $\frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} = 1$.

From the ellipse equation, $y_0^2 = b^2 \left(1 - \frac{x_0^2}{a^2}\right) = \frac{b^2}{a^2}(a^2 - x_0^2)$.

Since $y_0 > 0$, we have $y_0 = \frac{b}{a}\sqrt{a^2 - x_0^2}$.

For the vertices $P$ and $Q$ to be distinct from $A$ and form a triangle, $x_0$ must be in the interval $(-a, a)$ and $y_0 \neq 0$. The condition $y_0 \neq 0$ implies $x_0 \neq \pm a$. So, the domain for $x_0$ is $(-a, a)$.

The base of the triangle is the segment $PQ$. Its length is $2y_0$.

The altitude of the triangle from vertex $A=(a,0)$ to the base $PQ$ (which lies on the vertical line $x=x_0$) is the horizontal distance between $A$ and the line $x=x_0$. This distance is $|a - x_0|$. Since $x_0 \in (-a, a)$, $a - x_0 > 0$. So the altitude is $a - x_0$.

The area of the triangle $APQ$, denoted by $Area(x_0)$, is given by:

$Area(x_0) = \frac{1}{2} \times \text{base} \times \text{altitude}$

$Area(x_0) = \frac{1}{2} \times (2y_0) \times (a - x_0)$

$Area(x_0) = y_0 (a - x_0)$

Substitute the expression for $y_0$:

To find the maximum area, we can maximize the square of the area, which simplifies differentiation. Let $Z(x_0) = [Area(x_0)]^2$.

$Z(x_0) = \left(\frac{b}{a}\sqrt{a^2 - x_0^2} (a - x_0)\right)^2$

$Z(x_0) = \frac{b^2}{a^2} (a^2 - x_0^2) (a - x_0)^2$

$Z(x_0) = \frac{b^2}{a^2} (a - x_0)(a + x_0) (a - x_0)^2$

We need to find the maximum of $Z(x_0)$ for $x_0 \in (-a, a)$. We differentiate $Z(x_0)$ with respect to $x_0$ and set the derivative to zero.

$\frac{dZ}{dx_0} = \frac{d}{dx_0} \left[ \frac{b^2}{a^2} (a + x_0) (a - x_0)^3 \right]$

Using the product rule $(uv)' = u'v + uv'$ with $u = a+x_0$ and $v = (a-x_0)^3$:

$\frac{du}{dx_0} = 1$

$\frac{dv}{dx_0} = 3(a - x_0)^2 \cdot \frac{d}{dx_0}(a - x_0) = 3(a - x_0)^2 \cdot (-1) = -3(a - x_0)^2$

$\frac{dZ}{dx_0} = \frac{b^2}{a^2} \left[ 1 \cdot (a - x_0)^3 + (a + x_0) \cdot (-3)(a - x_0)^2 \right]$

Factor out the common term $(a - x_0)^2$:

$\frac{dZ}{dx_0} = \frac{b^2}{a^2} (a - x_0)^2 \left[ (a - x_0) - 3(a + x_0) \right]$

$\frac{dZ}{dx_0} = \frac{b^2}{a^2} (a - x_0)^2 \left[ a - x_0 - 3a - 3x_0 \right]$

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To find the critical points, we set $\frac{dZ}{dx_0} = 0$.

Since $\frac{2b^2}{a^2} \neq 0$ (as $a, b$ are semi-axes lengths), this implies $(a - x_0)^2 = 0$ or $(a + 2x_0) = 0$.

Case 1: $(a - x_0)^2 = 0 \implies x_0 = a$. This is an endpoint of the interval $(-a, a)$. At $x_0 = a$, $y_0 = \frac{b}{a}\sqrt{a^2 - a^2} = 0$, so $P = Q = (a, 0) = A$. This results in a degenerate triangle with area 0, which is a minimum.

Case 2: $a + 2x_0 = 0 \implies x_0 = -\frac{a}{2}$. This value is in the interval $(-a, a)$ since $a > 0$. This is a potential location for a maximum.

---

We can use the first derivative test to confirm if $x_0 = -\frac{a}{2}$ corresponds to a maximum. The sign of $\frac{dZ}{dx_0} = -\frac{2b^2}{a^2} (a - x_0)^2 (a + 2x_0)$ depends on the sign of $-(a + 2x_0)$ for $x_0 \in (-a, a)$, since $(a - x_0)^2 > 0$ and $-\frac{2b^2}{a^2} < 0$.

• If $x_0 < -\frac{a}{2}$ (e.g., $x_0 = -0.6a$), then $a + 2x_0 < 0$, so $-(a + 2x_0) > 0$. Thus $\frac{dZ}{dx_0} > 0$.
• If $x_0 > -\frac{a}{2}$ (e.g., $x_0 = -0.4a$), then $a + 2x_0 > 0$, so $-(a + 2x_0) < 0$. Thus $\frac{dZ}{dx_0} < 0$.

Since $\frac{dZ}{dx_0}$ changes from positive to negative as $x_0$ increases through $x_0 = -\frac{a}{2}$, $Z(x_0)$ has a local maximum at $x_0 = -\frac{a}{2}$. Consequently, $Area(x_0)$ also has a local maximum at $x_0 = -\frac{a}{2}$. Since there is only one critical point in the interval and the boundary points give minimum area (0), this local maximum is the absolute maximum.

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Now, we calculate the maximum area by substituting $x_0 = -\frac{a}{2}$ into the area formula (i):

$Area_{max} = \frac{b}{a}\sqrt{a^2 - (-\frac{a}{2})^2} (a - (-\frac{a}{2}))$

$Area_{max} = \frac{b}{a}\sqrt{a^2 - \frac{a^2}{4}} (a + \frac{a}{2})$

$Area_{max} = \frac{b}{a}\sqrt{\frac{4a^2 - a^2}{4}} (\frac{3a}{2})$

$Area_{max} = \frac{b}{a}\sqrt{\frac{3a^2}{4}} (\frac{3a}{2})$

Since $a > 0$, $\sqrt{a^2} = a$ and $\sqrt{4} = 2$:

$Area_{max} = \frac{b}{a} \cdot \frac{\sqrt{3}a}{2} \cdot \frac{3a}{2}$

$Area_{max} = \frac{b}{\cancel{a}} \cdot \frac{\sqrt{3}\cancel{a}}{2} \cdot \frac{3a}{2}$

If the major axis is along the y-axis ($b>a$) and the vertex is at $(0, \pm b)$, a similar calculation would yield the same result $\frac{3\sqrt{3}ab}{4}$.

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Answer:

The maximum area of the isosceles triangle inscribed in the ellipse with its vertex at one end of the major axis is $\mathbf{\frac{3\sqrt{3}}{4}ab}$.

Given:

Depth of the tank, $h = 2$ m.

Volume of the tank, $V = 8$ m$^3$.

Cost of building the base = $\textsf{₹} 70$ per sq metre.

Cost of building the sides = $\textsf{₹} 45$ per sq metre.

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To Find:

The cost of the least expensive tank.

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Solution:

Let the length, width, and depth of the rectangular tank be $l$, $w$, and $h$ respectively.

The volume of the tank is given by $V = lwh$.

We are given $h = 2$ m and $V = 8$ m$^3$.

From equation (i), we can express the width $w$ in terms of length $l$ (since $l > 0$):

The tank has an open top, so the total surface area to be constructed consists of the base and the four sides.

Area of the base $= lw$.

Area of the four sides $= 2(lh + wh)$.

The total cost of building the tank, $C$, is the sum of the cost of the base and the cost of the sides.

