Chapter 6 Application Of Derivatives

Given:

Radius of the circle is denoted by $r$.

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

The area of a circle is given by the formula $A = \pi r^2$.

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To Find:

The rate of change of the area of the circle per second with respect to its radius $r$. This is represented by $\frac{dA}{dr}$.

We need to evaluate $\frac{dA}{dr}$ when $r = 5$ cm.

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Solution:

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

$A = \pi r^2$

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

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

Using the power rule for differentiation ($\frac{d}{dr}(r^n) = nr^{n-1}$), we get:

$\frac{dA}{dr} = \pi \cdot 2r^{2-1}$

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

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

Substitute $r = 5$ cm into the expression for $\frac{dA}{dr}$:

$\left(\frac{dA}{dr}\right)_{r=5} = 2\pi (5)$

$\left(\frac{dA}{dr}\right)_{r=5} = 10\pi$

The units of the rate of change of area with respect to radius are the units of area divided by the units of radius, which is cm$^2$/cm.

So, the rate of change is $10\pi$ cm$^2$/cm.

The question phrasing "per second with respect to its radius" might be slightly confusing. The standard interpretation of "rate of change of area ... with respect to its radius" is $\frac{dA}{dr}$. The "per second" part usually indicates differentiation with respect to time $t$, i.e., $\frac{dA}{dt}$. However, the question explicitly asks for the rate of change "with respect to its radius r". Therefore, we calculate $\frac{dA}{dr}$. If it intended $\frac{dA}{dt}$, it would typically provide $\frac{dr}{dt}$. Assuming the question means $\frac{dA}{dr}$, the calculation is as shown above.

The rate of change of the area of a circle with respect to its radius $r$ when $r = 5$ cm is $10\pi$ cm$^2$/cm.

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Final Answer:

The rate of change of the area of a circle per second with respect to its radius when the radius is 5 cm is $10\pi$ cm$^2$/cm.

Given:

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

The volume of the cube is increasing at a rate of 9 cubic centimetres per second.

This means the rate of change of volume ($V$) with respect to time ($t$) is $\frac{dV}{dt} = 9 \$cm^3/s$.

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To Find:

How fast is the surface area ($A$) of the cube increasing when the length of an edge is 10 centimetres?

This means we need to find $\frac{dA}{dt}$ when $s = 10 \$cm$.

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Solution:

Let $V$ be the volume of the cube and $A$ be the surface area of the cube.

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

$V = s^3$

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

$A = 6s^2$

We are given that the volume is increasing at a rate of 9 \$cm$^3$/s. This is the rate of change of volume with respect to time $t$.

Differentiate the volume equation with respect to time $t$:

$\frac{dV}{dt} = \frac{d}{dt}(s^3)$

Using the chain rule, $\frac{d}{dt}(s^3) = 3s^2 \frac{ds}{dt}$.

So, $\frac{dV}{dt} = 3s^2 \frac{ds}{dt}$.

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

$9 = 3s^2 \frac{ds}{dt}$

Now, we can solve for $\frac{ds}{dt}$, which is the rate of change of the edge length with respect to time:

$\frac{ds}{dt} = \frac{9}{3s^2}$

$\frac{ds}{dt} = \frac{3}{s^2}$

Next, we need to find the rate of change of the surface area with respect to time, $\frac{dA}{dt}$. Differentiate the surface area equation with respect to time $t$:

$\frac{dA}{dt} = \frac{d}{dt}(6s^2)$

Using the chain rule, $\frac{d}{dt}(6s^2) = 6 \cdot 2s \frac{ds}{dt} = 12s \frac{ds}{dt}$.

So, $\frac{dA}{dt} = 12s \frac{ds}{dt}$.

Substitute the expression for $\frac{ds}{dt}$ that we found ($\frac{3}{s^2}$) into this equation:

$\frac{dA}{dt} = 12s \left(\frac{3}{s^2}\right)$

$\frac{dA}{dt} = \frac{36s}{s^2}$

$\frac{dA}{dt} = \frac{36}{s}$

We need to find the rate of change of the surface area when the length of an edge is 10 centimetres, i.e., when $s = 10 \$cm$. Substitute $s = 10$ into the expression for $\frac{dA}{dt}$:

$\left(\frac{dA}{dt}\right)_{s=10} = \frac{36}{10}$

$\left(\frac{dA}{dt}\right)_{s=10} = 3.6$

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

Therefore, the surface area is increasing at a rate of 3.6 \$cm$^2$/s when the edge length is 10 \$cm.

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Final Answer:

The surface area of the cube is increasing at a rate of 3.6 \$cm$^2$/s when the length of an edge is 10 \$cm.

Given:

Let $r$ be the radius of the circular wave and $A$ be the enclosed area.

The speed of the waves is the rate of change of the radius with respect to time.

Rate of change of radius: $\frac{dr}{dt} = 4$ cm/s.

We need to find the rate of change of the enclosed area at the instant when $r = 10$ cm.

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To Find:

The rate of change of the enclosed area with respect to time, $\frac{dA}{dt}$, when $r = 10$ cm.

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Solution:

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

$A = \pi r^2$

To find the rate at which the enclosed area is increasing, we need to differentiate the area $A$ with respect to time $t$.

Differentiate both sides of the area equation with respect to $t$ using the chain rule:

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

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

Using the chain rule, $\frac{d}{dt}(r^2) = 2r \frac{dr}{dt}$.

So, we have:

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

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

We are given that the speed of the waves (rate of change of radius) is 4 cm/s, so $\frac{dr}{dt} = 4$ cm/s.

We need to find the rate of change of the area at the instant when the radius is 10 cm, so $r = 10$ cm.

Substitute the values $r = 10$ and $\frac{dr}{dt} = 4$ into the expression for $\frac{dA}{dt}$:

$\left(\frac{dA}{dt}\right)_{r=10} = 2\pi (10)(4)$

$\left(\frac{dA}{dt}\right)_{r=10} = 80\pi$

The units for the rate of change of area are square centimetres per second (\$cm$^2$/s).

Therefore, the enclosed area is increasing at a rate of $80\pi$ \$cm$^2$/s when the radius is 10 cm.

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Final Answer:

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

Given:

Length of the rectangle is $x$.

Width of the rectangle is $y$.

Rate of decrease of length: $\frac{dx}{dt} = -3 \$cm/minute.

Rate of increase of width: $\frac{dy}{dt} = 2 \$cm/minute.

Instant of interest: $x = 10 \$cm$ and $y = 6 \$cm$.

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To Find:

(a) The rate of change of the perimeter of the rectangle when $x=10 \$cm$ and $y=6 \$cm$.

(b) The rate of change of the area of the rectangle when $x=10 \$cm$ and $y=6 \$cm$.

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Solution:

Let $P$ be the perimeter of the rectangle and $A$ be the area of the rectangle.

(a) Rate of change of the perimeter:

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

$P = 2x + 2y$

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

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

Using the sum rule and chain rule:

$\frac{dP}{dt} = 2\frac{dx}{dt} + 2\frac{dy}{dt}$

Substitute the given rates of change, $\frac{dx}{dt} = -3$ \$cm/minute and $\frac{dy}{dt} = 2$ \$cm/minute:

$\frac{dP}{dt} = 2(-3) + 2(2)$

$\frac{dP}{dt} = -6 + 4$

$\frac{dP}{dt} = -2$

The rate of change of the perimeter is $-2$ \$cm/minute. A negative rate indicates that the perimeter is decreasing.

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(b) Rate of change of the area:

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

$A = xy$

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

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

Using the product rule, $\frac{d}{dt}(uv) = u\frac{dv}{dt} + v\frac{du}{dt}$:

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

We need to find this rate when $x = 10 \$cm$ and $y = 6 \$cm$. Substitute these values and the given rates $\frac{dx}{dt} = -3$ \$cm/minute and $\frac{dy}{dt} = 2$ \$cm/minute into the equation:

$\left(\frac{dA}{dt}\right)_{\substack{x=10 \\ y=6}} = (10)(2) + (6)(-3)$

$\left(\frac{dA}{dt}\right)_{\substack{x=10 \\ y=6}} = 20 - 18$

$\left(\frac{dA}{dt}\right)_{\substack{x=10 \\ y=6}} = 2$

The rate of change of the area is $2$ \$cm$^2$/minute. A positive rate indicates that the area is increasing.

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Final Answer:

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

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

Given:

The total cost function in Rupees associated with the production of $x$ units is:

$C(x) = 0.005 x^3 – 0.02 x^2 + 30x + 5000$

Marginal cost is defined as the instantaneous rate of change of total cost at any level of output.

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To Find:

The marginal cost when 3 units are produced, i.e., when $x = 3$.

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Solution:

The marginal cost is the rate of change of the total cost $C(x)$ with respect to the number of units produced, $x$. Mathematically, this is given by the derivative of $C(x)$ with respect to $x$, denoted as $C'(x)$ or $\frac{dC}{dx}$.

We differentiate the given cost function $C(x)$ with respect to $x$:

$C(x) = 0.005 x^3 – 0.02 x^2 + 30x + 5000$

$\frac{dC}{dx} = \frac{d}{dx} (0.005 x^3 – 0.02 x^2 + 30x + 5000)$

Using the rules of differentiation (power rule and constant multiple rule):

$\frac{dC}{dx} = 0.005 \cdot \frac{d}{dx}(x^3) – 0.02 \cdot \frac{d}{dx}(x^2) + 30 \cdot \frac{d}{dx}(x) + \frac{d}{dx}(5000)$

$\frac{dC}{dx} = 0.005 \cdot (3x^{3-1}) – 0.02 \cdot (2x^{2-1}) + 30 \cdot (1x^{1-1}) + 0$

$\frac{dC}{dx} = 0.015 x^2 – 0.04 x + 30$

So, the marginal cost function is:

$C'(x) = 0.015 x^2 – 0.04 x + 30$

We need to find the marginal cost when 3 units are produced, which means we need to evaluate $C'(x)$ at $x = 3$.

Substitute $x = 3$ into the marginal cost function:

$C'(3) = 0.015 (3)^2 – 0.04 (3) + 30$

$C'(3) = 0.015 \cdot 9 – 0.12 + 30$

$C'(3) = 0.135 – 0.12 + 30$

$C'(3) = 0.015 + 30$

$C'(3) = 30.015$

The marginal cost is in Rupees per unit.

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Final Answer:

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

Given:

The total revenue function in Rupees from the sale of $x$ units is:

$R(x) = 3x^2 + 36x + 5$

Marginal revenue is the rate of change of total revenue with respect to the number of items sold.

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To Find:

The marginal revenue when $x = 5$ units are sold.

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Solution:

The marginal revenue is the instantaneous rate of change of the total revenue $R(x)$ with respect to the number of units sold, $x$. This is given by the derivative of $R(x)$ with respect to $x$, denoted as $R'(x)$ or $\frac{dR}{dx}$.

We differentiate the given total revenue function $R(x)$ with respect to $x$:

$R(x) = 3x^2 + 36x + 5$

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

Using the rules of differentiation (power rule, constant multiple rule, and sum rule):

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

$\frac{dR}{dx} = 3 \cdot \frac{d}{dx}(x^2) + 36 \cdot \frac{d}{dx}(x) + 0$

$\frac{dR}{dx} = 3 \cdot (2x^{2-1}) + 36 \cdot (1x^{1-1})$

$\frac{dR}{dx} = 6x + 36$

So, the marginal revenue function is:

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

We need to find the marginal revenue when $x = 5$. Substitute $x = 5$ into the marginal revenue function:

$R'(5) = 6(5) + 36$

$R'(5) = 30 + 36$

$R'(5) = 66$

The marginal revenue is in Rupees per unit.

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Final Answer:

The marginal revenue when $x = 5$ is $\textsf{₹} \$66$.

Given:

The area of a circle is $A = \pi r^2$, where $r$ is the radius.

We need to find the rate of change of the area with respect to the radius, which is $\frac{dA}{dr}$.

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To Find:

The rate of change of the area of the circle with respect to its radius $r$ at two specific values:

(a) when $r = 3$ cm

(b) when $r = 4$ cm

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Solution:

The area of a circle is given by the formula:

$A = \pi r^2$

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

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

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

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

$\frac{dA}{dr} = \pi \cdot (2r^{2-1})$

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

This expression, $2\pi r$, represents the rate of change of the area of the circle with respect to its radius $r$ for any value of $r$.

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(a) Rate of change when $r = 3$ cm:

Substitute $r = 3$ cm into the expression for $\frac{dA}{dr}$:

$\left(\frac{dA}{dr}\right)_{r=3} = 2\pi (3)$

$\left(\frac{dA}{dr}\right)_{r=3} = 6\pi$

The units for $\frac{dA}{dr}$ are the units of area divided by the units of radius, which are \$cm$^2$/cm.

So, the rate of change of the area when $r = 3$ cm is $6\pi$ \$cm$^2$/cm.

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(b) Rate of change when $r = 4$ cm:

Substitute $r = 4$ cm into the expression for $\frac{dA}{dr}$:

$\left(\frac{dA}{dr}\right)_{r=4} = 2\pi (4)$

$\left(\frac{dA}{dr}\right)_{r=4} = 8\pi$

The units are \$cm$^2$/cm.

So, the rate of change of the area when $r = 4$ cm is $8\pi$ \$cm$^2$/cm.

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Final Answer:

(a) The rate of change of the area of the circle with respect to its radius when $r = 3$ cm is $6\pi$ \$cm$^2$/cm.

(b) The rate of change of the area of the circle with respect to its radius when $r = 4$ cm is $8\pi$ \$cm$^2$/cm.

Given:

Let $s$ be the length of an edge of the cube in centimetres.

Let $V$ be the volume of the cube in cubic centimetres.

Let $A$ be the surface area of the cube in square centimetres.

The volume is increasing at the rate of 8 \$cm$^3$/s, so $\frac{dV}{dt} = 8 \$cm$^3$/s.

We are interested in the instant when the length of an edge is 12 \$cm, i.e., $s = 12 \$cm.

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To Find:

How fast is the surface area increasing when $s = 12$ \$cm? This means we need to find $\frac{dA}{dt}$ when $s = 12$ \$cm.

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Solution:

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

$V = s^3$

To relate the rate of change of volume to the rate of change of the edge length, we differentiate both sides of the volume equation with respect to time $t$:

$\frac{dV}{dt} = \frac{d}{dt}(s^3)$

Using the chain rule, $\frac{d}{dt}(s^3) = 3s^2 \frac{ds}{dt}$.

So, we have:

$\frac{dV}{dt} = 3s^2 \frac{ds}{dt}$

We are given $\frac{dV}{dt} = 8$. Substituting this value, we get:

$8 = 3s^2 \frac{ds}{dt}$

We can express the rate of change of the edge length $\frac{ds}{dt}$ in terms of $s$:

$\frac{ds}{dt} = \frac{8}{3s^2}$

Now, consider the surface area of the cube. The surface area $A$ of a cube with edge length $s$ is given by:

$A = 6s^2$

To find the rate of change of the surface area, we differentiate both sides of the surface area equation with respect to time $t$:

$\frac{dA}{dt} = \frac{d}{dt}(6s^2)$

Using the constant multiple rule and the chain rule, $\frac{d}{dt}(6s^2) = 6 \cdot 2s \frac{ds}{dt} = 12s \frac{ds}{dt}$.

So, we have:

$\frac{dA}{dt} = 12s \frac{ds}{dt}$

Now, substitute the expression for $\frac{ds}{dt}$ that we found earlier ($\frac{8}{3s^2}$) into this equation:

$\frac{dA}{dt} = 12s \left(\frac{8}{3s^2}\right)$

Simplify the expression:

$\frac{dA}{dt} = \frac{12 \cdot 8 \cdot s}{3 \cdot s^2}$

$\frac{dA}{dt} = \frac{\cancel{12}^4 \cdot 8 \cdot \cancel{s}}{\cancel{3} \cdot s^{\cancel{2}^1}}$

$\frac{dA}{dt} = \frac{4 \cdot 8}{s}$

$\frac{dA}{dt} = \frac{32}{s}$

We need to find the rate of change of the surface area when the length of an edge is 12 \$cm. Substitute $s = 12$ \$cm into the expression for $\frac{dA}{dt}$:

$\left(\frac{dA}{dt}\right)_{s=12} = \frac{32}{12}$

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 4:

$\frac{32}{12} = \frac{\cancel{32}^8}{\cancel{12}_3} = \frac{8}{3}$

The units for the rate of change of surface area are square centimetres per second (\$cm$^2$/s).

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

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Final Answer:

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

Given:

Let $r$ be the radius of the circle in centimetres.

Let $A$ be the area of the circle in square centimetres.

The radius is increasing uniformly at the rate of 3 cm/s. This means $\frac{dr}{dt} = 3 \$cm/s.

We are interested in the instant when the radius is 10 cm, i.e., $r = 10 \$cm.

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To Find:

The rate at which the area of the circle is increasing when $r = 10$ \$cm. This means we need to find $\frac{dA}{dt}$ when $r = 10$ \$cm.

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Solution:

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

$A = \pi r^2$

To find the rate at which the area is increasing, we need to differentiate both sides of the area equation with respect to time $t$. We use the chain rule since $r$ is a function of $t$:

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

Using the constant multiple rule and the chain rule, $\frac{d}{dt}(r^2) = 2r \frac{dr}{dt}$.

So, we have:

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

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

We are given that the rate of increase of the radius is $\frac{dr}{dt} = 3$ \$cm/s.

We need to find the rate of increase of the area when the radius is $r = 10$ \$cm.

Substitute $r = 10$ and $\frac{dr}{dt} = 3$ into the expression for $\frac{dA}{dt}$:

$\left(\frac{dA}{dt}\right)_{r=10} = 2\pi (10)(3)$

$\left(\frac{dA}{dt}\right)_{r=10} = 2\pi \cdot 30$

$\left(\frac{dA}{dt}\right)_{r=10} = 60\pi$

The units for the rate of change of area are square centimetres per second (\$cm$^2$/s).

Therefore, the area of the circle is increasing at a rate of $60\pi$ \$cm$^2$/s when the radius is 10 cm.

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Final Answer:

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

Given:

Let $s$ be the length of an edge of the cube in centimetres.

Let $V$ be the volume of the cube in cubic centimetres.

The edge of the cube is increasing at the rate of 3 cm/s. This means $\frac{ds}{dt} = 3 \$cm/s.

We are interested in the instant when the edge is 10 cm long, i.e., $s = 10 \$cm.

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To Find:

How fast is the volume of the cube increasing when the edge is 10 cm long? This means we need to find $\frac{dV}{dt}$ when $s = 10$ \$cm.

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Solution:

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

$V = s^3$

To find the rate at which the volume is increasing, we need to differentiate both sides of the volume equation with respect to time $t$. We use the chain rule since $s$ is a function of $t$:

$\frac{dV}{dt} = \frac{d}{dt}(s^3)$

Using the chain rule, $\frac{d}{dt}(s^3) = 3s^{3-1} \frac{ds}{dt} = 3s^2 \frac{ds}{dt}$.

So, we have:

$\frac{dV}{dt} = 3s^2 \frac{ds}{dt}$

We are given that the rate of increase of the edge length is $\frac{ds}{dt} = 3$ \$cm/s.

We need to find the rate of increase of the volume when the edge length is $s = 10$ \$cm.

Substitute $s = 10$ and $\frac{ds}{dt} = 3$ into the expression for $\frac{dV}{dt}$:

$\left(\frac{dV}{dt}\right)_{s=10} = 3(10)^2 (3)$

$\left(\frac{dV}{dt}\right)_{s=10} = 3(100)(3)$

$\left(\frac{dV}{dt}\right)_{s=10} = 900$

The units for the rate of change of volume are cubic centimetres per second (\$cm$^3$/s).

Therefore, the volume of the cube is increasing at a rate of 900 \$cm$^3$/s when the edge is 10 cm long.

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Final Answer:

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

Given:

Let $r$ be the radius of the circular wave in centimetres.

Let $A$ be the enclosed area of the circular wave in square centimetres.

The speed of the waves is the rate at which the radius is increasing with respect to time $t$.

Rate of change of radius: $\frac{dr}{dt} = 5 \$cm/s.

We are interested in the instant when the radius is 8 cm, i.e., $r = 8 \$cm.

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To Find:

How fast is the enclosed area increasing when $r = 8$ \$cm? This means we need to find $\frac{dA}{dt}$ when $r = 8$ \$cm.

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Solution:

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

$A = \pi r^2$

To find the rate at which the enclosed area is increasing with respect to time, we need to differentiate both sides of the area equation with respect to time $t$. Since $r$ is changing with time, we use the chain rule:

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

Using the constant multiple rule and the chain rule, $\frac{d}{dt}(r^2) = 2r \frac{dr}{dt}$.

So, we have:

$\frac{dA}{dt} = \pi \cdot \left(2r \frac{dr}{dt}\right)$

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

We are given that the rate of increase of the radius is $\frac{dr}{dt} = 5$ \$cm/s.

We need to find the rate of increase of the area at the instant when the radius is $r = 8$ \$cm.

Substitute $r = 8$ and $\frac{dr}{dt} = 5$ into the expression for $\frac{dA}{dt}$:

$\left(\frac{dA}{dt}\right)_{r=8} = 2\pi (8)(5)$

$\left(\frac{dA}{dt}\right)_{r=8} = 2\pi \cdot 40$

$\left(\frac{dA}{dt}\right)_{r=8} = 80\pi$

The units for the rate of change of area with respect to time are square centimetres per second (\$cm$^2$/s).

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

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Final Answer:

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

Given:

Let $r$ be the radius of the circle in centimetres.

Let $C$ be the circumference of the circle in centimetres.

The radius is increasing at the rate of 0.7 cm/s. This means $\frac{dr}{dt} = 0.7 \$cm/s.

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To Find:

The rate of increase of the circumference of the circle. This means we need to find $\frac{dC}{dt}$.

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Solution:

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

$C = 2\pi r$

To find the rate at which the circumference is increasing, we need to differentiate both sides of the circumference equation with respect to time $t$. Since $r$ is changing with time, we use the chain rule:

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

Using the constant multiple rule, $\frac{d}{dt}(2\pi r) = 2\pi \frac{d}{dt}(r)$.

So, we have:

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

We are given that the rate of increase of the radius is $\frac{dr}{dt} = 0.7$ \$cm/s.

Substitute the value of $\frac{dr}{dt}$ into the expression for $\frac{dC}{dt}$:

$\frac{dC}{dt} = 2\pi (0.7)$

$\frac{dC}{dt} = 1.4\pi$

The units for the rate of change of circumference with respect to time are centimetres per second (cm/s).

Therefore, the rate of increase of the circumference is $1.4\pi$ \$cm/s.

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Final Answer:

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

Given:

Let $x$ be the length of the rectangle in centimetres.

Let $y$ be the width of the rectangle in centimetres.

Length $x$ is decreasing at the rate of 5 cm/minute. This means $\frac{dx}{dt} = -5 \$cm/minute.

Width $y$ is increasing at the rate of 4 cm/minute. This means $\frac{dy}{dt} = 4 \$cm/minute.

We are interested in the instant when $x = 8 \$cm$ and $y = 6 \$cm.

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To Find:

(a) The rate of change of the perimeter of the rectangle when $x = 8 \$cm$ and $y = 6 \$cm$.

(b) The rate of change of the area of the rectangle 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:

$P = 2x + 2y$

To find the rate of change of the perimeter with respect to time $t$, we differentiate $P$ with respect to $t$. Since $x$ and $y$ are functions of $t$, we use the chain rule:

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

Using the sum rule and the constant multiple rule:

$\frac{dP}{dt} = 2\frac{dx}{dt} + 2\frac{dy}{dt}$

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

$\frac{dP}{dt} = 2(-5) + 2(4)$

$\frac{dP}{dt} = -10 + 8$

$\frac{dP}{dt} = -2$

The units for the rate of change of perimeter are centimetres per minute (cm/minute).

The rate of change of the perimeter is $-2$ \$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 with respect to time $t$, we differentiate $A$ with respect to $t$. Since $x$ and $y$ are functions of $t$, we use the product rule:

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

Using the product rule, $\frac{d}{dt}(uv) = u\frac{dv}{dt} + v\frac{du}{dt}$:

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

We need to find this rate at the instant when $x = 8 \$cm$ and $y = 6 \$cm$. Substitute these values and the given rates $\frac{dx}{dt} = -5$ \$cm/minute and $\frac{dy}{dt} = 4$ \$cm/minute into the equation:

$\left(\frac{dA}{dt}\right)_{\substack{x=8 \\ y=6}} = (8)(4) + (6)(-5)$

$\left(\frac{dA}{dt}\right)_{\substack{x=8 \\ y=6}} = 32 - 30$

$\left(\frac{dA}{dt}\right)_{\substack{x=8 \\ y=6}} = 2$

The units for the rate of change of area are square centimetres per minute (\$cm$^2$/minute).

The rate of change of the area is $2$ \$cm$^2$/minute. The positive sign indicates that the area is increasing.

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Final Answer:

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

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

Given:

Let $r$ be the radius of the spherical balloon in centimetres.

Let $V$ be the volume of the balloon in cubic centimetres.

The balloon is being inflated by pumping in gas at a rate of 900 \$cm$^3$/s.

This means the rate of change of volume with respect to time $t$ is $\frac{dV}{dt} = 900 \$cm$^3$/s.

We are interested in the instant when the radius is 15 cm, i.e., $r = 15 \$cm.

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To Find:

The rate at which the radius of the balloon increases when $r = 15$ \$cm. This means we need to find $\frac{dr}{dt}$ when $r = 15$ \$cm.

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Solution:

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

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

To relate the rate of change of volume to the rate of change of the radius, we differentiate both sides of the volume equation with respect to time $t$. Since $r$ is changing with time, we use the chain rule:

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

Using the constant multiple rule and the chain rule ($\frac{d}{dt}(r^3) = 3r^2 \frac{dr}{dt}$), we get:

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

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

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

We are given that the rate of increase of the volume is $\frac{dV}{dt} = 900$ \$cm$^3$/s.

We need to find the rate of increase of the radius, $\frac{dr}{dt}$, at the instant when the radius is $r = 15$ \$cm.

Substitute the given values $\frac{dV}{dt} = 900$ and $r = 15$ into the expression for $\frac{dV}{dt}$:

$900 = 4\pi (15)^2 \frac{dr}{dt}$

Calculate $(15)^2$:

$15^2 = 225$

Substitute this value back into the equation:

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

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

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

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

Simplify the fraction:

$\frac{dr}{dt} = \frac{1}{\pi}$

The units for the rate of change of radius are centimetres per second (cm/s).

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

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Final Answer:

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

Given:

Let $r$ be the radius of the spherical balloon in centimetres.

Let $V$ be the volume of the balloon in cubic centimetres.

The balloon is always spherical and has a variable radius $r$.

We need to find the rate at which its volume is increasing with the radius when $r = 10 \$cm$.

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To Find:

The rate of change of the volume of the balloon with respect to its radius $r$, i.e., $\frac{dV}{dr}$, when $r = 10$ \$cm.

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Solution:

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

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

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

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

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

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

$\frac{dV}{dr} = \frac{4}{3}\pi \cdot (3r^{3-1})$

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

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

This expression, $4\pi r^2$, represents the rate of change of the volume of the sphere with respect to its radius $r$ for any value of $r$.

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

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

$\left(\frac{dV}{dr}\right)_{r=10} = 4\pi (100)$

$\left(\frac{dV}{dr}\right)_{r=10} = 400\pi$

The units for the rate of change of volume with respect to radius are cubic centimetres per centimetre (\$cm$^3$/cm).

Therefore, 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.

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Final Answer:

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

Given:

Let $L$ be the length of the ladder, $x$ be the distance of the bottom of the ladder from the wall, and $y$ be the height of the top of the ladder on the wall.

Length of the ladder: $L = 5 \$m$.

The bottom of the ladder is pulled away from the wall at a rate: $\frac{dx}{dt} = 2 \$cm/s$.

Convert the rate of $x$ to meters per second to match the unit of ladder length:

$\frac{dx}{dt} = 2 \$cm/s = \frac{2}{100} \$m/s = 0.02 \$m/s$.

We are interested in the instant when the foot of the ladder is 4 m away from the wall, i.e., when $x = 4 \$m$.

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To Find:

How fast is its height on the wall decreasing when $x = 4 \$m$? This means we need to find $\frac{dy}{dt}$ when $x = 4 \$m$.

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Solution:

The ladder, the wall, and the ground form a right-angled triangle. By the Pythagorean theorem, the relationship between the length of the ladder, the distance from the wall ($x$), and the height on the wall ($y$) is:

$x^2 + y^2 = L^2$

Substitute the given length of the ladder, $L = 5 \$m$:

$x^2 + y^2 = 5^2$

$x^2 + y^2 = 25$

Since $x$ and $y$ are changing with time, we differentiate both sides of this equation with respect to time $t$ to find the relationship between their rates of change. We use the chain rule:

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

$\frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = 0$

Using the chain rule, $\frac{d}{dt}(x^2) = 2x \frac{dx}{dt}$ and $\frac{d}{dt}(y^2) = 2y \frac{dy}{dt}$.

So, the differentiated equation is:

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

We can divide the entire equation by 2:

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

We need to find $\frac{dy}{dt}$ when $x = 4 \$m$. At this specific instant, we need to find the corresponding value of $y$. We can find this value using the Pythagorean relationship $x^2 + y^2 = 25$:

When $x = 4 \$m$:

$4^2 + y^2 = 25$

$16 + y^2 = 25$

$y^2 = 25 - 16$

$y^2 = 9$

Since $y$ represents a height on the wall, it must be positive:

$y = \sqrt{9} = 3 \$m$

Now we have the values for $x$, $y$, and $\frac{dx}{dt}$ at the instant of interest: $x = 4 \$m$, $y = 3 \$m$, and $\frac{dx}{dt} = 0.02 \$m/s$. Substitute these values into the differentiated equation $x \frac{dx}{dt} + y \frac{dy}{dt} = 0$:

$(4)(0.02) + (3) \frac{dy}{dt} = 0$

$0.08 + 3 \frac{dy}{dt} = 0$

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

$3 \frac{dy}{dt} = -0.08$

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

The unit of $\frac{dy}{dt}$ is meters per second (m/s). The question provided the rate of change of $x$ in cm/s, so we should provide the answer for the rate of change of $y$ in cm/s. Convert the rate from m/s to cm/s by multiplying by 100:

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

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

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

The negative sign indicates that the height $y$ is decreasing, which is expected as the bottom of the ladder is being pulled away from the wall.

The question asks "How fast is its height on the wall decreasing?". This phrasing asks for the magnitude of the rate of decrease.

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Final Answer:

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 curve is given by the equation $6y = x^3 + 2$.

The rate of change of the y-coordinate is 8 times the rate of change of the x-coordinate.

This can be written as: $\frac{dy}{dt} = 8 \frac{dx}{dt}$.

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To Find:

The points $(x, y)$ on the curve where the given condition on the rates of change holds.

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Solution:

We are given the equation of the curve:

$6y = x^3 + 2$

Since $x$ and $y$ are changing with time $t$, we differentiate both sides of this equation with respect to $t$ to find the relationship between their rates of change $\frac{dx}{dt}$ and $\frac{dy}{dt}$.

Differentiating with respect to $t$:

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

Using the chain rule on the left side and the power rule and sum rule on the right side:

$6 \frac{dy}{dt} = 3x^{3-1} \frac{dx}{dt} + 0$

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

We are given the condition that the rate of change of the y-coordinate is 8 times the rate of change of the x-coordinate:

$\frac{dy}{dt} = 8 \frac{dx}{dt}$

Substitute this condition into the differentiated curve equation:

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

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

Rearrange the equation to solve for $x$:

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

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

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

This equation is satisfied if either $(48 - 3x^2) = 0$ or $\frac{dx}{dt} = 0$. If $\frac{dx}{dt} = 0$, then $\frac{dy}{dt} = 8 \times 0 = 0$, meaning the particle is momentarily stationary. However, the problem implies movement and a specific ratio between non-zero rates. Thus, we consider the case where the term in the parenthesis is zero:

$48 - 3x^2 = 0$

$3x^2 = 48$

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

$x^2 = 16$

Taking the square root of both sides gives the possible values for $x$:

$x = \pm \sqrt{16}$

$x = 4$ or $x = -4$

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

Case 1: When $x = 4$

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

$6y = 64 + 2$

$6y = 66$

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

$y = 11$

This gives the point $(4, 11)$.

Case 2: When $x = -4$

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

$6y = -64 + 2$

$6y = -62$

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

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

This gives the point $(-4, -\frac{31}{3})$.

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

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Final Answer:

The points on the curve are $(4, 11)$ and $(-4, -\frac{31}{3})$.

Given:

Let $r$ be the radius of the air bubble in centimetres.

Let $V$ be the volume of the air bubble in cubic centimetres.

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

We are interested in the instant when the radius is 1 cm, i.e., $r = 1 \$cm.

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To Find:

At what rate is the volume of the bubble increasing when the radius is 1 cm? This means we need to find $\frac{dV}{dt}$ when $r = 1$ \$cm.

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Solution:

An air bubble is spherical. The volume $V$ of a sphere with radius $r$ is given by the formula:

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

To find the rate at which the volume is increasing, we need to differentiate both sides of the volume equation with respect to time $t$. Since $r$ is changing with time, we use the chain rule:

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

Using the constant multiple rule and the chain rule ($\frac{d}{dt}(r^3) = 3r^2 \frac{dr}{dt}$), we get:

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

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

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

We are given that the rate of increase of the radius is $\frac{dr}{dt} = \frac{1}{2}$ \$cm/s.

We need to find the rate of increase of the volume, $\frac{dV}{dt}$, at the instant when the radius is $r = 1$ \$cm.

Substitute the given values $\frac{dr}{dt} = \frac{1}{2}$ and $r = 1$ into the expression for $\frac{dV}{dt}$:

$\left(\frac{dV}{dt}\right)_{r=1} = 4\pi (1)^2 \left(\frac{1}{2}\right)$

$\left(\frac{dV}{dt}\right)_{r=1} = 4\pi (1) \left(\frac{1}{2}\right)$

$\left(\frac{dV}{dt}\right)_{r=1} = 4\pi \cdot \frac{1}{2}$

$\left(\frac{dV}{dt}\right)_{r=1} = 2\pi$

The units for the rate of change of volume are cubic centimetres per second (\$cm$^3$/s).

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

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Final Answer:

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

Given:

The balloon always remains spherical.

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

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To Find:

The rate of change of the volume of the balloon with respect to $x$, which is $\frac{dV}{dx}$.

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Solution:

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

The radius $r$ is half of the diameter $D$:

$r = \frac{D}{2}$

Substitute the given expression for the diameter:

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

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

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

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

To find the rate of change of the volume with respect to $x$, we first express the volume in terms of $x$ by substituting the expression for $r$:

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

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

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

Now, simplify the constant term:

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

$V = \frac{\cancel{4}\pi \cdot \cancel{27}^9}{\cancel{3} \cdot \cancel{64}_{16}} (2x + 1)^3$

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

Now, we differentiate $V$ with respect to $x$ to find the rate of change of its volume with respect to $x$. We use the chain rule:

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

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

Let $u = 2x + 1$. Then $\frac{du}{dx} = \frac{d}{dx}(2x + 1) = 2$.

Using the chain rule, $\frac{d}{dx}(u^3) = 3u^2 \frac{du}{dx}$.

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

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

Combine the constants:

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

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

Simplify the fraction $\frac{54}{16}$ by dividing both numerator and denominator by 2:

$\frac{54}{16} = \frac{\cancel{54}^{27}}{\cancel{16}_8} = \frac{27}{8}$

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

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

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Final Answer:

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

Given:

Let $V$ be the volume of the sand cone in cubic centimetres.

Let $h$ be the height of the cone in centimetres.

Let $r$ be the radius of the base of the cone in centimetres.

Sand is pouring from a pipe at the rate of 12 \$cm$^3$/s. This is the rate of change of the volume of the cone with respect to time $t$.

So, $\frac{dV}{dt} = 12 \$cm$^3$/s.

The height of the cone is always one-sixth of the radius of the base. This gives the relationship between $h$ and $r$:

$h = \frac{1}{6} r$

This implies $r = 6h$.

We are interested in the instant when the height is 4 cm, i.e., $h = 4 \$cm.

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To Find:

How fast is the height of the sand cone increasing when $h = 4 \$cm$? This means we need to find $\frac{dh}{dt}$ when $h = 4 \$cm$.

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Solution:

The volume $V$ of a cone with radius $r$ and height $h$ is given by the formula:

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

The relationship between $h$ and $r$ is $r = 6h$. To express the volume solely in terms of the height $h$, substitute $r = 6h$ into the volume formula:

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

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

$V = \frac{1}{3}\pi \cdot 36 h^3$

$V = \frac{\cancel{36}^{12}}{3}\pi h^3$

$V = 12\pi h^3$

Now that the volume is expressed as a function of height $h$, we can find the rate of change of volume with respect to time $t$ by differentiating $V$ with respect to $t$. Since $h$ is changing with time, we use the chain rule:

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

Using the constant multiple rule and the chain rule ($\frac{d}{dt}(h^3) = 3h^2 \frac{dh}{dt}$), we get:

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

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

$\frac{dV}{dt} = 36\pi h^2 \frac{dh}{dt}$

We are given that the rate of change of the volume is $\frac{dV}{dt} = 12$ \$cm$^3$/s.

We need to find the rate of increase of the height, $\frac{dh}{dt}$, at the instant when the height is $h = 4$ \$cm.

Substitute the given values $\frac{dV}{dt} = 12$ and $h = 4$ into the differentiated equation:

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

Calculate $(4)^2$:

$4^2 = 16$

Substitute this value back into the equation:

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

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

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

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

Simplify the fraction $\frac{12}{576}$ by dividing both numerator and denominator by 12:

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

So, the rate of change of the height is:

$\frac{dh}{dt} = \frac{1}{48\pi}$

The units for the rate of change of height are centimetres per second (cm/s).

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

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Final Answer:

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 in Rupees associated with the production of $x$ units of an item is:

$C(x) = 0.007x^3 – 0.003x^2 + 15x + 4000$

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

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To Find:

The marginal cost when 17 units are produced.

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Solution:

The marginal cost is defined as the instantaneous rate of change of the total cost with respect to the number of units produced. This is given by the derivative of the total cost function $C(x)$ with respect to $x$, denoted by $\frac{dC}{dx}$ or $C'(x)$.

We are given the total cost function:

$C(x) = 0.007x^3 – 0.003x^2 + 15x + 4000$

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

$\frac{dC}{dx} = \frac{d}{dx} (0.007x^3 – 0.003x^2 + 15x + 4000)$

Using the power rule and constant multiple rule for differentiation:

$\frac{dC}{dx} = 0.007 \cdot \frac{d}{dx}(x^3) – 0.003 \cdot \frac{d}{dx}(x^2) + 15 \cdot \frac{d}{dx}(x) + \frac{d}{dx}(4000)$

$\frac{dC}{dx} = 0.007 \cdot (3x^{3-1}) – 0.003 \cdot (2x^{2-1}) + 15 \cdot (1x^{1-1}) + 0$

$\frac{dC}{dx} = 0.021 x^2 – 0.006 x + 15$

So, the marginal cost function is:

$C'(x) = 0.021 x^2 – 0.006 x + 15$

We need to find the marginal cost when 17 units are produced, which means we need to evaluate $C'(x)$ at $x = 17$.

Substitute $x = 17$ into the marginal cost function:

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

First, calculate $(17)^2$:

$17^2 = 289$

Now substitute this value back:

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

Perform the multiplications:

$0.021 \times 289 = 6.069$

$0.006 \times 17 = 0.102$

Substitute these values back into the expression for $C'(17)$:

$C'(17) = 6.069 – 0.102 + 15$

$C'(17) = 5.967 + 15$

$C'(17) = 20.967$

The units for marginal cost are Rupees per unit.

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Final Answer:

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

Given:

The total revenue function in Rupees received from the sale of $x$ units of a product is:

$R(x) = 13x^2 + 26x + 15$

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

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To Find:

The marginal revenue when $x = 7$.

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Solution:

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

We are given the total revenue function:

$R(x) = 13x^2 + 26x + 15$

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

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

Using the power rule, constant multiple rule, and sum rule for differentiation:

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

$\frac{dR}{dx} = 13 \cdot \frac{d}{dx}(x^2) + 26 \cdot \frac{d}{dx}(x) + 0$

$\frac{dR}{dx} = 13 \cdot (2x^{2-1}) + 26 \cdot (1x^{1-1})$

$\frac{dR}{dx} = 26x + 26$

So, the marginal revenue function is:

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

We need to find the marginal revenue when $x = 7$. Substitute $x = 7$ into the marginal revenue function:

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

We can calculate $26 \times 7$:

$\begin{array}{cc}& 2 & 6 \\ \times & & 7 \\ \hline 1 & 8 & 2 \\ \hline \end{array}$

So, $26 \times 7 = 182$.

Substitute this value back into the expression for $R'(7)$:

$R'(7) = 182 + 26$

Now perform the addition:

$\begin{array}{cc}& 1 & 8 & 2 \\ + & & 2 & 6 \\ \hline & 2 & 0 & 8 \\ \hline \end{array}$

$R'(7) = 208$

The units for marginal revenue are Rupees per unit sold.

---

Final Answer:

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

Given:

The area of a circle is $A$, and its radius is $r$.

We need to find the rate of change of the area with respect to the radius when $r = 6$ cm.

---

To Find:

The value of $\frac{dA}{dr}$ when $r = 6$ cm.

---

Solution:

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

$A = \pi r^2$

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

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

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

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

$\frac{dA}{dr} = \pi \cdot (2r^{2-1})$

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

This expression, $2\pi r$, represents the rate of change of the area of the circle with respect to its radius $r$ for any value of $r$.

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

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

$\left(\frac{dA}{dr}\right)_{r=6} = 12\pi$

The units for the rate of change of area with respect to radius are units of area divided by units of radius, which is \$cm$^2$/cm.

So, the rate of change of the area when $r = 6$ cm is $12\pi$ \$cm$^2$/cm.

Comparing this result with the given options:

(A) $10\pi$

(B) $12\pi$

(C) $8\pi$

(D) $11\pi$

The calculated rate of change matches option (B).

---

Final Answer:

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

Given:

The total revenue function in Rupees received from the sale of $x$ units of a product is:

$R(x) = 3x^2 + 36x + 5$

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

---

To Find:

The marginal revenue when $x = 15$.

---

Solution:

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

We are given the total revenue function:

$R(x) = 3x^2 + 36x + 5$

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

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

Using the power rule, constant multiple rule, and sum rule for differentiation:

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

$\frac{dR}{dx} = 3 \cdot (2x) + 36 \cdot (1) + 0$

$\frac{dR}{dx} = 6x + 36$

So, the marginal revenue function is:

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

We need to find the marginal revenue when $x = 15$. Substitute $x = 15$ into the marginal revenue function:

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

$R'(15) = 90 + 36$

$R'(15) = 126$

The units for marginal revenue are Rupees per unit sold.

Comparing the calculated value with the given options, we find that the value is 126.

---

Final Answer:

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

The correct option is (D) 126.