$C = (\text{Area of base}) \times (\text{Cost per sq metre for base}) + (\text{Area of sides}) \times (\text{Cost per sq metre for sides})$

$C = (lw) \times 70 + 2(lh + wh) \times 45$

Substitute the values $lw = 4$ (from i) and $h = 2$ into the cost function (iii):

$C(l, w) = 70(4) + 90(l(2) + w(2))$

$C(l, w) = 280 + 90(2l + 2w)$

$C(l, w) = 280 + 180l + 180w$

Now, substitute $w = \frac{4}{l}$ (from ii) to express the cost as a function of a single variable $l$:

The domain of $l$ is $(0, \infty)$ since length must be positive.

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To find the minimum cost, we need to find the value of $l$ that minimizes $C(l)$. We do this by finding the critical points, where the derivative $C'(l) = 0$ or is undefined. The derivative is undefined only at $l=0$, which is not in the domain.

Differentiate $C(l)$ with respect to $l$:

$C'(l) = \frac{d}{dl} \left( 280 + 180l + 720l^{-1} \right)$

$C'(l) = 0 + 180(1) + 720(-1)l^{-2}$

Set $C'(l) = 0$ to find critical points:

Since $l > 0$, we take the positive square root:

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Now, we use the second derivative test to confirm if $l=2$ corresponds to a minimum cost.

Differentiate $C'(l) = 180 - 720l^{-2}$ with respect to $l$:

$C''(l) = \frac{d}{dl} (180 - 720l^{-2})$

$C''(l) = 0 - 720(-2)l^{-3}$

Evaluate $C''(l)$ at $l=2$ (from equation vi):

$C''(2) = \frac{1440}{2^3} = \frac{1440}{8}$

Since $C''(2) = 180 > 0$, the cost function $C(l)$ has a local minimum at $l=2$. As this is the only critical point in the domain $(0, \infty)$, it corresponds to the absolute minimum cost.

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The dimensions that minimize the cost are $l=2$ m. Using equation (ii), the width is $w = \frac{4}{l} = \frac{4}{2} = 2$ m. The depth is $h=2$ m (given).

So, the tank dimensions for the least expensive tank are 2 m $\times$ 2 m $\times$ 2 m (a cube).

Now, calculate the minimum cost by substituting $l=2$ into the cost function $C(l)$ (equation iv):

$C_{min} = C(2) = 280 + 180(2) + \frac{720}{2}$

$C_{min} = 280 + 360 + 360$

$C_{min} = 280 + 720$

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Answer:

The cost of the least expensive tank is $\mathbf{\textsf{₹} 1000}$.

Given:

The sum of the perimeter of a circle and a square is a constant $k$.

To Prove:

The sum of their areas is least when the side of the square is double the radius of the circle.

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Solution:

Let $r$ be the radius of the circle and $s$ be the side of the square.

The perimeter of the circle is $P_c = 2\pi r$.

The perimeter of the square is $P_s = 4s$.

According to the given information, the sum of their perimeters is a constant $k$:

From equation (i), we can express $s$ in terms of $r$ and $k$ (since $4s = k - 2\pi r$ and $s>0$, we must have $k - 2\pi r > 0$, which implies $r < \frac{k}{2\pi}$):

The area of the circle is $A_c = \pi r^2$.

The area of the square is $A_s = s^2$.

The sum of their areas, $A$, is:

$A = A_c + A_s = \pi r^2 + s^2$

Substitute the expression for $s$ from equation (ii) into the area equation to get the total area as a function of $r$:

The domain for $r$ is $r > 0$. Also, for $s > 0$, $k - 2\pi r > 0$, so $r < \frac{k}{2\pi}$. Thus, the relevant domain for $r$ is $\left(0, \frac{k}{2\pi}\right)$.

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To find the value of $r$ for which the area is least, we differentiate $A(r)$ with respect to $r$ and set the derivative to zero.

$A'(r) = \frac{d}{dr}\left(\pi r^2 + \frac{1}{16}(k - 2\pi r)^2\right)$

$A'(r) = 2\pi r + \frac{1}{16} \cdot 2(k - 2\pi r) \cdot \frac{d}{dr}(k - 2\pi r)$

$A'(r) = 2\pi r + \frac{1}{8}(k - 2\pi r) \cdot (-2\pi)$

$A'(r) = 2\pi r - \frac{2\pi}{8}(k - 2\pi r)$

Set $A'(r) = 0$ to find the critical points:

Divide by $\pi$ (since $\pi \neq 0$):

This critical value $r = \frac{k}{2(4+\pi)}$ lies within the domain $\left(0, \frac{k}{2\pi}\right)$ because $4+\pi > \pi$, so $2(4+\pi) > 2\pi$, and thus $\frac{1}{2(4+\pi)} < \frac{1}{2\pi}$.

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To confirm that this value of $r$ minimizes the area, we find the second derivative of $A(r)$.

From equation (iv), $A'(r) = 2\pi r - \frac{\pi k}{4} + \frac{2\pi^2 r}{4} = 2\pi r - \frac{\pi k}{4} + \frac{\pi^2 r}{2}$.

$A''(r) = \frac{d}{dr}\left(2\pi r - \frac{\pi k}{4} + \frac{\pi^2 r}{2}\right)$

$A''(r) = 2\pi - 0 + \frac{\pi^2}{2}$

Since $\pi > 0$ and $2 + \frac{\pi}{2} > 0$, we have $A''(r) > 0$ for all $r$. This means the function $A(r)$ is concave up, and the critical point found corresponds to a global minimum.

So, the area is least when $r = \frac{k}{2(4 + \pi)}$.

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Now, we need to check the relationship between the side of the square $s$ and the radius of the circle $r$ at this minimum area.

Substitute $r = \frac{k}{2(4 + \pi)}$ (from equation v) into the expression for $s$ (equation ii):

$s = \frac{k - 2\pi \left(\frac{k}{2(4 + \pi)}\right)}{4}$

$s = \frac{k - \frac{\pi k}{4 + \pi}}{4}$

$s = \frac{\frac{k(4 + \pi) - \pi k}{4 + \pi}}{4}$

$s = \frac{\frac{4k + \pi k - \pi k}{4 + \pi}}{4}$

$s = \frac{\frac{4k}{4 + \pi}}{4}$

Now, let's find $2r$ using the value of $r$ from equation (v):

Comparing equation (vii) and equation (viii), we see that $s = \frac{k}{4 + \pi}$ and $2r = \frac{k}{4 + \pi}$.

Therefore, $s = 2r$ when the sum of the areas is least.

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Conclusion:

We have shown that the sum of the areas of the circle and the square is minimized when the radius of the circle is $r = \frac{k}{2(4 + \pi)}$, and at this radius, the side of the square $s = \frac{k}{4 + \pi}$. This proves that the side of the square is double the radius of the circle ($s = 2r$) when the sum of their areas is least.

Given:

A window is composed of a rectangle surmounted by a semicircle.

The total perimeter of the window is 10 m.

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To Find:

The dimensions of the rectangular part of the window (width and height) that maximize the total area of the window.

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Solution:

Let the width of the rectangular part of the window be $w$ metres and the height be $h$ metres. The semicircle is surmounted on the width, so the diameter of the semicircle is $w$, and its radius is $r = \frac{w}{2}$ metres.

The total perimeter of the window consists of the three sides of the rectangle (two vertical sides and the base) and the arc length of the semicircle.

Perimeter $P = h + w + h + (\text{Arc length of semicircle})$

$P = 2h + w + \pi r$

Substitute $r = \frac{w}{2}$ into the perimeter equation:

$P = 2h + w + \pi \left(\frac{w}{2}\right)$

$P = 2h + w \left(1 + \frac{\pi}{2}\right)$

We are given that the total perimeter is 10 m.