Given:

The function $f(x) = 7x – 3$.

The domain is $R$ (the set of real numbers).

---

To Show:

The function $f(x)$ is strictly increasing on $R$.

---

Solution:

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

We are given the function:

$f(x) = 7x – 3$

To find the derivative of $f(x)$ with respect to $x$, we differentiate term by term:

$f'(x) = \frac{d}{dx}(7x – 3)$

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

Using the constant multiple rule and the power rule ($\frac{d}{dx}(x) = 1$) and the rule for the derivative of a constant ($\frac{d}{dx}(c) = 0$):

$f'(x) = 7 \cdot \frac{d}{dx}(x) – 0$

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

$f'(x) = 7$

The derivative of the function is $f'(x) = 7$.

We observe the value of the derivative:

$f'(x) = 7$

Since $7 > 0$, the derivative $f'(x)$ is positive for all real numbers $x$.

According to the definition, if the derivative of a function is positive on an interval, the function is strictly increasing on that interval.

In this case, $f'(x) = 7 > 0$ for all $x \in R$.

Therefore, the function $f(x) = 7x – 3$ is strictly increasing on $R$.

---

Alternatively, using the definition of a strictly increasing function:

A function $f$ is strictly increasing on an interval $I$ if for any two numbers $x_1, x_2 \in I$, with $x_1 < x_2$, we have $f(x_1) < f(x_2)$.

Let $x_1, x_2$ be any two real numbers such that $x_1 < x_2$.

Consider the function $f(x) = 7x - 3$.

Evaluate the function at $x_1$ and $x_2$:

$f(x_1) = 7x_1 - 3$

$f(x_2) = 7x_2 - 3$

Since we assumed $x_1 < x_2$, we can multiply the inequality by 7 (a positive number), which preserves the inequality direction:

$7x_1 < 7x_2$

Now, subtract 3 from both sides of the inequality:

$7x_1 - 3 < 7x_2 - 3$

By the definition of $f(x)$, this means:

$f(x_1) < f(x_2)$

Since for any $x_1, x_2 \in R$ with $x_1 < x_2$, we have $f(x_1) < f(x_2)$, the function $f(x) = 7x – 3$ is strictly increasing on $R$ by definition.

Given:

The function $f(x) = x^3 – 3x^2 + 4x$.

The domain is $R$ (the set of real numbers).

---

To Show:

The function $f(x)$ is increasing on $R$.

---

Solution:

A function $f(x)$ is increasing on an interval if its derivative $f'(x)$ is greater than or equal to zero ($f'(x) \geq 0$) for all $x$ in that interval.

We are given the function:

$f(x) = x^3 – 3x^2 + 4x$

To find the derivative of $f(x)$ with respect to $x$, we differentiate term by term using the power rule and constant multiple rule:

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

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

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

$f'(x) = 3x^2 – 6x + 4$

So, the derivative of the function is $f'(x) = 3x^2 – 6x + 4$.

To show that the function is increasing on $R$, we need to show that $f'(x) \geq 0$ for all $x \in R$.

The expression for $f'(x)$ is a quadratic in $x$. We can analyze its sign by completing the square or by looking at its discriminant.

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

Take out the common factor 3 from the terms involving $x$:

$f'(x) = 3(x^2 – 2x) + 4$

Complete the square for the expression inside the parenthesis, $x^2 - 2x$. The term needed is $(\frac{-2}{2})^2 = (-1)^2 = 1$. Add and subtract 1 inside the parenthesis:

$f'(x) = 3(x^2 – 2x + 1 - 1) + 4$

$f'(x) = 3((x^2 – 2x + 1) - 1) + 4$

Recognize the perfect square trinomial $(x^2 – 2x + 1) = (x - 1)^2$:

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

Distribute the 3:

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

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

Now, let's analyze the sign of $f'(x) = 3(x - 1)^2 + 1$ for all $x \in R$.

For any real number $x$, the term $(x - 1)^2$ is the square of a real number, which is always greater than or equal to zero:

$(x - 1)^2 \geq 0$

Multiply the inequality by 3 (a positive number), which preserves the inequality direction:

$3(x - 1)^2 \geq 3 \cdot 0$

$3(x - 1)^2 \geq 0$

Now, add 1 to both sides of the inequality:

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

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

Since $1 > 0$, we have $3(x - 1)^2 + 1 \geq 1 > 0$ for all $x \in R$.

Thus, $f'(x) = 3(x - 1)^2 + 1 \geq 1$ for all $x \in R$.

This shows that $f'(x)$ is strictly positive for all $x \in R$, except possibly when $(x-1)^2=0$, which occurs when $x=1$. At $x=1$, $f'(1) = 3(1-1)^2+1 = 1$, which is still positive. So, $f'(x) > 0$ for all $x \in R$.

Since $f'(x) \geq 0$ for all $x \in R$ (specifically, $f'(x) > 0$ for all $x \in R$), the function $f(x)$ is increasing on $R$. (Note: A function is strictly increasing if $f'(x) > 0$. It is increasing if $f'(x) \geq 0$. Since $f'(x) > 0$ for all $x$, the function is actually strictly increasing on R, which also satisfies the condition of being increasing on R).

---

Alternatively, using the discriminant:

The derivative is $f'(x) = 3x^2 – 6x + 4$. This is a quadratic equation of the form $ax^2 + bx + c$, where $a=3$, $b=-6$, and $c=4$.

To determine the sign of the quadratic, we can check the sign of the leading coefficient and the discriminant. The leading coefficient is $a = 3$, which is positive ($3 > 0$).

The discriminant $\Delta$ is given by $\Delta = b^2 - 4ac$.

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

$\Delta = 36 - 48$

$\Delta = -12$

Since the discriminant $\Delta = -12$ is negative ($\Delta < 0$), and the leading coefficient $a = 3$ is positive ($a > 0$), the quadratic $ax^2 + bx + c$ is always positive for all real values of $x$.

Therefore, $f'(x) = 3x^2 – 6x + 4 > 0$ for all $x \in R$.

Since $f'(x) > 0$ for all $x \in R$, the function $f(x)$ is strictly increasing on $R$, which implies it is also increasing on $R$.

---

Conclusion:

Since the derivative $f'(x) = 3x^2 – 6x + 4 = 3(x-1)^2 + 1$, and $3(x-1)^2 \geq 0$ for all $x \in R$, it follows that $f'(x) \geq 1 > 0$ for all $x \in R$. Thus $f'(x) \geq 0$ for all $x \in R$.

Therefore, the function $f(x) = x^3 – 3x^2 + 4x$ is increasing on $R$.

Given:

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

The domain is $x \in R$.

---

To Prove:

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

(a) decreasing in $(0, \pi)$

(b) increasing in $(\pi, 2\pi)$

(c) neither increasing nor decreasing in $(0, 2\pi)$

---

Solution:

To determine the intervals where the function is increasing or decreasing, we find the derivative of the function $f(x)$ with respect to $x$ and analyze its sign in the given intervals.

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

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

$f'(x) = -\sin x$

---

(a) On the interval $(0, \pi)$:

In the interval $(0, \pi)$, the values of the sine function $\sin x$ are positive (i.e., $\sin x > 0$ for all $x \in (0, \pi)$).

Therefore, the derivative $f'(x) = -\sin x$ will be negative in this interval.

$-\sin x < 0$ for all $x \in (0, \pi)$.

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

---

(b) On the interval $(\pi, 2\pi)$:

In the interval $(\pi, 2\pi)$, the values of the sine function $\sin x$ are negative (i.e., $\sin x < 0$ for all $x \in (\pi, 2\pi)$).

Therefore, the derivative $f'(x) = -\sin x$ will be positive in this interval.

$-\sin x > 0$ for all $x \in (\pi, 2\pi)$.

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

---

(c) On the interval $(0, 2\pi)$:

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

From part (a), we found that $f'(x) = -\sin x < 0$ for all $x \in (0, \pi)$.

From part (b), we found that $f'(x) = -\sin x > 0$ for all $x \in (\pi, 2\pi)$.

Since the derivative $f'(x)$ is negative on a portion of the interval $(0, 2\pi)$ (specifically, on $(0, \pi)$) and positive on another portion of the interval $(0, 2\pi)$ (specifically, on $(\pi, 2\pi)$), the function $f(x) = \cos x$ changes its monotonic behaviour within the interval $(0, 2\pi)$.

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

---

Hence, it is proven that 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 $f(x) = x^2 – 4x + 6$.

The domain is $R$ (the set of real numbers).

---

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 the derivative of the function $f(x)$ with respect to $x$ and then determine the sign of the derivative.

The function is given by:

$f(x) = x^2 – 4x + 6$

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

$f'(x) = \frac{d}{dx}(x^2 – 4x + 6)$

Using the sum rule, constant multiple rule, and power rule:

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

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

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

Now, we need to find the critical points where the derivative is zero or undefined. $f'(x)$ is a polynomial, so it is defined for all real $x$. Set $f'(x) = 0$ to find the critical points:

$2x - 4 = 0$

$2x = 4$

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

$x = 2$

The critical point $x=2$ divides the real number line into two open intervals: $(-\infty, 2)$ and $(2, \infty)$. We analyze the sign of $f'(x)$ in each of these intervals.

---

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

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

Evaluate $f'(x)$ at $x = 0$:

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

Since $f'(0) = -4 < 0$, the derivative $f'(x)$ is negative for all $x$ in the interval $(-\infty, 2)$.

When $f'(x) < 0$ on an interval, the function is strictly decreasing on that interval.

Since $f(x)$ is continuous at $x=2$, it is decreasing on the closed interval $(-\infty, 2]$.

---

Interval 2: $(2, \infty)$

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

Evaluate $f'(x)$ at $x = 3$:

$f'(3) = 2(3) - 4 = 6 - 4 = 2$

Since $f'(3) = 2 > 0$, the derivative $f'(x)$ is positive for all $x$ in the interval $(2, \infty)$.

When $f'(x) > 0$ on an interval, the function is strictly increasing on that interval.

Since $f(x)$ is continuous at $x=2$, it is increasing on the closed interval $[2, \infty)$.

---

(a) The function is increasing on the interval where $f'(x) \geq 0$. This occurs for $x \geq 2$. So the interval is $[2, \infty)$.

(b) The function is decreasing on the interval where $f'(x) \leq 0$. This occurs for $x \leq 2$. So the interval is $(-\infty, 2]$.

---

Final Answer:

The function $f(x) = x^2 – 4x + 6$ is:

(a) increasing on $[2, \infty)$

(b) decreasing on $(-\infty, 2]$

Given:

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

The domain is $R$ (the set of real numbers).

---

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 the derivative of the function $f(x)$ with respect to $x$ and then determine the sign of the derivative.

The function is given by:

$f(x) = 4x^3 – 6x^2 – 72x + 30$

Differentiate $f(x)$ with respect to $x$ using the sum rule, constant multiple rule, and power rule:

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

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

$f'(x) = 4(3x^2) – 6(2x) – 72(1) + 0$

$f'(x) = 12x^2 – 12x – 72$

Now, we need to find the critical points where the derivative is zero. Set $f'(x) = 0$:

$12x^2 – 12x – 72 = 0$

Divide the entire equation by 12 to simplify:

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

$x^2 – x – 6 = 0$

This is a quadratic equation. We can factor it to find the roots:

We need two numbers that multiply to -6 and add to -1. These numbers are -3 and 2.

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

The roots are $x - 3 = 0$ or $x + 2 = 0$.

$x = 3$ or $x = -2$

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

We can use the factored form of the derivative $f'(x) = 12(x - 3)(x + 2)$ to easily determine the sign.

---

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

Choose a test value, e.g., $x = -3$.

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

Since $f'(-3) = 72 > 0$, the derivative $f'(x)$ is positive for all $x$ in the interval $(-\infty, -2)$.

---

Interval 2: $(-2, 3)$

Choose a test value, e.g., $x = 0$.

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

Since $f'(0) = -72 < 0$, the derivative $f'(x)$ is negative for all $x$ in the interval $(-2, 3)$.

---

Interval 3: $(3, \infty)$

Choose a test value, e.g., $x = 4$.

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

Since $f'(4) = 72 > 0$, the derivative $f'(x)$ is positive for all $x$ in the interval $(3, \infty)$.

---

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

(a) The function is increasing on the intervals where $f'(x) \geq 0$. This occurs in $(-\infty, -2)$ and $(3, \infty)$. Since the function is continuous at $x = -2$ and $x = 3$, we include these points in the intervals.

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

(b) The function is decreasing on the intervals where $f'(x) \leq 0$. This occurs in $(-2, 3)$. Since the function is continuous at $x = -2$ and $x = 3$, we include these points in the interval.

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

---

Final Answer:

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

(a) increasing on $(-\infty, -2] \cup [3, \infty)$

(b) decreasing on $[-2, 3]$

Given:

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

The domain is $x \in \left[ 0, \frac{\pi}{2} \right]$.

---

To Find:

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

(a) increasing

(b) decreasing

---

Solution:

To determine the intervals where the function is increasing or decreasing, we find the derivative of the function $f(x)$ with respect to $x$ and then analyze the sign of the derivative on the given domain $\left[ 0, \frac{\pi}{2} \right]$.

The function is given by:

$f(x) = \sin(3x)$

Differentiate $f(x)$ with respect to $x$ using the chain rule:

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

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

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

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

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

Now, we need to find the critical points in the interval $\left[ 0, \frac{\pi}{2} \right]$ where the derivative is zero or undefined. The derivative $f'(x) = 3\cos(3x)$ is defined for all $x$. We set $f'(x) = 0$:

$3\cos(3x) = 0$

$\cos(3x) = 0$

We need to find values of $x \in \left[ 0, \frac{\pi}{2} \right]$ such that $\cos(3x) = 0$.

Let $\theta = 3x$. As $x$ ranges from $0$ to $\frac{\pi}{2}$, $3x$ ranges from $3 \cdot 0 = 0$ to $3 \cdot \frac{\pi}{2} = \frac{3\pi}{2}$.

So, we need to find $\theta \in \left[ 0, \frac{3\pi}{2} \right]$ such that $\cos(\theta) = 0$. The values of $\theta$ for which $\cos(\theta) = 0$ in this range are $\theta = \frac{\pi}{2}$ and $\theta = \frac{3\pi}{2}$.

Now, we find the corresponding values of $x$ by substituting back $\theta = 3x$:

Case 1: $3x = \frac{\pi}{2}$

$x = \frac{\pi}{6}$

This value is in the interval $\left[ 0, \frac{\pi}{2} \right]$, since $0 < \frac{\pi}{6} < \frac{\pi}{2}$.

Case 2: $3x = \frac{3\pi}{2}$

$x = \frac{3\pi}{6} = \frac{\pi}{2}$

This value is the upper endpoint of the interval $\left[ 0, \frac{\pi}{2} \right]$.

The critical point in the interior of the interval is $x = \frac{\pi}{6}$. This point divides the interval $\left[ 0, \frac{\pi}{2} \right]$ into two sub-intervals: $\left[ 0, \frac{\pi}{6} \right]$ and $\left[ \frac{\pi}{6}, \frac{\pi}{2} \right]$. We analyze the sign of $f'(x) = 3\cos(3x)$ in these intervals.

---

Interval 1: $\left[ 0, \frac{\pi}{6} \right]$

For $x$ in the open interval $\left( 0, \frac{\pi}{6} \right)$, $3x$ is in the interval $\left( 0, \frac{\pi}{2} \right)$.

In the interval $\left( 0, \frac{\pi}{2} \right)$, $\cos(\theta) > 0$. Therefore, $\cos(3x) > 0$ for $x \in \left( 0, \frac{\pi}{6} \right)$.

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

At the endpoints $x=0$ and $x=\frac{\pi}{6}$, $f'(0) = 3\cos(0) = 3(1) = 3 > 0$ and $f'(\frac{\pi}{6}) = 3\cos(\frac{\pi}{2}) = 3(0) = 0$.

Since $f'(x) \geq 0$ for all $x \in \left[ 0, \frac{\pi}{6} \right]$, the function $f(x)$ is increasing on $\left[ 0, \frac{\pi}{6} \right]$.

---

Interval 2: $\left[ \frac{\pi}{6}, \frac{\pi}{2} \right]$

For $x$ in the open interval $\left( \frac{\pi}{6}, \frac{\pi}{2} \right)$, $3x$ is in the interval $\left( \frac{\pi}{2}, \frac{3\pi}{2} \right)$.

In the interval $\left( \frac{\pi}{2}, \frac{3\pi}{2} \right)$, $\cos(\theta) < 0$. Therefore, $\cos(3x) < 0$ for $x \in \left( \frac{\pi}{6}, \frac{\pi}{2} \right)$.

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

At the endpoints $x=\frac{\pi}{6}$ and $x=\frac{\pi}{2}$, $f'(\frac{\pi}{6}) = 0$ and $f'(\frac{\pi}{2}) = 3\cos(\frac{3\pi}{2}) = 3(0) = 0$.

Since $f'(x) \leq 0$ for all $x \in \left[ \frac{\pi}{6}, \frac{\pi}{2} \right]$, the function $f(x)$ is decreasing on $\left[ \frac{\pi}{6}, \frac{\pi}{2} \right]$.

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Final Answer:

The function $f(x) = \sin(3x)$, $x \in \left[ 0, \frac{\pi}{2} \right]$ is:

(a) increasing on $\left[ 0, \frac{\pi}{6} \right]$

(b) decreasing on $\left[ \frac{\pi}{6}, \frac{\pi}{2} \right]$

Given:

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

The domain is $0 \leq x \leq 2\pi$, i.e., $x \in [0, 2\pi]$.

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To Find:

The intervals in which the function $f(x)$ is increasing or decreasing on the domain $[0, 2\pi]$.

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Solution:

To find the intervals where the function is increasing or decreasing, we first find the derivative of $f(x)$ with respect to $x$ and then analyze the sign of the derivative on the domain $[0, 2\pi]$.

The function is given by:

$f(x) = \sin x + \cos x$

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

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

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

$f'(x) = \cos x - \sin x$

Now, we need to find the critical points in the interval $[0, 2\pi]$ where the derivative is zero or undefined. The derivative $f'(x) = \cos x - \sin x$ is defined for all $x$. We set $f'(x) = 0$:

$\cos x - \sin x = 0$

$\cos x = \sin x$

Divide both sides by $\cos x$, assuming $\cos x \neq 0$ (If $\cos x = 0$, then $\sin x = \pm 1$. If $\cos x = 0$ and $\sin x = 0$, this is impossible. So we can divide by $\cos x$).

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

$\tan x = 1$

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

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

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

For $n = 2$: $x = 2 \cdot \pi + \frac{\pi}{4} = \frac{8\pi + \pi}{4} = \frac{9\pi}{4}$. This is outside $[0, 2\pi]$.

The critical points in the interval $[0, 2\pi]$ are $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$. These points divide the interval $[0, 2\pi]$ into three 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 these intervals.

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

In this interval, $0 < x < \frac{\pi}{4}$. For these values of $x$, $\cos x > \sin x$.

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

Since $f'(x) > 0$ on this interval, $f(x)$ is strictly increasing on $\left( 0, \frac{\pi}{4} \right)$. Including the endpoints where $f'(x)$ could be zero, $f(x)$ is increasing on $\left[ 0, \frac{\pi}{4} \right]$.

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

In this interval, $\frac{\pi}{4} < x < \frac{5\pi}{4}$. For these values of $x$, $\cos x < \sin x$. (This interval spans from Quadrant I through Quadrant II and into Quadrant III. In Quadrant I after $\frac{\pi}{4}$, $\sin x > \cos x$. In Quadrants II and III, $\sin x$ can be positive or negative, and $\cos x$ is negative, but the inequality $\cos x < \sin x$ holds). For instance, at $x = \frac{\pi}{2}$, $\cos(\frac{\pi}{2}) = 0$, $\sin(\frac{\pi}{2}) = 1$, so $0 - 1 = -1 < 0$. At $x = \pi$, $\cos(\pi) = -1$, $\sin(\pi) = 0$, so $-1 - 0 = -1 < 0$.

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

Since $f'(x) < 0$ on this interval, $f(x)$ is strictly decreasing on $\left( \frac{\pi}{4}, \frac{5\pi}{4} \right)$. Including the endpoints, $f(x)$ is decreasing on $\left[ \frac{\pi}{4}, \frac{5\pi}{4} \right]$.

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

In this interval, $\frac{5\pi}{4} < x < 2\pi$. For these values of $x$ (in Quadrant IV after $\frac{5\pi}{4}$), $\cos x > \sin x$. For instance, at $x = \frac{3\pi}{2}$, $\cos(\frac{3\pi}{2}) = 0$, $\sin(\frac{3\pi}{2}) = -1$, so $0 - (-1) = 1 > 0$. At $x = 2\pi$, $\cos(2\pi) = 1$, $\sin(2\pi) = 0$, so $1 - 0 = 1 > 0$.

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

Since $f'(x) > 0$ on this interval, $f(x)$ is strictly increasing on $\left( \frac{5\pi}{4}, 2\pi \right)$. Including the endpoint $x=2\pi$, $f(x)$ is increasing on $\left[ \frac{5\pi}{4}, 2\pi \right]$.

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Combining the closed intervals:

The function is increasing on $\left[ 0, \frac{\pi}{4} \right] \cup \left[ \frac{5\pi}{4}, 2\pi \right]$.

The function is decreasing on $\left[ \frac{\pi}{4}, \frac{5\pi}{4} \right]$.

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Final Answer:

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

increasing on $\left[ 0, \frac{\pi}{4} \right] \cup \left[ \frac{5\pi}{4}, 2\pi \right]$

decreasing on $\left[ \frac{\pi}{4}, \frac{5\pi}{4} \right]$

Given:

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

The domain is $R$, the set of all real numbers.

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To Show:

The function $f(x) = 3x + 17$ is strictly increasing on $R$.

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Solution (using derivative):

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

We are given the function:

$f(x) = 3x + 17$

To find the derivative of $f(x)$ with respect to $x$, we differentiate term by term:

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

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

Using the constant multiple rule and the power rule ($\frac{d}{dx}(x) = 1$), and the derivative of a constant ($\frac{d}{dx}(c) = 0$):

$f'(x) = 3 \cdot 1 + 0$

$f'(x) = 3$

The derivative of the function is $f'(x) = 3$.

We observe the value of the derivative:

$f'(x) = 3$

Since $3 > 0$, the derivative $f'(x)$ is strictly positive for all real numbers $x$.

Therefore, based on the condition for strictly increasing functions, $f(x) = 3x + 17$ is strictly increasing on $R$.

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Alternate Solution (using definition):

A function $f$ is strictly increasing on an interval $I$ if for any two numbers $x_1, x_2 \in I$, with $x_1 < x_2$, we have $f(x_1) < f(x_2)$.

Let $x_1, x_2$ be any two real numbers in the domain $R$ such that $x_1 < x_2$.

Consider the values of the function at $x_1$ and $x_2$:

$f(x_1) = 3x_1 + 17$

$f(x_2) = 3x_2 + 17$

Since we assumed $x_1 < x_2$, we can multiply the inequality by 3 (a positive number). Multiplying by a positive number preserves the direction of the inequality:

$3x_1 < 3x_2$

Now, add 17 to both sides of the inequality. Adding a constant to both sides also preserves the direction of the inequality:

$3x_1 + 17 < 3x_2 + 17$

By the definition of $f(x)$, the left side is $f(x_1)$ and the right side is $f(x_2)$. So, the inequality becomes:

$f(x_1) < f(x_2)$

Thus, for any $x_1, x_2 \in R$ with $x_1 < x_2$, we have shown that $f(x_1) < f(x_2)$.

Therefore, by the definition of a strictly increasing function, $f(x) = 3x + 17$ is strictly increasing on $R$.

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Conclusion:

Using both the derivative test and the definition of a strictly increasing function, we have shown that the function $f(x) = 3x + 17$ is strictly increasing on $R$.

Given:

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

The domain is $R$, the set of all real numbers.

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To Show:

The function $f(x) = e^{2x}$ is strictly increasing on $R$.

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Solution (using derivative):

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

We are given the function:

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

To find the derivative of $f(x)$ with respect to $x$, we differentiate using the chain rule. Let $u = 2x$, so $\frac{du}{dx} = 2$.

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

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

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

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

Now, we need to examine the sign of the derivative $f'(x) = 2e^{2x}$ for all $x \in R$.

The exponential function $e^u$ is always positive for any real value of $u$. In this case, $u = 2x$, which is a real number for any $x \in R$.

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

Since $e^{2x}$ is always positive, multiplying it by a positive constant (2) also results in a positive value:

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

Thus, $f'(x) > 0$ for all $x \in R$.

According to the condition for strictly increasing functions, if the derivative of a function is strictly positive on an interval, the function is strictly increasing on that interval.

Since $f'(x) > 0$ for all $x \in R$, the function $f(x) = e^{2x}$ is strictly increasing on $R$.

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Alternate Solution (using definition):

A function $f$ is strictly increasing on an interval $I$ if for any two numbers $x_1, x_2 \in I$, with $x_1 < x_2$, we have $f(x_1) < f(x_2)$.

Let $x_1, x_2$ be any two real numbers in the domain $R$ such that $x_1 < x_2$.

Consider the values of the function at $x_1$ and $x_2$:

$f(x_1) = e^{2x_1}$

$f(x_2) = e^{2x_2}$

Since $x_1 < x_2$, multiply the inequality by 2:

$2x_1 < 2x_2$

Now consider the exponential function $e^u$. The base $e$ is approximately 2.718, which is greater than 1. The exponential function $e^u$ is a strictly increasing function for all real $u$. This means if $u_1 < u_2$, then $e^{u_1} < e^{u_2}$.

Applying this property to $2x_1 < 2x_2$, let $u_1 = 2x_1$ and $u_2 = 2x_2$. Since $u_1 < u_2$, we have:

$e^{2x_1} < e^{2x_2}$

By the definition of $f(x)$, the left side is $f(x_1)$ and the right side is $f(x_2)$. So, the inequality becomes:

$f(x_1) < f(x_2)$

Thus, for any $x_1, x_2 \in R$ with $x_1 < x_2$, we have shown that $f(x_1) < f(x_2)$.

Therefore, by the definition of a strictly increasing function, $f(x) = e^{2x}$ is strictly increasing on $R$.

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Conclusion:

Using both the derivative test and the definition of a strictly increasing function, we have shown that the function $f(x) = e^{2x}$ is strictly increasing on $R$.

Given:

The function $f(x) = \sin x$.

The domain under consideration for parts (a) and (b) is effectively restricted by the intervals provided. For part (c), the domain is specified as $(0, \pi)$.

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To Show:

The function $f(x) = \sin x$ is:

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

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

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

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Solution:

To determine the intervals where the function is increasing or decreasing, we find the first derivative of the function $f(x)$ with respect to $x$ and analyze its sign in the given intervals.

The function is given by:

$f(x) = \sin x$

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

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

$f'(x) = \cos x$

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(a) On the interval $\left( 0, \frac{\pi}{2} \right)$:

We need to determine the sign of $f'(x) = \cos x$ for $x \in \left( 0, \frac{\pi}{2} \right)$.

In the first quadrant, which corresponds to the interval $\left( 0, \frac{\pi}{2} \right)$, the cosine function is positive.

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

Since $f'(x) = \cos x > 0$ for all $x \in \left( 0, \frac{\pi}{2} \right)$, the function $f(x) = \sin x$ is strictly increasing on the interval $\left( 0, \frac{\pi}{2} \right)$. A strictly increasing function is also considered increasing.

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(b) On the interval $\left( \frac{\pi}{2}, \pi \right)$:

We need to determine the sign of $f'(x) = \cos x$ for $x \in \left( \frac{\pi}{2}, \pi \right)$.

In the second quadrant, which corresponds to the interval $\left( \frac{\pi}{2}, \pi \right)$, the cosine function is negative.

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

Since $f'(x) = \cos x < 0$ for all $x \in \left( \frac{\pi}{2}, \pi \right)$, the function $f(x) = \sin x$ is strictly decreasing on the interval $\left( \frac{\pi}{2}, \pi \right)$. A strictly decreasing function is also considered decreasing.

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(c) On the interval $(0, \pi)$:

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

From part (a), we know that $f'(x) = \cos x > 0$ for $x \in \left( 0, \frac{\pi}{2} \right)$.

From part (b), we know that $f'(x) = \cos x < 0$ for $x \in \left( \frac{\pi}{2}, \pi \right)$.

At the point $x = \frac{\pi}{2}$ within the interval $(0, \pi)$, the derivative is $f'(\frac{\pi}{2}) = \cos(\frac{\pi}{2}) = 0$.

Since the derivative $f'(x)$ is positive on part of the interval $(0, \pi)$ and negative on another part of the interval $(0, \pi)$, the function $f(x) = \sin x$ changes its monotonic behaviour within the interval $(0, \pi)$. It is increasing on $\left( 0, \frac{\pi}{2} \right)$ and decreasing on $\left( \frac{\pi}{2}, \pi \right)$.

Therefore, the function $f(x) = \sin x$ is neither increasing nor decreasing on the entire interval $(0, \pi)$.

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Hence, it is shown that the function $f(x) = \sin x$ is increasing in $\left( 0, \frac{\pi}{2} \right)$, decreasing in $\left( \frac{\pi}{2}, \pi \right)$, and neither increasing nor decreasing in $(0, \pi)$.

Given:

The function $f(x) = 2x^2 – 3x$.

The domain is $R$ (the set of real numbers).

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To Find:

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

(a) increasing

(b) decreasing

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Solution:

To find the intervals where the function is increasing or decreasing, we first find the first derivative of the function $f(x)$ with respect to $x$ and then analyze the sign of the derivative.

The function is given by:

$f(x) = 2x^2 – 3x$

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

$f'(x) = \frac{d}{dx}(2x^2 – 3x)$

Using the difference rule, constant multiple rule, and power rule:

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

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

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

Now, we need to find the critical points where the derivative is zero or undefined. The derivative $f'(x) = 4x - 3$ is a polynomial, so it is defined for all real numbers $x$. We set $f'(x) = 0$ to find the critical points:

$4x - 3 = 0$

$4x = 3$

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

The critical point $x = \frac{3}{4}$ divides the real number line into two open intervals: $\left(-\infty, \frac{3}{4}\right)$ and $\left(\frac{3}{4}, \infty\right)$. We analyze the sign of $f'(x)$ in each of these intervals.

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Interval 1: $\left(-\infty, \frac{3}{4}\right)$

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

Evaluate $f'(x)$ at $x = 0$:

$f'(0) = 4(0) - 3 = -3$

Since $f'(0) = -3 < 0$, the derivative $f'(x)$ is negative for all $x$ in the interval $\left(-\infty, \frac{3}{4}\right)$.

When $f'(x) < 0$ on an interval, the function is strictly decreasing on that interval.

Since $f(x)$ is continuous at $x=\frac{3}{4}$, it is decreasing on the closed interval $\left(-\infty, \frac{3}{4}\right]$.

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

Choose a test value in this interval, for example, $x = 1$.

Evaluate $f'(x)$ at $x = 1$:

$f'(1) = 4(1) - 3 = 4 - 3 = 1$

Since $f'(1) = 1 > 0$, the derivative $f'(x)$ is positive for all $x$ in the interval $\left(\frac{3}{4}, \infty\right)$.

When $f'(x) > 0$ on an interval, the function is strictly increasing on that interval.

Since $f(x)$ is continuous at $x=\frac{3}{4}$, it is increasing on the closed interval $\left[\frac{3}{4}, \infty\right)$.

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(a) The function is increasing on the interval where $f'(x) \geq 0$. This occurs for $x \geq \frac{3}{4}$. So the interval is $\left[\frac{3}{4}, \infty\right)$.

(b) The function is decreasing on the interval where $f'(x) \leq 0$. This occurs for $x \leq \frac{3}{4}$. So the interval is $\left(-\infty, \frac{3}{4}\right]$.

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Final Answer:

The function $f(x) = 2x^2 – 3x$ is:

(a) increasing on $\left[\frac{3}{4}, \infty\right)$

(b) decreasing on $\left(-\infty, \frac{3}{4}\right]$

Given:

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

The domain is $R$ (the set of real numbers).

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To Find:

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

(a) increasing

(b) decreasing

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Solution:

To determine the intervals where the function is increasing or decreasing, we find the first derivative of the function $f(x)$ with respect to $x$ and then analyze the sign of the derivative.

The function is given by:

$f(x) = 2x^3 – 3x^2 – 36x + 7$

Differentiate $f(x)$ with respect to $x$ using the sum rule, constant multiple rule, and power rule:

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

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

$f'(x) = 2(3x^2) – 3(2x) – 36(1) + 0$

$f'(x) = 6x^2 – 6x – 36$

Now, we need to find the critical points where the derivative is zero. Set $f'(x) = 0$:

$6x^2 – 6x – 36 = 0$

Divide the entire equation by 6 to simplify:

$x^2 – x – 6 = 0$

This is a quadratic equation. We can factor it to find the roots:

We look for two numbers that multiply to -6 and add to -1. These numbers are -3 and 2.

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

The roots are $x - 3 = 0$ or $x + 2 = 0$.

$x = 3$ or $x = -2$

The critical points are $x = -2$ and $x = 3$. These points divide the real number line into three open intervals: $(-\infty, -2)$, $(-2, 3)$, and $(3, \infty)$. We analyze the sign of $f'(x) = 6x^2 – 6x – 36 = 6(x - 3)(x + 2)$ in each of these intervals.

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

Choose a test value in this interval, e.g., $x = -3$.

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

Since $f'(-3) = 36 > 0$, the derivative $f'(x)$ is positive for all $x$ in the interval $(-\infty, -2)$.

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Interval 2: $(-2, 3)$

Choose a test value in this interval, e.g., $x = 0$.

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

Since $f'(0) = -72 < 0$, the derivative $f'(x)$ is negative for all $x$ in the interval $(-2, 3)$.

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Interval 3: $(3, \infty)$

Choose a test value in this interval, e.g., $x = 4$.

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

Since $f'(4) = 36 > 0$, the derivative $f'(x)$ is positive for all $x$ in the interval $(3, \infty)$.

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

(a) The function is increasing on the intervals where $f'(x) \geq 0$. This occurs on $(-\infty, -2)$ and $(3, \infty)$. Since $f(x)$ is a polynomial, it is continuous at the critical points, so we include them in the intervals.

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

(b) The function is decreasing on the interval where $f'(x) \leq 0$. This occurs on $(-2, 3)$. Since $f(x)$ is continuous at the critical points, we include them in the interval.

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

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Final Answer:

The function $f(x) = 2x^3 – 3x^2 – 36x + 7$ is:

(a) increasing on $(-\infty, -2] \cup [3, \infty)$

(b) decreasing on $[-2, 3]$

To find the intervals where the function is strictly increasing or decreasing, we find the first derivative of the function and determine the intervals where the derivative is strictly positive ($f'(x) > 0$) or strictly negative ($f'(x) < 0$).

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(a) $f(x) = x^2 + 2x – 5$

Solution:

The function is $f(x) = x^2 + 2x – 5$.

Find the derivative $f'(x)$:

$f'(x) = \frac{d}{dx}(x^2 + 2x – 5)$

$f'(x) = 2x + 2$

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

$2x + 2 = 0$

$2x = -2$

$x = -1$

The critical point $x = -1$ divides the real number line into two intervals: $(-\infty, -1)$ and $(-1, \infty)$.

Analyze the sign of $f'(x)$ in these intervals:

• For $x \in (-\infty, -1)$, choose a test value, e.g., $x = -2$. $f'(-2) = 2(-2) + 2 = -4 + 2 = -2$. Since $f'(-2) < 0$, $f'(x) < 0$ on $(-\infty, -1)$.
• For $x \in (-1, \infty)$, choose a test value, e.g., $x = 0$. $f'(0) = 2(0) + 2 = 2$. Since $f'(0) > 0$, $f'(x) > 0$ on $(-1, \infty)$.

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

$f(x)$ is strictly increasing when $f'(x) > 0$, which is on $(-1, \infty)$.

$f(x)$ is strictly decreasing when $f'(x) < 0$, which is on $(-\infty, -1)$.

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(b) $f(x) = 10 – 6x – 2x^2$

Solution:

The function is $f(x) = 10 – 6x – 2x^2$.

Find the derivative $f'(x)$:

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

$f'(x) = 0 - 6 - 2(2x)$

$f'(x) = -6 - 4x$

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

$-6 - 4x = 0$

$-4x = 6$

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

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

Analyze the sign of $f'(x)$ in these intervals:

• For $x \in (-\infty, -\frac{3}{2})$, choose a test value, e.g., $x = -2$. $f'(-2) = -6 - 4(-2) = -6 + 8 = 2$. Since $f'(-2) > 0$, $f'(x) > 0$ on $(-\infty, -\frac{3}{2})$.
• For $x \in (-\frac{3}{2}, \infty)$, choose a test value, e.g., $x = 0$. $f'(0) = -6 - 4(0) = -6$. Since $f'(0) < 0$, $f'(x) < 0$ on $(-\frac{3}{2}, \infty)$.

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

$f(x)$ is strictly increasing when $f'(x) > 0$, which is on $(-\infty, -\frac{3}{2})$.

$f(x)$ is strictly decreasing when $f'(x) < 0$, which is on $(-\frac{3}{2}, \infty)$.

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(c) $f(x) = –2x^3 – 9x^2 – 12x + 1$

Solution:

The function is $f(x) = –2x^3 – 9x^2 – 12x + 1$.

Find the derivative $f'(x)$:

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

$f'(x) = -2(3x^2) - 9(2x) - 12(1) + 0$

$f'(x) = -6x^2 - 18x - 12$

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

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

Divide by -6:

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

Factor the quadratic:

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

The critical points are $x = -1$ and $x = -2$. Arrange them in increasing order: $x = -2, x = -1$.

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

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

• For $x \in (-\infty, -2)$, choose a test value, e.g., $x = -3$. $f'(-3) = -6(-3+1)(-3+2) = -6(-2)(-1) = -12$. Since $f'(-3) < 0$, $f'(x) < 0$ on $(-\infty, -2)$.
• For $x \in (-2, -1)$, choose a test value, e.g., $x = -1.5$. $f'(-1.5) = -6(-1.5+1)(-1.5+2) = -6(-0.5)(0.5) = 1.5$. Since $f'(-1.5) > 0$, $f'(x) > 0$ on $(-2, -1)$.
• For $x \in (-1, \infty)$, choose a test value, e.g., $x = 0$. $f'(0) = -6(0+1)(0+2) = -6(1)(2) = -12$. Since $f'(0) < 0$, $f'(x) < 0$ on $(-1, \infty)$.

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

$f(x)$ is strictly increasing when $f'(x) > 0$, which is on $(-2, -1)$.

$f(x)$ is strictly decreasing when $f'(x) < 0$, which is on $(-\infty, -2)$ and $(-1, \infty)$.

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(d) $f(x) = 6 – 9x – x^2$

Solution:

The function is $f(x) = 6 – 9x – x^2$.

Find the derivative $f'(x)$:

$f'(x) = \frac{d}{dx}(6 – 9x – x^2)$

$f'(x) = 0 - 9 - 2x$

$f'(x) = -9 - 2x$

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

$-9 - 2x = 0$

$-2x = 9$

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

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

Analyze the sign of $f'(x)$ in these intervals:

• For $x \in (-\infty, -\frac{9}{2})$, choose a test value, e.g., $x = -5$. $f'(-5) = -9 - 2(-5) = -9 + 10 = 1$. Since $f'(-5) > 0$, $f'(x) > 0$ on $(-\infty, -\frac{9}{2})$.
• For $x \in (-\frac{9}{2}, \infty)$, choose a test value, e.g., $x = 0$. $f'(0) = -9 - 2(0) = -9$. Since $f'(0) < 0$, $f'(x) < 0$ on $(-\frac{9}{2}, \infty)$.

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

$f(x)$ is strictly increasing when $f'(x) > 0$, which is on $(-\infty, -\frac{9}{2})$.

$f(x)$ is strictly decreasing when $f'(x) < 0$, which is on $(-\frac{9}{2}, \infty)$.

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(e) $f(x) = (x + 1)^3 (x – 3)^3$

Solution:

The function is $f(x) = (x + 1)^3 (x – 3)^3$. This can be written as $f(x) = ((x + 1)(x – 3))^3 = (x^2 - 2x - 3)^3$.

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

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

$f'(x) = 3(x^2 - 2x - 3)^{3-1} \cdot \frac{d}{dx}(x^2 - 2x - 3)$

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

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

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

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

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

This equation is zero if any of the factors are zero:

• $(x+1)^2 = 0 \implies x + 1 = 0 \implies x = -1$
• $(x-3)^2 = 0 \implies x - 3 = 0 \implies x = 3$
• $x - 1 = 0 \implies x = 1$

The critical points are $x = -1, x = 1, x = 3$. Arrange them in increasing order: $x = -1, x = 1, x = 3$.

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

Analyze the sign of $f'(x) = 6(x+1)^2 (x-3)^2 (x - 1)$ in these intervals. Note that $6(x+1)^2 \geq 0$ and $(x-3)^2 \geq 0$ for all real $x$. The sign of $f'(x)$ is determined by the sign of the factor $(x-1)$.

• For $x \in (-\infty, -1)$, choose a test value, e.g., $x = -2$. $(x-1) = -2 - 1 = -3 < 0$. Since the other factors are non-negative, $f'(-2) = 6(-1)^2(-5)^2(-3) < 0$. $f'(x) < 0$ on $(-\infty, -1)$.
• For $x \in (-1, 1)$, choose a test value, e.g., $x = 0$. $(x-1) = 0 - 1 = -1 < 0$. Since the other factors are non-negative, $f'(0) = 6(1)^2(-3)^2(-1) < 0$. $f'(x) < 0$ on $(-1, 1)$.
• For $x \in (1, 3)$, choose a test value, e.g., $x = 2$. $(x-1) = 2 - 1 = 1 > 0$. Since the other factors are non-negative, $f'(2) = 6(3)^2(-1)^2(1) > 0$. $f'(x) > 0$ on $(1, 3)$.
• For $x \in (3, \infty)$, choose a test value, e.g., $x = 4$. $(x-1) = 4 - 1 = 3 > 0$. Since the other factors are non-negative, $f'(4) = 6(5)^2(1)^2(3) > 0$. $f'(x) > 0$ on $(3, \infty)$.

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

$f(x)$ is strictly increasing when $f'(x) > 0$, which is on $(1, 3)$ and $(3, \infty)$.

$f(x)$ is strictly decreasing when $f'(x) < 0$, which is on $(-\infty, -1)$ and $(-1, 1)$.

Given:

The function $y = \log (1 + x) - \frac{2x}{2 + x}$.

The domain is $x > -1$. For $\log(1+x)$ to be defined, $1+x > 0$, which means $x > -1$. For $\frac{2x}{2+x}$ to be defined, $2+x \neq 0$, which means $x \neq -2$. Since the domain is $x > -1$, $x \neq -2$ is satisfied within this domain.

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To Show:

The function $y$ is an increasing function of $x$ throughout its domain ($x > -1$).