From equation (i), we can express the height $h$ in terms of the width $w$:

$2h = 10 - w \left(\frac{2 + \pi}{2}\right)$

For the dimensions to be physically meaningful, $w > 0$ and $h \geq 0$.

$h \geq 0 \implies 5 - \frac{w(2 + \pi)}{4} \geq 0 \implies 5 \geq \frac{w(2 + \pi)}{4} \implies 20 \geq w(2 + \pi) \implies w \leq \frac{20}{2 + \pi}$.

Thus, the domain for the width $w$ is $0 < w \leq \frac{20}{2 + \pi}$.

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The total area of the window, $A$, which determines the amount of light admitted, is the sum of the area of the rectangle and the area of the semicircle.

$A = (\text{Area of rectangle}) + (\text{Area of semicircle})$

$A = wh + \frac{1}{2} \pi r^2$

Substitute $r = \frac{w}{2}$ into the area equation:

$A = wh + \frac{1}{2} \pi \left(\frac{w}{2}\right)^2 = wh + \frac{\pi w^2}{8}$

Now, substitute the expression for $h$ from equation (ii) into the area equation to express the area as a function of $w$:

$A(w) = w \left(5 - \frac{w(2 + \pi)}{4}\right) + \frac{\pi w^2}{8}$

$A(w) = 5w - \frac{w^2(2 + \pi)}{4} + \frac{\pi w^2}{8}$

To combine the $w^2$ terms, find a common denominator (8):

$A(w) = 5w - \frac{2w^2(2 + \pi)}{8} + \frac{\pi w^2}{8}$

$A(w) = 5w + \frac{w^2}{8} [-2(2 + \pi) + \pi]$

$A(w) = 5w + \frac{w^2}{8} [-4 - 2\pi + \pi]$

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To find the maximum area, we differentiate $A(w)$ with respect to $w$ and find the critical points by setting the derivative equal to zero.

$A'(w) = \frac{d}{dw}\left(5w - \frac{(4 + \pi)w^2}{8}\right)$

$A'(w) = 5 \cdot \frac{d}{dw}(w) - \frac{(4 + \pi)}{8} \cdot \frac{d}{dw}(w^2)$

$A'(w) = 5(1) - \frac{(4 + \pi)}{8} \cdot (2w)$

Set $A'(w) = 0$ to find the critical point(s):

This critical value $w = \frac{20}{4 + \pi}$ lies within the domain $0 < w \leq \frac{20}{2 + \pi}$, since $4 + \pi > 2 + \pi \implies \frac{1}{4 + \pi} < \frac{1}{2 + \pi} \implies \frac{20}{4 + \pi} < \frac{20}{2 + \pi}$.

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To confirm that this value of $w$ corresponds to a maximum area, we use the second derivative test.

Differentiate $A'(w) = 5 - \frac{(4 + \pi)}{4} w$ with respect to $w$:

$A''(w) = \frac{d}{dw} \left(5 - \frac{(4 + \pi)}{4} w \right)$

Since $\pi > 0$, $4 + \pi > 0$, and therefore $A''(w) = -\frac{4 + \pi}{4} < 0$ for all $w$. This indicates that the area function $A(w)$ is concave down everywhere, and the critical point $w = \frac{20}{4 + \pi}$ corresponds to an absolute maximum.

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Now, we find the dimensions of the window that maximize the area. The width of the rectangle is $w = \frac{20}{4 + \pi}$ m. The height $h$ is found by substituting this value of $w$ into equation (ii):

$h = 5 - \frac{\left(\frac{20}{4 + \pi}\right)(2 + \pi)}{4}$

$h = 5 - \frac{20(2 + \pi)}{4(4 + \pi)}$

$h = 5 - \frac{5(2 + \pi)}{4 + \pi}$

$h = \frac{5(4 + \pi)}{4 + \pi} - \frac{5(2 + \pi)}{4 + \pi}$

The radius of the semicircle at this width is $r = \frac{w}{2} = \frac{1}{2} \left(\frac{20}{4 + \pi}\right) = \frac{10}{4 + \pi}$. Notice that the height of the rectangle is equal to the radius of the semicircle ($h=r$).

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Answer:

To admit maximum light, the dimensions of the rectangular part of the window should be width $= \mathbf{\frac{20}{4 + \pi}}$ m and height $= \mathbf{\frac{10}{4 + \pi}}$ m.

Given:

A right-angled triangle. A point on the hypotenuse is at a distance $a$ from one leg and a distance $b$ from the other leg.

To Show:

The minimum length of the hypotenuse is $\left( a^{\frac{2}{3}} + b^{\frac{2}{3}} \right)^{\frac{3}{2}}$.

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Solution:

Let the right-angled triangle be placed in the first quadrant of a coordinate system, with the vertex containing the right angle at the origin $O(0,0)$. Let the legs of the triangle lie along the positive x and y axes. Let the vertices on the axes be $B(X, 0)$ and $A(0, Y)$, where $X > 0$ and $Y > 0$. The hypotenuse is the line segment connecting $A$ and $B$.

The equation of the line representing the hypotenuse is given by the intercept form:

Let $P(x_p, y_p)$ be the point on the hypotenuse. The distance of $P$ from the y-axis (the leg OA) is $|x_p|$. We are given this distance is $b$, so $x_p = b$ (since P is in the first quadrant). The distance of $P$ from the x-axis (the leg OB) is $|y_p|$. We are given this distance is $a$, so $y_p = a$.

Thus, the coordinates of the point on the hypotenuse are $P(b, a)$.

Since the point $P(b, a)$ lies on the hypotenuse, its coordinates must satisfy the equation of the hypotenuse (i):

This equation relates the lengths of the legs $X$ and $Y$ based on the position of the point $P(b, a)$. For a non-degenerate triangle with P on the hypotenuse between A and B, we must have $0 < b < X$ and $0 < a < Y$.

The length of the hypotenuse, $L$, is the distance between points $A(0, Y)$ and $B(X, 0)$:

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We need to minimize $L$. We can express $Y$ in terms of $X$ and the constants $a$ and $b$ from equation (ii):

$\frac{a}{Y} = 1 - \frac{b}{X} = \frac{X - b}{X}$

Substitute this expression for $Y$ into the formula for $L$:

$L(X) = \sqrt{X^2 + \left(\frac{aX}{X - b}\right)^2} = \sqrt{X^2 + \frac{a^2 X^2}{(X - b)^2}}$

$L(X) = \sqrt{X^2 \left(1 + \frac{a^2}{(X - b)^2}\right)} = X \sqrt{1 + \frac{a^2}{(X - b)^2}}$

Since $X > b$, $|X - b| = X - b$.

The domain for $X$ is $(b, \infty)$. To minimize $L(X)$, we can equivalently minimize $L^2(X)$. Let $f(X) = L^2(X)$.

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Differentiate $f(X)$ with respect to $X$ to find critical points:

$f'(X) = \frac{d}{dX} \left(X^2 + a^2 X^2 (X - b)^{-2}\right)$

$f'(X) = 2X + a^2 \left[ 2X (X - b)^{-2} + X^2 (-2)(X - b)^{-3} \cdot 1 \right]$

$f'(X) = 2X + 2a^2 X (X - b)^{-2} - 2a^2 X^2 (X - b)^{-3}$

$f'(X) = 2X \left[ 1 + a^2 (X - b)^{-2} - a^2 X (X - b)^{-3} \right]$

$f'(X) = 2X \left[ 1 + \frac{a^2}{(X - b)^2} - \frac{a^2 X}{(X - b)^3} \right]$

$f'(X) = 2X \left[ \frac{(X - b)^3 + a^2 (X - b) - a^2 X}{(X - b)^3} \right]$

Let's try factoring earlier:

$f'(X) = 2X + 2a^2 X (X-b)^{-3} [(X-b) - X]$

$f'(X) = 2X + 2a^2 X (X-b)^{-3} [-b]$

$f'(X) = 2X - \frac{2a^2 b X}{(X-b)^3}$

Set $f'(X) = 0$ to find the critical points. Since $X > b > 0$, $2X \neq 0$.

Taking the cube root of both sides:

To confirm this is a minimum, we find the second derivative $f''(X)$:

$f'(X) = 2X - 2a^2 b (X-b)^{-3}$

$f''(X) = \frac{d}{dX} (2X - 2a^2 b (X-b)^{-3})$

$f''(X) = 2 - 2a^2 b (-3)(X-b)^{-4} \cdot 1$

For $X > b$, $(X-b)^4 > 0$. Since $a > 0$ and $b > 0$, $6a^2 b > 0$. Thus, $f''(X) = 2 + \frac{6a^2 b}{(X-b)^4} > 0$ for all $X > b$. This confirms that the critical point found corresponds to an absolute minimum of $f(X)$, and therefore of $L(X)$.

---

Now, we find the minimum length of the hypotenuse by substituting the critical value of $X$ from equation (vii) into the expression for $L(X)$ from equation (iv):

$L_{min} = \frac{X}{X - b} \sqrt{(X - b)^2 + a^2}$

From equation (vii), we know $X - b = a^{2/3} b^{1/3}$. Substitute this into the expression for $L_{min}$:

$L_{min} = \frac{b + a^{2/3} b^{1/3}}{a^{2/3} b^{1/3}} \sqrt{(a^{2/3} b^{1/3})^2 + a^2}$

$L_{min} = \left(\frac{b}{a^{2/3} b^{1/3}} + \frac{a^{2/3} b^{1/3}}{a^{2/3} b^{1/3}}\right) \sqrt{a^{4/3} b^{2/3} + a^2}$

$L_{min} = \left(b^{1 - 1/3} a^{-2/3} + 1\right) \sqrt{a^2(a^{-2/3} b^{2/3} + 1)}$

$L_{min} = \left(a^{-2/3} b^{2/3} + 1\right) a \sqrt{a^{-2/3} b^{2/3} + 1}$

$L_{min} = a (a^{-2/3} b^{2/3} + 1)^{3/2}$

This is not the target form. Let's simplify the expression for $L^2$ at the critical point instead.

$L^2 = f(X) = X^2 + \frac{a^2 X^2}{(X-b)^2}$

We know $(X-b)^2 = (a^{2/3} b^{1/3})^2 = a^{4/3} b^{2/3}$.

$L^2 = X^2 + \frac{a^2 X^2}{a^{4/3} b^{2/3}} = X^2 + X^2 a^{2 - 4/3} b^{-2/3} = X^2 + X^2 a^{2/3} b^{-2/3}$

$L^2 = X^2 (1 + a^{2/3} b^{-2/3})$

Substitute $X = b + a^{2/3} b^{1/3} = b^{1/3}(b^{2/3} + a^{2/3})$:

$X^2 = (b^{1/3}(b^{2/3} + a^{2/3}))^2 = b^{2/3} (b^{2/3} + a^{2/3})^2$

$L^2 = b^{2/3} (a^{2/3} + b^{2/3})^2 \left(1 + \frac{a^{2/3}}{b^{2/3}}\right)$

$L^2 = b^{2/3} (a^{2/3} + b^{2/3})^2 \left(\frac{b^{2/3} + a^{2/3}}{b^{2/3}}\right)$

$L^2 = \cancel{b^{2/3}} (a^{2/3} + b^{2/3})^2 \frac{(a^{2/3} + b^{2/3})}{\cancel{b^{2/3}}}$

$L^2 = (a^{2/3} + b^{2/3})^2 (a^{2/3} + b^{2/3})$

Taking the square root:

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Conclusion:

We found that the length of the hypotenuse is minimized when $X = b + a^{2/3} b^{1/3}$, and the minimum length at this value of $X$ is $\left( a^{\frac{2}{3}} + b^{\frac{2}{3}} \right)^{\frac{3}{2}}$.

Thus, the minimum length of the hypotenuse is $\left( a^{\frac{2}{3}} + b^{\frac{2}{3}} \right)^{\frac{3}{2}}$.

Given: The function is $f(x) = (x – 2)^4 (x + 1)^3$.

To Find: The points of local maxima, local minima, and points of inflection.

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Solution:

To find the local extrema and points of inflection, we need the first and second derivatives of the function.

Given $f(x) = (x – 2)^4 (x + 1)^3$.

First, find the first derivative $f'(x)$ using the product rule $\frac{d}{dx}(uv) = u'v + uv'$ with $u = (x-2)^4$ and $v = (x+1)^3$.

$u' = \frac{d}{dx}(x-2)^4 = 4(x-2)^3 \cdot \frac{d}{dx}(x-2) = 4(x-2)^3 \cdot 1 = 4(x-2)^3$

$v' = \frac{d}{dx}(x+1)^3 = 3(x+1)^2 \cdot \frac{d}{dx}(x+1) = 3(x+1)^2 \cdot 1 = 3(x+1)^2$

$f'(x) = u'v + uv'$

$f'(x) = 4(x-2)^3 (x+1)^3 + (x-2)^4 \cdot 3(x+1)^2$

Factor out the common terms $(x-2)^3$ and $(x+1)^2$:

$f'(x) = (x-2)^3 (x+1)^2 [4(x+1) + 3(x-2)]$

$f'(x) = (x-2)^3 (x+1)^2 [4x + 4 + 3x - 6]$

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To find local extrema, we find the critical points by setting $f'(x) = 0$.

This equation is satisfied if any of the factors are zero:

• $(x-2)^3 = 0 \implies x - 2 = 0 \implies x = 2$
• $(x+1)^2 = 0 \implies x + 1 = 0 \implies x = -1$
• $7x - 2 = 0 \implies 7x = 2 \implies x = \frac{2}{7}$

The critical points are $x = -1$, $x = \frac{2}{7}$, and $x = 2$.

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Now we use the first derivative test to classify these critical points. We examine the sign of $f'(x) = (x-2)^3 (x+1)^2 (7x - 2)$ in intervals around the critical points. Note that $(x+1)^2 \geq 0$ for all real $x$. The sign of $f'(x)$ depends on the signs of $(x-2)^3$ and $(7x-2)$.

• For $x < -1$ (e.g., $x=-2$): $(x-2)^3$ is $(-)$, $(x+1)^2$ is $(+)$, $(7x-2)$ is $(-)$. $f'(x) \approx (-) (+) (-) = (+)$. $f$ is increasing.
• For $-1 < x < \frac{2}{7}$ (e.g., $x=0$): $(x-2)^3$ is $(-)$, $(x+1)^2$ is $(+)$, $(7x-2)$ is $(-)$. $f'(x) \approx (-) (+) (-) = (+)$. $f$ is increasing.
• At $x = -1$: $f'(-1) = 0$. $f'(x)$ does not change sign around $x=-1$. This indicates $x=-1$ is not a local extremum. It's a point of inflection with a horizontal tangent.
• For $\frac{2}{7} < x < 2$ (e.g., $x=1$): $(x-2)^3$ is $(-)$, $(x+1)^2$ is $(+)$, $(7x-2)$ is $(+)$. $f'(x) \approx (-) (+) (+) = (-)$. $f$ is decreasing.
• At $x = \frac{2}{7}$: $f'(\frac{2}{7}) = 0$. $f'(x)$ changes sign from $(+)$ to $(-)$ around $x = \frac{2}{7}$. This indicates a local maximum at $x = \frac{2}{7}$.
• For $x > 2$ (e.g., $x=3$): $(x-2)^3$ is $(+)$, $(x+1)^2$ is $(+)$, $(7x-2)$ is $(+)$. $f'(x) \approx (+) (+) (+) = (+)$. $f$ is increasing.
• At $x = 2$: $f'(2) = 0$. $f'(x)$ changes sign from $(-)$ to $(+)$ around $x = 2$. This indicates a local minimum at $x = 2$.