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Solution:

To show that the function is increasing throughout its domain, we need to find its derivative with respect to $x$ and show that it is greater than or equal to zero ($y' \geq 0$) for all $x > -1$. For a strictly increasing function, we would need to show $y' > 0$, but the question asks to show it is an increasing function, which typically means $y' \geq 0$. Let's find the derivative $y' = \frac{dy}{dx}$.

$y = \log (1 + x) - \frac{2x}{2 + x}$

Differentiate $y$ with respect to $x$:

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

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

For the first term, $\frac{d}{dx}(\log(1+x))$, we use the chain rule. Let $u = 1+x$, then $\frac{du}{dx} = 1$. $\frac{d}{dx}(\log u) = \frac{1}{u} \frac{du}{dx} = \frac{1}{1+x} \cdot 1 = \frac{1}{1+x}$.

For the second term, $\frac{d}{dx} \left( \frac{2x}{2 + x} \right)$, we use the quotient rule, $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$, where $u = 2x$ and $v = 2+x$.

$u' = \frac{d}{dx}(2x) = 2$

$v' = \frac{d}{dx}(2+x) = 1$

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

Now, substitute these derivatives back into the expression for $y'$:

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

To determine the sign of $y'$, we combine the terms into a single fraction:

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

Expand the numerator:

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

$4(1 + x) = 4 + 4x$

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

So, the derivative is:

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

Now we need to analyze the sign of $y'$ throughout the domain $x > -1$.

Consider the terms in the expression for $y'$:

• Numerator: $x^2$. The square of any real number is greater than or equal to zero. So, $x^2 \geq 0$ for all $x$. $x^2 = 0$ only when $x=0$.
• Denominator: $(1 + x)(2 + x)^2$.

Consider the terms in the denominator for $x > -1$:

• $(1 + x)$: Since $x > -1$, adding 1 to both sides gives $x + 1 > -1 + 1$, so $1 + x > 0$. This term is always positive on the domain.
• $(2 + x)^2$: Since $x > -1$, $2+x > 2-1 = 1$. The term $(2+x)^2$ is the square of a number greater than 1, so it is always positive. $(2+x)^2 > 0$ for all $x > -1$.

So, for the domain $x > -1$:

• The numerator $x^2$ is $\geq 0$.
• The denominator $(1 + x)(2 + x)^2$ is $(positive) \times (positive) = positive$.

Therefore, the fraction $\frac{x^2}{(1 + x)(2 + x)^2}$ is $\frac{\geq 0}{> 0}$, which means the derivative $y'$ is greater than or equal to zero for all $x > -1$.

$y' = \frac{x^2}{(1 + x)(2 + x)^2} \geq 0$ for all $x > -1$.

The derivative is equal to zero only when the numerator is zero, which happens when $x^2 = 0$, i.e., $x = 0$. At $x=0$, $y' = \frac{0^2}{(1+0)(2+0)^2} = 0$.

Since $y' \geq 0$ for all $x > -1$, the function $y$ is an increasing function of $x$ throughout its domain $x > -1$. Note that it is strictly increasing everywhere except at the single point $x=0$, where the derivative is zero. However, having the derivative equal to zero at isolated points does not prevent a function from being increasing.

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Conclusion:

Since $\frac{dy}{dx} = \frac{x^2}{(1 + x)(2 + x)^2} \geq 0$ for all $x > -1$, the function $y = \log (1 + x) - \frac{2x}{2 + x}$ is an increasing function of $x$ throughout its domain $x > -1$.

Given:

The function $y = [x(x – 2)]^2$.

The domain is $R$ (the set of real numbers).

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To Find:

The values of $x$ for which the function $y$ is increasing.

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Solution:

To find the intervals where the function is increasing, we find the first derivative $\frac{dy}{dx}$ and determine where it is greater than or equal to zero ($\frac{dy}{dx} \geq 0$).

The function is given by:

$y = [x(x – 2)]^2$

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

Differentiate $y$ with respect to $x$ using the chain rule. Let $u = x^2 - 2x$, then $\frac{du}{dx} = 2x - 2$.

$\frac{dy}{dx} = \frac{d}{dx}((x^2 - 2x)^2)$

$\frac{dy}{dx} = 2(x^2 - 2x)^{2-1} \cdot \frac{d}{dx}(x^2 - 2x)$

$\frac{dy}{dx} = 2(x^2 - 2x) \cdot (2x - 2)$

We can factor out $x$ from the first term and 2 from the second term:

$\frac{dy}{dx} = 2(x(x - 2)) \cdot 2(x - 1)$

$\frac{dy}{dx} = 4x(x - 2)(x - 1)$

Now, we need to find the critical points where the derivative is zero. Set $\frac{dy}{dx} = 0$:

$4x(x - 2)(x - 1) = 0$

This equation is zero if any of the factors are zero:

• $x = 0$
• $x - 2 = 0 \implies x = 2$
• $x - 1 = 0 \implies x = 1$

The critical points are $x = 0, x = 1, x = 2$. Arrange them in increasing order: $x = 0, x = 1, x = 2$.

These critical points divide the real number line into four open intervals: $(-\infty, 0)$, $(0, 1)$, $(1, 2)$, and $(2, \infty)$. We analyze the sign of $\frac{dy}{dx} = 4x(x - 2)(x - 1)$ in each of these intervals.

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Interval 1: $(-\infty, 0)$

Choose a test value, e.g., $x = -1$.

$\frac{dy}{dx} = 4(-1)(-1 - 2)(-1 - 1) = 4(-1)(-3)(-2) = 4(-1)(6) = -24$

Since $\frac{dy}{dx} < 0$, the function is strictly decreasing on $(-\infty, 0)$.

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Interval 2: $(0, 1)$

Choose a test value, e.g., $x = 0.5$.

$\frac{dy}{dx} = 4(0.5)(0.5 - 2)(0.5 - 1) = 4(0.5)(-1.5)(-0.5) = 4(0.5)(0.75) = 2(0.75) = 1.5$

Since $\frac{dy}{dx} > 0$, the function is strictly increasing on $(0, 1)$.

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Interval 3: $(1, 2)$

Choose a test value, e.g., $x = 1.5$.

$\frac{dy}{dx} = 4(1.5)(1.5 - 2)(1.5 - 1) = 4(1.5)(-0.5)(0.5) = 6(-0.25) = -1.5$

Since $\frac{dy}{dx} < 0$, the function is strictly decreasing on $(1, 2)$.

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Interval 4: $(2, \infty)$

Choose a test value, e.g., $x = 3$.

$\frac{dy}{dx} = 4(3)(3 - 2)(3 - 1) = 4(3)(1)(2) = 24$

Since $\frac{dy}{dx} > 0$, the function is strictly increasing on $(2, \infty)$.

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The function is increasing on the intervals where $\frac{dy}{dx} \geq 0$. These are the intervals where $\frac{dy}{dx} > 0$ along with the critical points where $\frac{dy}{dx} = 0$.

The intervals where $\frac{dy}{dx} > 0$ are $(0, 1)$ and $(2, \infty)$.

The critical points where $\frac{dy}{dx} = 0$ are $x = 0, x = 1, x = 2$.

Including the endpoints where the derivative is zero, the function is increasing on $[0, 1] \cup [2, \infty)$.

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Final Answer:

The function $y = [x(x – 2)]^2$ is increasing on the intervals $[0, 1] \cup [2, \infty)$.

Given:

The function $y = \frac{4 \sin \theta}{(2 + \cos\theta)} - \theta$.

The domain is $\left[ 0, \frac{\pi}{2} \right]$.

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To Prove:

The function $y$ is an increasing function of $\theta$ in $\left[ 0, \frac{\pi}{2} \right]$.

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Solution:

To prove that the function is increasing on the given interval, we need to find its derivative with respect to $\theta$ and show that it is greater than or equal to zero ($\frac{dy}{d\theta} \geq 0$) for all $\theta \in \left[ 0, \frac{\pi}{2} \right]$.

The function is given by:

$y = \frac{4 \sin \theta}{(2 + \cos\theta)} - \theta$

Differentiate $y$ with respect to $\theta$:

$\frac{dy}{d\theta} = \frac{d}{d\theta} \left( \frac{4 \sin \theta}{2 + \cos\theta} - \theta \right)$

$\frac{dy}{d\theta} = \frac{d}{d\theta} \left( \frac{4 \sin \theta}{2 + \cos\theta} \right) - \frac{d}{d\theta} (\theta)$

The derivative of the second term is $\frac{d}{d\theta}(\theta) = 1$.

For the first term, $\frac{d}{d\theta} \left( \frac{4 \sin \theta}{2 + \cos\theta} \right)$, 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 $v = 2 + \cos\theta$.

$u' = \frac{d}{d\theta}(4 \sin \theta) = 4 \cos \theta$

$v' = \frac{d}{d\theta}(2 + \cos\theta) = 0 - \sin \theta = -\sin \theta$

So, $\frac{d}{d\theta} \left( \frac{4 \sin \theta}{2 + \cos\theta} \right) = \frac{(4 \cos \theta)(2 + \cos\theta) - (4 \sin \theta)(-\sin \theta)}{(2 + \cos\theta)^2}$

Expand the numerator:

Numerator = $8 \cos \theta + 4 \cos^2 \theta - (-4 \sin^2 \theta)$

Numerator = $8 \cos \theta + 4 \cos^2 \theta + 4 \sin^2 \theta$

Factor out 4 from the last two terms and use the identity $\cos^2 \theta + \sin^2 \theta = 1$:

Numerator = $8 \cos \theta + 4(\cos^2 \theta + \sin^2 \theta)$

Numerator = $8 \cos \theta + 4(1)$

Numerator = $8 \cos \theta + 4$

So, the derivative of the first term is $\frac{8 \cos \theta + 4}{(2 + \cos\theta)^2}$.

Now, substitute this back into the expression for $\frac{dy}{d\theta}$:

$\frac{dy}{d\theta} = \frac{8 \cos \theta + 4}{(2 + \cos\theta)^2} - 1$

To determine the sign of $\frac{dy}{d\theta}$ on the interval $\left[ 0, \frac{\pi}{2} \right]$, we combine the terms into a single fraction:

$\frac{dy}{d\theta} = \frac{8 \cos \theta + 4 - 1 \cdot (2 + \cos\theta)^2}{(2 + \cos\theta)^2}$

Expand the term $(2 + \cos\theta)^2$ in the numerator:

$(2 + \cos\theta)^2 = 4 + 4 \cos \theta + \cos^2 \theta$

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:

$\frac{dy}{d\theta} = \frac{\cos \theta (4 - \cos \theta)}{(2 + \cos\theta)^2}$

Now we need to analyze the sign of $\frac{dy}{d\theta}$ on the domain $\left[ 0, \frac{\pi}{2} \right]$.

Consider the terms in the expression for $\frac{dy}{d\theta}$ for $\theta \in \left[ 0, \frac{\pi}{2} \right]$:

• $\cos \theta$: In the interval $\left[ 0, \frac{\pi}{2} \right]$, $\cos \theta$ ranges from $\cos(0) = 1$ to $\cos(\frac{\pi}{2}) = 0$. So, $\cos \theta \geq 0$ for $\theta \in \left[ 0, \frac{\pi}{2} \right]$. $\cos \theta = 0$ only at $\theta = \frac{\pi}{2}$.
• $4 - \cos \theta$: Since $0 \leq \cos \theta \leq 1$ in the interval $\left[ 0, \frac{\pi}{2} \right]$, the value of $4 - \cos \theta$ ranges from $4 - 1 = 3$ to $4 - 0 = 4$. So, $4 - \cos \theta \geq 3 > 0$. This term is always positive.
• $(2 + \cos\theta)^2$: Since $0 \leq \cos \theta \leq 1$, $2 + \cos \theta$ ranges from $2 + 0 = 2$ to $2 + 1 = 3$. So, $2 + \cos \theta \geq 2$. The term $(2 + \cos\theta)^2 \geq 2^2 = 4 > 0$. This term is always positive.

So, for $\theta \in \left[ 0, \frac{\pi}{2} \right]$:

• The numerator $\cos \theta (4 - \cos \theta)$ is $(\geq 0) \times (> 0)$, which is $\geq 0$. The numerator is 0 only when $\cos \theta = 0$, which occurs at $\theta = \frac{\pi}{2}$.
• The denominator $(2 + \cos\theta)^2$ is $> 0$.

Therefore, the fraction $\frac{\cos \theta (4 - \cos \theta)}{(2 + \cos\theta)^2}$ is $\frac{\geq 0}{> 0}$, which means the derivative $\frac{dy}{d\theta}$ is greater than or equal to zero for all $\theta \in \left[ 0, \frac{\pi}{2} \right]$.

$\frac{dy}{d\theta} \geq 0$ for all $\theta \in \left[ 0, \frac{\pi}{2} \right]$.

The derivative is equal to zero only at $\theta = \frac{\pi}{2}$. Since the derivative is non-negative throughout the interval, the function is increasing on $\left[ 0, \frac{\pi}{2} \right]$.

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Conclusion:

Since $\frac{dy}{d\theta} = \frac{\cos \theta (4 - \cos \theta)}{(2 + \cos\theta)^2} \geq 0$ 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 of $\theta$ in $\left[ 0, \frac{\pi}{2} \right]$.

Given:

The logarithmic function, which is typically represented as $f(x) = \log_b x$ where $b > 1$ is the base, or commonly, the natural logarithm $f(x) = \ln x$ (base $e \approx 2.718$). We will use $f(x) = \ln x$ for this proof.

The domain is $(0, \infty)$.

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To Prove:

The logarithmic function $f(x) = \ln x$ is increasing on the interval $(0, \infty)$.

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Solution:

A function $f(x)$ is increasing on an interval if its first derivative $f'(x)$ is greater than or equal to zero ($f'(x) \geq 0$) for all $x$ in that interval. To show it is strictly increasing, we would show $f'(x) > 0$. Since we will find that the derivative is strictly positive, the function is strictly increasing, which implies it is also increasing.

Let the logarithmic function be $f(x) = \ln x$.

To find the derivative of $f(x)$ with respect to $x$, we use the standard differentiation rule for the natural logarithm:

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

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

Now, we need to examine the sign of the derivative $f'(x) = \frac{1}{x}$ on the given domain $(0, \infty)$.

The domain $(0, \infty)$ includes all positive real numbers. For any value of $x$ in this domain, $x > 0$.

Consider the expression for the derivative, $f'(x) = \frac{1}{x}$.

If $x$ is a positive number (i.e., $x > 0$), then its reciprocal $\frac{1}{x}$ is also a positive number.

Thus, $f'(x) = \frac{1}{x} > 0$ for all $x \in (0, \infty)$.

Since the derivative $f'(x)$ is strictly positive ($f'(x) > 0$) for all $x$ in the interval $(0, \infty)$, the function $f(x) = \ln x$ is strictly increasing on this interval.

A strictly increasing function is also an increasing function.

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Alternate Solution (using definition):

A function $f$ is increasing on an interval $I$ if for any two numbers $x_1, x_2 \in I$, with $x_1 < x_2$, we have $f(x_1) \leq f(x_2)$. For strictly increasing, we need $f(x_1) < f(x_2)$. We will show it's strictly increasing.

Let $x_1, x_2$ be any two numbers in the domain $(0, \infty)$ such that $0 < x_1 < x_2$.

Consider the ratio $\frac{x_2}{x_1}$. Since $x_1 < x_2$ and both are positive, $\frac{x_2}{x_1} > 1$.

Now consider the function values $f(x_1) = \ln x_1$ and $f(x_2) = \ln x_2$.

We know a property of logarithms: $\ln \left(\frac{a}{b}\right) = \ln a - \ln b$.

Consider the difference between $f(x_2)$ and $f(x_1)$:

$f(x_2) - f(x_1) = \ln x_2 - \ln x_1$

$f(x_2) - f(x_1) = \ln \left(\frac{x_2}{x_1}\right)$

Since $0 < x_1 < x_2$, we have $\frac{x_2}{x_1} > 1$.

The natural logarithm function $\ln u$ has the property that $\ln u > 0$ if $u > 1$.

Applying this property with $u = \frac{x_2}{x_1} > 1$, we get:

$\ln \left(\frac{x_2}{x_1}\right) > 0$

So, $f(x_2) - f(x_1) > 0$.

This means $f(x_2) > f(x_1)$, or $f(x_1) < f(x_2)$.

Thus, for any $x_1, x_2 \in (0, \infty)$ with $x_1 < x_2$, we have shown that $f(x_1) < f(x_2)$.

Therefore, by the definition of a strictly increasing function, $f(x) = \ln x$ is strictly increasing on $(0, \infty)$, which implies it is also an increasing function on $(0, \infty)$.

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Conclusion:

Using both the derivative test and the definition of an increasing function, we have proven that the logarithmic function is increasing on $(0, \infty)$.

Given:

The function $f(x) = x^2 – x + 1$.

The interval is $(-1, 1)$.

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To Prove:

The function $f(x) = x^2 – x + 1$ is neither strictly increasing nor strictly decreasing on the interval $(– 1, 1)$.

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Solution:

To determine if a function is strictly increasing or strictly decreasing on an interval, we examine the sign of its first derivative on that interval. A function is strictly increasing if its derivative is strictly positive ($f'(x) > 0$), and strictly decreasing if its derivative is strictly negative ($f'(x) < 0$). If the derivative changes sign within the interval, the function is neither strictly increasing nor strictly decreasing on the entire interval.

We are given the function:

$f(x) = x^2 – x + 1$

Find the first derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(x^2 – x + 1)$

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

$f'(x) = 2x^{2-1} - 1x^{1-1} + 0$

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

Now, we need to analyze the sign of $f'(x) = 2x - 1$ on the interval $(-1, 1)$.

We find the point where the derivative is zero within this interval by setting $f'(x) = 0$:

$2x - 1 = 0$

$2x = 1$

$x = \frac{1}{2}$

The critical point $x = \frac{1}{2}$ lies within the interval $(-1, 1)$, because $-1 < \frac{1}{2} < 1$. This critical point divides the interval $(-1, 1)$ into two sub-intervals: $\left( -1, \frac{1}{2} \right)$ and $\left( \frac{1}{2}, 1 \right)$.

Let's examine the sign of $f'(x)$ in each sub-interval:

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Sub-interval 1: $\left( -1, \frac{1}{2} \right)$

For any $x$ in this interval, $x < \frac{1}{2}$.

Multiply the inequality by 2: $2x < 1$.

Subtract 1 from both sides: $2x - 1 < 0$.

Since $f'(x) = 2x - 1$, this means $f'(x) < 0$ for all $x \in \left( -1, \frac{1}{2} \right)$.

Thus, the function $f(x)$ is strictly decreasing on the interval $\left( -1, \frac{1}{2} \right)$.

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Sub-interval 2: $\left( \frac{1}{2}, 1 \right)$

For any $x$ in this interval, $x > \frac{1}{2}$.

Multiply the inequality by 2: $2x > 1$.

Subtract 1 from both sides: $2x - 1 > 0$.

Since $f'(x) = 2x - 1$, this means $f'(x) > 0$ for all $x \in \left( \frac{1}{2}, 1 \right)$.

Thus, the function $f(x)$ is strictly increasing on the interval $\left( \frac{1}{2}, 1 \right)$.

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Since the function $f(x)$ is strictly decreasing on the interval $\left( -1, \frac{1}{2} \right)$ and strictly increasing on the interval $\left( \frac{1}{2}, 1 \right)$, it changes its monotonic 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 $(-1, 1)$.

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Conclusion:

As the derivative $f'(x) = 2x - 1$ is negative on $\left( -1, \frac{1}{2} \right)$ and positive on $\left( \frac{1}{2}, 1 \right)$, the function $f(x)$ is strictly decreasing on $\left( -1, \frac{1}{2} \right]$ and strictly increasing on $\left[ \frac{1}{2}, 1 \right)$. Hence, it is neither strictly increasing nor strictly decreasing on $(-1, 1)$.

Given:

We are given four functions and the interval $\left( 0, \frac{\pi}{2} \right)$.

The functions are:

(A) $f(x) = \cos x$

(B) $f(x) = \cos 2x$

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

(D) $f(x) = \tan x$

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To Find:

Which of the given functions is decreasing on the interval $\left( 0, \frac{\pi}{2} \right)$.

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Solution:

A function $f(x)$ is decreasing on an open interval if its first derivative $f'(x)$ is less than or equal to zero ($f'(x) \leq 0$) for all $x$ in that interval. If $f'(x) < 0$ for all $x$ in the open interval, the function is strictly decreasing.

We will find the derivative for each function and examine its sign on the interval $\left( 0, \frac{\pi}{2} \right)$.

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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. Specifically, $\sin x \in (0, 1)$.

Therefore, $f'(x) = -\sin x$ is negative for all $x \in \left( 0, \frac{\pi}{2} \right)$.

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

Since $f'(x) < 0$, the function $f(x) = \cos x$ is strictly decreasing on $\left( 0, \frac{\pi}{2} \right)$. A strictly decreasing function is also a decreasing function.

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(B) $f(x) = \cos 2x$

The derivative is $f'(x) = \frac{d}{dx}(\cos 2x)$. Using the chain rule, $\frac{d}{dx}(\cos u) = -\sin u \frac{du}{dx}$. Let $u = 2x$, so $\frac{du}{dx} = 2$.

$f'(x) = -\sin(2x) \cdot 2 = -2\sin(2x)$.

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 $\sin \theta$ is positive. So, $\sin(2x) > 0$ for all $x \in \left( 0, \frac{\pi}{2} \right)$.

Therefore, $f'(x) = -2\sin(2x)$ is negative for all $x \in \left( 0, \frac{\pi}{2} \right)$.

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

Since $f'(x) < 0$, the function $f(x) = \cos 2x$ is strictly decreasing on $\left( 0, \frac{\pi}{2} \right)$. A strictly decreasing function is also a decreasing function.

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

The derivative is $f'(x) = \frac{d}{dx}(\cos 3x)$. Using the chain rule, let $u = 3x$, so $\frac{du}{dx} = 3$.

$f'(x) = -\sin(3x) \cdot 3 = -3\sin(3x)$.

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 $\left( 0, \frac{3\pi}{2} \right)$, the sign of $\sin(3x)$ changes:

• For $3x \in (0, \pi)$, $\sin(3x) > 0$. This corresponds to $x \in \left( 0, \frac{\pi}{3} \right)$. In this interval, $f'(x) = -3\sin(3x) < 0$.
• For $3x \in (\pi, \frac{3\pi}{2})$, $\sin(3x) < 0$. This corresponds to $x \in \left( \frac{\pi}{3}, \frac{\pi}{2} \right)$. In this interval, $f'(x) = -3\sin(3x) > 0$.

Since the derivative $f'(x)$ is negative on a part of the interval $\left( 0, \frac{\pi}{2} \right)$ and positive on another part, the function $f(x) = \cos 3x$ 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 $f'(x) = \frac{d}{dx}(\tan x) = \sec^2 x$.

For $x \in \left( 0, \frac{\pi}{2} \right)$, $\cos x$ is positive. $\cos x \in (0, 1)$.

$\sec x = \frac{1}{\cos x}$, so $\sec x$ is also positive. $\sec x \in (1, \infty)$.

The square of a positive number is positive. So, $f'(x) = \sec^2 x > 0$ for all $x \in \left( 0, \frac{\pi}{2} \right)$.

Since $f'(x) > 0$, the function $f(x) = \tan x$ is strictly increasing on $\left( 0, \frac{\pi}{2} \right)$.

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Summary of findings on the interval $\left( 0, \frac{\pi}{2} \right)$:

• (A) $f(x) = \cos x$: Strictly decreasing ($f'(x) < 0$)
• (B) $f(x) = \cos 2x$: Strictly decreasing ($f'(x) < 0$)
• (C) $f(x) = \cos 3x$: Neither increasing nor decreasing (derivative changes sign)
• (D) $f(x) = \tan x$: Strictly increasing ($f'(x) > 0$)

Both functions (A) and (B) are strictly decreasing on the interval $\left( 0, \frac{\pi}{2} \right)$, and thus they are both decreasing on this interval according to the standard definition ($f'(x) \leq 0$). Since the question format implies a single correct answer, and both (A) and (B) satisfy the condition, there might be an issue with the question as posed in a single-choice format, or a specific convention is assumed (e.g., selecting the simplest function that satisfies the property). Based on standard interpretations where one option must be chosen, $\cos x$ is a fundamental example of a function decreasing in the first quadrant.

Therefore, we select option (A).

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Final Answer:

Based on the analysis, both $\cos x$ and $\cos 2x$ are decreasing on $\left( 0, \frac{\pi}{2} \right)$. Assuming the question expects a single answer from the options, the most common and fundamental example of a function decreasing in the first quadrant is $\cos x$.

The correct option is (A) cos x.

Given:

The function $f(x) = x^{100} + \sin x –1$.

We need to find the intervals from the given options where the function is decreasing.

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To Find:

The interval among (A), (B), (C) where $f(x)$ is decreasing.

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Solution:

A function $f(x)$ is decreasing on an interval if its first derivative $f'(x)$ is less than or equal to zero ($f'(x) \leq 0$) for all $x$ in that interval. For strictly decreasing, we need $f'(x) < 0$ in the open interval.

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

$f(x) = x^{100} + \sin x –1$

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

$f'(x) = \frac{d}{dx}(x^{100} + \sin x –1)$

$f'(x) = \frac{d}{dx}(x^{100}) + \frac{d}{dx}(\sin x) - \frac{d}{dx}(1)$

$f'(x) = 100x^{100-1} + \cos x - 0$

$f'(x) = 100x^{99} + \cos x$

Now we examine the sign of $f'(x) = 100x^{99} + \cos x$ on each of the given intervals:

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(A) Interval $(0, 1)$

For $x \in (0, 1)$, we have $x > 0$.

The term $x^{99}$ is positive for $x > 0$. Thus, $100x^{99} > 0$ for $x \in (0, 1)$.

The interval $(0, 1)$ is a subset of $\left( 0, \frac{\pi}{2} \right)$ since $1$ radian $\approx 57.3^\circ$ and $\frac{\pi}{2}$ radians $\approx 90^\circ$. So $0 < x < 1 < \frac{\pi}{2}$.

In the interval $\left( 0, \frac{\pi}{2} \right)$, the cosine function $\cos x$ is positive. Thus, $\cos x > 0$ for $x \in (0, 1)$.

So, for $x \in (0, 1)$, $f'(x) = 100x^{99} + \cos x = (positive) + (positive) > 0$.

The function $f(x)$ is strictly increasing on $(0, 1)$.

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(B) Interval $\left( \frac{\pi}{2}, \pi \right)$

For $x \in \left( \frac{\pi}{2}, \pi \right)$, we have $x > \frac{\pi}{2} \approx 1.57$. Since $x > 0$, the term $x^{99}$ is positive. Thus, $100x^{99} > 0$ for $x \in \left( \frac{\pi}{2}, \pi \right)$. The smallest value of $x^{99}$ in this interval is $(\frac{\pi}{2})^{99}$, which is a large positive number as $\frac{\pi}{2} > 1$.

In the interval $\left( \frac{\pi}{2}, \pi \right)$, which is the second quadrant, the cosine function $\cos x$ is negative. The values of $\cos x$ in this interval are in $(-1, 0)$.

So, for $x \in \left( \frac{\pi}{2}, \pi \right)$, $f'(x) = 100x^{99} + \cos x = (large \$positive) + (negative \$between \$ -1 \$ and \$ 0)$.

The magnitude of $100x^{99}$ is very large for $x > \frac{\pi}{2}$. The magnitude of $\cos x$ is at most 1.

Thus, the large positive term $100x^{99}$ dominates the negative term $\cos x$.

For any $x \in \left( \frac{\pi}{2}, \pi \right)$, $x > \frac{\pi}{2} > 1$. So $x^{99} > 1$. $100x^{99} > 100$.

The minimum value of $f'(x)$ in this interval would be greater than $100 \cdot (\frac{\pi}{2})^{99} + (-1)$, which is clearly positive.

Therefore, $f'(x) = 100x^{99} + \cos x > 0$ on $\left( \frac{\pi}{2}, \pi \right)$.

The function $f(x)$ is strictly increasing on $\left( \frac{\pi}{2}, \pi \right)$.

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(C) Interval $\left( 0, \frac{\pi}{2} \right)$

For $x \in \left( 0, \frac{\pi}{2} \right)$, we have $x > 0$.

The term $x^{99}$ is positive for $x > 0$. Thus, $100x^{99} > 0$ for $x \in \left( 0, \frac{\pi}{2} \right)$.

In the interval $\left( 0, \frac{\pi}{2} \right)$, which is the first quadrant, the cosine function $\cos x$ is positive. Thus, $\cos x > 0$ for $x \in \left( 0, \frac{\pi}{2} \right)$.

So, for $x \in \left( 0, \frac{\pi}{2} \right)$, $f'(x) = 100x^{99} + \cos x = (positive) + (positive) > 0$.

The function $f(x)$ is strictly increasing on $\left( 0, \frac{\pi}{2} \right)$.

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Based on the analysis, the derivative $f'(x)$ is strictly positive on all the given intervals (A), (B), and (C). This means the function is strictly increasing on these intervals.

Therefore, the function is not decreasing on any of the intervals provided in options (A), (B), or (C).

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Final Answer:

The function is decreasing on none of the given intervals.

The correct option is (D) None of these.

Given:

The function $f(x) = x^2 + ax + 1$.

The interval is $(1, 2)$.

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

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To Find:

The least value of $a$ such that $f(x)$ is strictly increasing on $(1, 2)$.

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Solution:

A function $f(x)$ is strictly increasing on an open interval $(c, d)$ if its first derivative $f'(x)$ is strictly positive ($f'(x) > 0$) for all $x \in (c, d)$.

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

$f(x) = x^2 + ax + 1$

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

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

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

$f'(x) = 2x^{2-1} + a \cdot x^{1-1} + 0$

$f'(x) = 2x + a$

For $f(x)$ to be strictly increasing on the interval $(1, 2)$, we must have $f'(x) > 0$ for all $x \in (1, 2)$.

So, we require:

$2x + a > 0$ for all $x \in (1, 2)$.

Rearrange the inequality to isolate $a$:

$a > -2x$ for all $x \in (1, 2)$.

This means that $a$ must be greater than $-2x$ for every value of $x$ in the interval $(1, 2)$. To satisfy this condition, $a$ must be greater than the maximum value of $-2x$ on the interval $(1, 2)$.

Let's consider the function $g(x) = -2x$. We need $a > g(x)$ for all $x \in (1, 2)$.

On the interval $(1, 2)$, the values of $x$ satisfy $1 < x < 2$.

Multiply the inequality by -2. When multiplying an inequality by a negative number, the direction of the inequality sign reverses:

$1 < x < 2$

$-2 \cdot 1 > -2 \cdot x > -2 \cdot 2$

$-2 > -2x > -4$

Rewrite the inequality in standard order:

$-4 < -2x < -2$

So, the values of $-2x$ on the interval $(1, 2)$ are in the range $(-4, -2)$. The maximum value that $-2x$ approaches on this interval is -2 (as $x$ approaches 1 from the right). However, since the interval is open, $-2x$ never actually reaches -2. The supremum (least upper bound) of $-2x$ on $(1, 2)$ is -2.

For $a$ to be strictly greater than $-2x$ for all $x \in (1, 2)$, $a$ must be greater than the supremum of $-2x$ on that interval.

Thus, $a > -2$.

The condition for strict increase on an open interval is $f'(x) > 0$. If we consider the closed interval $[1, 2]$, the condition for increasing is $f'(x) \geq 0$. In some contexts, the boundary points are included when considering increasing/decreasing behaviour of the function itself, even if the derivative test strictly applies to the open interval. Let's check if including the boundary in the derivative condition is necessary for "strictly increasing" on an open interval.

If $a = -2$, then $f'(x) = 2x - 2 = 2(x - 1)$. On $(1, 2)$, $x > 1$, so $x - 1 > 0$, and $f'(x) = 2(x-1) > 0$. This satisfies the condition for strictly increasing on $(1, 2)$.

If $a = -2 - \epsilon$ for some small $\epsilon > 0$, then $f'(x) = 2x + (-2 - \epsilon) = 2x - 2 - \epsilon$. At $x=1.1$, $f'(1.1) = 2(1.1) - 2 - \epsilon = 2.2 - 2 - \epsilon = 0.2 - \epsilon$. If $\epsilon$ is small (e.g., $\epsilon = 0.1$), $f'(1.1) = 0.1 > 0$. However, we need $f'(x) > 0$ for *all* $x \in (1, 2)$.

Let's re-examine $f'(x) = 2x + a$. We need $2x + a > 0$ for all $x \in (1, 2)$.

$a > -2x$ for all $x \in (1, 2)$.

This inequality must hold for all $x$ strictly between 1 and 2. As $x$ gets closer to 1 from the right ($x \to 1^+$), $-2x$ gets closer to $-2(1) = -2$. Since $a$ must be strictly greater than $-2x$ for all $x$ in the interval, $a$ must be strictly greater than the supremum of $-2x$ in that interval. The supremum of $(-4, -2)$ is $-2$. Thus, $a > -2$.

The least value that $a$ can take while satisfying $a > -2$ is not a specific number, but the values approach -2. However, the question asks for the *least value* of $a$ such that $f(x)$ is strictly increasing. If $a > -2$, $f'(x) = 2x+a$. For $x \in (1,2)$, $2x \in (2, 4)$. So $f'(x) = 2x+a \in (2+a, 4+a)$. For $f'(x)$ to be strictly positive on $(1,2)$, we need the lower bound to be non-negative, and since the interval is open, strictly positive. So we need $2+a \geq 0$, which means $a \geq -2$. However, the derivative must be strictly positive on the open interval. If $a = -2$, $f'(x) = 2x - 2 = 2(x-1)$. For $x \in (1,2)$, $x > 1$, so $x-1 > 0$, which implies $f'(x) = 2(x-1) > 0$. So $a=-2$ makes the function strictly increasing on $(1,2)$. If $a < -2$, say $a = -2 - \epsilon$ for $\epsilon > 0$. Then $f'(x) = 2x - 2 - \epsilon$. As $x \to 1^+$, $f'(x) \to 2(1) - 2 - \epsilon = -\epsilon$. Since the limit is negative, there will be values of $x$ in $(1, 2)$ close to 1 for which $f'(x)$ is negative. Thus $a$ cannot be less than -2.

So, the condition $f'(x) > 0$ for all $x \in (1, 2)$ is equivalent to $\min_{x \in [1, 2]} f'(x) \geq 0$ and $f'(x) > 0$ at endpoints if included, but here the interval is open. The function $f'(x) = 2x+a$ is a linear function with a positive slope (2), so it is an increasing function of $x$. The minimum value of $f'(x)$ on the interval $(1, 2)$ is approached as $x$ approaches 1 from the right. The minimum value on the closed interval $[1, 2]$ occurs at $x=1$.

For $f(x)$ to be strictly increasing on $(1, 2)$, we need $f'(x) \geq 0$ for $x \in (1, 2)$ and $f'(x)$ cannot be identically zero on any subinterval.

If $f'(x) = 2x+a \geq 0$ for all $x \in (1, 2)$, then the minimum value of $2x+a$ for $x \in (1, 2)$ must be $\geq 0$. Since $2x+a$ is increasing in $x$, the minimum occurs as $x \to 1^+$. The value approaches $2(1)+a = 2+a$. So we need $2+a \geq 0$, which means $a \geq -2$.

If $a = -2$, then $f'(x) = 2x - 2$. For $x \in (1, 2)$, $x > 1$, so $x-1 > 0$, $f'(x) = 2(x-1) > 0$. Thus $f(x)$ is strictly increasing on $(1, 2)$.

If $a > -2$, then $f'(x) = 2x+a$. For $x \in (1, 2)$, $2x > 2$, so $f'(x) = 2x+a > 2+a$. Since $a > -2$, $2+a > 0$. So $f'(x) > 0$ on $(1, 2)$. Thus $f(x)$ is strictly increasing on $(1, 2)$.

The condition for strict increase on an open interval is often stated as $f'(x) > 0$. If $f'(x) = 0$ at isolated points, the function can still be strictly increasing. However, if $f'(x) \geq 0$ and $f'(x) = 0$ only at isolated points, the function is strictly increasing. If $f'(x) \geq 0$ and $f'(x)$ is zero on an entire subinterval, it is only increasing, not strictly increasing.

In this case, $f'(x) = 2x+a$. This is zero at most at one point $x = -a/2$. If this point is in $(1, 2)$ and $f'(-a/2) = 0$, but $f'(x)$ is positive or negative on either side, it can affect strict monotonicity. However, the requirement is $f'(x) > 0$ throughout $(1, 2)$. The minimum value of $2x+a$ on the interval is approached as $x \to 1^+$. This value is $2+a$. For $f'(x) > 0$ on $(1, 2)$, we need the minimum value on the closed interval $[1, 2]$ to be non-negative, and $f'(x)$ not identically zero. The minimum on $[1, 2]$ is $2(1)+a = 2+a$. So we need $2+a \geq 0$, or $a \geq -2$. If $a=-2$, $f'(x) = 2(x-1)$, which is 0 at $x=1$, but strictly positive on $(1, 2)$. If $a > -2$, $f'(x) = 2x+a$. At $x=1$, $f'(1) = 2+a > 0$. Since $f'(x)$ is increasing, $f'(x) > 0$ for all $x > 1$. So $a \geq -2$ ensures $f'(x) \geq 0$ for $x \geq 1$, and specifically $f'(x) > 0$ for $x > 1$ unless $2x+a$ is always zero (which is not possible here). Thus, the condition $a \geq -2$ ensures $f'(x) \geq 0$ on $[1, 2]$ and $f'(x) > 0$ on $(1, 2)$.

The smallest value of $a$ that satisfies $a \geq -2$ is $a = -2$.

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Final Answer:

The least value of $a$ is $-2$.

Given:

The function $f(x) = x + \frac{1}{x}$.

The interval $I$ is any interval disjoint from $[-1, 1]$.

This means the interval $I$ is either a subset of $(-\infty, -1)$ or a subset of $(1, \infty)$.

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To Prove:

The function $f(x)$ is increasing on $I$.

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Solution:

To prove that the function is increasing on the interval $I$, we need to find its derivative $f'(x)$ and show that $f'(x) \geq 0$ for all $x \in I$.

The function is given by:

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

Find the first derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(x + x^{-1})$

$f'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(x^{-1})$

$f'(x) = 1 \cdot x^{1-1} + (-1) \cdot x^{-1-1}$

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

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

To analyze the sign of $f'(x)$, we can write it as a single fraction:

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

Now, we need to examine the sign of $f'(x)$ on any interval $I$ that is disjoint from $[-1, 1]$. This means $I \subseteq (-\infty, -1)$ or $I \subseteq (1, \infty)$.

Case 1: $I \subseteq (-\infty, -1)$

If $x \in I$, then $x < -1$.

Consider the numerator $x^2 - 1$. If $x < -1$, then $x^2 > (-1)^2 = 1$. So, $x^2 - 1 > 0$.

Consider the denominator $x^2$. For any real number $x \neq 0$, $x^2 > 0$. Since $x < -1$, $x \neq 0$, so $x^2 > 0$.

Thus, for $x \in (-\infty, -1)$, $f'(x) = \frac{x^2 - 1}{x^2} = \frac{> 0}{> 0} > 0$.

Since $f'(x) > 0$ for all $x \in (-\infty, -1)$, the function $f(x)$ is strictly increasing on $(-\infty, -1)$. Any interval $I$ that is a subset of $(-\infty, -1)$ will also have $f'(x) > 0$, so $f(x)$ is strictly increasing (and thus increasing) on $I$.

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Case 2: $I \subseteq (1, \infty)$

If $x \in I$, then $x > 1$.

Consider the numerator $x^2 - 1$. If $x > 1$, then $x^2 > 1^2 = 1$. So, $x^2 - 1 > 0$.

Consider the denominator $x^2$. If $x > 1$, then $x \neq 0$, so $x^2 > 0$.

Thus, for $x \in (1, \infty)$, $f'(x) = \frac{x^2 - 1}{x^2} = \frac{> 0}{> 0} > 0$.

Since $f'(x) > 0$ for all $x \in (1, \infty)$, the function $f(x)$ is strictly increasing on $(1, \infty)$. Any interval $I$ that is a subset of $(1, \infty)$ will also have $f'(x) > 0$, so $f(x)$ is strictly increasing (and thus increasing) on $I$.

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In both cases, where $I$ is any interval disjoint from $[-1, 1]$, the derivative $f'(x) > 0$ for all $x \in I$.

Therefore, the function $f(x) = x + \frac{1}{x}$ is strictly increasing on any interval $I$ disjoint from $[-1, 1]$. A strictly increasing function is also increasing.

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Conclusion:

Since the derivative $f'(x) = 1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2}$ is positive for all $x$ such that $x^2 > 1$, i.e., for $x \in (-\infty, -1) \cup (1, \infty)$, and $I$ is a subset of this union, it means $f'(x) > 0$ for all $x \in I$. 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 the function requires $\sin x > 0$. This occurs in the intervals $(2n\pi, (2n+1)\pi)$ for integer $n$. We are considering the behaviour on $\left( 0, \frac{\pi}{2} \right)$ and $\left( \frac{\pi}{2}, \pi \right)$, which are subsets of $(0, \pi)$, where $\sin x > 0$.

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To Prove:

The function $f(x) = \log (\sin x)$ is:

(a) increasing on $\left( 0, \frac{\pi}{2} \right)$

(b) decreasing on $\left( \frac{\pi}{2}, \pi \right)$

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Solution:

To determine the intervals where the function is increasing or decreasing, we find the first derivative of the function $f(x)$ with respect to $x$ and then analyze the sign of the derivative on the given intervals.

The function is given by:

$f(x) = \log (\sin x)$

Let $u = \sin x$. Then $\frac{du}{dx} = \cos x$.

Using the chain rule, $\frac{d}{dx}(\log u) = \frac{1}{u} \frac{du}{dx}$.

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

$f'(x) = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x)$

$f'(x) = \frac{1}{\sin x} \cdot \cos x$

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

So, the derivative is $f'(x) = \cot x$.

Now, we analyze the sign of $f'(x) = \cot x$ on the given intervals $\left( 0, \frac{\pi}{2} \right)$ and $\left( \frac{\pi}{2}, \pi \right)$.

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(a) On the interval $\left( 0, \frac{\pi}{2} \right)$:

In the first quadrant, which corresponds to the interval $\left( 0, \frac{\pi}{2} \right)$, both $\sin x$ and $\cos x$ are positive.

The cotangent function $\cot x = \frac{\cos x}{\sin x}$ is the ratio of two positive numbers in this interval.

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

Since $f'(x) = \cot x > 0$ for all $x \in \left( 0, \frac{\pi}{2} \right)$, the function $f(x) = \log (\sin x)$ is strictly increasing on the interval $\left( 0, \frac{\pi}{2} \right)$. A strictly increasing function is also considered increasing.

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(b) On the interval $\left( \frac{\pi}{2}, \pi \right)$:

In the second quadrant, which corresponds to the interval $\left( \frac{\pi}{2}, \pi \right)$, $\sin x$ is positive and $\cos x$ is negative.

The cotangent function $\cot x = \frac{\cos x}{\sin x}$ is the ratio of a negative number to a positive number in this interval.

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

Since $f'(x) = \cot x < 0$ for all $x \in \left( \frac{\pi}{2}, \pi \right)$, the function $f(x) = \log (\sin x)$ is strictly decreasing on the interval $\left( \frac{\pi}{2}, \pi \right)$. A strictly decreasing function is also considered decreasing.

---

Hence, it is proven that 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 $f(x) = \log |\cos x|$.

We are considering the behaviour of the function on the intervals $\left( 0, \frac{\pi}{2} \right)$ and $\left( \frac{3\pi}{2}, 2 \pi \right)$. Note that for the function $\log |\cos x|$ to be defined, $|\cos x| > 0$, which means $\cos x \neq 0$. Both given intervals are within the domain of the function where $\cos x \neq 0$.

---

To Prove:

The function $f(x) = \log |\cos x|$ is:

(a) decreasing on $\left( 0, \frac{\pi}{2} \right)$

(b) increasing on $\left( \frac{3\pi}{2}, 2 \pi \right)$

---

Solution:

To determine the intervals where the function is increasing or decreasing, we find the first derivative of the function $f(x)$ with respect to $x$ and then analyze the sign of the derivative on the given intervals.

The function is given by:

$f(x) = \log |\cos x|$

To find the derivative of $f(x)$, we use the chain rule. Let $u = \cos x$. Then $\frac{du}{dx} = -\sin x$. The derivative of $\log |u|$ with respect to $u$ is $\frac{1}{u}$.

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

$f'(x) = \frac{1}{\cos x} \cdot \frac{d}{dx}(\cos x)$

$f'(x) = \frac{1}{\cos x} \cdot (-\sin x)$

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

Recall that $\frac{\sin x}{\cos x} = \tan x$. So, the derivative is:

$f'(x) = -\tan x$

Now, we analyze the sign of $f'(x) = -\tan x$ on the given intervals.

---

(a) On the interval $\left( 0, \frac{\pi}{2} \right)$:

This interval corresponds to the first quadrant.

For any $x$ in the interval $\left( 0, \frac{\pi}{2} \right)$, the value of $\tan x$ is positive ($\tan x > 0$).

For example, if $x = \frac{\pi}{4}$, $\tan(\frac{\pi}{4}) = 1 > 0$.

Since $\tan x > 0$ on $\left( 0, \frac{\pi}{2} \right)$, the derivative $f'(x) = -\tan x$ is negative on this interval.

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

Since $f'(x) < 0$ on $\left( 0, \frac{\pi}{2} \right)$, the function $f(x) = \log |\cos x|$ is strictly decreasing on this interval. A strictly decreasing function is also considered decreasing.

---

(b) On the interval $\left( \frac{3\pi}{2}, 2 \pi \right)$:

This interval corresponds to the fourth quadrant.

For any $x$ in the interval $\left( \frac{3\pi}{2}, 2 \pi \right)$, the value of $\tan x$ is negative ($\tan x < 0$).

For example, if $x = \frac{7\pi}{4}$, which is in this interval, $\tan(\frac{7\pi}{4}) = \tan(2\pi - \frac{\pi}{4}) = -\tan(\frac{\pi}{4}) = -1 < 0$.

Since $\tan x < 0$ on $\left( \frac{3\pi}{2}, 2 \pi \right)$, the derivative $f'(x) = -\tan x$ is positive on this interval.

$f'(x) = -(\text{negative}) = \text{positive}$.

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

Since $f'(x) > 0$ on $\left( \frac{3\pi}{2}, 2 \pi \right)$, the function $f(x) = \log |\cos x|$ is strictly increasing on this interval. A strictly increasing function is also considered increasing.

---

Hence, it is proven that 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 $f(x) = x^3 – 3x^2 + 3x – 100$.

The domain is $R$ (the set of real numbers).

---

To Prove:

The function $f(x)$ is increasing in $R$.

---

Solution:

To prove that the function is increasing in $R$, we need to find its first derivative $f'(x)$ and show that $f'(x) \geq 0$ for all $x \in R$. If we can show $f'(x) > 0$ except at isolated points where $f'(x) = 0$, then it's strictly increasing, which is a stronger condition than increasing.

The function is given by:

$f(x) = x^3 – 3x^2 + 3x – 100$

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

$f'(x) = \frac{d}{dx}(x^3 – 3x^2 + 3x – 100)$

$f'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(3x^2) + \frac{d}{dx}(3x) - \frac{d}{dx}(100)$

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

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

So, the derivative is $f'(x) = 3x^2 - 6x + 3$.

To analyze the sign of $f'(x)$, we can factor the expression:

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

The expression inside the parenthesis is a perfect square trinomial: $x^2 - 2x + 1 = (x - 1)^2$.

So, the derivative can be written as:

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

Now, we need to examine the sign of $f'(x) = 3(x - 1)^2$ for all $x \in R$.

Consider the term $(x - 1)^2$. The square of any real number is always greater than or equal to zero.

$(x - 1)^2 \geq 0$ for all $x \in R$.

The term $(x - 1)^2$ is equal to zero only when $x - 1 = 0$, which means $x = 1$. For all other real values of $x$, $(x - 1)^2 > 0$.

Now, consider $f'(x) = 3(x - 1)^2$. Since $3$ is a positive constant, the sign of $f'(x)$ is the same as the sign of $(x - 1)^2$.

So, $f'(x) \geq 0$ for all $x \in R$.

Specifically:

• If $x \neq 1$, then $(x - 1)^2 > 0$, so $f'(x) = 3(x - 1)^2 > 0$.
• If $x = 1$, then $(x - 1)^2 = 0$, so $f'(1) = 3(0) = 0$.

So, $f'(x) \geq 0$ for all $x \in R$, and $f'(x) = 0$ only at the isolated point $x = 1$.

According to the definition, if $f'(x) \geq 0$ on an interval, the function is increasing on that interval. Since this holds for all $x \in R$, the function is increasing on $R$. Furthermore, since $f'(x) = 0$ only at a single point, the function is actually strictly increasing on $R$. A strictly increasing function is also an increasing function.

---

Conclusion:

Since the derivative $f'(x) = 3(x - 1)^2 \geq 0$ for all $x \in R$, the function $f(x) = x^3 – 3x^2 + 3x – 100$ is increasing in $R$.

Given:

The function $y = x^2 e^{-x}$.

The domain is $R$ (the set of real numbers).

---

To Find:

The interval among the given options in which the function $y$ is increasing.

---

Solution:

To find the intervals where the function is increasing, we find the first derivative $\frac{dy}{dx}$ and determine where it is greater than or equal to zero ($\frac{dy}{dx} \geq 0$). For strictly increasing on an open interval, we check where $\frac{dy}{dx} > 0$.

The function is given by:

$y = x^2 e^{-x}$

Differentiate $y$ with respect to $x$ using the product rule, $\frac{d}{dx}(uv) = u'v + uv'$. Let $u = x^2$ and $v = e^{-x}$.

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

$v' = \frac{d}{dx}(e^{-x})$. Using the chain rule, let $w = -x$, so $\frac{dw}{dx} = -1$. $\frac{d}{dx}(e^w) = e^w \frac{dw}{dx} = e^{-x} (-1) = -e^{-x}$.

So, $\frac{dy}{dx} = u'v + uv'$:

$\frac{dy}{dx} = (2x)(e^{-x}) + (x^2)(-e^{-x})$

Factor out the common term $e^{-x}$:

$\frac{dy}{dx} = e^{-x}(2x - x^2)$

Factor the term in the parenthesis:

$\frac{dy}{dx} = e^{-x} x (2 - x)$

So, the derivative is $\frac{dy}{dx} = xe^{-x}(2 - x)$.

Now, we need to find the critical points where the derivative is zero or undefined. The derivative is defined for all real $x$. We set $\frac{dy}{dx} = 0$:

$xe^{-x}(2 - x) = 0$

This equation is zero if any of the factors are zero:

• $x = 0$
• $e^{-x} = 0$. The exponential function $e^{-x}$ is always positive ($e^{-x} > 0$) for all real $x$. So, this factor is never zero.
• $2 - x = 0 \implies x = 2$

The critical points are $x = 0$ and $x = 2$. These critical points divide the real number line into three open intervals: $(-\infty, 0)$, $(0, 2)$, and $(2, \infty)$. We analyze the sign of $\frac{dy}{dx} = xe^{-x}(2 - x)$ in each of these intervals. Remember that $e^{-x}$ is always positive.

The sign of $\frac{dy}{dx}$ is determined by the product of $x$ and $(2-x)$.

---

Interval 1: $(-\infty, 0)$

Choose a test value, e.g., $x = -1$.

$x = -1$ (negative)

$2 - x = 2 - (-1) = 3$ (positive)

$\frac{dy}{dx} = (\text{positive}) \times (\text{negative}) \times (\text{positive}) = \text{negative}$.

So, $\frac{dy}{dx} < 0$ on $(-\infty, 0)$. The function is strictly decreasing on $(-\infty, 0)$.

---

Interval 2: $(0, 2)$

Choose a test value, e.g., $x = 1$.

$x = 1$ (positive)

$2 - x = 2 - 1 = 1$ (positive)

$\frac{dy}{dx} = (\text{positive}) \times (\text{positive}) \times (\text{positive}) = \text{positive}$.

So, $\frac{dy}{dx} > 0$ on $(0, 2)$. The function is strictly increasing on $(0, 2)$.

---

Interval 3: $(2, \infty)$

Choose a test value, e.g., $x = 3$.

$x = 3$ (positive)

$2 - x = 2 - 3 = -1$ (negative)

$\frac{dy}{dx} = (\text{positive}) \times (\text{positive}) \times (\text{negative}) = \text{negative}$.

So, $\frac{dy}{dx} < 0$ on $(2, \infty)$. The function is strictly decreasing on $(2, \infty)$.

---

The function is increasing on the interval where $\frac{dy}{dx} \geq 0$. From the analysis of the open intervals, $\frac{dy}{dx} > 0$ on $(0, 2)$. At the critical points $x=0$ and $x=2$, $\frac{dy}{dx} = 0$. Since the function is continuous, it is increasing on the closed interval $[0, 2]$.

Now we compare this with the given options:

(A) $(– ∞, ∞)$: The function is not increasing on the entire real line as it decreases on $(-\infty, 0)$ and $(2, \infty)$.

(B) $(– 2, 0)$: This interval is a subset of $(-\infty, 0)$. On $(-\infty, 0)$, the function is strictly decreasing.

(C) $(2, ∞)$: On $(2, \infty)$, the function is strictly decreasing.

(D) $(0, 2)$: On $(0, 2)$, the function is strictly increasing.

The question asks for the interval where $y$ is increasing. The function is strictly increasing on $(0, 2)$. It is also increasing on the closed interval $[0, 2]$. The option (D) is $(0, 2)$, which matches one of the intervals where the function is strictly increasing.

---

Final Answer:

The interval in which $y = x^2 e^{-x}$ is increasing is $(0, 2)$.

The correct option is (D) $(0, 2)$.

Given:

The equation of the curve is $y = x^3 – x$.

We need to find the slope of the tangent at $x = 2$.

---

To Find:

The slope of the tangent to the curve $y = x^3 – x$ at $x = 2$.

---

Solution:

The slope of the tangent to a curve $y = f(x)$ at a point $(x_0, y_0)$ is given by the value of the derivative of the function at that point, i.e., $f'(x_0)$ or $\left(\frac{dy}{dx}\right)_{x=x_0}$.

We are given the curve $y = x^3 – x$.

First, find the derivative of $y$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(x^3 – x)$

Using the difference rule and the power rule:

$\frac{dy}{dx} = \frac{d}{dx}(x^3) - \frac{d}{dx}(x)$

$\frac{dy}{dx} = 3x^{3-1} - 1x^{1-1}$

$\frac{dy}{dx} = 3x^2 - 1$

Now, we need to find the slope of the tangent at $x = 2$. Substitute $x = 2$ into the expression for $\frac{dy}{dx}$:

Slope of the tangent at $x = 2$ = $\left(\frac{dy}{dx}\right)_{x=2} = 3(2)^2 - 1$

Calculate $(2)^2$:

$2^2 = 4$

Substitute this value back:

Slope = $3(4) - 1$

Slope = $12 - 1$

Slope = $11$

The slope of the tangent to the curve $y = x^3 – x$ at $x = 2$ is 11.

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Final Answer:

The slope of the tangent to the curve $y = x^3 – x$ at $x = 2$ is 11.

Given:

The equation of the curve is $y = \sqrt{4x-3} - 1$.

The slope of the tangent to the curve is given as $\frac{2}{3}$.

---

To Find:

The point $(x, y)$ on the curve where the slope of the tangent is $\frac{2}{3}$.

---

Solution:

The slope of the tangent to the curve $y = f(x)$ at any point $(x, y)$ is given by the derivative $\frac{dy}{dx}$.

We are given the curve $y = \sqrt{4x-3} - 1$. We can write $\sqrt{4x-3}$ as $(4x-3)^{1/2}$.

Find the derivative of $y$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}((4x-3)^{1/2} - 1)$

Using the difference rule, $\frac{dy}{dx} = \frac{d}{dx}((4x-3)^{1/2}) - \frac{d}{dx}(1)$.

The derivative of the constant term $-1$ is 0.

For the term $(4x-3)^{1/2}$, we use the chain rule. Let $u = 4x-3$. Then $\frac{du}{dx} = \frac{d}{dx}(4x-3) = 4$.

The derivative of $u^{1/2}$ with respect to $u$ is $\frac{1}{2} u^{1/2 - 1} = \frac{1}{2} u^{-1/2}$.

Applying the chain rule:

$\frac{d}{dx}((4x-3)^{1/2}) = \frac{1}{2} (4x-3)^{-1/2} \cdot \frac{d}{dx}(4x-3)$

$\frac{d}{dx}((4x-3)^{1/2}) = \frac{1}{2} (4x-3)^{-1/2} \cdot 4$

$\frac{d}{dx}((4x-3)^{1/2}) = 2 (4x-3)^{-1/2}$

$\frac{d}{dx}((4x-3)^{1/2}) = \frac{2}{\sqrt{4x-3}}$

So, the derivative of the function is:

$\frac{dy}{dx} = \frac{2}{\sqrt{4x-3}}$

We are given that the slope of the tangent is $\frac{2}{3}$. Set the derivative equal to this value:

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

Divide both sides by 2 (since $2 \neq 0$):

$\frac{1}{\sqrt{4x-3}} = \frac{1}{3}$

Taking the reciprocal of both sides:

$\sqrt{4x-3} = 3$

Square both sides of the equation to eliminate the square root:

$(\sqrt{4x-3})^2 = 3^2$

$4x-3 = 9$

Add 3 to both sides:

$4x = 9 + 3$

$4x = 12$

Divide by 4:

$x = \frac{12}{4}$

$x = 3$

This value of $x$ ($x=3$) is in the domain of the function and its derivative, as $4(3) - 3 = 12 - 3 = 9 \geq 0$.

Now, find the corresponding $y$-coordinate by substituting $x = 3$ into the original curve equation $y = \sqrt{4x-3} - 1$:

$y = \sqrt{4(3)-3} - 1$

$y = \sqrt{12-3} - 1$

$y = \sqrt{9} - 1$

$y = 3 - 1$

$y = 2$

The point at which the tangent to the curve has a slope of $\frac{2}{3}$ is $(3, 2)$.

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Final Answer:

The point is $(3, 2)$.

Given:

The equation of the curve is $y + \frac{2}{x-3}= 0$.

The slope of the tangent line is $m = 2$.

---

To Find:

The equation of all lines having a slope of 2 and being tangent to the given curve.

---

Solution:

First, rewrite the equation of the curve to express $y$ as a function of $x$:

$y = -\frac{2}{x-3}$

To find the slope of the tangent to the curve at any point $(x, y)$, we need to find the first derivative $\frac{dy}{dx}$.

We can write $y = -2(x-3)^{-1}$.

Differentiate $y$ with respect to $x$ using the chain rule:

$\frac{dy}{dx} = \frac{d}{dx}(-2(x-3)^{-1})$

Let $u = x-3$. Then $\frac{du}{dx} = 1$.

$\frac{dy}{dx} = -2 \cdot (-1)(x-3)^{-1-1} \cdot \frac{d}{dx}(x-3)$

$\frac{dy}{dx} = 2(x-3)^{-2} \cdot 1$

$\frac{dy}{dx} = \frac{2}{(x-3)^2}$

We are given that the slope of the tangent is 2. So, we set the derivative equal to 2:

$\frac{2}{(x-3)^2} = 2$

Divide both sides by 2:

$\frac{1}{(x-3)^2} = 1$

$(x-3)^2 = 1$

Take the square root of both sides:

$x-3 = \pm \sqrt{1}$

$x-3 = \pm 1$

This gives two possible values for the x-coordinates of the points of tangency:

Case 1: $x-3 = 1$

$x = 1 + 3 = 4$

Case 2: $x-3 = -1$

$x = -1 + 3 = 2$

Now, find the corresponding y-coordinates for each x-value by substituting them back into the original curve equation $y = -\frac{2}{x-3}$.

For $x = 4$:

$y = -\frac{2}{4-3} = -\frac{2}{1} = -2$

The first point of tangency is $(4, -2)$.

For $x = 2$:

$y = -\frac{2}{2-3} = -\frac{2}{-1} = 2$

The second point of tangency is $(2, 2)$.

Now, we find the equation of the tangent line for each point using the point-slope form of a linear equation: $y - y_1 = m(x - x_1)$, where $m=2$.

Equation of the first tangent line (at $(4, -2)$ with $m=2$):

$y - (-2) = 2(x - 4)$

$y + 2 = 2x - 8$

$y = 2x - 8 - 2$

$y = 2x - 10$

Equation of the second tangent line (at $(2, 2)$ with $m=2$):

$y - 2 = 2(x - 2)$

$y - 2 = 2x - 4$

$y = 2x - 4 + 2$

$y = 2x - 2$

There are two lines tangent to the curve with a slope of 2.

---

Final Answer:

The equations of the tangent lines are $y = 2x - 10$ and $y = 2x - 2$.

These can also be written as $2x - y - 10 = 0$ and $2x - y - 2 = 0$.

Given:

The equation of the curve is $\frac{x^{2}}{4} + \frac{y^{2}}{25} = 1$. This is the equation of an ellipse.

---

To Find:

The points on the curve where the tangents are:

(i) parallel to the x-axis

(ii) parallel to the y-axis

---

Solution:

To find the slope of the tangent at any point $(x, y)$ on the curve, we need to find the derivative $\frac{dy}{dx}$ using implicit differentiation.

The equation of the curve is:

$\frac{x^{2}}{4} + \frac{y^{2}}{25} = 1$

Differentiate both sides of the equation with respect to $x$:

$\frac{d}{dx}\left(\frac{x^{2}}{4} + \frac{y^{2}}{25}\right) = \frac{d}{dx}(1)$

$\frac{1}{4} \frac{d}{dx}(x^{2}) + \frac{1}{25} \frac{d}{dx}(y^{2}) = 0$

$\frac{1}{4} (2x) + \frac{1}{25} (2y \frac{dy}{dx}) = 0$

$\frac{x}{2} + \frac{2y}{25} \frac{dy}{dx} = 0$

Now, solve for $\frac{dy}{dx}$:

$\frac{2y}{25} \frac{dy}{dx} = -\frac{x}{2}$

$\frac{dy}{dx} = -\frac{x}{2} \cdot \frac{25}{2y}$

$\frac{dy}{dx} = -\frac{25x}{4y}$

This is the general expression for the slope of the tangent at any point $(x, y)$ on the curve, provided $y \neq 0$.

---

(i) Tangents parallel to the x-axis:

A line parallel to the x-axis has a slope of 0. So, we set $\frac{dy}{dx} = 0$:

$-\frac{25x}{4y} = 0$

For this fraction to be zero, the numerator must be zero (assuming the denominator is not zero):

$-25x = 0$

$x = 0$

Now, find the corresponding y-coordinates by substituting $x = 0$ back into the original curve equation $\frac{x^{2}}{4} + \frac{y^{2}}{25} = 1$:

$\frac{0^{2}}{4} + \frac{y^{2}}{25} = 1$

$0 + \frac{y^{2}}{25} = 1$

$\frac{y^{2}}{25} = 1$

$y^{2} = 25$

$y = \pm \sqrt{25}$

$y = \pm 5$

The points where the tangent is parallel to the x-axis are $(0, 5)$ and $(0, -5)$.

At these points, $y = \pm 5 \neq 0$, so our assumption that the denominator $4y$ is not zero was valid.

---

(ii) Tangents parallel to the y-axis:

A line parallel to the y-axis has an infinite slope. This occurs when the denominator of the derivative $\frac{dy}{dx}$ is zero, provided the numerator is non-zero at that point.

We set the denominator of $\frac{dy}{dx} = -\frac{25x}{4y}$ to zero:

$4y = 0$

$y = 0$

Now, find the corresponding x-coordinates by substituting $y = 0$ back into the original curve equation $\frac{x^{2}}{4} + \frac{y^{2}}{25} = 1$:

$\frac{x^{2}}{4} + \frac{0^{2}}{25} = 1$

$\frac{x^{2}}{4} + 0 = 1$

$\frac{x^{2}}{4} = 1$

$x^{2} = 4$

$x = \pm \sqrt{4}$

$x = \pm 2$

The points where the tangent's slope is undefined are $(2, 0)$ and $(-2, 0)$.

At these points, the numerator of $\frac{dy}{dx}$ is $-25x$, which is $-25(\pm 2) = \mp 50 \neq 0$. Since the numerator is non-zero and the denominator is zero, the slope is indeed infinite (vertical tangent), meaning the tangent is parallel to the y-axis.

---

Final Answer:

The points on the curve $\frac{x^{2}}{4} + \frac{y^{2}}{25} = 1$ are:

(i) parallel to the x-axis at $(0, 5)$ and $(0, -5)$.

(ii) parallel to the y-axis at $(2, 0)$ and $(-2, 0)$.

Given:

The equation of the curve is $y = \frac{x - 7}{(x - 2) (x - 3)}$.

We need to find the equation of the tangent at the point where the curve cuts the x-axis.

---

To Find:

The equation of the tangent line to the curve $y = \frac{x - 7}{(x - 2) (x - 3)}$ at its x-intercept.

---

Solution:

First, we need to find the coordinates of the point where the curve cuts the x-axis. A curve cuts the x-axis when the y-coordinate is zero.

Set $y = 0$ in the equation of the curve:

$0 = \frac{x - 7}{(x - 2) (x - 3)}$

For a fraction to be zero, the numerator must be zero, provided the denominator is not zero.

Numerator: $x - 7 = 0$

$x = 7$

Check the denominator at $x = 7$: $(7 - 2)(7 - 3) = (5)(4) = 20$. Since the denominator is not zero, $x = 7$ is a valid solution.

So, the curve cuts the x-axis at the point where $x = 7$. The corresponding y-coordinate is $y=0$.

The point of tangency is $(x_1, y_1) = (7, 0)$.

Next, we need to find the slope of the tangent line at this point. The slope of the tangent is given by the value of the derivative $\frac{dy}{dx}$ evaluated at $x = 7$.

The function is $y = \frac{x - 7}{(x - 2) (x - 3)}$.

We can rewrite the denominator by expanding it:

$(x - 2) (x - 3) = x^2 - 3x - 2x + 6 = x^2 - 5x + 6$

So, $y = \frac{x - 7}{x^2 - 5x + 6}$.

Now, we find the derivative $\frac{dy}{dx}$ using the quotient rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$.

Let $u = x - 7$ and $v = x^2 - 5x + 6$.

The derivative of $u$ is $u' = \frac{d}{dx}(x - 7) = 1$.

The derivative of $v$ is $v' = \frac{d}{dx}(x^2 - 5x + 6) = 2x - 5$.

Apply the quotient rule:

$\frac{dy}{dx} = \frac{(1)(x^2 - 5x + 6) - (x - 7)(2x - 5)}{(x^2 - 5x + 6)^2}$

Expand the terms in the numerator:

$(x - 7)(2x - 5) = x(2x) + x(-5) - 7(2x) - 7(-5) = 2x^2 - 5x - 14x + 35 = 2x^2 - 19x + 35$

Substitute this back into the numerator:

Numerator = $(x^2 - 5x + 6) - (2x^2 - 19x + 35)$

Numerator = $x^2 - 5x + 6 - 2x^2 + 19x - 35$

Numerator = $-x^2 + 14x - 29$

So, the derivative is:

$\frac{dy}{dx} = \frac{-x^2 + 14x - 29}{(x^2 - 5x + 6)^2}$

Now, evaluate the slope of the tangent at the point where $x = 7$:

Slope $m = \left(\frac{dy}{dx}\right)_{x=7} = \frac{-(7)^2 + 14(7) - 29}{((7)^2 - 5(7) + 6)^2}$

$m = \frac{-49 + 98 - 29}{(49 - 35 + 6)^2}$

$m = \frac{49 - 29}{(14 + 6)^2}$

$m = \frac{20}{(20)^2}$

$m = \frac{20}{400}$

Simplify the fraction:

$m = \frac{\cancel{20}^1}{\cancel{400}_{20}}$

$m = \frac{1}{20}$

The slope of the tangent at the point $(7, 0)$ is $\frac{1}{20}$.

Finally, we find the equation of the tangent line using the point-slope form: $y - y_1 = m(x - x_1)$.

Substitute $(x_1, y_1) = (7, 0)$ and $m = \frac{1}{20}$:

$y - 0 = \frac{1}{20}(x - 7)$

$y = \frac{1}{20}(x - 7)$

Multiply both sides by 20 to clear the fraction:

$20y = x - 7$

Rearrange the equation to the standard form $Ax + By + C = 0$:

$x - 20y - 7 = 0$

---

Final Answer:

The equation of the tangent to the curve at the point where it cuts the x-axis is $y = \frac{1}{20}(x - 7)$, or $x - 20y - 7 = 0$.

Given:

The equation of the curve is $x^{\frac{2}{3}} + y^{\frac{2}{3}} = 2$.

The point of tangency is $(1, 1)$. We can verify that this point is on the curve: $1^{\frac{2}{3}} + 1^{\frac{2}{3}} = 1 + 1 = 2$, which satisfies the equation.

---

To Find:

The equations of the tangent line and the normal line to the curve at the point $(1, 1)$.

---

Solution:

First, we find the slope of the tangent to the curve at any point $(x, y)$ by finding the derivative $\frac{dy}{dx}$ using implicit differentiation.

The equation of the curve is:

$x^{\frac{2}{3}} + y^{\frac{2}{3}} = 2$

Differentiate both sides of the equation with respect to $x$:

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

$\frac{d}{dx}(x^{\frac{2}{3}}) + \frac{d}{dx}(y^{\frac{2}{3}}) = 0$

Using the power rule and the chain rule for $y^{\frac{2}{3}}$:

$\frac{2}{3} x^{\frac{2}{3}-1} + \frac{2}{3} y^{\frac{2}{3}-1} \frac{dy}{dx} = 0$

$\frac{2}{3} x^{-\frac{1}{3}} + \frac{2}{3} y^{-\frac{1}{3}} \frac{dy}{dx} = 0$

$\frac{2}{3x^{\frac{1}{3}}} + \frac{2}{3y^{\frac{1}{3}}} \frac{dy}{dx} = 0$

Multiply the entire equation by $\frac{3}{2}$ to simplify:

$\frac{1}{x^{\frac{1}{3}}} + \frac{1}{y^{\frac{1}{3}}} \frac{dy}{dx} = 0$

Solve for $\frac{dy}{dx}$:

$\frac{1}{y^{\frac{1}{3}}} \frac{dy}{dx} = -\frac{1}{x^{\frac{1}{3}}}$

$\frac{dy}{dx} = -\frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}}}$

$\frac{dy}{dx} = -\left(\frac{y}{x}\right)^{\frac{1}{3}}$

This is the general expression for the slope of the tangent at any point $(x, y)$ on the curve, provided $x \neq 0$ and $y \neq 0$.

Now, evaluate the slope of the tangent at the given point $(1, 1)$:

Slope of the tangent $m_t = \left(\frac{dy}{dx}\right)_{(1,1)} = -\left(\frac{1}{1}\right)^{\frac{1}{3}}$

$m_t = -(1)^{\frac{1}{3}} = -1$

The slope of the tangent at $(1, 1)$ is $-1$.

The equation of the tangent line at the point $(x_1, y_1) = (1, 1)$ with slope $m_t = -1$ can be found using the point-slope form: $y - y_1 = m_t(x - x_1)$.

$y - 1 = -1(x - 1)$

$y - 1 = -x + 1$

$y = -x + 1 + 1$

$y = -x + 2$

This can also be written as $x + y - 2 = 0$.

---

Next, find the equation of the normal line. The normal line to the curve at a point is perpendicular to the tangent line at that point.

The slope of the normal line, $m_n$, is the negative reciprocal of the slope of the tangent line, $m_t$, provided $m_t \neq 0$ and $m_t$ is finite.

$m_n = -\frac{1}{m_t}$

Since $m_t = -1$, the slope of the normal is:

$m_n = -\frac{1}{-1} = 1$

The equation of the normal line at the point $(x_1, y_1) = (1, 1)$ with slope $m_n = 1$ can be found using the point-slope form: $y - y_1 = m_n(x - x_1)$.

$y - 1 = 1(x - 1)$

$y - 1 = x - 1$

$y = x - 1 + 1$

$y = x$

This can also be written as $x - y = 0$.

---

Final Answer:

The equation of the tangent to the curve $x^{\frac{2}{3}} + y^{\frac{2}{3}} = 2$ at $(1, 1)$ is $y = -x + 2$ or $x + y - 2 = 0$.

The equation of the normal to the curve at $(1, 1)$ is $y = x$ or $x - y = 0$.

Given:

The parametric equations of the curve are:

We need to find the equation of the tangent at the point where $t = \frac{\pi}{2}$.

---

To Find:

The equation of the tangent to the curve at $t = \frac{\pi}{2}$.

---

Solution:

First, we find the coordinates of the point of tangency by substituting $t = \frac{\pi}{2}$ into the given parametric equations.

Substitute $t = \frac{\pi}{2}$ into Equation (1):

$x = a \sin^3 \left(\frac{\pi}{2}\right)$

Since $\sin \left(\frac{\pi}{2}\right) = 1$, we have:

$x = a (1)^3 = a$

Substitute $t = \frac{\pi}{2}$ into Equation (2):

$y = b \cos^3 \left(\frac{\pi}{2}\right)$

Since $\cos \left(\frac{\pi}{2}\right) = 0$, we have:

$y = b (0)^3 = 0$

The point of tangency is $(x_1, y_1) = (a, 0)$.

Next, we find the slope of the tangent to the curve at any point. For a parametric curve, the slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.

Differentiate $x$ with respect to $t$:

$\frac{dx}{dt} = \frac{d}{dt}(a \sin^3 t)$

Using the constant multiple rule and the chain rule ($u = \sin t$, $\frac{du}{dt} = \cos t$):

$\frac{dx}{dt} = a \cdot 3 \sin^{3-1} t \cdot \frac{d}{dt}(\sin t)$

$\frac{dx}{dt} = 3a \sin^2 t \cos t$

Differentiate $y$ with respect to $t$:

$\frac{dy}{dt} = \frac{d}{dt}(b \cos^3 t)$

Using the constant multiple rule and the chain rule ($v = \cos t$, $\frac{dv}{dt} = -\sin t$):

$\frac{dy}{dt} = b \cdot 3 \cos^{3-1} t \cdot \frac{d}{dt}(\cos t)$

$\frac{dy}{dt} = 3b \cos^2 t (-\sin t)$

$\frac{dy}{dt} = -3b \cos^2 t \sin t$

Now, find the slope $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-3b \cos^2 t \sin t}{3a \sin^2 t \cos t}$

Simplify the expression (for $\cos t \neq 0$ and $\sin t \neq 0$):

$\frac{dy}{dx} = -\frac{3b}{3a} \frac{\cos^2 t \sin t}{\sin^2 t \cos t} = -\frac{b}{a} \frac{\cos t}{\sin t} = -\frac{b}{a} \cot t$

Now, evaluate the slope at $t = \frac{\pi}{2}$. We need to find the limit as $t \to \frac{\pi}{2}$ if the direct substitution results in an indeterminate form, or evaluate $\frac{dy}{dt}$ and $\frac{dx}{dt}$ at $t=\frac{\pi}{2}$.

At $t = \frac{\pi}{2}$, $\cos t = 0$ and $\sin t = 1$.

$\left(\frac{dx}{dt}\right)_{t=\frac{\pi}{2}} = 3a \sin^2 \left(\frac{\pi}{2}\right) \cos \left(\frac{\pi}{2}\right) = 3a (1)^2 (0) = 0$

$\left(\frac{dy}{dt}\right)_{t=\frac{\pi}{2}} = -3b \cos^2 \left(\frac{\pi}{2}\right) \sin \left(\frac{\pi}{2}\right) = -3b (0)^2 (1) = 0$

Since both $\frac{dx}{dt}$ and $\frac{dy}{dt}$ are 0 at $t=\frac{\pi}{2}$, we consider the limit of the slope expression:

Slope $m = \lim_{t \to \frac{\pi}{2}} -\frac{b}{a} \cot t = \lim_{t \to \frac{\pi}{2}} -\frac{b}{a} \frac{\cos t}{\sin t}$

As $t \to \frac{\pi}{2}$, $\cos t \to 0$ and $\sin t \to 1$.

$m = -\frac{b}{a} \frac{0}{1} = 0$

The slope of the tangent at $t = \frac{\pi}{2}$ is $m = 0$.

The equation of the tangent line at the point $(x_1, y_1) = (a, 0)$ with slope $m = 0$ can be found using the point-slope form: $y - y_1 = m(x - x_1)$.

$y - 0 = 0(x - a)$

$y = 0$

This is the equation of the tangent line, which is a horizontal line.

---

Final Answer:

The equation of the tangent to the curve at the point where $t = \frac{\pi}{2}$ is $y = 0$.

Given:

The equation of the curve is $y = 3x^4 – 4x$.

We need to find the slope of the tangent at $x = 4$.

---

To Find:

The slope of the tangent to the curve $y = 3x^4 – 4x$ at $x = 4$.

---

Solution:

The slope of the tangent to a curve $y = f(x)$ at a point specified by its x-coordinate $x_0$ is given by the value of the derivative $\frac{dy}{dx}$ evaluated at $x = x_0$.

We are given the curve $y = 3x^4 – 4x$.

First, find the derivative of $y$ with respect to $x$. Using the difference rule, constant multiple rule, and power rule for differentiation:

$\frac{dy}{dx} = \frac{d}{dx}(3x^4 – 4x)$

$\frac{dy}{dx} = \frac{d}{dx}(3x^4) - \frac{d}{dx}(4x)$

$\frac{dy}{dx} = 3 \cdot (4x^{4-1}) - 4 \cdot (1x^{1-1})$

$\frac{dy}{dx} = 12x^3 - 4$

Now, we need to find the slope of the tangent at $x = 4$. Substitute $x = 4$ into the expression for $\frac{dy}{dx}$:

Slope of the tangent at $x = 4$ = $\left(\frac{dy}{dx}\right)_{x=4} = 12(4)^3 - 4$

Calculate $(4)^3$:

$4^3 = 4 \times 4 \times 4 = 16 \times 4 = 64$

Substitute this value back:

Slope = $12(64) - 4$

Perform the multiplication $12 \times 64$:

$\begin{array}{cc}& & 6 & 4 \\ \times & & 1 & 2 \\ \hline & 1 & 2 & 8 \\ 6 & 4 & \times & \\ \hline 7 & 6 & 8 \\ \hline \end{array}$

$12 \times 64 = 768$

Substitute this value back:

Slope = $768 - 4$

Slope = $764$

The slope of the tangent to the curve $y = 3x^4 – 4x$ at $x = 4$ is 764.

---

Final Answer:

The slope of the tangent to the curve at $x = 4$ is 764.

Given:

The equation of the curve is $y = \frac{x - 1}{x - 2}$, where $x \neq 2$.

We need to find the slope of the tangent at $x = 10$.

---

To Find:

The slope of the tangent to the curve $y = \frac{x - 1}{x - 2}$ at $x = 10$.

---

Solution:

The slope of the tangent to a curve $y = f(x)$ at a specific x-value $x_0$ is given by the value of the derivative $\frac{dy}{dx}$ evaluated at $x = x_0$.

We are given the curve $y = \frac{x - 1}{x - 2}$.

First, find the derivative of $y$ with respect to $x$. We use the quotient rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$.

Let $u = x - 1$ and $v = x - 2$.

The derivative of $u$ is $u' = \frac{d}{dx}(x - 1) = 1$.

The derivative of $v$ is $v' = \frac{d}{dx}(x - 2) = 1$.

Apply the quotient rule:

$\frac{dy}{dx} = \frac{(1)(x - 2) - (x - 1)(1)}{(x - 2)^2}$

Simplify the numerator:

Numerator = $(x - 2) - (x - 1)$

Numerator = $x - 2 - x + 1$

Numerator = $-1$

So, the derivative is:

$\frac{dy}{dx} = \frac{-1}{(x - 2)^2}$

Now, we need to find the slope of the tangent at $x = 10$. Substitute $x = 10$ into the expression for $\frac{dy}{dx}$:

Slope of the tangent at $x = 10$ = $\left(\frac{dy}{dx}\right)_{x=10} = \frac{-1}{(10 - 2)^2}$

$\left(\frac{dy}{dx}\right)_{x=10} = \frac{-1}{(8)^2}$

Calculate $(8)^2$:

$8^2 = 64$

Substitute this value back:

Slope = $\frac{-1}{64}$

The slope of the tangent to the curve $y = \frac{x - 1}{x - 2}$ at $x = 10$ is $-\frac{1}{64}$.

---

Final Answer:

The slope of the tangent to the curve at $x = 10$ is $-\frac{1}{64}$.

Given:

The equation of the curve is $y = x^3 – x + 1$.

We need to find the slope of the tangent at the point where the x-coordinate is 2, i.e., at $x = 2$.

---

To Find:

The slope of the tangent to the curve $y = x^3 – x + 1$ at $x = 2$.

---

Solution:

The slope of the tangent to a curve $y = f(x)$ at a point specified by its x-coordinate $x_0$ is given by the value of the derivative $\frac{dy}{dx}$ evaluated at $x = x_0$.

We are given the curve $y = x^3 – x + 1$.

First, find the derivative of $y$ with respect to $x$. Using the sum/difference rule and the power rule for differentiation:

$\frac{dy}{dx} = \frac{d}{dx}(x^3 – x + 1)$

$\frac{dy}{dx} = \frac{d}{dx}(x^3) - \frac{d}{dx}(x) + \frac{d}{dx}(1)$

$\frac{dy}{dx} = 3x^{3-1} - 1x^{1-1} + 0$

$\frac{dy}{dx} = 3x^2 - 1$

Now, we need to find the slope of the tangent at $x = 2$. Substitute $x = 2$ into the expression for $\frac{dy}{dx}$:

Slope of the tangent at $x = 2$ = $\left(\frac{dy}{dx}\right)_{x=2} = 3(2)^2 - 1$

Calculate $(2)^2$:

$2^2 = 4$

Substitute this value back:

Slope = $3(4) - 1$

Slope = $12 - 1$

Slope = $11$

The slope of the tangent to the curve $y = x^3 – x + 1$ at the point whose x-coordinate is 2 is 11.

---

Final Answer:

The slope of the tangent to the curve at the point whose x-coordinate is 2 is 11.

Given:

The equation of the curve is $y = x^3 – 3x + 2$.

We need to find the slope of the tangent at the point where the x-coordinate is 3, i.e., at $x = 3$.

---

To Find:

The slope of the tangent to the curve $y = x^3 – 3x + 2$ at $x = 3$.

---

Solution:

The slope of the tangent to a curve $y = f(x)$ at a point specified by its x-coordinate $x_0$ is given by the value of the derivative $\frac{dy}{dx}$ evaluated at $x = x_0$.

We are given the curve $y = x^3 – 3x + 2$.

First, find the derivative of $y$ with respect to $x$. Using the sum/difference rule and the power rule for differentiation:

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

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

$\frac{dy}{dx} = 3x^{3-1} - 3(1x^{1-1}) + 0$

$\frac{dy}{dx} = 3x^2 - 3$

Now, we need to find the slope of the tangent at $x = 3$. Substitute $x = 3$ into the expression for $\frac{dy}{dx}$:

Slope of the tangent at $x = 3$ = $\left(\frac{dy}{dx}\right)_{x=3} = 3(3)^2 - 3$

Calculate $(3)^2$:

$3^2 = 9$

Substitute this value back:

Slope = $3(9) - 3$

Slope = $27 - 3$

Slope = $24$

The slope of the tangent to the curve $y = x^3 – 3x + 2$ at the point whose x-coordinate is 3 is 24.

---

Final Answer:

The slope of the tangent to the curve at the point whose x-coordinate is 3 is 24.

Given:

The parametric equations of the curve are:

$x = a \cos^3 \theta$

$y = a \sin^3 \theta$

We need to find the slope of the normal to the curve at $\theta = \frac{\pi}{4}$.

---

To Find:

The slope of the normal to the curve at $\theta = \frac{\pi}{4}$.

---

Solution:

First, we find the slope of the tangent to the curve at any point. For a parametric curve, the slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$.

Differentiate $x$ with respect to $\theta$:

$\frac{dx}{d\theta} = \frac{d}{d\theta}(a \cos^3 \theta)$

Using the constant multiple rule and the chain rule ($u = \cos \theta$, $\frac{du}{d\theta} = -\sin \theta$):

$\frac{dx}{d\theta} = a \cdot 3 \cos^{3-1} \theta \cdot \frac{d}{d\theta}(\cos \theta)$

$\frac{dx}{d\theta} = 3a \cos^2 \theta (-\sin \theta)$

$\frac{dx}{d\theta} = -3a \cos^2 \theta \sin \theta$

Differentiate $y$ with respect to $\theta$:

$\frac{dy}{d\theta} = \frac{d}{d\theta}(a \sin^3 \theta)$

Using the constant multiple rule and the chain rule ($v = \sin \theta$, $\frac{dv}{d\theta} = \cos \theta$):

$\frac{dy}{d\theta} = a \cdot 3 \sin^{3-1} \theta \cdot \frac{d}{d\theta}(\sin \theta)$

$\frac{dy}{d\theta} = 3a \sin^2 \theta \cos \theta$

Now, find the slope of the tangent $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta}$

Simplify the expression (for $\sin \theta \neq 0$ and $\cos \theta \neq 0$):

$\frac{dy}{dx} = -\frac{3a}{3a} \frac{\sin^2 \theta \cos \theta}{\cos^2 \theta \sin \theta} = -1 \cdot \frac{\sin \theta}{\cos \theta} = -\tan \theta$

This is the slope of the tangent at any parameter value $\theta$ (where $\sin \theta \neq 0$ and $\cos \theta \neq 0$).