(i) Local maxima: The function has a local maximum at $x = \frac{2}{7}$.

(ii) Local minima: The function has a local minimum at $x = 2$.

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(iii) Points of inflection: Points of inflection occur where the concavity of the function changes. This happens where $f''(x) = 0$ or $f''(x)$ is undefined, and the sign of $f''(x)$ changes around that point.

Find the second derivative $f''(x)$ by differentiating $f'(x) = (x-2)^3 (x+1)^2 (7x - 2)$. We can use the product rule $(uvw)' = u'vw + uv'w + uvw'$ with $u=(x-2)^3$, $v=(x+1)^2$, and $w=(7x-2)$.

$u' = 3(x-2)^2$

$v' = 2(x+1)$

$w' = 7$

$f''(x) = [3(x-2)^2](x+1)^2 (7x-2) + (x-2)^3 [2(x+1)](7x-2) + (x-2)^3 (x+1)^2 [7]$

Factor out $(x-2)^2 (x+1)$:

$f''(x) = (x-2)^2 (x+1) [3(x+1)(7x-2) + 2(x-2)(7x-2) + 7(x-2)(x+1)]

Expand the terms inside the bracket:

• $3(7x^2 - 2x + 7x - 2) = 3(7x^2 + 5x - 2) = 21x^2 + 15x - 6$
• $2(7x^2 - 14x - 2x + 4) = 2(7x^2 - 16x + 4) = 14x^2 - 32x + 8$
• $7(x^2 + x - 2x - 2) = 7(x^2 - x - 2) = 7x^2 - 7x - 14$

Summing the terms inside the bracket:

$(21x^2 + 14x^2 + 7x^2) + (15x - 32x - 7x) + (-6 + 8 - 14) = 42x^2 - 24x - 12$

So, $f''(x) = (x-2)^2 (x+1) (42x^2 - 24x - 12)$.

Factor out 6 from the quadratic term:

Potential points of inflection are where $f''(x) = 0$.

This gives potential points at $x = 2$, $x = -1$, and the roots of $7x^2 - 4x - 2 = 0$.

Using the quadratic formula for $7x^2 - 4x - 2 = 0$:

$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(7)(-2)}}{2(7)} = \frac{4 \pm \sqrt{16 + 56}}{14} = \frac{4 \pm \sqrt{72}}{14}$

$x = \frac{4 \pm 6\sqrt{2}}{14} = \frac{2 \pm 3\sqrt{2}}{7}$

The potential inflection points are $x = -1$, $x = \frac{2 - 3\sqrt{2}}{7}$, $x = \frac{2 + 3\sqrt{2}}{7}$, and $x = 2$.

We check for a sign change in $f''(x) = 6(x-2)^2 (x+1) (7x^2 - 4x - 2)$ around these points. The term $6(x-2)^2$ is always non-negative (and zero only at $x=2$), so the sign of $f''(x)$ is determined by $(x+1)(7x^2 - 4x - 2)$. The roots of $7x^2 - 4x - 2 = 0$ are $x_1 = \frac{2 - 3\sqrt{2}}{7}$ and $x_2 = \frac{2 + 3\sqrt{2}}{7}$. Numerically, $x_1 \approx -0.32$ and $x_2 \approx 0.89$. The roots of $f''(x)$ in increasing order are approximately $-1, -0.32, 0.89, 2$.

• Around $x = -1$: The sign of $(x+1)$ changes from $-$ to $+$ as $x$ crosses $-1$. The quadratic term $(7x^2 - 4x - 2)$ is positive around $-1$ (since $-1 < x_1$ and $-1 < x_2$). Thus, the sign of $f''(x)$ changes from $-(+) = -$ to $(+)(+) = +$ as $x$ crosses $-1$. So, $x=-1$ is a point of inflection.

$f''(-1) = 0$

(Verified above)

• Around $x = \frac{2 - 3\sqrt{2}}{7}$: The sign of $(x+1)$ is positive around $x_1 \approx -0.32$. The sign of the quadratic term $(7x^2 - 4x - 2)$ changes from $+$ to $-$ as $x$ crosses $x_1$. Thus, the sign of $f''(x)$ changes from $(+)(+) = +$ to $(+)(-) = -$ as $x$ crosses $x_1$. So, $x = \frac{2 - 3\sqrt{2}}{7}$ is a point of inflection.

$f''\left(\frac{2 - 3\sqrt{2}}{7}\right) = 0$

[Root of quadratic term]

• Around $x = \frac{2 + 3\sqrt{2}}{7}$: The sign of $(x+1)$ is positive around $x_2 \approx 0.89$. The sign of the quadratic term $(7x^2 - 4x - 2)$ changes from $-$ to $+$ as $x$ crosses $x_2$. Thus, the sign of $f''(x)$ changes from $(+)(-) = -$ to $(+)(+) = +$ as $x$ crosses $x_2$. So, $x = \frac{2 + 3\sqrt{2}}{7}$ is a point of inflection.

$f''\left(\frac{2 + 3\sqrt{2}}{7}\right) = 0$

[Root of quadratic term]

• Around $x = 2$: The term $(x-2)^2$ does not change sign (it's positive on both sides of 2). The term $(x+1)$ is positive around 2. The quadratic term $(7x^2 - 4x - 2)$ is positive around 2 (since $2 > x_2$). Thus, the sign of $f''(x)$ does not change as $x$ crosses 2 (it's positive on both sides). So, $x=2$ is not a point of inflection, even though $f''(2)=0$.

$f''(2) = 0$

[Root of $(x-2)^2$]

(iii) Points of inflection: The function has points of inflection at $x = -1$, $x = \frac{2 - 3\sqrt{2}}{7}$, and $x = \frac{2 + 3\sqrt{2}}{7}$.

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Summary of Answer:

(i) Local maxima at $\mathbf{x = \frac{2}{7}}$.

(ii) Local minima at $\mathbf{x = 2}$.

(iii) Points of inflection at $\mathbf{x = -1}$, $\mathbf{x = \frac{2 - 3\sqrt{2}}{7}}$, and $\mathbf{x = \frac{2 + 3\sqrt{2}}{7}}$.

Given: The function is $f(x) = \cos^2 x + \sin x$ on the closed interval $x \in [0, \pi]$.

To Find: The absolute maximum and minimum values of the function $f(x)$ on the given interval.

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Solution:

The given function is $f(x) = \cos^2 x + \sin x$.

Using the identity $\cos^2 x = 1 - \sin^2 x$, we can rewrite the function in terms of $\sin x$:

To find the absolute extrema on the closed interval $[0, \pi]$, we need to evaluate the function at the critical points within the interval and at the endpoints of the interval.

First, find the critical points by finding the first derivative $f'(x)$ and setting it to zero.