Now, evaluate the slope of the tangent at $\theta = \frac{\pi}{4}$:

Slope of the tangent $m_t = \left(\frac{dy}{dx}\right)_{\theta=\frac{\pi}{4}} = -\tan \left(\frac{\pi}{4}\right)$

Since $\tan \left(\frac{\pi}{4}\right) = 1$, the slope of the tangent is:

$m_t = -1$

The slope of the normal line, $m_n$, is the negative reciprocal of the slope of the tangent line, $m_t$, provided $m_t \neq 0$ and is finite.

$m_n = -\frac{1}{m_t}$

Since $m_t = -1$, the slope of the normal is:

$m_n = -\frac{1}{-1} = 1$

The slope of the normal to the curve at $\theta = \frac{\pi}{4}$ is 1.

Note: At $\theta = \frac{\pi}{4}$, $\sin \theta = \frac{1}{\sqrt{2}}$ and $\cos \theta = \frac{1}{\sqrt{2}}$, which are non-zero. The values $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$ are also non-zero at this point: $\frac{dx}{d\theta} = -3a (\frac{1}{\sqrt{2}})^2 (\frac{1}{\sqrt{2}}) = -3a \frac{1}{2\sqrt{2}} \neq 0$ and $\frac{dy}{d\theta} = 3a (\frac{1}{\sqrt{2}})^2 (\frac{1}{\sqrt{2}}) = 3a \frac{1}{2\sqrt{2}} \neq 0$. So the slope is well-defined.

---

Final Answer:

The slope of the normal to the curve at $\theta = \frac{\pi}{4}$ is 1.

Given:

The parametric equations of the curve are:

$x = 1 - a \sin \theta$

$y = b \cos^2 \theta$

We need to find the slope of the normal to this curve at the point where $\theta = \frac{\pi}{2}$.

---

Solution:

First, we find the derivatives of $x$ and $y$ with respect to $\theta$.

Differentiating $x$ with respect to $\theta$:

$\frac{dx}{d\theta} = \frac{d}{d\theta} (1 - a \sin \theta)$

$\frac{dx}{d\theta} = 0 - a (\cos \theta)$

$\frac{dx}{d\theta} = -a \cos \theta$

Differentiating $y$ with respect to $\theta$:

$\frac{dy}{d\theta} = \frac{d}{d\theta} (b \cos^2 \theta)$

Using the chain rule:

$\frac{dy}{d\theta} = b \cdot (2 \cos \theta) \cdot \frac{d}{d\theta}(\cos \theta)$

$\frac{dy}{d\theta} = 2b \cos \theta \cdot (-\sin \theta)$

$\frac{dy}{d\theta} = -2b \sin \theta \cos \theta$

---

The slope of the tangent to the curve is given by $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$.

$\frac{dy}{dx} = \frac{-2b \sin \theta \cos \theta}{-a \cos \theta}$

Assuming $\cos \theta \neq 0$, we can simplify:

$\frac{dy}{dx} = \frac{2b \sin \theta}{a}$

---

Now, we need to find the slope of the tangent at $\theta = \frac{\pi}{2}$.

Let's evaluate the derivatives at $\theta = \frac{\pi}{2}$:

$\frac{dx}{d\theta}\Big|_{\theta=\pi/2} = -a \cos\left(\frac{\pi}{2}\right) = -a(0) = 0$

$\frac{dy}{d\theta}\Big|_{\theta=\pi/2} = -2b \sin\left(\frac{\pi}{2}\right) \cos\left(\frac{\pi}{2}\right) = -2b(1)(0) = 0$

Since both derivatives are zero, we have an indeterminate form $\frac{0}{0}$. We evaluate the limit of the simplified expression for $\frac{dy}{dx}$ as $\theta \to \frac{\pi}{2}$ (or use the simplified expression directly as the cancellation of $\cos \theta$ represents the limit process):

Slope of the tangent at $\theta = \frac{\pi}{2}$ is:

$m_{tangent} = \frac{dy}{dx}\Big|_{\theta=\pi/2} = \frac{2b \sin(\pi/2)}{a}$

$m_{tangent} = \frac{2b(1)}{a} = \frac{2b}{a}$

---

The slope of the normal is the negative reciprocal of the slope of the tangent.

Slope of the normal, $m_{normal} = -\frac{1}{m_{tangent}}$

$m_{normal} = -\frac{1}{\left(\frac{2b}{a}\right)}$

$m_{normal} = -\frac{a}{2b}$

---

Therefore, the slope of the normal to the curve $x = 1 - a \sin \theta$, $y = b \cos^2 \theta$ at $\theta = \frac{\pi}{2}$ is $-\frac{a}{2b}$ (assuming $a \neq 0$ and $b \neq 0$).

Given:

The equation of the curve is $y = x^3 - 3x^2 - 9x + 7$.

---

To Find:

The points on the curve where the tangent is parallel to the x-axis.

---

Solution:

The slope of the tangent to the curve at any point $(x, y)$ is given by the derivative $\frac{dy}{dx}$.

First, differentiate the equation of the curve with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(x^3 - 3x^2 - 9x + 7)$

$\frac{dy}{dx} = 3x^2 - 3(2x) - 9(1) + 0$

$\frac{dy}{dx} = 3x^2 - 6x - 9$

---

A tangent line is parallel to the x-axis if its slope is zero.

Therefore, we set $\frac{dy}{dx} = 0$:

$3x^2 - 6x - 9 = 0$

Divide the equation by 3:

$x^2 - 2x - 3 = 0$

Factor the quadratic equation:

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

$x(x - 3) + 1(x - 3) = 0$

$(x - 3)(x + 1) = 0$

This gives us two possible values for $x$: $x = 3$ or $x = -1$.

---

Now, we find the corresponding $y$-coordinates for these values of $x$ using the original equation of the curve, $y = x^3 - 3x^2 - 9x + 7$.

Case 1: When $x = 3$

$y = (3)^3 - 3(3)^2 - 9(3) + 7$

$y = 27 - 3(9) - 27 + 7$

$y = 27 - 27 - 27 + 7$

$y = -20$

So, one point is (3, -20).

Case 2: When $x = -1$

$y = (-1)^3 - 3(-1)^2 - 9(-1) + 7$

$y = -1 - 3(1) + 9 + 7$

$y = -1 - 3 + 9 + 7$

$y = -4 + 16$

$y = 12$

So, the other point is (-1, 12).

---

The points on the curve $y = x^3 - 3x^2 - 9x + 7$ where the tangent is parallel to the x-axis are (3, -20) and (-1, 12).

Given:

The equation of the curve is $y = (x - 2)^2$.

The chord joins the points A(2, 0) and B(4, 4).

---

To Find:

A point on the curve where the tangent is parallel to the chord AB.

---

Solution:

First, find the slope of the chord joining the points A(2, 0) and B(4, 4).

The slope of the chord ($m_{chord}$) is given by the formula:

$m_{chord} = \frac{y_2 - y_1}{x_2 - x_1}$

$m_{chord} = \frac{4 - 0}{4 - 2} = \frac{4}{2} = 2$

---

Next, find the slope of the tangent to the curve $y = (x - 2)^2$ at any point $(x, y)$. The slope of the tangent ($m_{tangent}$) is the derivative of the curve's equation with respect to $x$.

$\frac{dy}{dx} = \frac{d}{dx}((x - 2)^2)$

Using the chain rule:

$\frac{dy}{dx} = 2(x - 2)^{2-1} \cdot \frac{d}{dx}(x - 2)$

$\frac{dy}{dx} = 2(x - 2) \cdot (1)$

$m_{tangent} = \frac{dy}{dx} = 2(x - 2)$

---

We are looking for a point on the curve where the tangent is parallel to the chord AB. This means the slope of the tangent must be equal to the slope of the chord.

$m_{tangent} = m_{chord}$

$2(x - 2) = 2$

Divide both sides by 2:

$x - 2 = 1$

Add 2 to both sides:

$x = 1 + 2$

$x = 3$

---

Now, find the corresponding $y$-coordinate of the point on the curve when $x = 3$. Substitute $x = 3$ into the equation of the curve $y = (x - 2)^2$.

$y = (3 - 2)^2$

$y = (1)^2$

$y = 1$

---

Therefore, the point on the curve $y = (x - 2)^2$ at which the tangent is parallel to the chord joining (2, 0) and (4, 4) is (3, 1).

Given:

The equation of the curve is $y = x^3 - 11x + 5$.

The equation of the tangent line is $y = x - 11$.

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To Find:

The point on the curve where the tangent line is $y = x - 11$.

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Solution:

First, find the slope of the given tangent line $y = x - 11$.

Comparing this equation with the slope-intercept form $y = mx + c$, where $m$ is the slope, we get:

Slope of the tangent line, $m = 1$.

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Next, find the slope of the tangent to the curve $y = x^3 - 11x + 5$ by differentiating with respect to $x$.

$\frac{dy}{dx} = \frac{d}{dx}(x^3 - 11x + 5)$

$\frac{dy}{dx} = 3x^2 - 11$

The slope of the tangent to the curve at a point $(x, y)$ is $3x^2 - 11$.

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Since the tangent line $y = x - 11$ touches the curve at the required point, the slope of the tangent to the curve at that point must be equal to the slope of the line $y = x - 11$.

Therefore, we set $\frac{dy}{dx} = 1$:

$3x^2 - 11 = 1$

$3x^2 = 1 + 11$

$3x^2 = 12$

$x^2 = \frac{12}{3}$

$x^2 = 4$

$x = \pm \sqrt{4}$

$x = 2$ or $x = -2$.

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Now, we need to find the corresponding $y$-coordinates for these $x$-values using the equation of the curve $y = x^3 - 11x + 5$.

Case 1: When $x = 2$

$y = (2)^3 - 11(2) + 5$

$y = 8 - 22 + 5$

$y = 13 - 22$

$y = -9$

So, one potential point is $(2, -9)$.

Case 2: When $x = -2$

$y = (-2)^3 - 11(-2) + 5$

$y = -8 + 22 + 5$

$y = 14 + 5$

$y = 19$

So, the other potential point is $(-2, 19)$.

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The point of tangency must lie on both the curve and the tangent line $y = x - 11$. Let's check which of these points satisfies the equation of the tangent line.

Check point $(2, -9)$:

Substitute $x = 2$ and $y = -9$ into $y = x - 11$.

$-9 = 2 - 11$

$-9 = -9$ (True)

Check point $(-2, 19)$:

Substitute $x = -2$ and $y = 19$ into $y = x - 11$.

$19 = -2 - 11$

$19 = -13$ (False)

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Therefore, the only point that lies on both the curve and the tangent line $y = x - 11$ is (2, -9).

Given:

The equation of the curve is $y = \frac{1}{x-1}$, where $x \neq 1$.

The required slope of the tangent lines is $m = -1$.

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To Find:

The equations of all tangent lines to the curve with slope $-1$.

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Solution:

First, find the slope of the tangent to the curve at any point $(x, y)$ by differentiating the equation of the curve with respect to $x$.

We can write $y = (x-1)^{-1}$.

Using the power rule and chain rule:

$\frac{dy}{dx} = \frac{d}{dx}((x-1)^{-1})$

$\frac{dy}{dx} = -1 \cdot (x-1)^{-1-1} \cdot \frac{d}{dx}(x-1)$

$\frac{dy}{dx} = -1 \cdot (x-1)^{-2} \cdot (1)$

$\frac{dy}{dx} = -\frac{1}{(x-1)^2}$

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We are given that the slope of the tangent lines is $-1$. Therefore, we set the derivative equal to $-1$:

$\frac{dy}{dx} = -1$

$-\frac{1}{(x-1)^2} = -1$

Multiply both sides by $-1$:

$\frac{1}{(x-1)^2} = 1$

Take the reciprocal of both sides (or cross-multiply):

$(x-1)^2 = 1$

Take the square root of both sides:

$x-1 = \pm \sqrt{1}$

$x-1 = \pm 1$

This gives two possible values for $x$:

Case 1: $x-1 = 1 \implies x = 1 + 1 \implies x = 2$

Case 2: $x-1 = -1 \implies x = -1 + 1 \implies x = 0$

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Now, find the corresponding $y$-coordinates for these $x$-values using the equation of the curve $y = \frac{1}{x-1}$.

For $x = 2$:

$y = \frac{1}{2-1} = \frac{1}{1} = 1$. The point of tangency is $(2, 1)$.

For $x = 0$:

$y = \frac{1}{0-1} = \frac{1}{-1} = -1$. The point of tangency is $(0, -1)$.

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Finally, find the equations of the tangent lines using the point-slope form $y - y_1 = m(x - x_1)$, with slope $m = -1$.

Tangent at (2, 1):

$y - 1 = -1(x - 2)$

$y - 1 = -x + 2$

$y = -x + 2 + 1$

$y = -x + 3$

Or, rearranging the terms: $x + y - 3 = 0$

Tangent at (0, -1):

$y - (-1) = -1(x - 0)$

$y + 1 = -x$

$y = -x - 1$

Or, rearranging the terms: $x + y + 1 = 0$

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The equations of the lines having slope $-1$ that are tangents to the curve $y = \frac{1}{x-1}$ are $x + y - 3 = 0$ and $x + y + 1 = 0$.

Given:

The equation of the curve is $y = \frac{1}{x-3}$, where $x \neq 3$.

The required slope of the tangent lines is $m = 2$.

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To Find:

The equations of all tangent lines to the curve with slope $2$.

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Solution:

First, find the slope of the tangent to the curve at any point $(x, y)$ by differentiating the equation of the curve with respect to $x$.

We can write $y = (x-3)^{-1}$.

Using the power rule and chain rule:

$\frac{dy}{dx} = \frac{d}{dx}((x-3)^{-1})$

$\frac{dy}{dx} = -1 \cdot (x-3)^{-1-1} \cdot \frac{d}{dx}(x-3)$

$\frac{dy}{dx} = -1 \cdot (x-3)^{-2} \cdot (1)$

$\frac{dy}{dx} = -\frac{1}{(x-3)^2}$

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We are given that the slope of the tangent lines must be $2$. Therefore, we set the derivative equal to $2$:

$\frac{dy}{dx} = 2$

$-\frac{1}{(x-3)^2} = 2$

Multiply both sides by $-1$:

$\frac{1}{(x-3)^2} = -2$

Rearrange the equation:

$(x-3)^2 = -\frac{1}{2}$

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The square of any real number, $(x-3)^2$, must be non-negative (greater than or equal to zero).

However, the equation $(x-3)^2 = -\frac{1}{2}$ requires the square of a real number to be negative, which is impossible.

There are no real values of $x$ that satisfy this equation.

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Therefore, there are no points on the curve $y = \frac{1}{x-3}$ where the slope of the tangent is 2.

Consequently, there are no tangent lines to the given curve with a slope of 2.

Given:

The equation of the curve is $y = \frac{1}{x^2 - 2x + 3}$.

The required slope of the tangent lines is $m = 0$.

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To Find:

The equations of all tangent lines to the curve with slope $0$.

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Solution:

First, find the slope of the tangent to the curve at any point $(x, y)$ by differentiating the equation of the curve with respect to $x$.

We can write $y = (x^2 - 2x + 3)^{-1}$.

Using the chain rule:

$\frac{dy}{dx} = \frac{d}{dx}((x^2 - 2x + 3)^{-1})$

$\frac{dy}{dx} = -1 \cdot (x^2 - 2x + 3)^{-1-1} \cdot \frac{d}{dx}(x^2 - 2x + 3)$

$\frac{dy}{dx} = -1 \cdot (x^2 - 2x + 3)^{-2} \cdot (2x - 2)$

$\frac{dy}{dx} = -\frac{2x - 2}{(x^2 - 2x + 3)^2}$

$\frac{dy}{dx} = \frac{2 - 2x}{(x^2 - 2x + 3)^2}$

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We are given that the slope of the tangent lines must be $0$. Therefore, we set the derivative equal to $0$:

$\frac{dy}{dx} = 0$

$\frac{2 - 2x}{(x^2 - 2x + 3)^2} = 0$

A fraction is equal to zero only when its numerator is zero and its denominator is non-zero.

$2 - 2x = 0$

$2 = 2x$

$x = 1$

We should check if the denominator is non-zero at $x=1$.

Denominator: $(x^2 - 2x + 3)^2 = ((1)^2 - 2(1) + 3)^2 = (1 - 2 + 3)^2 = (2)^2 = 4$.

Since the denominator is $4 \neq 0$ at $x=1$, the value $x=1$ is valid.

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Now, find the corresponding $y$-coordinate for $x=1$ using the equation of the curve $y = \frac{1}{x^2 - 2x + 3}$.

$y = \frac{1}{(1)^2 - 2(1) + 3}$

$y = \frac{1}{1 - 2 + 3}$

$y = \frac{1}{2}$

So, the point of tangency is $(1, \frac{1}{2})$.

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Finally, find the equation of the tangent line using the point-slope form $y - y_1 = m(x - x_1)$, with the point $(1, \frac{1}{2})$ and slope $m = 0$.

$y - \frac{1}{2} = 0(x - 1)$

$y - \frac{1}{2} = 0$

$y = \frac{1}{2}$

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There is only one such line. The equation of the tangent line to the curve with slope 0 is $y = \frac{1}{2}$.

Given:

The equation of the curve (ellipse) is $\frac{x^2}{9} + \frac{y^2}{16} = 1$.

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To Find:

The points on the curve where the tangents are:

(i) parallel to the x-axis

(ii) parallel to the y-axis

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Solution:

First, differentiate the equation of the curve with respect to $x$ using implicit differentiation to find the slope of the tangent, $\frac{dy}{dx}$.

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

$\frac{1}{9} \frac{d}{dx}(x^2) + \frac{1}{16} \frac{d}{dx}(y^2) = 0$

$\frac{1}{9}(2x) + \frac{1}{16}(2y \frac{dy}{dx}) = 0$

$\frac{2x}{9} + \frac{2y}{16} \frac{dy}{dx} = 0$

$\frac{2x}{9} + \frac{y}{8} \frac{dy}{dx} = 0$

Now, solve for $\frac{dy}{dx}$:

$\frac{y}{8} \frac{dy}{dx} = -\frac{2x}{9}$

$\frac{dy}{dx} = -\frac{2x}{9} \cdot \frac{8}{y}$

$\frac{dy}{dx} = -\frac{16x}{9y}$

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(i) Tangents parallel to x-axis

A tangent is parallel to the x-axis when its slope is zero.

$\frac{dy}{dx} = 0$

$-\frac{16x}{9y} = 0$

For a fraction to be zero, the numerator must be zero (assuming the denominator is non-zero).

$-16x = 0$

$x = 0$

Now, substitute $x=0$ into the equation of the ellipse to find the corresponding $y$-values:

$\frac{(0)^2}{9} + \frac{y^2}{16} = 1$

$0 + \frac{y^2}{16} = 1$

$y^2 = 16$

$y = \pm \sqrt{16}$

$y = \pm 4$

The points where the tangent is parallel to the x-axis are (0, 4) and (0, -4).

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(ii) Tangents parallel to y-axis

A tangent is parallel to the y-axis (vertical tangent) when its slope $\frac{dy}{dx}$ is undefined.

The slope $\frac{dy}{dx} = -\frac{16x}{9y}$ is undefined when the denominator is zero (assuming the numerator is non-zero).

$9y = 0$

$y = 0$

Now, substitute $y=0$ into the equation of the ellipse to find the corresponding $x$-values:

$\frac{x^2}{9} + \frac{(0)^2}{16} = 1$

$\frac{x^2}{9} + 0 = 1$

$x^2 = 9$

$x = \pm \sqrt{9}$

$x = \pm 3$

The points where the tangent is parallel to the y-axis are (3, 0) and (-3, 0).

(i) y = x⁴ – 6x³ + 13x² – 10x + 5 at (0, 5)

Solution:

Given curve: $y = x^4 - 6x^3 + 13x^2 - 10x + 5$.

Differentiate with respect to $x$ to find the slope of the tangent:

$\frac{dy}{dx} = \frac{d}{dx}(x^4 - 6x^3 + 13x^2 - 10x + 5)$

$\frac{dy}{dx} = 4x^3 - 18x^2 + 26x - 10$

The slope of the tangent at the point $(0, 5)$ is found by substituting $x=0$ into $\frac{dy}{dx}$:

$m_{tangent} = \frac{dy}{dx}\Big|_{x=0} = 4(0)^3 - 18(0)^2 + 26(0) - 10 = -10$.

The slope of the normal is the negative reciprocal of the slope of the tangent:

$m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{-10} = \frac{1}{10}$.

Equation of the tangent at $(0, 5)$ using point-slope form $y - y_1 = m(x - x_1)$:

$y - 5 = -10(x - 0)$

$y - 5 = -10x$

$10x + y - 5 = 0$ (Equation of the tangent)

Equation of the normal at $(0, 5)$:

$y - 5 = \frac{1}{10}(x - 0)$

$10(y - 5) = x$

$10y - 50 = x$

$x - 10y + 50 = 0$ (Equation of the normal)

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(ii) y = x⁴ – 6x³ + 13x² – 10x + 5 at (1, 3)

Solution:

Given curve: $y = x^4 - 6x^3 + 13x^2 - 10x + 5$.

The derivative is $\frac{dy}{dx} = 4x^3 - 18x^2 + 26x - 10$.

The slope of the tangent at the point $(1, 3)$ is found by substituting $x=1$ into $\frac{dy}{dx}$:

$m_{tangent} = \frac{dy}{dx}\Big|_{x=1} = 4(1)^3 - 18(1)^2 + 26(1) - 10$

$m_{tangent} = 4 - 18 + 26 - 10 = 30 - 28 = 2$.

The slope of the normal is:

$m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{2}$.

Equation of the tangent at $(1, 3)$:

$y - 3 = 2(x - 1)$

$y - 3 = 2x - 2$

$2x - y + 1 = 0$ (Equation of the tangent)

Equation of the normal at $(1, 3)$:

$y - 3 = -\frac{1}{2}(x - 1)$

$2(y - 3) = -(x - 1)$

$2y - 6 = -x + 1$

$x + 2y - 7 = 0$ (Equation of the normal)

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(iii) y = x³ at (1, 1)

Solution:

Given curve: $y = x^3$.

Differentiate with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(x^3) = 3x^2$.

The slope of the tangent at the point $(1, 1)$ is:

$m_{tangent} = \frac{dy}{dx}\Big|_{x=1} = 3(1)^2 = 3$.

The slope of the normal is:

$m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{3}$.

Equation of the tangent at $(1, 1)$:

$y - 1 = 3(x - 1)$

$y - 1 = 3x - 3$

$3x - y - 2 = 0$ (Equation of the tangent)

Equation of the normal at $(1, 1)$:

$y - 1 = -\frac{1}{3}(x - 1)$

$3(y - 1) = -(x - 1)$

$3y - 3 = -x + 1$

$x + 3y - 4 = 0$ (Equation of the normal)

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(iv) y = x² at (0, 0)

Solution:

Given curve: $y = x^2$.

Differentiate with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(x^2) = 2x$.

The slope of the tangent at the point $(0, 0)$ is:

$m_{tangent} = \frac{dy}{dx}\Big|_{x=0} = 2(0) = 0$.

Since the slope of the tangent is 0, the tangent line is horizontal.

Equation of the tangent at $(0, 0)$:

$y - 0 = 0(x - 0)$

$y = 0$ (Equation of the tangent - the x-axis)

Since the tangent is horizontal, the normal line is vertical. A vertical line passing through $(0, 0)$ has the equation $x = 0$.

Alternatively, the slope of the normal $m_{normal} = -1/0$, which is undefined, indicating a vertical line.

$x = 0$ (Equation of the normal - the y-axis)

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(v) x = cos t, y = sin t at t = $\frac{\pi}{4}$

Solution:

Given parametric equations: $x = \cos t$, $y = \sin t$.

First, find the coordinates of the point when $t = \frac{\pi}{4}$:

$x_1 = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

$y_1 = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

The point is $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$.

Next, find the slope $\frac{dy}{dx}$ using parametric differentiation:

$\frac{dx}{dt} = \frac{d}{dt}(\cos t) = -\sin t$

$\frac{dy}{dt} = \frac{d}{dt}(\sin t) = \cos t$

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\cos t}{-\sin t} = -\cot t$

The slope of the tangent at $t = \frac{\pi}{4}$ is:

$m_{tangent} = \frac{dy}{dx}\Big|_{t=\pi/4} = -\cot\left(\frac{\pi}{4}\right) = -1$.

The slope of the normal is:

$m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{-1} = 1$.

Equation of the tangent at $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$:

$y - \frac{\sqrt{2}}{2} = -1 \left( x - \frac{\sqrt{2}}{2} \right)$

$y - \frac{\sqrt{2}}{2} = -x + \frac{\sqrt{2}}{2}$

$x + y = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}$

$x + y = \frac{2\sqrt{2}}{2}$

$x + y = \sqrt{2}$ or $x + y - \sqrt{2} = 0$ (Equation of the tangent)

Equation of the normal at $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$:

$y - \frac{\sqrt{2}}{2} = 1 \left( x - \frac{\sqrt{2}}{2} \right)$

$y - \frac{\sqrt{2}}{2} = x - \frac{\sqrt{2}}{2}$

$y = x$

$x - y = 0$ (Equation of the normal)

Given:

The equation of the curve is $y = x^2 - 2x + 7$.

Line L1: $2x - y + 9 = 0$

Line L2: $5y - 15x = 13$

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To Find:

(a) The equation of the tangent line to the curve which is parallel to line L1.

(b) The equation of the tangent line to the curve which is perpendicular to line L2.

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Solution:

First, find the slope of the tangent to the curve $y = x^2 - 2x + 7$ at any point $(x, y)$ by differentiating with respect to $x$.

$\frac{dy}{dx} = \frac{d}{dx}(x^2 - 2x + 7)$

$\frac{dy}{dx} = 2x - 2$

So, the slope of the tangent at any point $(x, y)$ on the curve is $m_{tangent} = 2x - 2$.

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(a) Tangent parallel to the line 2x – y + 9 = 0

Find the slope of the line $2x - y + 9 = 0$. Rewrite it in slope-intercept form ($y = mx + c$):

$y = 2x + 9$

The slope of this line is $m_{L1} = 2$.

Since the tangent line is parallel to this line, their slopes must be equal.

$m_{tangent} = m_{L1}$

$2x - 2 = 2$

$2x = 4$

$x = 2$

Now, find the $y$-coordinate of the point of tangency by substituting $x=2$ into the curve's equation:

$y = (2)^2 - 2(2) + 7$

$y = 4 - 4 + 7$

$y = 7$

The point of tangency is $(2, 7)$.

The equation of the tangent line with slope $m=2$ passing through $(2, 7)$ is given by the point-slope form $y - y_1 = m(x - x_1)$:

$y - 7 = 2(x - 2)$

$y - 7 = 2x - 4$

$2x - y - 4 + 7 = 0$

$2x - y + 3 = 0$

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(b) Tangent perpendicular to the line 5y – 15x = 13

Find the slope of the line $5y - 15x = 13$. Rewrite it in slope-intercept form ($y = mx + c$):

$5y = 15x + 13$

$y = \frac{15}{5}x + \frac{13}{5}$

$y = 3x + \frac{13}{5}$

The slope of this line is $m_{L2} = 3$.

Since the tangent line is perpendicular to this line, the product of their slopes is $-1$.

$m_{tangent} \times m_{L2} = -1$

$m_{tangent} \times 3 = -1$

$m_{tangent} = -\frac{1}{3}$

Now, set the slope of the tangent to the curve equal to $-\frac{1}{3}$:

$\frac{dy}{dx} = -\frac{1}{3}$

$2x - 2 = -\frac{1}{3}$

$2x = 2 - \frac{1}{3}$

$2x = \frac{6 - 1}{3}$

$2x = \frac{5}{3}$

$x = \frac{5}{3 \times 2}$

$x = \frac{5}{6}$

Now, find the $y$-coordinate of the point of tangency by substituting $x=\frac{5}{6}$ into the curve's equation:

$y = \left(\frac{5}{6}\right)^2 - 2\left(\frac{5}{6}\right) + 7$

$y = \frac{25}{36} - \frac{10}{6} + 7$

$y = \frac{25}{36} - \frac{10 \times 6}{6 \times 6} + \frac{7 \times 36}{1 \times 36}$

$y = \frac{25 - 60 + 252}{36}$

$y = \frac{277 - 60}{36}$

$y = \frac{217}{36}$

The point of tangency is $\left(\frac{5}{6}, \frac{217}{36}\right)$.

The equation of the tangent line with slope $m=-\frac{1}{3}$ passing through $\left(\frac{5}{6}, \frac{217}{36}\right)$ is:

$y - \frac{217}{36} = -\frac{1}{3}\left(x - \frac{5}{6}\right)$

Multiply the entire equation by 36 to clear the denominators:

$36y - 36 \left(\frac{217}{36}\right) = 36 \left(-\frac{1}{3}\right) \left(x - \frac{5}{6}\right)$

$36y - 217 = -12 \left(x - \frac{5}{6}\right)$

$36y - 217 = -12x + 12 \left(\frac{5}{6}\right)$

$36y - 217 = -12x + 10$

$12x + 36y - 217 - 10 = 0$

$12x + 36y - 227 = 0$

Given:

The equation of the curve is $y = 7x^3 + 11$.

We need to consider the points on the curve where $x = 2$ and $x = -2$.

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To Prove:

The tangents to the curve at the points where $x = 2$ and $x = -2$ are parallel.

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Proof:

Two lines (or tangents) are parallel if their slopes are equal.

The slope of the tangent to the curve at any point $(x, y)$ is given by the derivative $\frac{dy}{dx}$.

First, differentiate the equation of the curve with respect to $x$:

$y = 7x^3 + 11$

$\frac{dy}{dx} = \frac{d}{dx}(7x^3 + 11)$

$\frac{dy}{dx} = 7(3x^2) + 0$

$\frac{dy}{dx} = 21x^2$

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Now, find the slope of the tangent at the point where $x = 2$. Let this slope be $m_1$.

$m_1 = \frac{dy}{dx}\Big|_{x=2} = 21(2)^2$

$m_1 = 21(4)$

$m_1 = 84$

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Next, find the slope of the tangent at the point where $x = -2$. Let this slope be $m_2$.

$m_2 = \frac{dy}{dx}\Big|_{x=-2} = 21(-2)^2$

$m_2 = 21(4)$

$m_2 = 84$

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Comparing the slopes, we find that $m_1 = 84$ and $m_2 = 84$.

Since $m_1 = m_2$, the slopes of the tangents at $x=2$ and $x=-2$ are equal.

Therefore, the tangents to the curve $y = 7x^3 + 11$ at the points where $x = 2$ and $x = -2$ are parallel.

Hence proved.

Given:

The equation of the curve is $y = x^3$.

The condition is that the slope of the tangent at a point $(x, y)$ on the curve is equal to the y-coordinate of that point.

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To Find:

The points on the curve satisfying the given condition.

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Solution:

First, find the slope of the tangent to the curve $y = x^3$ at any point $(x, y)$ by differentiating with respect to $x$.

$\frac{dy}{dx} = \frac{d}{dx}(x^3)$

$\frac{dy}{dx} = 3x^2$

So, the slope of the tangent at $(x, y)$ is $m = 3x^2$.

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According to the problem, the slope of the tangent ($\frac{dy}{dx}$) is equal to the y-coordinate ($y$) of the point.

Therefore, we have the condition:

$\frac{dy}{dx} = y$

$3x^2 = y$

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The point $(x, y)$ must also lie on the curve $y = x^3$. So we have a system of two equations:

Substitute the expression for $y$ from equation (ii) into equation (i):

$x^3 = 3x^2$

Rearrange the equation to solve for $x$:

$x^3 - 3x^2 = 0$

Factor out $x^2$:

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

This equation gives two possible values for $x$:

$x^2 = 0 \implies x = 0$

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

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Now, find the corresponding $y$-coordinates using the equation of the curve, $y = x^3$.

Case 1: When $x = 0$

$y = (0)^3 = 0$

The point is $(0, 0)$. Let's check the condition: Slope = $3x^2 = 3(0)^2 = 0$. Y-coordinate = $0$. Condition $0 = 0$ is satisfied.

Case 2: When $x = 3$

$y = (3)^3 = 27$

The point is $(3, 27)$. Let's check the condition: Slope = $3x^2 = 3(3)^2 = 3(9) = 27$. Y-coordinate = $27$. Condition $27 = 27$ is satisfied.

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Therefore, the points on the curve $y = x^3$ at which the slope of the tangent is equal to the y-coordinate of the point are (0, 0) and (3, 27).

Given:

The equation of the curve is $y = 4x^3 - 2x^5$.

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To Find:

All points $(x_0, y_0)$ on the curve such that the tangent line at $(x_0, y_0)$ passes through the origin $(0, 0)$.

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Solution:

Let $(x_0, y_0)$ be a point on the curve. Then its coordinates must satisfy the curve's equation:

First, find the slope of the tangent to the curve at any point $(x, y)$ by differentiating $y$ with respect to $x$.

$\frac{dy}{dx} = \frac{d}{dx}(4x^3 - 2x^5)$

$\frac{dy}{dx} = 4(3x^2) - 2(5x^4)$

$\frac{dy}{dx} = 12x^2 - 10x^4$

The slope of the tangent at the point $(x_0, y_0)$ is:

$m = \frac{dy}{dx}\Big|_{x=x_0} = 12x_0^2 - 10x_0^4$

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The equation of the tangent line at the point $(x_0, y_0)$ using the point-slope form $Y - y_1 = m(X - x_1)$ is:

$Y - y_0 = (12x_0^2 - 10x_0^4)(X - x_0)$

We are given that this tangent line passes through the origin $(0, 0)$. Substitute $X=0$ and $Y=0$ into the tangent equation:

$0 - y_0 = (12x_0^2 - 10x_0^4)(0 - x_0)$

$-y_0 = (12x_0^2 - 10x_0^4)(-x_0)$

$y_0 = x_0(12x_0^2 - 10x_0^4)$

$y_0 = 12x_0^3 - 10x_0^5$

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Now we have two expressions for $y_0$ (from equation (i) and equation (ii)). Equating them:

$4x_0^3 - 2x_0^5 = 12x_0^3 - 10x_0^5$

Bring all terms to one side:

$(10x_0^5 - 2x_0^5) + (4x_0^3 - 12x_0^3) = 0$

$8x_0^5 - 8x_0^3 = 0$

Factor out $8x_0^3$:

$8x_0^3(x_0^2 - 1) = 0$

$8x_0^3(x_0 - 1)(x_0 + 1) = 0$

This equation holds if $x_0^3 = 0$ or $x_0 - 1 = 0$ or $x_0 + 1 = 0$.

So, the possible values for $x_0$ are $x_0 = 0$, $x_0 = 1$, and $x_0 = -1$.

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Find the corresponding $y_0$-coordinates using the equation of the curve, $y_0 = 4x_0^3 - 2x_0^5$.

Case 1: If $x_0 = 0$

$y_0 = 4(0)^3 - 2(0)^5 = 0 - 0 = 0$.

The point is $(0, 0)$.

Case 2: If $x_0 = 1$

$y_0 = 4(1)^3 - 2(1)^5 = 4(1) - 2(1) = 4 - 2 = 2$.

The point is $(1, 2)$.

Case 3: If $x_0 = -1$

$y_0 = 4(-1)^3 - 2(-1)^5 = 4(-1) - 2(-1) = -4 - (-2) = -4 + 2 = -2$.

The point is $(-1, -2)$.

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The points on the curve $y = 4x^3 - 2x^5$ at which the tangent passes through the origin are (0, 0), (1, 2), and (-1, -2).

Given:

The equation of the curve is $x^2 + y^2 - 2x - 3 = 0$.

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To Find:

The points on the curve where the tangents are parallel to the x-axis.

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Solution:

A tangent line is parallel to the x-axis if its slope is zero. The slope of the tangent to the curve at any point $(x, y)$ is given by the derivative $\frac{dy}{dx}$.

We differentiate the equation of the curve $x^2 + y^2 - 2x - 3 = 0$ with respect to $x$ using implicit differentiation:

$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - \frac{d}{dx}(2x) - \frac{d}{dx}(3) = \frac{d}{dx}(0)$

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

$2y \frac{dy}{dx} = 2 - 2x$

$\frac{dy}{dx} = \frac{2 - 2x}{2y}$

$\frac{dy}{dx} = \frac{2(1 - x)}{2y}$

$\frac{dy}{dx} = \frac{1 - x}{y}$

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For the tangent to be parallel to the x-axis, its slope must be zero.

$\frac{dy}{dx} = 0$

$\frac{1 - x}{y} = 0$

This condition holds true when the numerator is zero and the denominator is non-zero.

$1 - x = 0$

$x = 1$

We also require $y \neq 0$.

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Now, substitute $x = 1$ into the original equation of the curve to find the corresponding $y$-coordinates:

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

$(1)^2 + y^2 - 2(1) - 3 = 0$

$1 + y^2 - 2 - 3 = 0$

$y^2 - 4 = 0$

$y^2 = 4$

$y = \pm \sqrt{4}$

$y = 2$ or $y = -2$

Both $y=2$ and $y=-2$ satisfy the condition $y \neq 0$.

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Therefore, the points on the curve $x^2 + y^2 - 2x - 3 = 0$ at which the tangents are parallel to the x-axis are (1, 2) and (1, -2).

(Note: The curve is a circle $(x-1)^2 + y^2 = 4$, centered at $(1,0)$ with radius 2. The points $(1,2)$ and $(1,-2)$ are the top and bottom points of the circle, where the tangents are indeed horizontal.)

Given:

The equation of the curve is $ay^2 = x^3$.

The point on the curve is $P(x_1, y_1) = (am^2, am^3)$.

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To Find:

The equation of the normal to the curve at the point $(am^2, am^3)$.

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Solution:

First, we need to find the slope of the tangent to the curve at the point $(am^2, am^3)$. We differentiate the equation of the curve $ay^2 = x^3$ with respect to $x$ using implicit differentiation.

$\frac{d}{dx}(ay^2) = \frac{d}{dx}(x^3)$

$a \cdot 2y \frac{dy}{dx} = 3x^2$

$2ay \frac{dy}{dx} = 3x^2$

Solving for $\frac{dy}{dx}$ (the slope of the tangent):

$\frac{dy}{dx} = \frac{3x^2}{2ay}$ (Provided $a \neq 0, y \neq 0$)

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Now, evaluate the slope of the tangent at the given point $(x_1, y_1) = (am^2, am^3)$.

$m_{tangent} = \frac{dy}{dx}\Big|_{(am^2, am^3)} = \frac{3(am^2)^2}{2a(am^3)}$

$m_{tangent} = \frac{3a^2m^4}{2a^2m^3}$

Assuming $a \neq 0$ and $m \neq 0$ (so $y \neq 0$):

$m_{tangent} = \frac{3\cancel{a^2}m^{\cancel{4}}}{\cancel{2}\cancel{a^2}\cancel{m^3}}$

$m_{tangent} = \frac{3m}{2}$

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The slope of the normal ($m_{normal}$) is the negative reciprocal of the slope of the tangent.

$m_{normal} = -\frac{1}{m_{tangent}}$

$m_{normal} = -\frac{1}{\frac{3m}{2}}$

$m_{normal} = -\frac{2}{3m}$ (for $m \neq 0$)

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Now, use the point-slope form of a line, $Y - y_1 = m_{normal}(X - x_1)$, to find the equation of the normal at the point $(am^2, am^3)$ with slope $-\frac{2}{3m}$.

$Y - am^3 = -\frac{2}{3m}(X - am^2)$

Multiply both sides by $3m$ to eliminate the fraction:

$3m(Y - am^3) = -2(X - am^2)$

$3mY - 3am^4 = -2X + 2am^2$

Rearrange the terms to the standard form $Ax + By + C = 0$:

$2X + 3mY - 2am^2 - 3am^4 = 0$

$2X + 3mY - am^2(2 + 3m^2) = 0$

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Case Check (m=0):

If $m=0$, the point is $(a(0)^2, a(0)^3) = (0, 0)$.

At $(0,0)$, the slope of the tangent $\frac{dy}{dx} = \frac{3x^2}{2ay} = \frac{3(0)^2}{2a(0)}$ is indeterminate in this form. However, calculating the limit or observing the tangent slope formula $m_{tangent} = \frac{3m}{2}$ gives $m_{tangent}=0$ when $m=0$.

A tangent with slope 0 is horizontal ($Y=0$). The normal line must be vertical and pass through $(0,0)$, so its equation is $X=0$.

Let's check if our derived normal equation gives $X=0$ when $m=0$:

$2X + 3(0)Y - a(0)^2(2 + 3(0)^2) = 0$

$2X + 0 - 0 = 0$

$2X = 0 \implies X = 0$.

The formula holds for $m=0$ as well.

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The equation of the normal to the curve $ay^2 = x^3$ at the point $(am^2, am^3)$ is:

$2x + 3my - am^2(2 + 3m^2) = 0$

Or equivalently:

$2x + 3my - 2am^2 - 3am^4 = 0$

Given:

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

The line to which the normals are parallel is $x + 14y + 4 = 0$.

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To Find:

The equations of the normal lines to the curve that are parallel to the given line.

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Solution:

First, find the slope of the given line $x + 14y + 4 = 0$. We can rewrite it in the slope-intercept form $y = mx + c$.

$14y = -x - 4$

$y = -\frac{1}{14}x - \frac{4}{14}$

$y = -\frac{1}{14}x - \frac{2}{7}$

The slope of this line is $m_{line} = -\frac{1}{14}$.

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Since the required normal lines are parallel to the given line, the slope of the normals ($m_{normal}$) must be equal to the slope of the given line.

$m_{normal} = m_{line} = -\frac{1}{14}$.

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Next, find the slope of the tangent ($m_{tangent}$) to the curve $y = x^3 + 2x + 6$. The slope of the tangent is given by the derivative $\frac{dy}{dx}$.

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

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

So, $m_{tangent} = 3x^2 + 2$.

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The slope of the normal is the negative reciprocal of the slope of the tangent:

$m_{normal} = -\frac{1}{m_{tangent}}$

We know $m_{normal} = -\frac{1}{14}$, so:

$-\frac{1}{14} = -\frac{1}{3x^2 + 2}$

Equating the denominators:

$14 = 3x^2 + 2$

$3x^2 = 14 - 2$

$3x^2 = 12$

$x^2 = \frac{12}{3}$

$x^2 = 4$

$x = \pm \sqrt{4}$

$x = 2$ or $x = -2$.

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Now, find the corresponding $y$-coordinates for these $x$-values using the equation of the curve $y = x^3 + 2x + 6$.

Case 1: When $x = 2$

$y = (2)^3 + 2(2) + 6 = 8 + 4 + 6 = 18$.

The point on the curve is $(2, 18)$.

Case 2: When $x = -2$

$y = (-2)^3 + 2(-2) + 6 = -8 - 4 + 6 = -6$.

The point on the curve is $(-2, -6)$.

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We need to find the equations of the normal lines at these points. The slope of the normal is $m_{normal} = -\frac{1}{14}$. Use the point-slope form $Y - y_1 = m(X - x_1)$.