$f'(x) = \frac{d}{dx}(1 + \sin x - \sin^2 x)$

$f'(x) = 0 + \cos x - 2\sin x \frac{d}{dx}(\sin x)$

Factor out $\cos x$:

Set $f'(x) = 0$ to find the critical points:

This equation holds if $\cos x = 0$ or $1 - 2\sin x = 0$ (i.e., $\sin x = \frac{1}{2}$).

For the interval $[0, \pi]$:

• If $\cos x = 0$, then $x = \frac{\pi}{2}$. This point is in the interval $[0, \pi]$.
• If $\sin x = \frac{1}{2}$, then $x = \frac{\pi}{6}$ or $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$. Both these points are in the interval $[0, \pi]$.

The critical points in the interval $[0, \pi]$ are $x = \frac{\pi}{6}$, $x = \frac{\pi}{2}$, and $x = \frac{5\pi}{6}$.

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Now, evaluate the function $f(x) = 1 + \sin x - \sin^2 x$ at the critical points and the endpoints of the interval $[0, \pi]$.

Endpoints: $x = 0$ and $x = \pi$.

• At $x = 0$: $f(0) = 1 + \sin 0 - \sin^2 0 = 1 + 0 - 0^2 = 1$.
• At $x = \pi$: $f(\pi) = 1 + \sin \pi - \sin^2 \pi = 1 + 0 - 0^2 = 1$.

Critical points: $x = \frac{\pi}{6}$, $x = \frac{\pi}{2}$, and $x = \frac{5\pi}{6}$.

• At $x = \frac{\pi}{6}$: $\sin(\frac{\pi}{6}) = \frac{1}{2}$. $f(\frac{\pi}{6}) = 1 + \sin(\frac{\pi}{6}) - \sin^2(\frac{\pi}{6}) = 1 + \frac{1}{2} - \left(\frac{1}{2}\right)^2 = 1 + \frac{1}{2} - \frac{1}{4} = \frac{4+2-1}{4} = \frac{5}{4}$.
• At $x = \frac{\pi}{2}$: $\sin(\frac{\pi}{2}) = 1$. $f(\frac{\pi}{2}) = 1 + \sin(\frac{\pi}{2}) - \sin^2(\frac{\pi}{2}) = 1 + 1 - (1)^2 = 1 + 1 - 1 = 1$.
• At $x = \frac{5\pi}{6}$: $\sin(\frac{5\pi}{6}) = \sin(\pi - \frac{\pi}{6}) = \sin(\frac{\pi}{6}) = \frac{1}{2}$. $f(\frac{5\pi}{6}) = 1 + \sin(\frac{5\pi}{6}) - \sin^2(\frac{5\pi}{6}) = 1 + \frac{1}{2} - \left(\frac{1}{2}\right)^2 = 1 + \frac{1}{2} - \frac{1}{4} = \frac{4+2-1}{4} = \frac{5}{4}$.

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Comparing the values of the function at the critical points and endpoints:

• $f(0) = 1$
• $f(\pi) = 1$
• $f(\frac{\pi}{6}) = \frac{5}{4} = 1.25$
• $f(\frac{\pi}{2}) = 1$
• $f(\frac{5\pi}{6}) = \frac{5}{4} = 1.25$

The highest value among these is $\frac{5}{4}$, and the lowest value is $1$.

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Answer:

The absolute maximum value of $f(x)$ on $[0, \pi]$ is $\mathbf{\frac{5}{4}}$, which occurs at $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$.

The absolute minimum value of $f(x)$ on $[0, \pi]$ is $\mathbf{1}$, which occurs at $x = 0$, $x = \frac{\pi}{2}$, and $x = \pi$.

Given:

A sphere of radius $r$. A right circular cone is inscribed in the sphere.

To Show:

The altitude of the cone of maximum volume is $\frac{4r}{3}$.

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Solution:

Let the radius of the sphere be $r$. Let the altitude of the inscribed right circular cone be $h$ and the radius of its base be $r_c$.

Consider a cross-section of the sphere and the cone through the axis of the cone. This cross-section is a circle of radius $r$ with an inscribed isosceles triangle. The altitude of the triangle is the altitude of the cone, $h$. The base of the triangle is the diameter of the cone's base, $2r_c$.

Let the center of the sphere be at the origin $(0,0)$. Let the vertex of the cone be at $(0, r)$ (one end of a vertical diameter). The base of the cone is a horizontal circle at some height $y = r - h$ below the center of the sphere. The radius of the cone's base, $r_c$, is the x-coordinate of the point on the sphere at height $y = r - h$.

The equation of the sphere in this cross-section is $x^2 + y^2 = r^2$.

At the base of the cone, $x = r_c$ and $y = r - h$. Substituting these into the sphere equation:

Solve for $r_c^2$:

For a valid cone inscribed in the sphere, the altitude $h$ must be greater than 0 and at most the diameter of the sphere $2r$. So the domain for $h$ is $0 < h \leq 2r$.

The volume of a right circular cone is given by $V = \frac{1}{3} \pi r_c^2 h$.

Substitute the expression for $r_c^2$ from equation (i) into the volume formula:

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To find the maximum volume, we find the critical points by differentiating $V(h)$ with respect to $h$ and setting the derivative to zero.

$V'(h) = \frac{d}{dh} \left(\frac{\pi}{3} (2rh^2 - h^3)\right)$

$V'(h) = \frac{\pi}{3} (2r \cdot 2h - 3h^2)$

Set $V'(h) = 0$:

Since $\frac{\pi}{3} \neq 0$, we have:

Factor out $h$:

This gives two possible values for $h$: $h = 0$ or $4r - 3h = 0$.

• $h = 0$: This corresponds to a degenerate cone with zero volume, which is the minimum possible volume.
• $4r - 3h = 0 \implies 3h = 4r \implies h = \frac{4r}{3}$.

This critical value $h = \frac{4r}{3}$ is within the domain $0 < h \leq 2r$ since $\frac{4}{3} < 2$.

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To confirm that $h = \frac{4r}{3}$ yields a maximum volume, we use the second derivative test.

Differentiate $V'(h) = \frac{\pi}{3} (4rh - 3h^2)$ with respect to $h$:

$V''(h) = \frac{d}{dh} \left(\frac{\pi}{3} (4rh - 3h^2)\right)$

$V''(h) = \frac{\pi}{3} (4r \cdot 1 - 6h)$

Evaluate $V''(h)$ at the critical point $h = \frac{4r}{3}$ (from equation iv):

$V''\left(\frac{4r}{3}\right) = \frac{\pi}{3} \left(4r - 6\left(\frac{4r}{3}\right)\right)$

$V''\left(\frac{4r}{3}\right) = \frac{\pi}{3} \left(4r - \frac{24r}{3}\right)$

$V''\left(\frac{4r}{3}\right) = \frac{\pi}{3} (4r - 8r)$

Since $r$ is the radius of the sphere, $r > 0$. Thus, $V''\left(\frac{4r}{3}\right) = -\frac{4\pi r}{3} < 0$.

According to the second derivative test, if $V'(c)=0$ and $V''(c) < 0$ at a critical point $c$, then there is a local maximum at $c$. Since $h = \frac{4r}{3}$ is the only critical point within the domain $(0, 2r)$ that yields a positive volume, it corresponds to the absolute maximum volume.

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Conclusion:

The altitude of the right circular cone of maximum volume inscribed in a sphere of radius $r$ is indeed $\mathbf{\frac{4r}{3}}$.

Given:

A function $f$ is defined on the closed interval $[a, b]$.

The derivative of the function, $f'(x)$, is strictly positive for all $x$ in the open interval $(a, b)$.

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To Prove:

The function $f$ is an increasing function on the interval $(a, b)$.