Normal at (2, 18):

$Y - 18 = -\frac{1}{14}(X - 2)$

$14(Y - 18) = -(X - 2)$

$14Y - 252 = -X + 2$

$X + 14Y - 252 - 2 = 0$

$X + 14Y - 254 = 0$

Normal at (-2, -6):

$Y - (-6) = -\frac{1}{14}(X - (-2))$

$Y + 6 = -\frac{1}{14}(X + 2)$

$14(Y + 6) = -(X + 2)$

$14Y + 84 = -X - 2$

$X + 14Y + 84 + 2 = 0$

$X + 14Y + 86 = 0$

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The equations of the normals to the curve $y = x^3 + 2x + 6$ which are parallel to the line $x + 14y + 4 = 0$ are $x + 14y - 254 = 0$ and $x + 14y + 86 = 0$.

Given:

The equation of the parabola is $y^2 = 4ax$.

The point on the parabola is $(x_1, y_1) = (at^2, 2at)$. This is the parametric form of a point on the parabola.

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To Find:

The equation of the tangent line to the parabola at the point $(at^2, 2at)$.

The equation of the normal line to the parabola at the point $(at^2, 2at)$.

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Solution:

We need to find the slope of the tangent, $\frac{dy}{dx}$, at the given point.

Method 1: Implicit Differentiation

Differentiate $y^2 = 4ax$ with respect to $x$:

$\frac{d}{dx}(y^2) = \frac{d}{dx}(4ax)$

$2y \frac{dy}{dx} = 4a$

$\frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y}$

At the point $(at^2, 2at)$, the slope of the tangent is:

$m_{tangent} = \frac{dy}{dx}\Big|_{(at^2, 2at)} = \frac{2a}{2at} = \frac{1}{t}$ (assuming $a \neq 0, t \neq 0$)

Method 2: Parametric Differentiation

We have $x = at^2$ and $y = 2at$.

Differentiate $x$ and $y$ with respect to the parameter $t$:

$\frac{dx}{dt} = \frac{d}{dt}(at^2) = 2at$

$\frac{dy}{dt} = \frac{d}{dt}(2at) = 2a$

The slope of the tangent is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:

$m_{tangent} = \frac{2a}{2at} = \frac{1}{t}$ (assuming $a \neq 0, t \neq 0$)

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Equation of the Tangent:

Use the point-slope form $Y - y_1 = m(X - x_1)$ with the point $(at^2, 2at)$ and slope $m_{tangent} = \frac{1}{t}$.

$Y - 2at = \frac{1}{t}(X - at^2)$

Multiply by $t$ (assuming $t \neq 0$):

$t(Y - 2at) = X - at^2$

$tY - 2at^2 = X - at^2$

Rearrange the terms:

$X - tY + 2at^2 - at^2 = 0$

$X - tY + at^2 = 0$ or $tY = X + at^2$ (Equation of the tangent)

(Note: If $t=0$, the point is $(0,0)$. The tangent slope $1/t$ is undefined, indicating a vertical tangent $X=0$. The formula $tY = X + at^2$ gives $0 = X + 0$, i.e., $X=0$. So the formula holds for $t=0$ as well.)

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Equation of the Normal:

The slope of the normal is the negative reciprocal of the slope of the tangent.

$m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{(1/t)} = -t$ (assuming $t \neq 0$)

Use the point-slope form $Y - y_1 = m(X - x_1)$ with the point $(at^2, 2at)$ and slope $m_{normal} = -t$.

$Y - 2at = -t(X - at^2)$

$Y - 2at = -tX + at^3$

Rearrange the terms:

$tX + Y - 2at - at^3 = 0$

$tX + Y = 2at + at^3$ or $Y + tX = 2at + at^3$ (Equation of the normal)

(Note: If $t=0$, the point is $(0,0)$. The tangent is vertical ($X=0$). The normal must be horizontal ($Y=0$). The formula $Y + tX = 2at + at^3$ gives $Y + 0 = 0 + 0$, i.e., $Y=0$. So the formula holds for $t=0$ as well.)

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Final Answer:

Equation of the tangent at $(at^2, 2at)$ is: $t y = x + a t^2$

Equation of the normal at $(at^2, 2at)$ is: $y + t x = 2 a t + a t^3$

Given:

Curve $C_1$: $x = y^2$

Curve $C_2$: $xy = k$

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To Prove:

The curves $C_1$ and $C_2$ cut at right angles if $8k^2 = 1$.

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Proof:

Two curves cut at right angles (or orthogonally) if the tangents to the curves at their point of intersection are perpendicular to each other. This means the product of the slopes of their tangents at the intersection point must be $-1$.

First, find the point(s) of intersection of the two curves. Substitute $x = y^2$ from $C_1$ into $C_2$:

$(y^2)y = k$

$y^3 = k$

$y = k^{1/3}$

Now substitute this value of $y$ back into $x = y^2$:

$x = (k^{1/3})^2 = k^{2/3}$

So, the point of intersection is $P(k^{2/3}, k^{1/3})$.

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Next, find the slope of the tangent for each curve at the point P.

For curve $C_1: x = y^2$

Differentiate implicitly with respect to $x$:

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

$1 = 2y \frac{dy}{dx}$

$\frac{dy}{dx} = \frac{1}{2y}$

Let the slope of the tangent to $C_1$ at P be $m_1$. Evaluate $\frac{dy}{dx}$ at $y = k^{1/3}$:

$m_1 = \frac{1}{2(k^{1/3})} = \frac{1}{2k^{1/3}}$

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For curve $C_2: xy = k$

Differentiate implicitly with respect to $x$:

$\frac{d}{dx}(xy) = \frac{d}{dx}(k)$

Using the product rule:

$x \frac{dy}{dx} + y \frac{d}{dx}(x) = 0$

$x \frac{dy}{dx} + y(1) = 0$

$x \frac{dy}{dx} = -y$

$\frac{dy}{dx} = -\frac{y}{x}$

Let the slope of the tangent to $C_2$ at P be $m_2$. Evaluate $\frac{dy}{dx}$ at $x = k^{2/3}$ and $y = k^{1/3}$:

$m_2 = -\frac{k^{1/3}}{k^{2/3}} = -k^{(1/3 - 2/3)} = -k^{-1/3} = -\frac{1}{k^{1/3}}$

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The curves cut at right angles if the product of their slopes at the point of intersection is $-1$.

$m_1 \times m_2 = -1$

Substitute the values of $m_1$ and $m_2$:

$\left( \frac{1}{2k^{1/3}} \right) \times \left( -\frac{1}{k^{1/3}} \right) = -1$

$-\frac{1}{2k^{1/3} \cdot k^{1/3}} = -1$

$-\frac{1}{2k^{(1/3 + 1/3)}} = -1$

$-\frac{1}{2k^{2/3}} = -1$

Multiply both sides by $-1$:

$\frac{1}{2k^{2/3}} = 1$

$1 = 2k^{2/3}$

To eliminate the fractional exponent, cube both sides of the equation:

$(1)^3 = (2k^{2/3})^3$

$1 = 2^3 (k^{2/3})^3$

$1 = 8 k^{(2/3) \times 3}$

$1 = 8 k^2$

Or, $8k^2 = 1$.

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Thus, the condition for the curves $x = y^2$ and $xy = k$ to cut at right angles is $8k^2 = 1$.

Hence proved.

Given:

The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

The point on the hyperbola is $(x_0, y_0)$. This means the point satisfies the equation:

$\frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} = 1$.

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To Find:

The equation of the tangent line to the hyperbola at $(x_0, y_0)$.

The equation of the normal line to the hyperbola at $(x_0, y_0)$.

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Solution:

First, find the slope of the tangent to the hyperbola by differentiating its equation with respect to $x$ using implicit differentiation.

$\frac{d}{dx}\left(\frac{x^2}{a^2} - \frac{y^2}{b^2}\right) = \frac{d}{dx}(1)$

$\frac{1}{a^2}\frac{d}{dx}(x^2) - \frac{1}{b^2}\frac{d}{dx}(y^2) = 0$

$\frac{1}{a^2}(2x) - \frac{1}{b^2}(2y \frac{dy}{dx}) = 0$

$\frac{2x}{a^2} - \frac{2y}{b^2}\frac{dy}{dx} = 0$

$\frac{2y}{b^2}\frac{dy}{dx} = \frac{2x}{a^2}$

Assuming $y \neq 0$:

$\frac{dy}{dx} = \frac{2x}{a^2} \cdot \frac{b^2}{2y}$

$\frac{dy}{dx} = \frac{b^2 x}{a^2 y}$

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The slope of the tangent at the point $(x_0, y_0)$ is:

$m_{tangent} = \frac{dy}{dx}\Big|_{(x_0, y_0)} = \frac{b^2 x_0}{a^2 y_0}$ (assuming $y_0 \neq 0$)

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Equation of the Tangent:

Using the point-slope form $Y - y_1 = m(X - x_1)$ with point $(x_0, y_0)$ and slope $m_{tangent}$:

$Y - y_0 = \frac{b^2 x_0}{a^2 y_0}(X - x_0)$

Multiply both sides by $a^2 y_0$:

$a^2 y_0 (Y - y_0) = b^2 x_0 (X - x_0)$

$a^2 y_0 Y - a^2 y_0^2 = b^2 x_0 X - b^2 x_0^2$

$b^2 x_0^2 - a^2 y_0^2 = b^2 x_0 X - a^2 y_0 Y$

Divide both sides by $a^2 b^2$:

$\frac{b^2 x_0^2}{a^2 b^2} - \frac{a^2 y_0^2}{a^2 b^2} = \frac{b^2 x_0 X}{a^2 b^2} - \frac{a^2 y_0 Y}{a^2 b^2}$

$\frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} = \frac{x_0 X}{a^2} - \frac{y_0 Y}{b^2}$

Since $(x_0, y_0)$ lies on the hyperbola, $\frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} = 1$. Substituting this:

$1 = \frac{x_0 X}{a^2} - \frac{y_0 Y}{b^2}$

Thus, the equation of the tangent at $(x_0, y_0)$ is:

$\frac{x x_0}{a^2} - \frac{y y_0}{b^2} = 1$

(This equation also holds for the case $y_0=0$, where the tangent is vertical $x=\pm a$.)

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Equation of the Normal:

The slope of the normal is the negative reciprocal of the slope of the tangent (assuming $m_{tangent}$ is non-zero and defined, i.e., $x_0 \neq 0$ and $y_0 \neq 0$):

$m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{\frac{b^2 x_0}{a^2 y_0}} = -\frac{a^2 y_0}{b^2 x_0}$

Using the point-slope form $Y - y_1 = m(X - x_1)$ with point $(x_0, y_0)$ and slope $m_{normal}$:

$Y - y_0 = -\frac{a^2 y_0}{b^2 x_0}(X - x_0)$

Multiply both sides by $b^2 x_0$:

$b^2 x_0 (Y - y_0) = -a^2 y_0 (X - x_0)$

$b^2 x_0 Y - b^2 x_0 y_0 = -a^2 y_0 X + a^2 x_0 y_0$

$a^2 y_0 X + b^2 x_0 Y = a^2 x_0 y_0 + b^2 x_0 y_0$

$a^2 y_0 X + b^2 x_0 Y = (a^2 + b^2)x_0 y_0$

Alternatively, dividing by $x_0 y_0$ (if $x_0 \neq 0, y_0 \neq 0$):

$\frac{a^2 X}{x_0} + \frac{b^2 Y}{y_0} = a^2 + b^2$

Another form is obtained by rearranging the equation before the last step:

$a^2 y_0 (X - x_0) + b^2 x_0 (Y - y_0) = 0$

Let's use the form $a^2 y_0 X + b^2 x_0 Y = (a^2 + b^2)x_0 y_0$.

Thus, the equation of the normal at $(x_0, y_0)$ is:

$a^2 y_0 (x - x_0) + b^2 x_0 (y - y_0) = 0$

or

$\frac{a^2 x}{x_0} + \frac{b^2 y}{y_0} = a^2 + b^2$ (if $x_0 \neq 0, y_0 \neq 0$)

(The first form holds even if $y_0=0$, giving $b^2 x_0(y-0)=0$, which simplifies to $y=0$, the correct normal equation.)

Given:

The equation of the curve is $y = \sqrt{3x-2}$.

The line to which the tangent is parallel is $4x - 2y + 5 = 0$.

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To Find:

The equation of the tangent line to the curve that is parallel to the given line.

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Solution:

First, find the slope of the given line $4x - 2y + 5 = 0$. We can rewrite it in the slope-intercept form $y = mx + c$.

$2y = 4x + 5$

$y = \frac{4}{2}x + \frac{5}{2}$

$y = 2x + \frac{5}{2}$

The slope of this line is $m_{line} = 2$.

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Since the required tangent line is parallel to the given line, the slope of the tangent ($m_{tangent}$) must be equal to the slope of the line.

$m_{tangent} = m_{line} = 2$.

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Next, find the slope of the tangent to the curve $y = \sqrt{3x-2}$ by differentiating with respect to $x$.

$y = (3x-2)^{1/2}$

Using the chain rule:

$\frac{dy}{dx} = \frac{1}{2}(3x-2)^{(1/2) - 1} \cdot \frac{d}{dx}(3x-2)$

$\frac{dy}{dx} = \frac{1}{2}(3x-2)^{-1/2} \cdot (3)$

$\frac{dy}{dx} = \frac{3}{2(3x-2)^{1/2}} = \frac{3}{2\sqrt{3x-2}}$

The slope of the tangent at any point $(x, y)$ on the curve is $\frac{3}{2\sqrt{3x-2}}$.

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We set the slope of the tangent equal to the required slope, which is 2:

$\frac{3}{2\sqrt{3x-2}} = 2$

Multiply both sides by $2\sqrt{3x-2}$:

$3 = 4\sqrt{3x-2}$

Square both sides:

$9 = 16(3x-2)$

$9 = 48x - 32$

$48x = 9 + 32$

$48x = 41$

$x = \frac{41}{48}$

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Now, find the corresponding $y$-coordinate of the point of tangency by substituting $x = \frac{41}{48}$ into the equation of the curve $y = \sqrt{3x-2}$.

$y = \sqrt{3\left(\frac{41}{48}\right) - 2}$

$y = \sqrt{\frac{41}{16} - 2}$

$y = \sqrt{\frac{41 - 32}{16}}$

$y = \sqrt{\frac{9}{16}} = \frac{3}{4}$

The point of tangency is $\left(\frac{41}{48}, \frac{3}{4}\right)$.

---

Finally, find the equation of the tangent line using the point-slope form $Y - y_1 = m(X - x_1)$, with the point $\left(\frac{41}{48}, \frac{3}{4}\right)$ and slope $m = 2$.

$Y - \frac{3}{4} = 2\left(X - \frac{41}{48}\right)$

$Y - \frac{3}{4} = 2X - \frac{41}{24}$

Multiply the entire equation by 24 to eliminate the fractions:

$24Y - 24\left(\frac{3}{4}\right) = 24(2X) - 24\left(\frac{41}{24}\right)$

$24Y - 18 = 48X - 41$

Rearrange the terms into the form $Ax + By + C = 0$:

$48X - 24Y - 41 + 18 = 0$

$48X - 24Y - 23 = 0$

---

The equation of the required tangent line is $48x - 24y - 23 = 0$.

Given:

The equation of the curve is $y = 2x^2 + 3 \sin x$.

We need to find the slope of the normal at $x = 0$.

---

Solution:

First, find the slope of the tangent to the curve by differentiating the equation with respect to $x$.

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

$\frac{dy}{dx} = 4x + 3 \cos x$

---

Now, evaluate the slope of the tangent ($m_{tangent}$) at $x = 0$ by substituting $x=0$ into the derivative.

$m_{tangent} = \frac{dy}{dx}\Big|_{x=0} = 4(0) + 3 \cos(0)$

$m_{tangent} = 0 + 3(1)$

$m_{tangent} = 3$

---

The slope of the normal ($m_{normal}$) is the negative reciprocal of the slope of the tangent.

$m_{normal} = -\frac{1}{m_{tangent}}$

$m_{normal} = -\frac{1}{3}$

---

Comparing this result with the given options:

(A) 3

(B) $\frac{1}{3}$

(C) -3

(D) $-\frac{1}{3}$

The calculated slope of the normal is $-\frac{1}{3}$, which corresponds to option (D).

Hence, the correct answer is (D).

Given:

The equation of the curve is $y^2 = 4x$.

The equation of the tangent line is $y = x + 1$.

---

To Find:

The point of tangency.

---

Solution:

Method 1: Using Slopes

1. Find the slope of the given tangent line $y = x + 1$. Comparing with $y = mx + c$, the slope is $m_{tangent} = 1$.

2. Find the slope of the tangent to the curve $y^2 = 4x$ by differentiating implicitly with respect to $x$.

$2y \frac{dy}{dx} = 4$

$\frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y}$

3. At the point of tangency, the slope of the curve's tangent must equal the slope of the given line.

$\frac{dy}{dx} = 1$

$\frac{2}{y} = 1 \implies y = 2$

4. Substitute $y=2$ into the equation of the curve to find the x-coordinate.

$y^2 = 4x$

$(2)^2 = 4x$

$4 = 4x \implies x = 1$

5. The point of tangency is $(1, 2)$.

---

Method 2: Solving Equations

1. The point of tangency must satisfy both the equation of the curve and the equation of the line.

2. Substitute $y = x + 1$ into the curve equation $y^2 = 4x$.

$(x+1)^2 = 4x$

$x^2 + 2x + 1 = 4x$

$x^2 - 2x + 1 = 0$

$(x-1)^2 = 0$

$x = 1$

3. Substitute $x=1$ back into the line equation $y = x + 1$.

$y = 1 + 1 = 2$

4. The point of intersection is $(1, 2)$. Since the quadratic equation for $x$ yielded only one solution, the line is indeed tangent to the curve at this point.

---

Comparing the result $(1, 2)$ with the given options:

(A) (1, 2)

(B) (2, 1)

(C) (1, -2)

(D) (-1, 2)

The calculated point of tangency is $(1, 2)$, which corresponds to option (A).

Hence, the correct answer is (A).

To Find:

Approximate value of $\sqrt{36.6}$ using differentials.

---

Solution:

Let the function be $y = f(x) = \sqrt{x}$.

We want to approximate $\sqrt{36.6}$. We choose a value of $x$ close to 36.6 for which $\sqrt{x}$ is known.

Let $x = 36$.

Then the change in $x$, denoted by $\Delta x$, is:

$\Delta x = 36.6 - x = 36.6 - 36 = 0.6$.

---

Now, we find the differential $dy$, which approximates the change in $y$ ($\Delta y$).

The formula for the differential is $dy = f'(x) \Delta x$ or $dy = \left(\frac{dy}{dx}\right) \Delta x$.

First, find the derivative of $y = \sqrt{x} = x^{1/2}$:

$\frac{dy}{dx} = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$

---

Evaluate the derivative at $x = 36$:

$\frac{dy}{dx}\Big|_{x=36} = \frac{1}{2\sqrt{36}} = \frac{1}{2 \times 6} = \frac{1}{12}$.

---

Now calculate the differential $dy$:

$dy = \left(\frac{dy}{dx}\Big|_{x=36}\right) \Delta x$

$dy = \left(\frac{1}{12}\right) (0.6)$

$dy = \frac{0.6}{12} = \frac{6}{120} = \frac{1}{20} = 0.05$.

---

The approximation formula using differentials is:

$f(x + \Delta x) \approx f(x) + dy$

$\sqrt{x + \Delta x} \approx \sqrt{x} + dy$

Substituting the values $x=36$ and $dy=0.05$:

$\sqrt{36.6} \approx \sqrt{36} + 0.05$

$\sqrt{36.6} \approx 6 + 0.05$

$\sqrt{36.6} \approx 6.05$.

---

Therefore, the approximate value of $\sqrt{36.6}$ using differentials is 6.05.

To Find:

Approximate value of $25^{1/3}$ using differentials.

---

Solution:

Let the function be $y = f(x) = x^{1/3}$.

We want to approximate $f(25)$. We need to choose a value of $x$ close to 25 for which the cube root $x^{1/3}$ is known.

The nearest perfect cube to 25 is 27 ($3^3=27$).

Let $x = 27$.

Then the change in $x$, denoted by $\Delta x$, is:

$\Delta x = 25 - x = 25 - 27 = -2$.

---

Now, we find the differential $dy$, which approximates the change in $y$ ($\Delta y$).

The formula for the differential is $dy = f'(x) \Delta x$ or $dy = \left(\frac{dy}{dx}\right) \Delta x$.

First, find the derivative of $y = x^{1/3}$:

$\frac{dy}{dx} = \frac{d}{dx}(x^{1/3}) = \frac{1}{3}x^{(1/3) - 1} = \frac{1}{3}x^{-2/3}$

$\frac{dy}{dx} = \frac{1}{3x^{2/3}}$

---

Evaluate the derivative at $x = 27$:

$\frac{dy}{dx}\Big|_{x=27} = \frac{1}{3(27)^{2/3}}$

Since $(27)^{2/3} = (27^{1/3})^2 = (3)^2 = 9$,

$\frac{dy}{dx}\Big|_{x=27} = \frac{1}{3 \times 9} = \frac{1}{27}$.

---

Now calculate the differential $dy$:

$dy = \left(\frac{dy}{dx}\Big|_{x=27}\right) \Delta x$

$dy = \left(\frac{1}{27}\right) (-2)$

$dy = -\frac{2}{27}$.

---

The approximation formula using differentials is:

$f(x + \Delta x) \approx f(x) + dy$

$(x + \Delta x)^{1/3} \approx x^{1/3} + dy$

Substituting the values $x=27$ and $dy = -\frac{2}{27}$:

$25^{1/3} \approx 27^{1/3} + \left(-\frac{2}{27}\right)$

$25^{1/3} \approx 3 - \frac{2}{27}$

$25^{1/3} \approx \frac{3 \times 27}{27} - \frac{2}{27} = \frac{81 - 2}{27} = \frac{79}{27}$

Calculating the decimal value:

$2/27 \approx 0.074$

$25^{1/3} \approx 3 - 0.074 = 2.926$.

---

Therefore, the approximate value of $25^{1/3}$ using differentials is $\frac{79}{27}$ or approximately 2.926.

Given:

The function is $f(x) = 3x^2 + 5x + 3$.

---

To Find:

The approximate value of $f(3.02)$ using differentials.

---

Solution:

We want to approximate $f(3.02)$. Let's choose a value of $x$ close to 3.02 where $f(x)$ is easy to calculate.

Let $x = 3$.

The change in $x$ is $\Delta x = 3.02 - x = 3.02 - 3 = 0.02$.

---

Calculate the value of the function at $x=3$:

$f(3) = 3(3)^2 + 5(3) + 3$

$f(3) = 3(9) + 15 + 3$

$f(3) = 27 + 15 + 3 = 45$.

---

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

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

$f'(x) = 6x + 5$.

---

Evaluate the derivative at $x=3$:

$f'(3) = 6(3) + 5 = 18 + 5 = 23$.

---

The differential $\Delta y$ (or $df$) is approximated by $dy$, where:

$dy = f'(x) \Delta x$

$dy = f'(3) \times (0.02)$

$dy = 23 \times 0.02 = 0.46$.

---

The approximation formula using differentials is:

$f(x + \Delta x) \approx f(x) + dy$

$f(3.02) \approx f(3) + 0.46$

$f(3.02) \approx 45 + 0.46$

$f(3.02) \approx 45.46$.

---

Therefore, the approximate value of $f(3.02)$ is 45.46.

Given:

A cube with side length $x$ meters.

The side length is increased by 2%.

---

To Find:

The approximate change in the volume (V) of the cube using differentials.

---

Solution:

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

$V = x^3$

---

The increase in the side length is 2% of $x$. Let $\Delta x$ be the change in the side length.

$\Delta x = 2\% \times x = \frac{2}{100} \times x = 0.02x$

---

We want to find the approximate change in volume, $\Delta V$, which can be approximated by the differential $dV$.

The differential $dV$ is given by the formula:

$dV = \left(\frac{dV}{dx}\right) \Delta x$

First, find the derivative of the volume function $V = x^3$ with respect to $x$:

$\frac{dV}{dx} = \frac{d}{dx}(x^3) = 3x^2$

---

Now, substitute the derivative and the change $\Delta x$ into the formula for $dV$:

$dV = (3x^2) (\Delta x)$

$dV = (3x^2) (0.02x)$

$dV = 3 \times 0.02 \times x^2 \times x$

$dV = 0.06 x^3$

---

The approximate change in the volume of the cube is $dV = 0.06x^3$ cubic meters.

This means that increasing the side length by 2% causes an approximate increase in volume of $0.06x^3 \text{ m}^3$.

Alternatively, since $V=x^3$, the approximate change is $0.06 V$. This represents a 6% approximate increase in volume.

The approximate change in volume is $0.06x^3 \text{ m}^3$.

Given:

Measured radius of the sphere, $r = 9$ cm.

Error in measuring the radius, $\Delta r = 0.03$ cm.

---

To Find:

The approximate error in calculating the volume of the sphere.

---

Solution:

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

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

---

We need to find the approximate error in volume, $\Delta V$. This can be approximated by the differential $dV$.

The formula for the differential $dV$ is:

$dV = \left(\frac{dV}{dr}\right) \Delta r$

First, find the derivative of the volume function $V$ with respect to the radius $r$:

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

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

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

---

Now, evaluate the derivative at the measured radius $r = 9$ cm:

$\frac{dV}{dr}\Big|_{r=9} = 4\pi (9)^2$

$\frac{dV}{dr}\Big|_{r=9} = 4\pi (81) = 324\pi$

---

Calculate the approximate error in volume, $dV$, using the formula $dV = \left(\frac{dV}{dr}\right) \Delta r$ with $\Delta r = 0.03$ cm:

$dV = (324\pi) (\Delta r)$

$dV = (324\pi) (0.03)$

$dV = (324 \times 0.03) \pi$

$dV = 9.72 \pi$

---

Therefore, the approximate error in calculating the volume is $9.72 \pi$ cm³.

General Approach:

Let $y = f(x)$. We use the approximation $f(x + \Delta x) \approx f(x) + \Delta y$, where $\Delta y \approx dy = \left(\frac{dy}{dx}\right) \Delta x$.

---

(i) $\sqrt{25.3}$

Let $y = f(x) = \sqrt{x}$. Let $x=25$ and $\Delta x = 0.3$.

$f(x) = \sqrt{25} = 5$.

$\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$.

At $x=25$, $\frac{dy}{dx} = \frac{1}{2\sqrt{25}} = \frac{1}{10} = 0.1$.

$dy = \left(\frac{dy}{dx}\right) \Delta x = (0.1)(0.3) = 0.03$.

$\sqrt{25.3} \approx f(x) + dy = 5 + 0.03 = 5.03$.

Approximate value: 5.030.

---

(ii) $\sqrt{49.5}$

Let $y = f(x) = \sqrt{x}$. Let $x=49$ and $\Delta x = 0.5$.

$f(x) = \sqrt{49} = 7$.

$\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$.

At $x=49$, $\frac{dy}{dx} = \frac{1}{2\sqrt{49}} = \frac{1}{14}$.

$dy = \left(\frac{dy}{dx}\right) \Delta x = \left(\frac{1}{14}\right)(0.5) = \frac{1}{28}$.

$\sqrt{49.5} \approx f(x) + dy = 7 + \frac{1}{28} \approx 7 + 0.0357...$

Approximate value: 7.036.

---

(iii) $\sqrt{0.6}$

Let $y = f(x) = \sqrt{x}$. Let $x=0.64$ and $\Delta x = 0.6 - 0.64 = -0.04$.

$f(x) = \sqrt{0.64} = 0.8$.

$\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$.

At $x=0.64$, $\frac{dy}{dx} = \frac{1}{2\sqrt{0.64}} = \frac{1}{2(0.8)} = \frac{1}{1.6} = \frac{10}{16} = 0.625$.

$dy = \left(\frac{dy}{dx}\right) \Delta x = (0.625)(-0.04) = -0.025$.

$\sqrt{0.6} \approx f(x) + dy = 0.8 - 0.025 = 0.775$.

Approximate value: 0.775.

---

(iv) $\left( 0.009 \right)^{\frac{1}{3}}$

Let $y = f(x) = x^{1/3}$. Let $x=0.008$ and $\Delta x = 0.009 - 0.008 = 0.001$.

$f(x) = (0.008)^{1/3} = 0.2$.

$\frac{dy}{dx} = \frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}}$.

At $x=0.008$, $\frac{dy}{dx} = \frac{1}{3(0.008)^{2/3}} = \frac{1}{3(0.2)^2} = \frac{1}{3(0.04)} = \frac{1}{0.12} = \frac{100}{12} = \frac{25}{3}$.

$dy = \left(\frac{dy}{dx}\right) \Delta x = \left(\frac{25}{3}\right)(0.001) = \frac{0.025}{3}$.

$(0.009)^{1/3} \approx f(x) + dy = 0.2 + \frac{0.025}{3} \approx 0.2 + 0.00833...$

Approximate value: 0.208.

---

(v) $\left( 0.999 \right)^{\frac{1}{10}}$

Let $y = f(x) = x^{1/10}$. Let $x=1$ and $\Delta x = 0.999 - 1 = -0.001$.

$f(x) = (1)^{1/10} = 1$.

$\frac{dy}{dx} = \frac{1}{10}x^{-9/10} = \frac{1}{10x^{9/10}}$.

At $x=1$, $\frac{dy}{dx} = \frac{1}{10(1)^{9/10}} = \frac{1}{10} = 0.1$.

$dy = \left(\frac{dy}{dx}\right) \Delta x = (0.1)(-0.001) = -0.0001$.

$(0.999)^{1/10} \approx f(x) + dy = 1 - 0.0001 = 0.9999$.

Approximate value: 1.000 (rounded to 3 decimal places).

---

(vi) $\left( 15 \right)^{\frac{1}{4}}$

Let $y = f(x) = x^{1/4}$. Let $x=16$ and $\Delta x = 15 - 16 = -1$.

$f(x) = (16)^{1/4} = 2$.

$\frac{dy}{dx} = \frac{1}{4}x^{-3/4} = \frac{1}{4x^{3/4}}$.

At $x=16$, $\frac{dy}{dx} = \frac{1}{4(16)^{3/4}} = \frac{1}{4(2)^3} = \frac{1}{32}$.

$dy = \left(\frac{dy}{dx}\right) \Delta x = \left(\frac{1}{32}\right)(-1) = -\frac{1}{32}$.

$(15)^{1/4} \approx f(x) + dy = 2 - \frac{1}{32} = 2 - 0.03125 = 1.96875$.

Approximate value: 1.969.

---

(vii) $\left( 26 \right)^{\frac{1}{3}}$

Let $y = f(x) = x^{1/3}$. Let $x=27$ and $\Delta x = 26 - 27 = -1$.

$f(x) = (27)^{1/3} = 3$.

$\frac{dy}{dx} = \frac{1}{3x^{2/3}}$.

At $x=27$, $\frac{dy}{dx} = \frac{1}{3(27)^{2/3}} = \frac{1}{3(3)^2} = \frac{1}{27}$.

$dy = \left(\frac{dy}{dx}\right) \Delta x = \left(\frac{1}{27}\right)(-1) = -\frac{1}{27}$.

$(26)^{1/3} \approx f(x) + dy = 3 - \frac{1}{27} \approx 3 - 0.037037... = 2.96296...$

Approximate value: 2.963.

---

(viii) $\left( 255 \right)^{\frac{1}{4}}$

Let $y = f(x) = x^{1/4}$. Let $x=256$ and $\Delta x = 255 - 256 = -1$.

$f(x) = (256)^{1/4} = 4$.

$\frac{dy}{dx} = \frac{1}{4x^{3/4}}$.

At $x=256$, $\frac{dy}{dx} = \frac{1}{4(256)^{3/4}} = \frac{1}{4(4)^3} = \frac{1}{256}$.

$dy = \left(\frac{dy}{dx}\right) \Delta x = \left(\frac{1}{256}\right)(-1) = -\frac{1}{256}$.

$(255)^{1/4} \approx f(x) + dy = 4 - \frac{1}{256} \approx 4 - 0.003906... = 3.99609...$

Approximate value: 3.996.

---

(ix) $\left( 82 \right)^{\frac{1}{4}}$

Let $y = f(x) = x^{1/4}$. Let $x=81$ and $\Delta x = 82 - 81 = 1$.

$f(x) = (81)^{1/4} = 3$.

$\frac{dy}{dx} = \frac{1}{4x^{3/4}}$.

At $x=81$, $\frac{dy}{dx} = \frac{1}{4(81)^{3/4}} = \frac{1}{4(3)^3} = \frac{1}{108}$.

$dy = \left(\frac{dy}{dx}\right) \Delta x = \left(\frac{1}{108}\right)(1) = \frac{1}{108}$.

$(82)^{1/4} \approx f(x) + dy = 3 + \frac{1}{108} \approx 3 + 0.009259... = 3.009259...$

Approximate value: 3.009.

---

(x) $\left( 401 \right)^{\frac{1}{2}}$

Let $y = f(x) = \sqrt{x}$. Let $x=400$ and $\Delta x = 401 - 400 = 1$.

$f(x) = \sqrt{400} = 20$.

$\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$.

At $x=400$, $\frac{dy}{dx} = \frac{1}{2\sqrt{400}} = \frac{1}{40} = 0.025$.

$dy = \left(\frac{dy}{dx}\right) \Delta x = (0.025)(1) = 0.025$.

$\sqrt{401} \approx f(x) + dy = 20 + 0.025 = 20.025$.

Approximate value: 20.025.

---

(xi) $\left( 0.0037 \right)^{\frac{1}{2}}$

Let $y = f(x) = \sqrt{x}$. Let $x=0.0036$ and $\Delta x = 0.0037 - 0.0036 = 0.0001$.

$f(x) = \sqrt{0.0036} = 0.06$.

$\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$.

At $x=0.0036$, $\frac{dy}{dx} = \frac{1}{2\sqrt{0.0036}} = \frac{1}{2(0.06)} = \frac{1}{0.12} = \frac{100}{12} = \frac{25}{3}$.

$dy = \left(\frac{dy}{dx}\right) \Delta x = \left(\frac{25}{3}\right)(0.0001) = \frac{0.0025}{3}$.

$\sqrt{0.0037} \approx f(x) + dy = 0.06 + \frac{0.0025}{3} \approx 0.06 + 0.000833... = 0.060833...$

Approximate value: 0.061.

---

(xii) $\left( 26.57 \right)^{\frac{1}{3}}$

Let $y = f(x) = x^{1/3}$. Let $x=27$ and $\Delta x = 26.57 - 27 = -0.43$.

$f(x) = (27)^{1/3} = 3$.

$\frac{dy}{dx} = \frac{1}{3x^{2/3}}$.

At $x=27$, $\frac{dy}{dx} = \frac{1}{3(27)^{2/3}} = \frac{1}{27}$.

$dy = \left(\frac{dy}{dx}\right) \Delta x = \left(\frac{1}{27}\right)(-0.43) = -\frac{0.43}{27}$.

$(26.57)^{1/3} \approx f(x) + dy = 3 - \frac{0.43}{27} \approx 3 - 0.015925... = 2.98407...$

Approximate value: 2.984.

---

(xiii) $\left( 81.5 \right)^{\frac{1}{4}}$

Let $y = f(x) = x^{1/4}$. Let $x=81$ and $\Delta x = 81.5 - 81 = 0.5$.

$f(x) = (81)^{1/4} = 3$.

$\frac{dy}{dx} = \frac{1}{4x^{3/4}}$.

At $x=81$, $\frac{dy}{dx} = \frac{1}{4(81)^{3/4}} = \frac{1}{108}$.

$dy = \left(\frac{dy}{dx}\right) \Delta x = \left(\frac{1}{108}\right)(0.5) = \frac{1}{216}$.

$(81.5)^{1/4} \approx f(x) + dy = 3 + \frac{1}{216} \approx 3 + 0.004629... = 3.004629...$

Approximate value: 3.005.

---

(xiv) $\left( 3.968 \right)^{\frac{3}{2}}$

Let $y = f(x) = x^{3/2}$. Let $x=4$ and $\Delta x = 3.968 - 4 = -0.032$.

$f(x) = (4)^{3/2} = (4^{1/2})^3 = 2^3 = 8$.

$\frac{dy}{dx} = \frac{3}{2}x^{1/2} = \frac{3}{2}\sqrt{x}$.

At $x=4$, $\frac{dy}{dx} = \frac{3}{2}\sqrt{4} = \frac{3}{2}(2) = 3$.

$dy = \left(\frac{dy}{dx}\right) \Delta x = (3)(-0.032) = -0.096$.

$(3.968)^{3/2} \approx f(x) + dy = 8 - 0.096 = 7.904$.

Approximate value: 7.904.

---

(xv) $\left( 32.15 \right)^{\frac{1}{5}}$

Let $y = f(x) = x^{1/5}$. Let $x=32$ and $\Delta x = 32.15 - 32 = 0.15$.

$f(x) = (32)^{1/5} = 2$.

$\frac{dy}{dx} = \frac{1}{5}x^{-4/5} = \frac{1}{5x^{4/5}}$.

At $x=32$, $\frac{dy}{dx} = \frac{1}{5(32)^{4/5}} = \frac{1}{5(2)^4} = \frac{1}{80}$.

$dy = \left(\frac{dy}{dx}\right) \Delta x = \left(\frac{1}{80}\right)(0.15) = \frac{0.15}{80} = \frac{15}{8000} = \frac{3}{1600}$.

$(32.15)^{1/5} \approx f(x) + dy = 2 + \frac{3}{1600} = 2 + 0.001875 = 2.001875$.

Approximate value: 2.002.

Given:

The function is $f(x) = 4x^2 + 5x + 2$.

---

To Find:

The approximate value of $f(2.01)$ using differentials.

---

Solution:

We want to approximate $f(2.01)$. Let's choose a value of $x$ close to 2.01 where $f(x)$ is easy to calculate.

Let $x = 2$.

The change in $x$ is $\Delta x = 2.01 - x = 2.01 - 2 = 0.01$.

---

Calculate the value of the function at $x=2$:

$f(2) = 4(2)^2 + 5(2) + 2$

$f(2) = 4(4) + 10 + 2$

$f(2) = 16 + 10 + 2 = 28$.

---

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

$f'(x) = \frac{d}{dx}(4x^2 + 5x + 2)$

$f'(x) = 8x + 5$.

---

Evaluate the derivative at $x=2$:

$f'(2) = 8(2) + 5 = 16 + 5 = 21$.

---

The differential $\Delta y$ (or $df$) is approximated by $dy$, where:

$dy = f'(x) \Delta x$

$dy = f'(2) \times (0.01)$

$dy = 21 \times 0.01 = 0.21$.

---

The approximation formula using differentials is:

$f(x + \Delta x) \approx f(x) + dy$

$f(2.01) \approx f(2) + 0.21$

$f(2.01) \approx 28 + 0.21$

$f(2.01) \approx 28.21$.

---

Therefore, the approximate value of $f(2.01)$ is 28.21.

Given:

The function is $f(x) = x^3 - 7x^2 + 15$.

---

To Find:

The approximate value of $f(5.001)$ using differentials.

---

Solution:

We want to approximate $f(5.001)$. Let's choose a value of $x$ close to 5.001 where $f(x)$ is easy to calculate.

Let $x = 5$.

The change in $x$ is $\Delta x = 5.001 - x = 5.001 - 5 = 0.001$.

---

Calculate the value of the function at $x=5$:

$f(5) = (5)^3 - 7(5)^2 + 15$

$f(5) = 125 - 7(25) + 15$

$f(5) = 125 - 175 + 15$

$f(5) = 140 - 175 = -35$.

---

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

$f'(x) = \frac{d}{dx}(x^3 - 7x^2 + 15)$

$f'(x) = 3x^2 - 14x$.

---

Evaluate the derivative at $x=5$:

$f'(5) = 3(5)^2 - 14(5)$

$f'(5) = 3(25) - 70$

$f'(5) = 75 - 70 = 5$.

---

The differential $\Delta y$ (or $df$) is approximated by $dy$, where:

$dy = f'(x) \Delta x$

$dy = f'(5) \times (0.001)$

$dy = 5 \times 0.001 = 0.005$.

---

The approximation formula using differentials is:

$f(x + \Delta x) \approx f(x) + dy$

$f(5.001) \approx f(5) + 0.005$

$f(5.001) \approx -35 + 0.005$

$f(5.001) \approx -34.995$.

---

Therefore, the approximate value of $f(5.001)$ is -34.995.

Given:

A cube with side length $x$ meters.

The side length is increased by 1%.

---

To Find:

The approximate change in the volume (V) of the cube using differentials.

---

Solution:

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

$V = x^3$

---

The increase in the side length is 1% of $x$. Let $\Delta x$ be the change in the side length.

$\Delta x = 1\% \times x = \frac{1}{100} \times x = 0.01x$

---

We want to find the approximate change in volume, $\Delta V$. This can be approximated by the differential $dV$.

The formula for the differential $dV$ is given by:

$dV = \left(\frac{dV}{dx}\right) \Delta x$

First, find the derivative of the volume function $V = x^3$ with respect to $x$:

$\frac{dV}{dx} = \frac{d}{dx}(x^3) = 3x^2$

---

Now, substitute the derivative and the change $\Delta x$ into the formula for $dV$:

$dV = (3x^2) (\Delta x)$

$dV = (3x^2) (0.01x)$

$dV = 3 \times 0.01 \times x^2 \times x$

$dV = 0.03 x^3$

---

The approximate change in the volume of the cube is $dV = 0.03x^3$ cubic meters.

This means that increasing the side length by 1% causes an approximate increase in volume of $0.03x^3 \text{ m}^3$.

(Alternatively, since $V=x^3$, the approximate change is $0.03 V$. This represents a 3% approximate increase in volume.)

Given:

A cube with side length $x$ meters.

The side length is decreased by 1%.

---

To Find:

The approximate change in the surface area (S) of the cube using differentials.

---

Solution:

The surface area S of a cube with side length $x$ is given by the formula (since a cube has 6 square faces, each with area $x^2$):

$S = 6x^2$

---

The decrease in the side length is 1% of $x$. Let $\Delta x$ be the change in the side length.

Since it's a decrease, the change is negative:

$\Delta x = -1\% \times x = -\frac{1}{100} \times x = -0.01x$

---

We want to find the approximate change in surface area, $\Delta S$. This can be approximated by the differential $dS$.

The formula for the differential $dS$ is given by:

$dS = \left(\frac{dS}{dx}\right) \Delta x$

First, find the derivative of the surface area function $S = 6x^2$ with respect to $x$:

$\frac{dS}{dx} = \frac{d}{dx}(6x^2) = 6(2x) = 12x$

---

Now, substitute the derivative and the change $\Delta x$ into the formula for $dS$:

$dS = (12x) (\Delta x)$

$dS = (12x) (-0.01x)$

$dS = - (12 \times 0.01) \times x \times x$

$dS = -0.12 x^2$

---

The approximate change in the surface area of the cube is $dS = -0.12x^2$ square meters.

The negative sign indicates a decrease in the surface area.

Therefore, the approximate change (decrease) in surface area is $0.12x^2 \text{ m}^2$.

(Alternatively, since $S=6x^2$, the approximate change is $dS = -0.12x^2 = -\frac{0.12}{6} (6x^2) = -0.02 S$. This represents a 2% approximate decrease in surface area.)

Given:

Measured radius of the sphere, $r = 7$ m.

Error in measuring the radius, $\Delta r = 0.02$ m.

---

To Find:

The approximate error in calculating the volume (V) of the sphere.

---

Solution:

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

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

---

We need to find the approximate error in volume, $\Delta V$. This can be approximated by the differential $dV$.

The formula for the differential $dV$ is:

$dV = \left(\frac{dV}{dr}\right) \Delta r$

First, find the derivative of the volume function $V$ with respect to the radius $r$:

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

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

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

---

Now, evaluate the derivative at the measured radius $r = 7$ m:

$\frac{dV}{dr}\Big|_{r=7} = 4\pi (7)^2$

$\frac{dV}{dr}\Big|_{r=7} = 4\pi (49)$

$\frac{dV}{dr}\Big|_{r=7} = 196\pi$

---

Calculate the approximate error in volume, $dV$, using the formula $dV = \left(\frac{dV}{dr}\right) \Delta r$ with $\Delta r = 0.02$ m:

$dV = (196\pi) (\Delta r)$

$dV = (196\pi) (0.02)$

$dV = (196 \times 0.02) \pi$

$dV = 3.92 \pi$

---

Therefore, the approximate error in calculating the volume is $3.92 \pi$ m³.

Given:

Measured radius of the sphere, $r = 9$ m.

Error in measuring the radius, $\Delta r = 0.03$ m.

---

To Find:

The approximate error in calculating the surface area (S) of the sphere.

---

Solution:

The surface area S of a sphere with radius $r$ is given by the formula:

$S = 4\pi r^2$

---

We need to find the approximate error in surface area, $\Delta S$. This can be approximated by the differential $dS$.

The formula for the differential $dS$ is:

$dS = \left(\frac{dS}{dr}\right) \Delta r$

First, find the derivative of the surface area function $S$ with respect to the radius $r$:

$\frac{dS}{dr} = \frac{d}{dr}\left(4\pi r^2\right)$

$\frac{dS}{dr} = 4\pi (2r)$

$\frac{dS}{dr} = 8\pi r$

---

Now, evaluate the derivative at the measured radius $r = 9$ m:

$\frac{dS}{dr}\Big|_{r=9} = 8\pi (9)$

$\frac{dS}{dr}\Big|_{r=9} = 72\pi$

---

Calculate the approximate error in surface area, $dS$, using the formula $dS = \left(\frac{dS}{dr}\right) \Delta r$ with $\Delta r = 0.03$ m:

$dS = (72\pi) (\Delta r)$

$dS = (72\pi) (0.03)$

$dS = (72 \times 0.03) \pi$

$dS = 2.16 \pi$

---

Therefore, the approximate error in calculating the surface area is $2.16 \pi$ m².

Given:

The function is $f(x) = 3x^2 + 15x + 5$.

---

To Find:

The approximate value of $f(3.02)$ using differentials.

---

Solution:

We use the approximation formula $f(x + \Delta x) \approx f(x) + dy$, where $dy = f'(x) \Delta x$.

We want to approximate $f(3.02)$. Let $x = 3$.

Then $\Delta x = 3.02 - 3 = 0.02$.

---

Calculate the value of $f(x)$ at $x=3$:

$f(3) = 3(3)^2 + 15(3) + 5$

$f(3) = 3(9) + 45 + 5$

$f(3) = 27 + 45 + 5 = 77$.

---

Calculate the derivative $f'(x)$:

$f'(x) = \frac{d}{dx}(3x^2 + 15x + 5)$

$f'(x) = 6x + 15$.

---

Evaluate the derivative $f'(x)$ at $x=3$:

$f'(3) = 6(3) + 15 = 18 + 15 = 33$.

---

Calculate the differential $dy$:

$dy = f'(3) \Delta x$

$dy = 33 \times 0.02 = 0.66$.

---

Apply the approximation formula:

$f(3.02) \approx f(3) + dy$

$f(3.02) \approx 77 + 0.66$

$f(3.02) \approx 77.66$.

---

Comparing this result with the given options:

(A) 47.66

(B) 57.66

(C) 67.66

(D) 77.66

The calculated approximate value is 77.66, which matches option (D).

Hence, the correct answer is (D).

Given:

A cube with side length $x$ meters.

The side length is increased by 3%.

---

To Find:

The approximate change in the volume (V) of the cube using differentials.

---

Solution:

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

$V = x^3$

---

The increase in the side length is 3% of $x$. Let $\Delta x$ be the change in the side length.

$\Delta x = 3\% \times x = \frac{3}{100} \times x = 0.03x$

---

We want to find the approximate change in volume, $\Delta V$. This can be approximated by the differential $dV$.

The formula for the differential $dV$ is given by:

$dV = \left(\frac{dV}{dx}\right) \Delta x$

First, find the derivative of the volume function $V = x^3$ with respect to $x$:

$\frac{dV}{dx} = \frac{d}{dx}(x^3) = 3x^2$

---

Now, substitute the derivative and the change $\Delta x$ into the formula for $dV$:

$dV = (3x^2) (\Delta x)$

$dV = (3x^2) (0.03x)$

$dV = 3 \times 0.03 \times x^2 \times x$

$dV = 0.09 x^3$

---

The approximate change in the volume of the cube is $dV = 0.09x^3$ cubic meters.

Comparing this result with the given options:

(A) 0.06 x³ m³

(B) 0.6 x³ m³

(C) 0.09 x³ m³

(D) 0.9 x³ m³

The calculated approximate change in volume is $0.09 x^3$ m³, which matches option (C).

Hence, the correct answer is (C).

Given:

The function is $f(x) = x^2$, where the domain is all real numbers ($x \in \mathbf{R}$).

---

To Find:

The maximum and minimum values of the function $f(x)$, if they exist.

---

Solution:

Consider the function $f(x) = x^2$.

Minimum Value:

We know that the square of any real number is always non-negative.

That is, $x^2 \ge 0$ for all $x \in \mathbf{R}$.

The smallest possible value of $x^2$ occurs when $x=0$.

At $x=0$, $f(0) = (0)^2 = 0$.

Since $f(x) = x^2 \ge 0$ for all $x \in \mathbf{R}$, the minimum value of the function is 0, and it is attained at $x=0$.

---

Maximum Value:

As the value of $x$ increases without bound (approaches $\infty$) or decreases without bound (approaches $-\infty$), the value of $x^2$ also increases without bound.

For example, if we take $x = 10$, $f(10) = 100$. If we take $x = 100$, $f(100) = 10000$. If we take $x = -10$, $f(-10) = 100$. If we take $x = -100$, $f(-100) = 10000$.

We can always find a value of $x$ such that $f(x)$ is greater than any given large number.

Therefore, the function $f(x) = x^2$ does not have a maximum value on the set of real numbers.

---

Conclusion:

The function $f(x) = x^2$ has a minimum value of 0 at $x = 0$.

The function $f(x) = x^2$ has no maximum value.

Given:

The function is $f(x) = |x|$, where the domain is all real numbers ($x \in \mathbf{R}$).

---

To Find:

The maximum and minimum values of the function $f(x)$, if they exist.

---

Solution:

Consider the function $f(x) = |x|$. The absolute value of a number is its distance from zero, which is always non-negative.

Minimum Value:

By definition of the absolute value:

$f(x) = |x| = \begin{cases} x & \text{, if } x \ge 0 \\ -x & \text{, if } x < 0 \end{cases}$

If $x \ge 0$, then $f(x) = x \ge 0$.

If $x < 0$, then $-x > 0$, so $f(x) = -x > 0$.

In all cases, $f(x) = |x| \ge 0$ for all $x \in \mathbf{R}$.

The smallest possible value of $|x|$ occurs when $x=0$.

At $x=0$, $f(0) = |0| = 0$.

Since $f(x) = |x| \ge 0$ for all $x \in \mathbf{R}$, the minimum value of the function is 0, and it is attained at $x=0$.

---

Maximum Value:

As the value of $x$ increases towards $\infty$, the value of $|x| = x$ also increases towards $\infty$.

As the value of $x$ decreases towards $-\infty$, the value of $|x| = -x$ increases towards $\infty$.

For example, if $x = 10$, $f(10) = |10| = 10$. If $x = -100$, $f(-100) = |-100| = 100$.

We can always find a value of $x$ such that $f(x) = |x|$ is greater than any given large number.

Therefore, the function $f(x) = |x|$ does not have a maximum value on the set of real numbers.

---

Conclusion:

The function $f(x) = |x|$ has a minimum value of 0 at $x = 0$.

The function $f(x) = |x|$ has no maximum value.

Given:

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

The domain of the function is the open interval $(0, 1)$, which means $0 < x < 1$.

---

To Find:

The maximum and minimum values of the function $f(x)$ on the interval $(0, 1)$, if they exist.

---

Solution:

Consider the function $f(x) = x$ defined on the open interval $(0, 1)$.

Minimum Value:

As $x$ takes values closer and closer to 0 from the right side (e.g., 0.1, 0.01, 0.001, ...), the value of $f(x) = x$ also gets closer and closer to 0.

However, since the interval is open, $x$ can never be exactly 0. For any value $c > 0$ (no matter how small), we can always find an $x$ in the interval $(0, 1)$ such that $0 < x < c$. For example, $x = c/2$.

This means that $f(x)$ can be made arbitrarily close to 0, but it never actually reaches 0 within the domain $(0, 1)$.

Therefore, the function $f(x) = x$ has no minimum value in the interval $(0, 1)$. The greatest lower bound (infimum) is 0, but it is not attained.

---

Maximum Value:

Similarly, as $x$ takes values closer and closer to 1 from the left side (e.g., 0.9, 0.99, 0.999, ...), the value of $f(x) = x$ also gets closer and closer to 1.

However, since the interval is open, $x$ can never be exactly 1. For any value $c < 1$ (no matter how close to 1), we can always find an $x$ in the interval $(0, 1)$ such that $c < x < 1$. For example, $x = (c+1)/2$.

This means that $f(x)$ can be made arbitrarily close to 1, but it never actually reaches 1 within the domain $(0, 1)$.

Therefore, the function $f(x) = x$ has no maximum value in the interval $(0, 1)$. The least upper bound (supremum) is 1, but it is not attained.

---

Conclusion:

The function $f(x) = x$ defined on the interval $(0, 1)$ has neither a maximum value nor a minimum value.

Given:

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

---

To Find:

All points of local maxima and local minima of the function $f(x)$.

---

Solution:

To find the points of local maxima and minima, we first find the critical points by finding the first derivative and setting it to zero.

Step 1: Find the first derivative, $f'(x)$.

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

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

---

Step 2: Find the critical points.

Set $f'(x) = 0$ and solve for $x$.

$3x^2 - 3 = 0$

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

$x^2 - 1 = 0$

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

The critical points are $x = 1$ and $x = -1$.

---

Step 3: Use the Second Derivative Test to classify the critical points.

Find the second derivative, $f''(x)$.

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

$f''(x) = 6x$

Now, evaluate $f''(x)$ at each critical point:

• At $x = 1$:

  $f''(1) = 6(1) = 6$.

  Since $f''(1) > 0$, the function $f(x)$ has a local minimum at $x = 1$.

  The value of the local minimum is $f(1) = (1)^3 - 3(1) + 3 = 1 - 3 + 3 = 1$.

  So, the point of local minimum is $(1, 1)$.

• At $x = -1$:

  $f''(-1) = 6(-1) = -6$.

  Since $f''(-1) < 0$, the function $f(x)$ has a local maximum at $x = -1$.

  The value of the local maximum is $f(-1) = (-1)^3 - 3(-1) + 3 = -1 + 3 + 3 = 5$.

  So, the point of local maximum is $(-1, 5)$.

---

Conclusion:

The function $f(x) = x^3 - 3x + 3$ has:

A point of local maximum at $x = -1$ (the point is $(-1, 5)$ and the local maximum value is 5).

A point of local minimum at $x = 1$ (the point is $(1, 1)$ and the local minimum value is 1).

Given:

The function is $f(x) = 2x^3 - 6x^2 + 6x + 5$.

---

To Find:

All points of local maxima and local minima of the function $f(x)$.

---

Solution:

To find the points of local maxima and minima, we first find the critical points by finding the first derivative and setting it to zero.

Step 1: Find the first derivative, $f'(x)$.

$f'(x) = \frac{d}{dx}(2x^3 - 6x^2 + 6x + 5)$

$f'(x) = 6x^2 - 12x + 6$

---

Step 2: Find the critical points.

Set $f'(x) = 0$ and solve for $x$.

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

Divide the equation by 6:

$x^2 - 2x + 1 = 0$

Factor the quadratic expression:

$(x - 1)^2 = 0$

This gives $x = 1$ as the only critical point.

---

Step 3: Use the Second Derivative Test (or First Derivative Test) to classify the critical point.

Find the second derivative, $f''(x)$.

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

$f''(x) = 12x - 12$

Now, evaluate $f''(x)$ at the critical point $x = 1$:

$f''(1) = 12(1) - 12 = 0$.

Since $f''(1) = 0$, the Second Derivative Test fails. We must use the First Derivative Test.

---

First Derivative Test:

We examine the sign of $f'(x) = 6x^2 - 12x + 6 = 6(x-1)^2$ around the critical point $x=1$.

• For $x < 1$: Choose a value slightly less than 1, say $x=0$.

  $f'(0) = 6(0-1)^2 = 6(1) = 6 > 0$.

• For $x > 1$: Choose a value slightly greater than 1, say $x=2$.

  $f'(2) = 6(2-1)^2 = 6(1)^2 = 6 > 0$.

Since the sign of $f'(x)$ does not change as $x$ passes through the critical point $x=1$ (it is positive on both sides), the point $x=1$ is neither a local maximum nor a local minimum.

Such a point, where the derivative is zero but does not change sign, is called a point of inflection.

---

Conclusion:

The function $f(x) = 2x^3 - 6x^2 + 6x + 5$ has no points of local maxima or local minima.

The point $x=1$ is a point of inflection.

Given:

The function is $f(x) = 3 + |x|$, where the domain is all real numbers ($x \in \mathbf{R}$).

---

To Find:

The local minimum value of the function $f(x)$.

---

Solution:

Consider the function $f(x) = 3 + |x|$.

We know that the absolute value function $|x|$ has the property:

$|x| \ge 0$ for all $x \in \mathbf{R}$.

The minimum value of $|x|$ is 0, which occurs when $x=0$.

---

Adding 3 to both sides of the inequality $|x| \ge 0$:

$3 + |x| \ge 3 + 0$

$f(x) \ge 3$ for all $x \in \mathbf{R}$.

---

This means that the value of the function $f(x)$ is always greater than or equal to 3.

The smallest value the function can attain is 3.

This minimum value is attained when $|x|$ is minimum, i.e., when $x=0$.

Let's calculate $f(0)$:

$f(0) = 3 + |0| = 3 + 0 = 3$.

---

Since $f(x) \ge f(0)$ for all $x \in \mathbf{R}$ (and thus for all $x$ in any open interval containing 0), the function $f(x)$ has a local minimum at $x=0$.

The value of this local minimum is $f(0) = 3$.

(Note: $f(x)$ is not differentiable at $x=0$, which is a critical point where the minimum occurs.)

---

Conclusion:

The local minimum value of the function $f(x) = 3 + |x|$ is 3, which occurs at $x=0$.

Given:

The function is $f(x) = 3x^4 + 4x^3 - 12x^2 + 12$.

---

To Find:

The local maximum and local minimum values of the function $f(x)$.

---

Solution:

To find the local maximum and minimum values, we use the first and second derivative tests.

Step 1: Find the first derivative, $f'(x)$.

$f'(x) = \frac{d}{dx}(3x^4 + 4x^3 - 12x^2 + 12)$

$f'(x) = 12x^3 + 12x^2 - 24x$

---

Step 2: Find the critical points.

Set $f'(x) = 0$ and solve for $x$.

$12x^3 + 12x^2 - 24x = 0$

Factor out $12x$:

$12x(x^2 + x - 2) = 0$

Factor the quadratic expression:

$12x(x+2)(x-1) = 0$

The critical points are $x = 0$, $x = -2$, and $x = 1$.

---

Step 3: Find the second derivative, $f''(x)$.

$f''(x) = \frac{d}{dx}(12x^3 + 12x^2 - 24x)$

$f''(x) = 36x^2 + 24x - 24$

---

Step 4: Use the Second Derivative Test to classify the critical points.

Evaluate $f''(x)$ at each critical point:

• At $x = 0$:

  $f''(0) = 36(0)^2 + 24(0) - 24 = -24$.

  Since $f''(0) < 0$, the function $f(x)$ has a local maximum at $x = 0$.

• At $x = 1$:

  $f''(1) = 36(1)^2 + 24(1) - 24 = 36 + 24 - 24 = 36$.

  Since $f''(1) > 0$, the function $f(x)$ has a local minimum at $x = 1$.

• At $x = -2$:

  $f''(-2) = 36(-2)^2 + 24(-2) - 24 = 36(4) - 48 - 24 = 144 - 72 = 72$.

  Since $f''(-2) > 0$, the function $f(x)$ has a local minimum at $x = -2$.

---

Step 5: Calculate the local maximum and minimum values.

• Local maximum value at $x = 0$:

  $f(0) = 3(0)^4 + 4(0)^3 - 12(0)^2 + 12 = 0 + 0 - 0 + 12 = 12$.

• Local minimum value at $x = 1$:

  $f(1) = 3(1)^4 + 4(1)^3 - 12(1)^2 + 12 = 3 + 4 - 12 + 12 = 7$.

• Local minimum value at $x = -2$:

  $f(-2) = 3(-2)^4 + 4(-2)^3 - 12(-2)^2 + 12$

  $f(-2) = 3(16) + 4(-8) - 12(4) + 12$

  $f(-2) = 48 - 32 - 48 + 12 = -20$.

---

Conclusion:

The function $f(x) = 3x^4 + 4x^3 - 12x^2 + 12$ has:

A local maximum value of 12 at $x = 0$.

A local minimum value of 7 at $x = 1$.

A local minimum value of -20 at $x = -2$.

Given:

The function is $f(x) = 2x^3 - 6x^2 + 6x + 5$.

---

To Find:

All points of local maxima and local minima of the function $f(x)$.

---

Solution:

To find the points of local maxima and minima, we first find the critical points by finding the first derivative and setting it to zero.

Step 1: Find the first derivative, $f'(x)$.

$f'(x) = \frac{d}{dx}(2x^3 - 6x^2 + 6x + 5)$

$f'(x) = 6x^2 - 12x + 6$

---

Step 2: Find the critical points.

Set $f'(x) = 0$ and solve for $x$.

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

Divide the equation by 6:

$x^2 - 2x + 1 = 0$

Factor the quadratic expression:

$(x - 1)^2 = 0$

This gives $x = 1$ as the only critical point.

---

Step 3: Use the Second Derivative Test (or First Derivative Test) to classify the critical point.

Find the second derivative, $f''(x)$.

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

$f''(x) = 12x - 12$

Now, evaluate $f''(x)$ at the critical point $x = 1$:

$f''(1) = 12(1) - 12 = 0$.

Since $f''(1) = 0$, the Second Derivative Test fails. We must use the First Derivative Test.

---

First Derivative Test:

We examine the sign of $f'(x) = 6x^2 - 12x + 6 = 6(x-1)^2$ around the critical point $x=1$.

• For $x < 1$: Choose a value slightly less than 1, say $x=0$.

  $f'(0) = 6(0-1)^2 = 6(1) = 6 > 0$.

• For $x > 1$: Choose a value slightly greater than 1, say $x=2$.

  $f'(2) = 6(2-1)^2 = 6(1)^2 = 6 > 0$.

Since the sign of $f'(x)$ does not change as $x$ passes through the critical point $x=1$ (it is positive on both sides), the point $x=1$ is neither a local maximum nor a local minimum.

Such a point, where the derivative is zero but does not change sign, is called a point of inflection.

---

Conclusion:

The function $f(x) = 2x^3 - 6x^2 + 6x + 5$ has no points of local maxima or local minima.

The point $x=1$ is a point of inflection.

Given:

Two positive numbers, let them be $x$ and $y$.

Condition 1: The numbers are positive, i.e., $x > 0$ and $y > 0$.

Condition 2: The sum of the numbers is 15, i.e., $x + y = 15$.

---

To Find:

The values of $x$ and $y$ such that the sum of their squares, $S = x^2 + y^2$, is minimum.

---

Solution:

Let the two positive numbers be $x$ and $y$.

We are given $x > 0$, $y > 0$ and $x + y = 15$.

We want to minimize the sum of squares, $S = x^2 + y^2$.

From the constraint $x + y = 15$, we can express $y$ in terms of $x$:

$y = 15 - x$.

Substitute this into the expression for $S$ to get $S$ as a function of $x$:

$S(x) = x^2 + (15 - x)^2$

Expand the expression:

$S(x) = x^2 + (225 - 30x + x^2)$

$S(x) = 2x^2 - 30x + 225$.

Since $x > 0$ and $y = 15 - x > 0$, we have $0 < x < 15$. The domain for $x$ is the interval $(0, 15)$.

---

To find the minimum value of $S(x)$, we first find the critical points by calculating the first derivative and setting it to zero.

$S'(x) = \frac{d}{dx}(2x^2 - 30x + 225)$

$S'(x) = 4x - 30$.

Set $S'(x) = 0$:

$4x - 30 = 0$

$4x = 30$

$x = \frac{30}{4} = \frac{15}{2} = 7.5$.

The critical point $x = 7.5$ lies within the domain $(0, 15)$.

---

Now, we use the second derivative test to determine if this critical point corresponds to a minimum.

$S''(x) = \frac{d}{dx}(4x - 30)$

$S''(x) = 4$.

Evaluate the second derivative at the critical point $x = 7.5$:

$S''(7.5) = 4$.

Since $S''(7.5) = 4 > 0$, the function $S(x)$ has a local minimum at $x = 7.5$. As the second derivative is always positive, this local minimum is the absolute minimum value for $S(x)$ on the interval $(0, 15)$.

---

Now, find the value of $y$ when $x = 7.5$:

$y = 15 - x$

$y = 15 - 7.5 = 7.5$.

Both $x = 7.5$ and $y = 7.5$ are positive, satisfying the initial condition.

---

Therefore, the two positive numbers whose sum is 15 and the sum of whose squares is minimum are 7.5 and 7.5.

Given:

The parabola $y = x^2$.

The point $P(0, c)$.

The constraint $\frac{1}{2} \le c \le 5$.

---

To Find:

The shortest distance between the point $P(0, c)$ and the parabola $y = x^2$.

---

Solution:

Let $Q(x, y)$ be any point on the parabola $y = x^2$. So, the coordinates of Q are $(x, x^2)$.

The distance $D$ between the point $P(0, c)$ and the point $Q(x, x^2)$ is given by the distance formula:

$D = \sqrt{(x - 0)^2 + (x^2 - c)^2}$

$D = \sqrt{x^2 + (x^2 - c)^2}$

Minimizing the distance $D$ is equivalent to minimizing the square of the distance, $S = D^2$.

$S(x) = x^2 + (x^2 - c)^2$

To find the minimum value of $S(x)$, we first find its critical points by calculating the derivative $S'(x)$ and setting it to zero.

$S'(x) = \frac{d}{dx} [x^2 + (x^2 - c)^2]$

$S'(x) = 2x + 2(x^2 - c) \cdot (2x)$

$S'(x) = 2x + 4x(x^2 - c)$

$S'(x) = 2x (1 + 2(x^2 - c))$

$S'(x) = 2x (1 + 2x^2 - 2c)$

Set $S'(x) = 0$:

$2x (2x^2 - 2c + 1) = 0$

This implies either $2x = 0$ or $2x^2 - 2c + 1 = 0$.

Case 1: $x = 0$.

Case 2: $2x^2 = 2c - 1 \implies x^2 = c - \frac{1}{2}$.

Since we are given $\frac{1}{2} \le c$, we have $c - \frac{1}{2} \ge 0$. Thus, $x^2 = c - \frac{1}{2}$ has real solutions $x = \pm \sqrt{c - \frac{1}{2}}$.

The critical points are $x = 0$ and $x = \pm \sqrt{c - \frac{1}{2}}$ (if $c \ge 1/2$).

---

Now, we use the second derivative test to classify these critical points.

$S''(x) = \frac{d}{dx}[4x^3 + (2 - 4c)x]$

$S''(x) = 12x^2 + 2 - 4c$

Evaluate $S''(x)$ at the critical points:

• At $x = 0$: $S''(0) = 12(0)^2 + 2 - 4c = 2 - 4c$. Since $c \ge 1/2$, $4c \ge 2$, so $2 - 4c \le 0$. If $c > 1/2$, $S''(0) < 0$, indicating a local maximum at $x=0$. If $c = 1/2$, $S''(0) = 0$, the test is inconclusive. We check the first derivative $S'(x) = 2x(2x^2) = 4x^3$. Since $S'(x)$ changes from negative to positive at $x=0$, it's a local minimum when $c=1/2$.
• At $x = \pm \sqrt{c - \frac{1}{2}}$ (when $c > 1/2$): $x^2 = c - \frac{1}{2}$. $S''\left(\pm \sqrt{c - \frac{1}{2}}\right) = 12\left(c - \frac{1}{2}\right) + 2 - 4c$ $S'' = 12c - 6 + 2 - 4c = 8c - 4$. Since $c > 1/2$, $8c > 4$, so $8c - 4 > 0$. This indicates local minima at $x = \pm \sqrt{c - \frac{1}{2}}$.

---

The minimum value of $S(x)$ occurs at the local minima.

If $c=1/2$, the minimum occurs at $x=0$. $S_{min} = S(0) = 0^2 + (0^2 - 1/2)^2 = (-1/2)^2 = 1/4$. The shortest distance is $D = \sqrt{1/4} = 1/2$.

If $c > 1/2$, the minimum occurs at $x = \pm \sqrt{c - \frac{1}{2}}$. Let's calculate $S$ at these points.

$S_{min} = x^2 + (x^2 - c)^2 = \left(c - \frac{1}{2}\right) + \left(\left(c - \frac{1}{2}\right) - c\right)^2$

$S_{min} = c - \frac{1}{2} + \left(-\frac{1}{2}\right)^2$

$S_{min} = c - \frac{1}{2} + \frac{1}{4}$

$S_{min} = c - \frac{1}{4}$.

The shortest distance is $D = \sqrt{c - \frac{1}{4}}$.

Note that this formula also works for $c=1/2$: $D = \sqrt{1/2 - 1/4} = \sqrt{1/4} = 1/2$.

The shortest distance depends only on $c$, and the condition $c \le 5$ doesn't change the method or the expression for the shortest distance, it just limits the possible values of the shortest distance.

---

Therefore, the shortest distance from the point $(0, c)$ to the parabola $y=x^2$ for $c \ge 1/2$ is $\sqrt{c - \frac{1}{4}}$.

Given:

Two vertical poles AP and BQ are located at points A and B respectively.

Height of pole AP = 16 m.

Height of pole BQ = 22 m.

Distance between the bases, AB = 20 m.

R is a point on the line segment AB.

---

To Find:

The distance of point R from point A (i.e., AR) such that the sum $RP^2 + RQ^2$ is minimum.

---

Solution:

Let's set up a coordinate system. Let point A be the origin (0, 0). Since AB lies along a straight line and AB = 20 m, the coordinates of point B are (20, 0).

Since AP is a vertical pole of height 16 m at A, the coordinates of point P are (0, 16).

Since BQ is a vertical pole of height 22 m at B, the coordinates of point Q are (20, 22).

Let R be a point on the line segment AB. Let the distance AR be $x$ meters. Since R lies on the line segment AB (which we've placed on the x-axis), the coordinates of R are (x, 0).

Since R lies on the segment AB, the possible values for $x$ are $0 \le x \le 20$.

The distance RB is AB - AR = $20 - x$ meters.

Now, we need to find the distances RP and RQ. We can use the distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$, which is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. The square of the distance is $(x_2 - x_1)^2 + (y_2 - y_1)^2$.

Calculating $RP^2$ using R(x, 0) and P(0, 16):

$RP^2 = (x - 0)^2 + (0 - 16)^2$

$RP^2 = x^2 + (-16)^2$

$RP^2 = x^2 + 256$

Calculating $RQ^2$ using R(x, 0) and Q(20, 22):

$RQ^2 = (x - 20)^2 + (0 - 22)^2$

$RQ^2 = (x - 20)^2 + (-22)^2$

$RQ^2 = (x - 20)^2 + 484$

We want to minimize the sum $S = RP^2 + RQ^2$. Let $S(x)$ represent this sum as a function of $x$.

$S(x) = RP^2 + RQ^2$

$S(x) = (x^2 + 256) + ((x - 20)^2 + 484)$

Expand $(x - 20)^2 = x^2 - 40x + 400$.

$S(x) = x^2 + 256 + x^2 - 40x + 400 + 484$

$S(x) = 2x^2 - 40x + (256 + 400 + 484)$

$S(x) = 2x^2 - 40x + 1140$

To find the value of $x$ that minimizes $S(x)$, we can use calculus. We find the first derivative of $S(x)$ with respect to $x$ and set it equal to zero.

$S'(x) = \frac{d}{dx}(2x^2 - 40x + 1140)$

$S'(x) = 4x - 40$

Set $S'(x) = 0$ to find critical points:

$4x - 40 = 0$

$4x = 40$

$x = \frac{40}{4}$

$x = 10$

To determine if this critical point corresponds to a minimum, we use the second derivative test. Find the second derivative $S''(x)$.

$S''(x) = \frac{d}{dx}(4x - 40)$

$S''(x) = 4$

Since $S''(x) = 4$, which is positive, the function $S(x)$ has a local minimum at $x = 10$.

The value $x=10$ is within the valid range for R on AB, i.e., $0 \le 10 \le 20$. Thus, the minimum value of $RP^2 + RQ^2$ occurs when $x = 10$.

The distance of the point R from A is $AR = x$.

$AR = 10$ m.

---

Therefore, the distance of the point R on AB from the point A such that $RP^2 + RQ^2$ is minimum is 10 m.

Given:

A trapezium with three sides of equal length, 10 cm each. Let these be the two non-parallel sides and one of the parallel sides (say, the shorter one).

---

To Find:

The maximum area of this trapezium.

---

Solution:

Let the trapezium be ABCD, where AB is parallel to DC.

Let AD = BC = DC = 10 cm (three equal sides).

Let AB be the longer parallel base.

Draw perpendiculars from D and C to the base AB, meeting AB at points P and Q respectively. Let the height of the trapezium be $h = DP = CQ$.

In $\triangle ADP$ and $\triangle BCQ$:

• AD = BC = 10 cm (Given)
• DP = CQ = $h$ (Height)
• $\angle APD = \angle BQC = 90^\circ$ (Construction)

So, $\triangle ADP \cong \triangle BCQ$ by RHS congruence (or by Pythagoras theorem, $AP = BQ$). Let $AP = BQ = x$.

Also, PQCD is a rectangle, so PQ = DC = 10 cm.

The length of the longer base AB is $AB = AP + PQ + QB = x + 10 + x = 10 + 2x$.

Let $\theta$ be the angle that the non-parallel side AD makes with the base AP (i.e., $\angle DAP = \theta$). Similarly, $\angle CBQ = \theta$. For a non-degenerate trapezium, $0 < \theta \le \pi/2$.

In the right-angled triangle $\triangle ADP$:

$x = AP = AD \cos \theta = 10 \cos \theta$

$h = DP = AD \sin \theta = 10 \sin \theta$

The area of the trapezium ABCD, denoted by $A$, is given by:

$A = \frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{Height}$

$A = \frac{1}{2} (DC + AB) \times h$

Substitute the expressions in terms of $\theta$:

$A(\theta) = \frac{1}{2} (10 + (10 + 2x)) \times h$

$A(\theta) = \frac{1}{2} (20 + 2x) h$

$A(\theta) = (10 + x) h$

Now substitute $x = 10 \cos \theta$ and $h = 10 \sin \theta$:

$A(\theta) = (10 + 10 \cos \theta) (10 \sin \theta)$

$A(\theta) = 10(1 + \cos \theta) \times 10 \sin \theta$

$A(\theta) = 100 (1 + \cos \theta) \sin \theta$

$A(\theta) = 100 (\sin \theta + \sin \theta \cos \theta)$

To find the maximum area, we need to find the value of $\theta$ for which $A(\theta)$ is maximum. We use calculus.

Differentiate $A(\theta)$ with respect to $\theta$:

$A'(\theta) = \frac{d}{d\theta} [100 (\sin \theta + \sin \theta \cos \theta)]

$A'(\theta) = 100 [\frac{d}{d\theta}(\sin \theta) + \frac{d}{d\theta}(\sin \theta \cos \theta)]$

Using the product rule for the second term: $\frac{d}{d\theta}(\sin \theta \cos \theta) = (\cos \theta)(\cos \theta) + (\sin \theta)(-\sin \theta) = \cos^2 \theta - \sin^2 \theta = \cos(2\theta)$.

$A'(\theta) = 100 [\cos \theta + \cos^2 \theta - \sin^2 \theta]$

Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$:

$A'(\theta) = 100 [\cos \theta + \cos^2 \theta - (1 - \cos^2 \theta)]$

$A'(\theta) = 100 [\cos \theta + \cos^2 \theta - 1 + \cos^2 \theta]$

$A'(\theta) = 100 [2 \cos^2 \theta + \cos \theta - 1]$

Set $A'(\theta) = 0$ to find critical points:

$100 [2 \cos^2 \theta + \cos \theta - 1] = 0$

$2 \cos^2 \theta + \cos \theta - 1 = 0$

This is a quadratic equation in $\cos \theta$. Let $y = \cos \theta$.

$2y^2 + y - 1 = 0$

Factorizing the quadratic:

$(2y - 1)(y + 1) = 0$

So, $2y - 1 = 0$ or $y + 1 = 0$.

$y = 1/2$ or $y = -1$.

Substitute back $y = \cos \theta$:

$\cos \theta = 1/2$ or $\cos \theta = -1$.

Since $\theta$ is the angle in a right triangle formed by the height ($0 < \theta \le \pi/2$), $\cos \theta$ must be non-negative. Also, $\cos \theta = -1$ corresponds to $\theta = \pi$, which doesn't form a trapezium.

Thus, we consider $\cos \theta = 1/2$.

This gives $\theta = \pi/3$ or $60^\circ$.

Now, we use the second derivative test to confirm if this value gives a maximum.

$A''(\theta) = \frac{d}{d\theta} [100 (2 \cos^2 \theta + \cos \theta - 1)]$

$A''(\theta) = 100 [2 \cdot 2 \cos \theta (-\sin \theta) - \sin \theta]$

$A''(\theta) = 100 [-4 \cos \theta \sin \theta - \sin \theta]$

$A''(\theta) = -100 \sin \theta (4 \cos \theta + 1)$

Evaluate $A''(\theta)$ at $\theta = \pi/3$:

$\sin(\pi/3) = \frac{\sqrt{3}}{2}$

$\cos(\pi/3) = \frac{1}{2}$

$A''(\pi/3) = -100 \left(\frac{\sqrt{3}}{2}\right) \left(4 \left(\frac{1}{2}\right) + 1\right)$

$A''(\pi/3) = -50\sqrt{3} (2 + 1)$

$A''(\pi/3) = -50\sqrt{3} (3) = -150\sqrt{3}$

Since $A''(\pi/3) < 0$, the area $A(\theta)$ is maximum at $\theta = \pi/3$.

Now, calculate the maximum area by substituting $\theta = \pi/3$ into the area formula $A(\theta) = 100 (\sin \theta + \sin \theta \cos \theta)$.

$A_{max} = 100 \left(\sin\left(\frac{\pi}{3}\right) + \sin\left(\frac{\pi}{3}\right) \cos\left(\frac{\pi}{3}\right)\right)$

$A_{max} = 100 \left(\frac{\sqrt{3}}{2} + \left(\frac{\sqrt{3}}{2}\right) \left(\frac{1}{2}\right)\right)$

$A_{max} = 100 \left(\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{4}\right)$

$A_{max} = 100 \left(\frac{2\sqrt{3} + \sqrt{3}}{4}\right)$

$A_{max} = 100 \left(\frac{3\sqrt{3}}{4}\right)$

$A_{max} = \frac{300\sqrt{3}}{4}$

$A_{max} = 75\sqrt{3}$

---

The maximum area of the trapezium is $75\sqrt{3}$ cm$^2$.

Given:

A right circular cone with fixed dimensions (let its radius be $R$ and height be $H$).

A right circular cylinder inscribed within this cone.

---

To Prove:

The radius of the inscribed cylinder with the greatest curved surface area is half the radius of the cone ($R/2$).

---

Solution:

Let the radius of the cone be $R$ and the height of the cone be $H$. These are constants.

Let the radius of the inscribed cylinder be $r$ and its height be $h$. These are variables.

Consider a vertical cross-section of the cone and the inscribed cylinder through the apex of the cone. This cross-section consists of an isosceles triangle (from the cone) with a rectangle (from the cylinder) inscribed in it.

Let the cone's apex be O, the center of its base be C, and a point on the circumference of the base be B. The triangle formed by the cross-section is $\triangle OAB$, where A is diametrically opposite to B. The height OC = $H$ and the radius CB = $R$. Consider the right-angled triangle $\triangle OCB$.

The inscribed cylinder has radius $r$ and height $h$. Let the top face of the cylinder intersect the slant height OB at point D. Let the point vertically below D on the base of the cone be E. Then DE is parallel to OC and DE = $h$. The radius of the cylinder is CE = $r$.

Consider the right-angled triangle $\triangle OCB$ and the smaller right-angled triangle $\triangle DEB$ (or the similar triangle above the cylinder with vertex O). Using similar triangles $\triangle OQC'$ and $\triangle OCB$, where Q is the center of the top face of the cylinder and C' is a point on the circumference of the top face (such that QC' = $r$), we have:

The height of the small cone above the cylinder is $H-h$. Its radius is $r$.

We need to express the height $h$ of the cylinder in terms of its radius $r$ and the cone's dimensions $R, H$.

$rH = R(H-h)$

$rH = RH - Rh$

$Rh = RH - rH$

The curved surface area (CSA) of the cylinder is given by the formula $S = 2 \pi r h$.

Let $S(r)$ be the CSA as a function of the cylinder's radius $r$. Substitute the expression for $h$ from (i):

$S(r) = 2 \pi r \left[ H \left( 1 - \frac{r}{R} \right) \right]$

$S(r) = 2 \pi H \left( r - \frac{r^2}{R} \right)$

To find the radius $r$ that maximizes the CSA, we need to find the derivative of $S(r)$ with respect to $r$ and set it to zero.

$\frac{dS}{dr} = \frac{d}{dr} \left[ 2 \pi H \left( r - \frac{r^2}{R} \right) \right]$

$\frac{dS}{dr} = 2 \pi H \left[ \frac{d}{dr}(r) - \frac{d}{dr}\left(\frac{r^2}{R}\right) \right]$

$\frac{dS}{dr} = 2 \pi H \left( 1 - \frac{2r}{R} \right)$

Set the derivative equal to zero to find critical points:

$\frac{dS}{dr} = 0$

$2 \pi H \left( 1 - \frac{2r}{R} \right) = 0$

Since $H > 0$ (for a valid cone), $2 \pi H \neq 0$. Therefore:

$1 - \frac{2r}{R} = 0$

$1 = \frac{2r}{R}$

$R = 2r$

$r = \frac{R}{2}$

Now, we need to verify that this value of $r$ corresponds to a maximum CSA using the second derivative test.

$\frac{d^2S}{dr^2} = \frac{d}{dr} \left[ 2 \pi H \left( 1 - \frac{2r}{R} \right) \right]$

$\frac{d^2S}{dr^2} = 2 \pi H \left[ \frac{d}{dr}(1) - \frac{d}{dr}\left(\frac{2r}{R}\right) \right]$

$\frac{d^2S}{dr^2} = 2 \pi H \left( 0 - \frac{2}{R} \right)$

$\frac{d^2S}{dr^2} = - \frac{4 \pi H}{R}$

Since $R > 0$ and $H > 0$, the second derivative $\frac{d^2S}{dr^2}$ is always negative.

This indicates that the curved surface area $S(r)$ is maximum when $r = \frac{R}{2}$.

---

Hence, it is proved 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.

Given:

The function $f(x) = 2x^3 - 15x^2 + 36x + 1$.

The interval $[1, 5]$.

---

To Find:

The absolute maximum and absolute minimum values of the function $f(x)$ on the interval $[1, 5]$.

---

Solution:

We are given the function $f(x) = 2x^3 - 15x^2 + 36x + 1$.

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.

Step 1: Find the critical points.

First, find the derivative of $f(x)$:

$f'(x) = \frac{d}{dx}(2x^3 - 15x^2 + 36x + 1)$

$f'(x) = 6x^2 - 30x + 36$

Next, set the derivative equal to zero to find the critical points:

$f'(x) = 0$

$6x^2 - 30x + 36 = 0$

Divide the equation by 6:

$x^2 - 5x + 6 = 0$

Factor the quadratic equation:

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

The critical points are $x = 2$ and $x = 3$.

Step 2: Check if critical points are within the interval $[1, 5]$.

Both critical points, $x = 2$ and $x = 3$, lie within the interval $[1, 5]$.

Step 3: Evaluate the function at the critical points and endpoints.

We need to evaluate $f(x)$ at $x = 1$, $x = 2$, $x = 3$, and $x = 5$.

At the endpoint $x = 1$:

$f(1) = 2(1)^3 - 15(1)^2 + 36(1) + 1 = 2 - 15 + 36 + 1 = 24$

At the critical point $x = 2$:

$f(2) = 2(2)^3 - 15(2)^2 + 36(2) + 1 = 2(8) - 15(4) + 72 + 1 = 16 - 60 + 72 + 1 = 29$

At the critical point $x = 3$:

$f(3) = 2(3)^3 - 15(3)^2 + 36(3) + 1 = 2(27) - 15(9) + 108 + 1 = 54 - 135 + 108 + 1 = 28$

At the endpoint $x = 5$:

$f(5) = 2(5)^3 - 15(5)^2 + 36(5) + 1 = 2(125) - 15(25) + 180 + 1 = 250 - 375 + 180 + 1 = 56$

Step 4: Determine the absolute maximum and minimum values.

Comparing the values calculated:

$f(1) = 24$

$f(2) = 29$

$f(3) = 28$

$f(5) = 56$

The largest value is 56, which occurs at $x = 5$.

The smallest value is 24, which occurs at $x = 1$.

---

Therefore, the absolute maximum value of the function $f(x)$ on the interval $[1, 5]$ is 56, and the absolute minimum value is 24.

Given:

The function $f(x) = 12x^{4/3} - 6x^{1/3}$.

The interval $x \in [-1, 1]$.

---

To Find:

The absolute maximum and absolute minimum values of the function $f(x)$ on the interval $[-1, 1]$.

---

Solution:

We are given the function $f(x) = 12x^{4/3} - 6x^{1/3}$.

To find the absolute maximum and minimum values on the closed interval $[-1, 1]$, we first find the critical points of $f(x)$ by calculating its derivative $f'(x)$.