An increasing function on an interval $(a, b)$ means that for any two points $x_1, x_2 \in (a, b)$ with $x_1 < x_2$, we have $f(x_1) < f(x_2)$.

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Proof:

Let $x_1$ and $x_2$ be any two arbitrary points in the interval $(a, b)$ such that $x_1 < x_2$.

Since $f'(x)$ exists for all $x \in (a, b)$, the function $f$ is differentiable on $(a, b)$.

A function that is differentiable on an open interval is continuous on that open interval. Furthermore, if a function is differentiable on $(a, b)$ and defined on $[a, b]$, it is continuous on $[a, b]$.

Consider the subinterval $[x_1, x_2]$, which is a closed interval within $[a, b]$. Since $a < x_1 < x_2 < b$, the interval $[x_1, x_2]$ is contained within $(a, b)$.

Thus, $f$ is continuous on the closed interval $[x_1, x_2]$.

Also, $f$ is differentiable on the open interval $(x_1, x_2)$.

Now, we can apply the Mean Value Theorem (MVT) to the function $f$ on the interval $[x_1, x_2]$.

The Mean Value Theorem states that if a function $f$ is continuous on $[p, q]$ and differentiable on $(p, q)$, then there exists at least one point $c \in (p, q)$ such that $f'(c) = \frac{f(q) - f(p)}{q - p}$.

Applying the MVT to $f$ on the interval $[x_1, x_2]$, there exists some point $c \in (x_1, x_2)$ such that:

From the given information (equation i), we know that $f'(x) > 0$ for all $x \in (a, b)$. Since $c \in (x_1, x_2)$ and $(x_1, x_2) \subset (a, b)$, it must be true that $f'(c) > 0$.

Substitute this into equation (ii):

We chose $x_1 < x_2$, which means $x_2 - x_1 > 0$.

Multiplying both sides of the inequality by the positive quantity $(x_2 - x_1)$, the inequality direction does not change:

We started by choosing any two points $x_1, x_2 \in (a, b)$ such that $x_1 < x_2$ and concluded that $f(x_1) < f(x_2)$. This is the definition of an increasing function on the interval $(a, b)$.

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Conclusion:

Therefore, the function $f$ is an increasing function on the interval $(a, b)$.

Given:

A sphere of radius $R$. A right circular cylinder is inscribed in the sphere.

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To Show:

The height of the cylinder of maximum volume is $\frac{2R}{\sqrt{3}}$.

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To Find:

The maximum volume of the cylinder.

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Solution:

Let the height of the inscribed right circular cylinder be $h$ and the radius of its base be $r_c$.

Consider a cross-section of the sphere and the cylinder through the center of the sphere and parallel to the axis of the cylinder. This cross-section is a circle of radius $R$ with an inscribed rectangle. The height of the rectangle is the height of the cylinder, $h$, and the width of the rectangle is the diameter of the cylinder's base, $2r_c$.

The diagonal of this inscribed rectangle is the diameter of the sphere, $2R$.

By the Pythagorean theorem, relating the sides of the rectangle and the diagonal:

We can express $r_c^2$ in terms of $h$ and $R$ from equation (i):

For a valid cylinder inscribed in the sphere, the height $h$ must be greater than 0 and at most the diameter of the sphere $2R$. So the domain for $h$ is $0 < h \leq 2R$. Also, $r_c^2 > 0$, so $R^2 - \frac{h^2}{4} > 0 \implies R^2 > \frac{h^2}{4} \implies 4R^2 > h^2 \implies 2R > h$. Thus, the practical domain for $h$ for a non-degenerate cylinder is $0 < h < 2R$.

The volume of a right circular cylinder is given by $V = \pi r_c^2 h$.

Substitute the expression for $r_c^2$ from equation (ii) into the volume formula to get the volume as a function of $h$:

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To find the maximum volume, we differentiate $V(h)$ with respect to $h$ and set the derivative to zero to find the critical points.

$V'(h) = \frac{d}{dh} \left[\pi \left(R^2 h - \frac{h^3}{4}\right)\right]$

$V'(h) = \pi \left(R^2 \cdot 1 - \frac{1}{4} \cdot 3h^2\right)$

Set $V'(h) = 0$:

Since $\pi \neq 0$, we have:

Since $h > 0$, we take the positive square root:

This critical value $h = \frac{2R}{\sqrt{3}}$ is within the domain $(0, 2R)$ since $\frac{2}{\sqrt{3}} = \frac{2 \times 1.732...}{3} < 2$.

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To confirm that $h = \frac{2R}{\sqrt{3}}$ yields a maximum volume, we use the second derivative test.

Differentiate $V'(h) = \pi \left(R^2 - \frac{3h^2}{4}\right)$ with respect to $h$:

$V''(h) = \frac{d}{dh} \left[\pi \left(R^2 - \frac{3h^2}{4}\right)\right]$

$V''(h) = \pi \left(0 - \frac{3}{4} \cdot 2h\right)$

Evaluate $V''(h)$ at the critical point $h = \frac{2R}{\sqrt{3}}$ (from equation vi):

$V''\left(\frac{2R}{\sqrt{3}}\right) = -\frac{3\pi}{2} \left(\frac{2R}{\sqrt{3}}\right)$

$V''\left(\frac{2R}{\sqrt{3}}\right) = -\frac{3\pi \cdot 2R}{2\sqrt{3}}$

Since $R$ is the radius of the sphere, $R > 0$. Thus, $V''\left(\frac{2R}{\sqrt{3}}\right) = -\pi R\sqrt{3} < 0$.

According to the second derivative test, if $V'(c)=0$ and $V''(c) < 0$ at a critical point $c$, then there is a local maximum at $c$. Since $h = \frac{2R}{\sqrt{3}}$ is the only critical point within the domain $(0, 2R)$, it corresponds to the absolute maximum volume.

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Show that the height is $\frac{2R}{\sqrt{3}}$:

We have found that the altitude (height) $h$ that maximizes the volume of the inscribed cylinder is $h = \frac{2R}{\sqrt{3}}$. This matches the value we were asked to show.

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Find the maximum volume:

The maximum volume occurs at $h = \frac{2R}{\sqrt{3}}$. We can find the corresponding radius squared $r_c^2$ using equation (ii):

$r_c^2 = R^2 - \frac{h^2}{4}$

Substitute $h = \frac{2R}{\sqrt{3}}$:

$r_c^2 = R^2 - \frac{\left(\frac{2R}{\sqrt{3}}\right)^2}{4} = R^2 - \frac{\frac{4R^2}{3}}{4} = R^2 - \frac{4R^2}{3 \cdot 4}$

Now, substitute the values of $h$ (from v) and $r_c^2$ (from vii) into the volume formula $V = \pi r_c^2 h$:

$V_{max} = \pi \left(\frac{2R^2}{3}\right) \left(\frac{2R}{\sqrt{3}}\right)$

$V_{max} = \frac{\pi \cdot 2R^2 \cdot 2R}{3 \cdot \sqrt{3}}$

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{3}$:

$V_{max} = \frac{4\pi R^3}{3\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{4\pi R^3 \sqrt{3}}{3 \cdot 3}$

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Answer:

The height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is $\mathbf{\frac{2R}{\sqrt{3}}}$.

The maximum volume of the cylinder is $\mathbf{\frac{4\pi R^3 \sqrt{3}}{9}}$.

Given:

A right circular cone with height $h$ and semi-vertical angle $\alpha$.

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To Show:

The height of the inscribed cylinder of maximum volume is $\frac{h}{3}$.

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To Find:

The maximum volume of the inscribed cylinder.

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Solution:

Let the height of the inscribed right circular cylinder be $H$ and the radius of its base be $r_c$.

Consider a vertical cross-section of the cone and the inscribed cylinder through their common axis. This cross-section shows an isosceles triangle (representing the cone) with height $h$ and a rectangle inscribed within it (representing the cylinder) with height $H$ and width $2r_c$.

Let the vertex of the cone be at the top. The radius of the cone's base is $R_{cone}$. From the semi-vertical angle $\alpha$, we have:

Assume the base of the cylinder rests on the base of the cone. The top circular base of the cylinder is at a height $H$ from the cone's base.

Consider the smaller right triangle formed by the cone's axis, the cylinder's radius, and the part of the cone's slant height above the cylinder's top surface. The height of this smaller triangle is the distance from the cone's apex to the top surface of the cylinder, which is $h - H$. The base of this smaller triangle is the radius of the cylinder, $r_c$.

This smaller triangle is similar to the larger right triangle formed by the cone's axis, the cone's base radius, and the cone's slant height (with height $h$ and base $R_{cone}$).

Using the property of similar triangles, the ratio of the base to height is constant:

Substitute $R_{cone} = h \tan \alpha$ from equation (i):

The height of the cylinder $H$ must be positive and less than the height of the cone $h$. So the domain for $H$ is $(0, h)$.

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The volume of a cylinder is given by $V = \pi r_c^2 H$.

Substitute the expression for $r_c$ from equation (ii) into the volume formula to get the volume as a function of $H$:

Since $h$ and $\alpha$ are constants, $\pi \tan^2 \alpha$ is a constant. Expand the expression:

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To find the maximum volume, we differentiate $V(H)$ with respect to $H$ and find the critical points by setting the derivative equal to zero.

$V'(H) = \frac{d}{dH} \left[ \pi \tan^2 \alpha (h^2 H - 2h H^2 + H^3) \right]$

$V'(H) = \pi \tan^2 \alpha \frac{d}{dH} (h^2 H - 2h H^2 + H^3)$

$V'(H) = \pi \tan^2 \alpha (h^2 \cdot 1 - 2h \cdot 2H + 3H^2)$

Set $V'(H) = 0$ to find the critical points:

Since $\pi \tan^2 \alpha \neq 0$ (assuming $0 < \alpha < \frac{\pi}{2}$), we must have:

This is a quadratic equation in $H$. We can factor it:

$(3H - h)(H - h) = 0$

This gives two possible values for $H$:

• $3H - h = 0 \implies H = \frac{h}{3}$
• $H - h = 0 \implies H = h$

The critical points are $H = \frac{h}{3}$ and $H = h$.

When $H = h$, the height of the cylinder is equal to the height of the cone. From equation (ii), $r_c = (h - h) \tan \alpha = 0$. This corresponds to a degenerate cylinder with volume 0, which is the minimum possible volume.

The critical point $H = \frac{h}{3}$ is in the domain $(0, h)$ and is likely the height that maximizes the volume.

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To confirm that $H = \frac{h}{3}$ yields a maximum volume, we use the second derivative test.

Differentiate $V'(H) = \pi \tan^2 \alpha (h^2 - 4hH + 3H^2)$ with respect to $H$:

$V''(H) = \frac{d}{dH} \left[ \pi \tan^2 \alpha (h^2 - 4hH + 3H^2) \right]$

$V''(H) = \pi \tan^2 \alpha (0 - 4h \cdot 1 + 3 \cdot 2H)$

Evaluate $V''(H)$ at the critical point $H = \frac{h}{3}$ (from equation v):

$V''\left(\frac{h}{3}\right) = \pi \tan^2 \alpha \left(6\left(\frac{h}{3}\right) - 4h\right)$

$V''\left(\frac{h}{3}\right) = \pi \tan^2 \alpha \left(\frac{\cancel{6}^{2} h}{\cancel{3}_{1}} - 4h\right)$

$V''\left(\frac{h}{3}\right) = \pi \tan^2 \alpha (2h - 4h)$

Since $h > 0$ and $\alpha \in (0, \frac{\pi}{2})$ implies $\tan^2 \alpha > 0$, we have $V''\left(\frac{h}{3}\right) < 0$.

According to the second derivative test, this confirms that the volume $V(H)$ has a local maximum at $H = \frac{h}{3}$. Since $H=h$ and $H \to 0^+$ correspond to minimum volume (0), this local maximum is the absolute maximum in the domain $(0, h)$.

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Show that the height is $\frac{h}{3}$:

We have found that the height $H$ that maximizes the volume of the inscribed cylinder is $H = \frac{h}{3}$. This means the height of the cylinder of greatest volume is one-third that of the cone.

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Find the maximum volume:

The maximum volume occurs at $H = \frac{h}{3}$. Substitute this value into the volume formula $V(H) = \pi \tan^2 \alpha (h-H)^2 H$ (equation iii):

$V_{max} = V\left(\frac{h}{3}\right) = \pi \tan^2 \alpha \left(h - \frac{h}{3}\right)^2 \left(\frac{h}{3}\right)$

$V_{max} = \pi \tan^2 \alpha \left(\frac{3h - h}{3}\right)^2 \left(\frac{h}{3}\right)$

$V_{max} = \pi \tan^2 \alpha \left(\frac{2h}{3}\right)^2 \left(\frac{h}{3}\right)$

$V_{max} = \pi \tan^2 \alpha \left(\frac{4h^2}{9}\right) \left(\frac{h}{3}\right)$

$V_{max} = \pi \tan^2 \alpha \frac{4h^3}{27}$

This matches the required expression for the maximum volume.

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Conclusion:

We have shown that the height of the cylinder of maximum volume inscribed in a right circular cone of height $h$ and semi-vertical angle $\alpha$ is $\mathbf{\frac{h}{3}}$.

The greatest volume of such a cylinder is $\mathbf{\frac{4}{27} \pi h^3 \tan^2 \alpha}$.

Given:

Radius of the cylindrical tank, $r = 10$ m.

Rate of filling the tank with wheat (rate of change of volume), $\frac{dV}{dt} = 314$ m$^3$/hour.

We can use $\pi \approx 3.14$ based on the numerical value of $\frac{dV}{dt}$.

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To Find:

The rate at which the depth of the wheat is increasing, $\frac{dh}{dt}$, where $h$ is the depth of the wheat.

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Solution:

Let $V$ be the volume of wheat in the cylindrical tank and $h$ be the depth of the wheat at time $t$.

The volume of a cylinder is given by the formula $V = \pi r^2 h$.

Since the radius of the tank is constant ($r=10$ m), we can substitute this value into the volume formula:

We are given the rate at which the volume is changing, $\frac{dV}{dt}$, and we need to find the rate at which the depth is changing, $\frac{dh}{dt}$. We can find this by differentiating equation (i) with respect to time $t$.

Differentiate both sides of equation (i) with respect to $t$:

$\frac{dV}{dt} = \frac{d}{dt}(100\pi h)$

Since $100\pi$ is a constant, we can pull it out of the differentiation:

Now, substitute the given value of $\frac{dV}{dt} = 314$ m$^3$/hour into equation (ii):

Solve for $\frac{dh}{dt}$:

Using the approximation $\pi \approx 3.14$:

The unit for the rate of change of depth is meters per hour (m/h).

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Comparing this result with the given options:

(A) 1 m/h

(B) 0.1 m/h

(C) 1.1 m/h

(D) 0.5 m/h

The calculated rate matches option (A).

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Answer:

The depth of the wheat is increasing at the rate of $\mathbf{1}$ m/h.

The correct option is (A).