$f'(x) = \frac{d}{dx} \left( 12x^{4/3} - 6x^{1/3} \right)$

$f'(x) = 12 \cdot \frac{4}{3} x^{(4/3 - 1)} - 6 \cdot \frac{1}{3} x^{(1/3 - 1)}$

$f'(x) = 16 x^{1/3} - 2 x^{-2/3}$

We can rewrite $f'(x)$ as:

$f'(x) = 16x^{1/3} - \frac{2}{x^{2/3}}$

$f'(x) = \frac{16x^{1/3} \cdot x^{2/3} - 2}{x^{2/3}}$

$f'(x) = \frac{16x - 2}{x^{2/3}}$

Critical points occur where $f'(x) = 0$ or where $f'(x)$ is undefined.

Set $f'(x) = 0$:

$\frac{16x - 2}{x^{2/3}} = 0$

This implies the numerator must be zero (provided the denominator is non-zero):

$16x - 2 = 0$

$16x = 2$

$x = \frac{2}{16} = \frac{1}{8}$

Now, find where $f'(x)$ is undefined. This occurs when the denominator is zero:

$x^{2/3} = 0$

This implies $x = 0$.

So, the critical points are $x = \frac{1}{8}$ and $x = 0$.

Both critical points $x = 0$ and $x = 1/8$ lie within the given interval $[-1, 1]$.

Now, we evaluate the function $f(x)$ at the critical points and the endpoints of the interval ($x = -1$ and $x = 1$).

At the endpoint $x = -1$:

$f(-1) = 12(-1)^{4/3} - 6(-1)^{1/3}$

$f(-1) = 12\left( ((-1)^4)^{1/3} \right) - 6(-1)$

$f(-1) = 12(1^{1/3}) + 6$

$f(-1) = 12(1) + 6 = 18$

At the critical point $x = 0$:

$f(0) = 12(0)^{4/3} - 6(0)^{1/3} = 12(0) - 6(0) = 0$

At the critical point $x = 1/8$:

$f(1/8) = 12(1/8)^{4/3} - 6(1/8)^{1/3}$

$f(1/8) = 12\left( (1/8)^{1/3} \right)^4 - 6(1/2)$

$f(1/8) = 12(1/2)^4 - 3$

$f(1/8) = 12(1/16) - 3$

$f(1/8) = \frac{12}{16} - 3 = \frac{3}{4} - 3 = \frac{3 - 12}{4} = -\frac{9}{4}$

At the endpoint $x = 1$:

$f(1) = 12(1)^{4/3} - 6(1)^{1/3}$

$f(1) = 12(1) - 6(1) = 12 - 6 = 6$

Comparing the values obtained: $f(-1) = 18$, $f(0) = 0$, $f(1/8) = -9/4$, $f(1) = 6$.

The largest value is 18.

The smallest value is $-9/4$.

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Therefore, the absolute maximum value of the function $f(x)$ on the interval $[-1, 1]$ is 18, which occurs at $x = -1$.

The absolute minimum value is -9/4, which occurs at $x = 1/8$.

Given:

The path of the Apache helicopter is given by the curve $y = x^2 + 7$.

The position of the soldier is at the point $(3, 7)$.

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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 path of the helicopter. Since the path is given by $y = x^2 + 7$, the coordinates of P can be written as $(x, x^2 + 7)$.

Let S be the position of the soldier, S$(3, 7)$.

The distance between the soldier S and the helicopter at point P is given by the distance formula:

$D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$D = \sqrt{(x - 3)^2 + ((x^2 + 7) - 7)^2}$

$D = \sqrt{(x - 3)^2 + (x^2)^2}$

$D = \sqrt{x^2 - 6x + 9 + x^4}$

To find the nearest distance, we need to minimize the distance $D$. Minimizing $D$ is equivalent to minimizing $D^2$. Let $Z = D^2$.

$Z(x) = (x - 3)^2 + x^4$

$Z(x) = x^2 - 6x + 9 + x^4$

$Z(x) = x^4 + x^2 - 6x + 9$

To find the minimum value of $Z(x)$, we first find its derivative with respect to $x$ and set it to zero.

$Z'(x) = \frac{d}{dx} (x^4 + x^2 - 6x + 9)$

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

Set $Z'(x) = 0$ to find critical points:

$4x^3 + 2x - 6 = 0$

Divide by 2:

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

We can observe by inspection that $x = 1$ is a solution, since $2(1)^3 + (1) - 3 = 2 + 1 - 3 = 0$.

To check if there are other real roots, we can factor the cubic polynomial. Since $x=1$ is a root, $(x-1)$ is a factor.

$(x - 1)(2x^2 + 2x + 3) = 2x^3 + 2x^2 + 3x - 2x^2 - 2x - 3 = 2x^3 + x - 3$.

So, the equation becomes $(x - 1)(2x^2 + 2x + 3) = 0$.

This gives $x - 1 = 0 \implies x = 1$.

For the quadratic factor $2x^2 + 2x + 3 = 0$, we check the discriminant $\Delta = b^2 - 4ac$.

$\Delta = (2)^2 - 4(2)(3) = 4 - 24 = -20$.

Since the discriminant is negative ($\Delta < 0$), the quadratic equation $2x^2 + 2x + 3 = 0$ has no real roots.

Thus, the only real critical point is $x = 1$.

Now, we use the second derivative test to determine if $x = 1$ corresponds to a minimum distance.

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

$Z''(x) = 12x^2 + 2$

Evaluate $Z''(x)$ at $x = 1$:

$Z''(1) = 12(1)^2 + 2 = 12 + 2 = 14$.

Since $Z''(1) = 14 > 0$, the function $Z(x)$ has a local minimum at $x = 1$. As this is the only critical point, this corresponds to the absolute minimum value of $Z(x)$.

The minimum value of $Z = D^2$ occurs at $x = 1$.

$Z_{min} = (1)^4 + (1)^2 - 6(1) + 9 = 1 + 1 - 6 + 9 = 5$.

The nearest distance $D_{min}$ is the square root of $Z_{min}$.

$D_{min} = \sqrt{Z_{min}} = \sqrt{5}$.

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Therefore, the nearest distance between the soldier and the helicopter is $\sqrt{5}$ units.

(i) f(x) = (2x – 1)² + 3

We are given the function $f(x) = (2x - 1)^2 + 3$.

The term $(2x - 1)^2$ is the square of a real number. The square of any real number is always non-negative.

So, $(2x - 1)^2 \ge 0$ for all real $x$.

The minimum value of $(2x - 1)^2$ is 0, which occurs when $2x - 1 = 0$, or $x = 1/2$.

Therefore, the minimum value of $f(x) = (2x - 1)^2 + 3$ is $0 + 3 = 3$.

Since $(2x - 1)^2$ can be made arbitrarily large (e.g., as $x \to \infty$ or $x \to -\infty$), the value of $f(x)$ can also be made arbitrarily large.

Thus, the function $f(x)$ does not have a maximum value.

Minimum value = 3. No maximum value.

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(ii) f(x) = 9x² + 12x + 2

We are given the function $f(x) = 9x^2 + 12x + 2$.

This is a quadratic function. We can find the minimum or maximum value by completing the square or using calculus.

Method 1: Completing the Square

$f(x) = 9(x^2 + \frac{12}{9}x) + 2$

$f(x) = 9(x^2 + \frac{4}{3}x) + 2$

$f(x) = 9\left(x^2 + \frac{4}{3}x + \left(\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2\right) + 2$

$f(x) = 9\left( \left(x + \frac{2}{3}\right)^2 - \frac{4}{9} \right) + 2$

$f(x) = 9\left(x + \frac{2}{3}\right)^2 - 9 \cdot \frac{4}{9} + 2$

$f(x) = 9\left(x + \frac{2}{3}\right)^2 - 4 + 2$

$f(x) = 9\left(x + \frac{2}{3}\right)^2 - 2$

Since $\left(x + \frac{2}{3}\right)^2 \ge 0$, the term $9\left(x + \frac{2}{3}\right)^2 \ge 0$.

The minimum value of $9\left(x + \frac{2}{3}\right)^2$ is 0, which occurs when $x + \frac{2}{3} = 0$, or $x = -2/3$.

Therefore, the minimum value of $f(x)$ is $0 - 2 = -2$.

Since the term $9\left(x + \frac{2}{3}\right)^2$ can grow indefinitely, $f(x)$ can be made arbitrarily large.

Thus, the function $f(x)$ does not have a maximum value.

Method 2: Using Calculus

$f'(x) = \frac{d}{dx}(9x^2 + 12x + 2) = 18x + 12$.

Set $f'(x) = 0$: $18x + 12 = 0 \implies 18x = -12 \implies x = -12/18 = -2/3$.

$f''(x) = \frac{d}{dx}(18x + 12) = 18$.

Since $f''(x) = 18 > 0$, the function has a local minimum at $x = -2/3$. Since it's a parabola opening upwards, this is the absolute minimum.

Minimum value $f(-2/3) = 9(-2/3)^2 + 12(-2/3) + 2 = 9(4/9) - 8 + 2 = 4 - 8 + 2 = -2$.

Minimum value = -2. No maximum value.

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(iii) f(x) = – (x – 1)² + 10

We are given the function $f(x) = -(x - 1)^2 + 10$.

The term $(x - 1)^2 \ge 0$ for all real $x$.

Therefore, $-(x - 1)^2 \le 0$. The maximum value of $-(x - 1)^2$ is 0, which occurs when $x - 1 = 0$, or $x = 1$.

The maximum value of $f(x) = -(x - 1)^2 + 10$ is $0 + 10 = 10$.

Since $(x - 1)^2$ can be made arbitrarily large, $-(x - 1)^2$ can be made arbitrarily negative (e.g., as $x \to \infty$ or $x \to -\infty$).

Thus, the function $f(x)$ does not have a minimum value.

Maximum value = 10. No minimum value.

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(iv) g(x) = x³ + 1

We are given the function $g(x) = x^3 + 1$.

Consider the behavior of the function as $x$ approaches positive and negative infinity.

As $x \to \infty$, $x^3 \to \infty$, so $g(x) \to \infty$.

As $x \to -\infty$, $x^3 \to -\infty$, so $g(x) \to -\infty$.

Since the function can take arbitrarily large positive and negative values, it does not have an absolute maximum or an absolute minimum value.

Alternatively, using calculus:

$g'(x) = \frac{d}{dx}(x^3 + 1) = 3x^2$.

Set $g'(x) = 0$: $3x^2 = 0 \implies x = 0$.

$g''(x) = \frac{d}{dx}(3x^2) = 6x$.

$g''(0) = 6(0) = 0$. The second derivative test is inconclusive.

However, note that $g'(x) = 3x^2 \ge 0$ for all $x$. This means the function is always non-decreasing. It does not turn around to create a local maximum or minimum, except for the point of horizontal tangency at $x = 0$, which is an inflection point.

Neither a maximum value nor a minimum value exists.

(i) f(x) = |x + 2| – 1

We know that the absolute value function $|y|$ is always non-negative, i.e., $|y| \ge 0$ for any real number $y$.

So, for our function, $|x + 2| \ge 0$ for all real $x$.

The minimum value of $|x + 2|$ is 0, which occurs when $x + 2 = 0$, or $x = -2$.

Now consider $f(x) = |x + 2| - 1$.

Since $|x + 2| \ge 0$, we have:

$f(x) = |x + 2| - 1 \ge 0 - 1$

$f(x) \ge -1$

The minimum value of $f(x)$ is -1, achieved at $x = -2$.

As $x$ becomes very large (positive or negative), $|x + 2|$ becomes arbitrarily large. Therefore, $f(x) = |x + 2| - 1$ can also become arbitrarily large.

Thus, the function has no maximum value.

Minimum value = -1. No maximum value.

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(ii) g(x) = – |x + 1| + 3

We know that $|x + 1| \ge 0$ for all real $x$.

Multiplying by -1 reverses the inequality:

$-|x + 1| \le 0$

The maximum value of $-|x + 1|$ is 0, which occurs when $x + 1 = 0$, or $x = -1$.

Now consider $g(x) = -|x + 1| + 3$.

Since $-|x + 1| \le 0$, we have:

$g(x) = -|x + 1| + 3 \le 0 + 3$

$g(x) \le 3$

The maximum value of $g(x)$ is 3, achieved at $x = -1$.

As $x$ becomes very large (positive or negative), $|x + 1|$ becomes arbitrarily large, so $-|x + 1|$ becomes arbitrarily large negative.

Thus, $g(x) = -|x + 1| + 3$ can become arbitrarily large negative (approaches $-\infty$).

Therefore, the function has no minimum value.

Maximum value = 3. No minimum value.

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(iii) h(x) = sin(2x) + 5

The range of the sine function, $\sin(\theta)$, is $[-1, 1]$.

This means that for any real number $x$, the value of $\sin(2x)$ must satisfy:

$-1 \le \sin(2x) \le 1$

To find the range of $h(x) = \sin(2x) + 5$, we add 5 to all parts of the inequality:

$-1 + 5 \le \sin(2x) + 5 \le 1 + 5$

$4 \le h(x) \le 6$

The minimum value of $h(x)$ is 4 (occurs when $\sin(2x) = -1$, e.g., when $2x = 3\pi/2 \implies x=3\pi/4$).

The maximum value of $h(x)$ is 6 (occurs when $\sin(2x) = 1$, e.g., when $2x = \pi/2 \implies x=\pi/4$).

Minimum value = 4. Maximum value = 6.

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(iv) f(x) = |sin 4x + 3|

We know the range of the sine function:

$-1 \le \sin(4x) \le 1$

Add 3 to all parts of the inequality:

$-1 + 3 \le \sin(4x) + 3 \le 1 + 3$

$2 \le \sin(4x) + 3 \le 4$

Since the value of the expression $\sin(4x) + 3$ is always between 2 and 4, it is always positive.

Therefore, the absolute value function does not change the expression:

$f(x) = |\sin(4x) + 3| = \sin(4x) + 3$

So, the range of $f(x)$ is the same as the range of $\sin(4x) + 3$.

$2 \le f(x) \le 4$

The minimum value of $f(x)$ is 2 (occurs when $\sin(4x) = -1$).

The maximum value of $f(x)$ is 4 (occurs when $\sin(4x) = 1$).

Minimum value = 2. Maximum value = 4.

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(v) h(x) = x + 1, x ∈ (– 1, 1)

The function is $h(x) = x + 1$.

The domain is the open interval $(-1, 1)$, which means $x$ can get arbitrarily close to -1 and 1, but $x$ cannot be equal to -1 or 1.

Consider the value of $h(x)$ as $x$ approaches the endpoints of the interval.

As $x$ approaches -1 (from the right side), $h(x)$ approaches $-1 + 1 = 0$. Since $x > -1$, $h(x) = x + 1 > -1 + 1 = 0$. So, $h(x)$ can get arbitrarily close to 0, but never reaches it.

As $x$ approaches 1 (from the left side), $h(x)$ approaches $1 + 1 = 2$. Since $x < 1$, $h(x) = x + 1 < 1 + 1 = 2$. So, $h(x)$ can get arbitrarily close to 2, but never reaches it.

The range of the function $h(x)$ on the interval $(-1, 1)$ is $(0, 2)$.

Since the interval for the range is open, the function does not attain a minimum value (it never reaches 0) and does not attain a maximum value (it never reaches 2).

Neither a maximum value nor a minimum value exists.

(i) f(x) = x²

Find the first derivative:

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

Set $f'(x) = 0$ to find critical points:

$2x = 0 \implies x = 0$

Find the second derivative:

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

Evaluate $f''(x)$ at the critical point $x=0$:

$f''(0) = 2 > 0$

Since the second derivative is positive, $x=0$ is a point of local minimum.

The local minimum value is $f(0) = (0)^2 = 0$.

There is no point of local maximum.

Local Minimum Point: $x = 0$. Local Minimum Value = 0. No local maximum.

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(ii) g(x) = x³ – 3x

Find the first derivative:

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

Set $g'(x) = 0$ to find critical points:

$3x^2 - 3 = 0$

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

$x^2 = 1 \implies x = 1$ or $x = -1$

Find the second derivative:

$g''(x) = \frac{d}{dx}(3x^2 - 3) = 6x$

Evaluate $g''(x)$ at the critical points:

At $x = 1$: $g''(1) = 6(1) = 6 > 0$. This indicates a local minimum at $x=1$. The local minimum value is $g(1) = (1)^3 - 3(1) = 1 - 3 = -2$.

At $x = -1$: $g''(-1) = 6(-1) = -6 < 0$. This indicates a local maximum at $x=-1$. The local maximum value is $g(-1) = (-1)^3 - 3(-1) = -1 + 3 = 2$.

Local Minimum Point: $x = 1$. Local Minimum Value = -2.

Local Maximum Point: $x = -1$. Local Maximum Value = 2.

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(iii) h(x) = sin x + cos x, 0 < x < $\frac{\pi}{2}$

Find the first derivative:

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

Set $h'(x) = 0$ to find critical points:

$\cos x - \sin x = 0$

$\cos x = \sin x$

Divide by $\cos x$ (which is non-zero in the interval $(0, \pi/2)$):

$\tan x = 1$

The solution in the interval $0 < x < \pi/2$ is $x = \frac{\pi}{4}$.

Find the second derivative:

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

Evaluate $h''(x)$ at the critical point $x = \pi/4$:

$h''(\pi/4) = -\sin(\pi/4) - \cos(\pi/4) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2}$

Since $h''(\pi/4) < 0$, the function has a local maximum at $x = \pi/4$.

The local maximum value is $h(\pi/4) = \sin(\pi/4) + \cos(\pi/4) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

There is no local minimum in the given interval.

Local Maximum Point: $x = \pi/4$. Local Maximum Value = $\sqrt{2}$. No local minimum.

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(iv) f(x) = sin x – cos x, 0 < x < 2π

Find the first derivative:

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

Set $f'(x) = 0$ to find critical points:

$\cos x + \sin x = 0$

$\sin x = -\cos x$

$\tan x = -1$

The solutions for $\tan x = -1$ in the interval $0 < x < 2\pi$ are in the second and fourth quadrants. They are $x = \frac{3\pi}{4}$ and $x = \frac{7\pi}{4}$.

Find the second derivative:

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

Evaluate $f''(x)$ at the critical points:

At $x = 3\pi/4$: $f''(3\pi/4) = -\sin(3\pi/4) + \cos(3\pi/4) = -\frac{1}{\sqrt{2}} + (-\frac{1}{\sqrt{2}}) = -\frac{2}{\sqrt{2}} = -\sqrt{2} < 0$. This indicates a local maximum at $x = 3\pi/4$.

The local maximum value is $f(3\pi/4) = \sin(3\pi/4) - \cos(3\pi/4) = \frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}}) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

At $x = 7\pi/4$: $f''(7\pi/4) = -\sin(7\pi/4) + \cos(7\pi/4) = -(-\frac{1}{\sqrt{2}}) + \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} > 0$. This indicates a local minimum at $x = 7\pi/4$.

The local minimum value is $f(7\pi/4) = \sin(7\pi/4) - \cos(7\pi/4) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2}$.

Local Maximum Point: $x = 3\pi/4$. Local Maximum Value = $\sqrt{2}$.

Local Minimum Point: $x = 7\pi/4$. Local Minimum Value = $-\sqrt{2}$.

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(v) f(x) = x³ – 6x² + 9x + 15

Find the first derivative:

$f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 9x + 15) = 3x^2 - 12x + 9$

Set $f'(x) = 0$ to find critical points:

$3x^2 - 12x + 9 = 0$

$3(x^2 - 4x + 3) = 0$

$x^2 - 4x + 3 = 0$

$(x - 1)(x - 3) = 0$

The critical points are $x = 1$ and $x = 3$.

Find the second derivative:

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

Evaluate $f''(x)$ at the critical points:

At $x = 1$: $f''(1) = 6(1) - 12 = 6 - 12 = -6 < 0$. This indicates a local maximum at $x = 1$. The local maximum value is $f(1) = (1)^3 - 6(1)^2 + 9(1) + 15 = 1 - 6 + 9 + 15 = 19$.

At $x = 3$: $f''(3) = 6(3) - 12 = 18 - 12 = 6 > 0$. This indicates a local minimum at $x = 3$. The local minimum value is $f(3) = (3)^3 - 6(3)^2 + 9(3) + 15 = 27 - 6(9) + 27 + 15 = 27 - 54 + 27 + 15 = 15$.

Local Maximum Point: $x = 1$. Local Maximum Value = 19.

Local Minimum Point: $x = 3$. Local Minimum Value = 15.

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(vi) g(x) = $\frac{x}{2} + \frac{2}{x}$, x > 0

Find the first derivative:

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

Set $g'(x) = 0$ to find critical points:

$\frac{1}{2} - \frac{2}{x^2} = 0$

$\frac{1}{2} = \frac{2}{x^2}$

$x^2 = 4$

Since $x > 0$, the only critical point is $x = 2$.

Find the second derivative:

$g''(x) = \frac{d}{dx}\left(\frac{1}{2} - 2x^{-2}\right) = 0 - 2(-2)x^{-3} = 4x^{-3} = \frac{4}{x^3}$

Evaluate $g''(x)$ at the critical point $x = 2$:

$g''(2) = \frac{4}{(2)^3} = \frac{4}{8} = \frac{1}{2} > 0$.

Since the second derivative is positive, the function has a local minimum at $x = 2$.

The local minimum value is $g(2) = \frac{2}{2} + \frac{2}{2} = 1 + 1 = 2$.

There is no local maximum.

Local Minimum Point: $x = 2$. Local Minimum Value = 2. No local maximum.

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(vii) g(x) = $\frac{1}{x^{2}+2}$

Find the first derivative using the chain rule or quotient rule. Let $g(x) = (x^2 + 2)^{-1}$.

$g'(x) = -1(x^2 + 2)^{-2} \cdot \frac{d}{dx}(x^2 + 2) = -(x^2 + 2)^{-2} (2x) = \frac{-2x}{(x^2 + 2)^2}$

Set $g'(x) = 0$ to find critical points:

$\frac{-2x}{(x^2 + 2)^2} = 0$

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

The denominator $(x^2 + 2)^2$ is never zero, so $x=0$ is the only critical point.

Find the second derivative using the quotient rule: $u = -2x, v = (x^2+2)^2$. $u' = -2, v' = 2(x^2+2)(2x) = 4x(x^2+2)$.

$g''(x) = \frac{v u' - u v'}{v^2} = \frac{(x^2 + 2)^2(-2) - (-2x)(4x(x^2+2))}{((x^2 + 2)^2)^2}$

$g''(x) = \frac{-2(x^2 + 2)^2 + 8x^2(x^2+2)}{(x^2 + 2)^4}$

Factor out $(x^2 + 2)$ from the numerator:

$g''(x) = \frac{(x^2 + 2)[-2(x^2 + 2) + 8x^2]}{(x^2 + 2)^4} = \frac{-2x^2 - 4 + 8x^2}{(x^2 + 2)^3} = \frac{6x^2 - 4}{(x^2 + 2)^3}$

Evaluate $g''(x)$ at the critical point $x = 0$:

$g''(0) = \frac{6(0)^2 - 4}{(0^2 + 2)^3} = \frac{-4}{(2)^3} = \frac{-4}{8} = -\frac{1}{2} < 0$.

Since the second derivative is negative, the function has a local maximum at $x = 0$.

The local maximum value is $g(0) = \frac{1}{0^2 + 2} = \frac{1}{2}$.

There is no local minimum.

Local Maximum Point: $x = 0$. Local Maximum Value = 1/2. No local minimum.

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(viii) f(x) = x$\sqrt{1-x}$, 0 < x < 1

Find the first derivative using the product rule: $u = x, v = \sqrt{1-x} = (1-x)^{1/2}$.

$u' = 1$

$v' = \frac{1}{2}(1-x)^{-1/2}(-1) = -\frac{1}{2\sqrt{1-x}}$

$f'(x) = u'v + uv' = 1 \cdot \sqrt{1-x} + x \left(-\frac{1}{2\sqrt{1-x}}\right)$

$f'(x) = \sqrt{1-x} - \frac{x}{2\sqrt{1-x}}$

Combine terms with a common denominator $2\sqrt{1-x}$:

$f'(x) = \frac{\sqrt{1-x} \cdot 2\sqrt{1-x} - x}{2\sqrt{1-x}} = \frac{2(1-x) - x}{2\sqrt{1-x}} = \frac{2 - 2x - x}{2\sqrt{1-x}} = \frac{2 - 3x}{2\sqrt{1-x}}$

Set $f'(x) = 0$ to find critical points:

$\frac{2 - 3x}{2\sqrt{1-x}} = 0$

$2 - 3x = 0 \implies 3x = 2 \implies x = \frac{2}{3}$.

The critical point $x = 2/3$ is within the domain $0 < x < 1$. The derivative is undefined at $x=1$, but that is outside the domain.

We can use the first derivative test or the second derivative test. Let's use the first derivative test.

Consider the sign of $f'(x) = \frac{2 - 3x}{2\sqrt{1-x}}$ around $x = 2/3$. The denominator $2\sqrt{1-x}$ is positive for $0 < x < 1$. The sign depends on the numerator $2 - 3x$.

If $0 < x < 2/3$, let $x = 1/2$. $2 - 3(1/2) = 2 - 1.5 = 0.5 > 0$. So $f'(x) > 0$.

If $2/3 < x < 1$, let $x = 3/4$. $2 - 3(3/4) = 2 - 9/4 = (8-9)/4 = -1/4 < 0$. So $f'(x) < 0$.

Since $f'(x)$ changes sign from positive to negative at $x = 2/3$, the function has a local maximum at $x = 2/3$.

The local maximum value is $f(2/3) = \frac{2}{3} \sqrt{1 - \frac{2}{3}} = \frac{2}{3} \sqrt{\frac{1}{3}} = \frac{2}{3} \frac{1}{\sqrt{3}} = \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{9}$.

There is no local minimum in the given interval.

Local Maximum Point: $x = 2/3$. Local Maximum Value = $\frac{2\sqrt{3}}{9}$. No local minimum.

To prove that a function does not have local maxima or minima, we can show that its first derivative is never equal to zero or that it doesn't change sign.

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(i) f(x) = e^x

Proof:

The given function is $f(x) = e^x$.

Find the first derivative:

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

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

$e^x = 0$

The exponential function $e^x$ is always positive for all real values of $x$ ($e^x > 0$). Therefore, $e^x$ can never be equal to zero.

Since there are no critical points where the derivative is zero, and the derivative $f'(x) = e^x$ is always positive, the function $f(x)$ is strictly increasing for all $x$.

Thus, $f(x) = e^x$ does not have any local maximum or local minimum values.

---

(ii) g(x) = log x

Proof:

The given function is $g(x) = \log x$. The domain of this function is $x > 0$.

Find the first derivative:

$g'(x) = \frac{d}{dx}(\log x) = \frac{1}{x}$

To find critical points, we set $g'(x) = 0$.

$\frac{1}{x} = 0$

This equation has no solution for $x$.

Furthermore, the derivative $g'(x) = \frac{1}{x}$ is defined and positive for all $x$ in the domain $(0, \infty)$.

Since there are no critical points and the derivative $g'(x)$ is always positive in its domain, the function $g(x)$ is strictly increasing on its domain $(0, \infty)$.

Thus, $g(x) = \log x$ does not have any local maximum or local minimum values.

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(iii) h(x) = x³ + x² + x + 1

Proof:

The given function is $h(x) = x^3 + x^2 + x + 1$.

Find the 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. Let's examine its discriminant, $\Delta = b^2 - 4ac$.

Here, $a = 3$, $b = 2$, $c = 1$.

$\Delta = (2)^2 - 4(3)(1) = 4 - 12 = -8$

Since the discriminant $\Delta = -8 < 0$, the quadratic equation $3x^2 + 2x + 1 = 0$ has no real roots.

Therefore, $h'(x)$ is never equal to zero for any real value of $x$.

Also, since the leading coefficient $a = 3$ is positive and the discriminant is negative, the quadratic $3x^2 + 2x + 1$ is always positive for all real $x$.

Since the derivative $h'(x)$ is always positive, the function $h(x)$ is strictly increasing for all $x$.

Thus, $h(x) = x^3 + x^2 + x + 1$ does not have any local maximum or local minimum values.

To find the absolute maximum and minimum values of a continuous function $f(x)$ on a closed interval $[a, b]$:

1. Find all critical points of $f(x)$ in the open interval $(a, b)$. (Critical points are where $f'(x) = 0$ or $f'(x)$ is undefined).
2. Evaluate $f(x)$ at these critical points.
3. Evaluate $f(x)$ at the endpoints of the interval, i.e., find $f(a)$ and $f(b)$.
4. The largest value among those found in steps 2 and 3 is the absolute maximum value, and the smallest value is the absolute minimum value.

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(i) f(x) = x³, x ∈ [-2, 2]

Given function: $f(x) = x^3$ on the interval $[-2, 2]$.

Step 1: Find critical points.

Find the derivative: $f'(x) = 3x^2$.

Set $f'(x) = 0$: $3x^2 = 0 \implies x = 0$.

The critical point $x=0$ is within the interval $[-2, 2]$.

Step 2: Evaluate function at critical points and endpoints.

Evaluate $f(x)$ at $x = -2, 0, 2$.

$f(-2) = (-2)^3 = -8$

$f(0) = (0)^3 = 0$

$f(2) = (2)^3 = 8$

Step 3: Compare values.

The values are -8, 0, 8.

The absolute maximum value is 8.

The 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]

Given function: $f(x) = \sin x + \cos x$ on the interval $[0, \pi]$.

Step 1: Find critical points.

Find the derivative: $f'(x) = \cos x - \sin x$.

Set $f'(x) = 0$: $\cos x - \sin x = 0 \implies \cos x = \sin x$.

Dividing by $\cos x$ (assuming $\cos x \neq 0$), we get $\tan x = 1$.

In the interval $[0, \pi]$, the solution is $x = \frac{\pi}{4}$. ($\cos(\pi/4) \neq 0$, so the division was valid).

The critical point $x=\pi/4$ is within the interval $[0, \pi]$.

Step 2: Evaluate function at critical points and endpoints.

Evaluate $f(x)$ at $x = 0, \pi/4, \pi$.

$f(0) = \sin(0) + \cos(0) = 0 + 1 = 1$

$f(\pi/4) = \sin(\pi/4) + \cos(\pi/4) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$

$f(\pi) = \sin(\pi) + \cos(\pi) = 0 + (-1) = -1$

Step 3: Compare values.

The values are 1, $\sqrt{2}$, and -1. Since $\sqrt{2} \approx 1.414$, the largest value is $\sqrt{2}$ and the smallest is -1.

The absolute maximum value is $\sqrt{2}$.

The 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 ∈ $\left[-2, \frac{9}{2}\right]$

Given function: $f(x) = 4x - \frac{1}{2}x^2$ on the interval $[-2, 9/2]$.

Step 1: Find critical points.

Find the derivative: $f'(x) = 4 - \frac{1}{2}(2x) = 4 - x$.

Set $f'(x) = 0$: $4 - x = 0 \implies x = 4$.

The critical point $x=4$ is within the interval $[-2, 9/2]$ (since $9/2 = 4.5$).

Step 2: Evaluate function at critical points and endpoints.

Evaluate $f(x)$ at $x = -2, 4, 9/2$.

$f(-2) = 4(-2) - \frac{1}{2}(-2)^2 = -8 - \frac{1}{2}(4) = -8 - 2 = -10$

$f(4) = 4(4) - \frac{1}{2}(4)^2 = 16 - \frac{1}{2}(16) = 16 - 8 = 8$

$f(9/2) = 4(9/2) - \frac{1}{2}(9/2)^2 = 18 - \frac{1}{2}\left(\frac{81}{4}\right) = 18 - \frac{81}{8} = \frac{144 - 81}{8} = \frac{63}{8}$

Step 3: Compare values.

The values are -10, 8, and $63/8$. Since $63/8 = 7.875$, the largest value is 8 and the smallest is -10.

The absolute maximum value is 8.

The 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]

Given function: $f(x) = (x - 1)^2 + 3$ on the interval $[-3, 1]$.

Step 1: Find critical points.

Find the derivative: $f'(x) = 2(x - 1)(1) + 0 = 2(x - 1)$.

Set $f'(x) = 0$: $2(x - 1) = 0 \implies x = 1$.

The critical point $x=1$ is an endpoint of the interval $[-3, 1]$. There are no critical points strictly inside the interval $(-3, 1)$.

Step 2: Evaluate function at endpoints (which includes the critical point).

Evaluate $f(x)$ at $x = -3$ and $x = 1$.

$f(-3) = (-3 - 1)^2 + 3 = (-4)^2 + 3 = 16 + 3 = 19$

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

Step 3: Compare values.

The values are 19 and 3.

The absolute maximum value is 19.

The absolute minimum value is 3.

Absolute Maximum Value = 19. Absolute Minimum Value = 3.

Given:

The profit function is given by $p(x) = 41 - 24x - 18x^2$.

---

To Find:

The maximum profit the company can make.

---

Solution:

The profit function is $p(x) = 41 - 24x - 18x^2$.

To find the maximum profit, we need to find the maximum value of the function $p(x)$. We can use calculus for this.

Step 1: Find the first derivative of $p(x)$.

$p'(x) = \frac{d}{dx} (41 - 24x - 18x^2)$

$p'(x) = 0 - 24 - 18(2x)$

$p'(x) = -24 - 36x$

Step 2: Find the critical points by setting $p'(x) = 0$.

$-24 - 36x = 0$

$-36x = 24$

$x = \frac{24}{-36}$

$x = -\frac{2 \times \cancel{12}}{3 \times \cancel{12}}$

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

Step 3: Use the second derivative test to determine if this critical point corresponds to a maximum.

Find the second derivative $p''(x)$:

$p''(x) = \frac{d}{dx} (-24 - 36x)$

$p''(x) = -36$

Since $p''(x) = -36$, which is negative, the function $p(x)$ has a local maximum at the critical point $x = -2/3$.

Because $p(x)$ is a quadratic function with a negative coefficient for the $x^2$ term (a parabola opening downwards), this local maximum is also the absolute maximum.

Step 4: Calculate the maximum profit by evaluating $p(x)$ at $x = -2/3$.

$p\left(-\frac{2}{3}\right) = 41 - 24\left(-\frac{2}{3}\right) - 18\left(-\frac{2}{3}\right)^2$

$p\left(-\frac{2}{3}\right) = 41 - \left(\frac{-48}{3}\right) - 18\left(\frac{4}{9}\right)$

$p\left(-\frac{2}{3}\right) = 41 - (-16) - \left(\frac{18 \times 4}{9}\right)$

$p\left(-\frac{2}{3}\right) = 41 + 16 - \left(\frac{\cancel{18}^2 \times 4}{\cancel{9}_1}\right)$

$p\left(-\frac{2}{3}\right) = 41 + 16 - 8$

$p\left(-\frac{2}{3}\right) = 57 - 8$

$p\left(-\frac{2}{3}\right) = 49$

---

Therefore, the maximum profit that the company can make is 49.

Given:

The function $f(x) = 3x^4 - 8x^3 + 12x^2 - 48x + 25$.

The interval $[0, 3]$.

---

To Find:

The absolute maximum value and the absolute minimum value of the function $f(x)$ on the interval $[0, 3]$.

---

Solution:

To find the absolute maximum and minimum values of a continuous function on a closed interval, we evaluate the function at its critical points within the interval and at the endpoints of the interval.

Step 1: Find the critical points.

First, find the derivative of $f(x)$:

$f'(x) = \frac{d}{dx}(3x^4 - 8x^3 + 12x^2 - 48x + 25)$

$f'(x) = 12x^3 - 24x^2 + 24x - 48$

Next, set the derivative equal to zero to find the critical points:

$f'(x) = 0$

$12x^3 - 24x^2 + 24x - 48 = 0$

Divide the equation by 12:

$x^3 - 2x^2 + 2x - 4 = 0$

Factor by grouping:

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

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

This gives two possibilities:

1) $x^2 + 2 = 0 \implies x^2 = -2$. This equation has no real solutions.

2) $x - 2 = 0 \implies x = 2$.

The only real critical point is $x = 2$.

Step 2: Check if critical points are within the interval $[0, 3]$.

The critical point $x = 2$ lies within the interval $[0, 3]$.

Step 3: Evaluate the function at the critical point(s) and endpoints.

We need to evaluate $f(x)$ at $x = 0$, $x = 2$, and $x = 3$.

At the endpoint $x = 0$:

$f(0) = 3(0)^4 - 8(0)^3 + 12(0)^2 - 48(0) + 25 = 0 - 0 + 0 - 0 + 25 = 25$

At the critical point $x = 2$:

$f(2) = 3(2)^4 - 8(2)^3 + 12(2)^2 - 48(2) + 25$

$f(2) = 3(16) - 8(8) + 12(4) - 96 + 25$

$f(2) = 48 - 64 + 48 - 96 + 25$

$f(2) = (48 + 48 + 25) - (64 + 96)$

$f(2) = 121 - 160 = -39$

At the endpoint $x = 3$:

$f(3) = 3(3)^4 - 8(3)^3 + 12(3)^2 - 48(3) + 25$

$f(3) = 3(81) - 8(27) + 12(9) - 144 + 25$

$f(3) = 243 - 216 + 108 - 144 + 25$

$f(3) = (243 + 108 + 25) - (216 + 144)$

$f(3) = 376 - 360 = 16$

Step 4: Determine the absolute maximum and minimum values.

Comparing the values calculated: $f(0) = 25$, $f(2) = -39$, and $f(3) = 16$.

The largest value is 25.

The smallest value is -39.

---

Therefore, on the interval $[0, 3]$:

The absolute maximum value of the function is 25 (which occurs at $x=0$).

The absolute minimum value of the function is -39 (which occurs at $x=2$).

Given:

The function $f(x) = \sin(2x)$.

The interval $[0, 2\pi]$.

---

To Find:

The points $x$ in the interval $[0, 2\pi]$ where the function $f(x) = \sin(2x)$ attains its maximum value.

---

Solution:

The function given is $f(x) = \sin(2x)$.

We know that the maximum value of the sine function, $\sin(\theta)$, is 1.

Therefore, the maximum value of $f(x) = \sin(2x)$ is 1.

We need to find the values of $x$ in the interval $[0, 2\pi]$ for which $\sin(2x) = 1$.

The general solution for $\sin(\theta) = 1$ is given by $\theta = \frac{\pi}{2} + 2n\pi$, where $n$ is any integer.

In our case, $\theta = 2x$. So, we have:

$2x = \frac{\pi}{2} + 2n\pi$

Divide by 2 to solve for $x$:

$x = \frac{\pi}{4} + n\pi$, where $n$ is an integer.

Now, we need to find the values of $n$ such that $x$ lies in the interval $[0, 2\pi]$.

$0 \le x \le 2\pi$

$0 \le \frac{\pi}{4} + n\pi \le 2\pi$

Subtract $\frac{\pi}{4}$ from all parts of the inequality:

$-\frac{\pi}{4} \le n\pi \le 2\pi - \frac{\pi}{4}$

$-\frac{\pi}{4} \le n\pi \le \frac{8\pi - \pi}{4}$

$-\frac{\pi}{4} \le n\pi \le \frac{7\pi}{4}$

Divide all parts by $\pi$ (since $\pi > 0$, the inequality signs remain the same):

$-\frac{1}{4} \le n \le \frac{7}{4}$

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

For $n = 0$:

$x = \frac{\pi}{4} + (0)\pi = \frac{\pi}{4}$

For $n = 1$:

$x = \frac{\pi}{4} + (1)\pi = \frac{\pi}{4} + \pi = \frac{\pi + 4\pi}{4} = \frac{5\pi}{4}$

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

---

Therefore, the function $\sin(2x)$ attains its maximum value at the points $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$ in the interval $[0, 2\pi]$.

Given:

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

---

To Find:

The maximum value of the function $f(x)$.

---

Solution:

We want to find the maximum value of $f(x) = \sin x + \cos x$.

We can rewrite the expression $\sin x + \cos x$ in the form $R \cos(x - \alpha)$ or $R \sin(x + \alpha)$. Let's use the form $R \cos(x - \alpha)$.

We know that $R \cos(x - \alpha) = R (\cos x \cos \alpha + \sin x \sin \alpha) = (R \sin \alpha) \sin x + (R \cos \alpha) \cos x$.

Comparing $f(x) = 1 \cdot \sin x + 1 \cdot \cos x$ with $(R \sin \alpha) \sin x + (R \cos \alpha) \cos x$, we must have:

Square equations (i) and (ii) and add them:

$(R \sin \alpha)^2 + (R \cos \alpha)^2 = 1^2 + 1^2$

$R^2 \sin^2 \alpha + R^2 \cos^2 \alpha = 1 + 1$

$R^2 (\sin^2 \alpha + \cos^2 \alpha) = 2$

Since $\sin^2 \alpha + \cos^2 \alpha = 1$, we have:

$R^2 (1) = 2$

$R^2 = 2$

$R = \sqrt{2}$ (We take the positive value for the amplitude R)

Now, divide equation (i) by equation (ii):

$\frac{R \sin \alpha}{R \cos \alpha} = \frac{1}{1}$

$\tan \alpha = 1$

Since both $\sin \alpha = 1/\sqrt{2}$ and $\cos \alpha = 1/\sqrt{2}$ are positive, $\alpha$ lies in the first quadrant. Thus, $\alpha = \frac{\pi}{4}$.

So, we can write the function as:

$f(x) = \sin x + \cos x = \sqrt{2} \cos\left(x - \frac{\pi}{4}\right)$

We know that the range of the cosine function is $[-1, 1]$.

$-1 \le \cos\left(x - \frac{\pi}{4}\right) \le 1$

Multiply the inequality by $\sqrt{2}$:

$-\sqrt{2} \le \sqrt{2} \cos\left(x - \frac{\pi}{4}\right) \le \sqrt{2}$

$-\sqrt{2} \le f(x) \le \sqrt{2}$

The maximum value of $f(x)$ is $\sqrt{2}$.

---

Alternate Solution using Calculus:

Let $f(x) = \sin x + \cos x$.

Find the first derivative:

$f'(x) = \cos x - \sin x$.

Set $f'(x) = 0$ to find critical points:

$\cos x - \sin x = 0 \implies \cos x = \sin x \implies \tan x = 1$.

The critical points occur when $x = \frac{\pi}{4} + n\pi$ for any integer $n$.

Find the second derivative:

$f''(x) = -\sin x - \cos x$.

Evaluate the second derivative at the critical points:

If $x = \frac{\pi}{4}$ (for $n=0$): $f''(\pi/4) = -\sin(\pi/4) - \cos(\pi/4) = -1/\sqrt{2} - 1/\sqrt{2} = -\sqrt{2}$. Since $f''(\pi/4) < 0$, this corresponds to a local maximum.

If $x = \frac{5\pi}{4}$ (for $n=1$): $f''(5\pi/4) = -\sin(5\pi/4) - \cos(5\pi/4) = -(-1/\sqrt{2}) - (-1/\sqrt{2}) = \sqrt{2}$. Since $f''(5\pi/4) > 0$, this corresponds to a local minimum.

The local maximum value occurs at $x = \frac{\pi}{4}$ (and points $2\pi$ apart).

$f(\pi/4) = \sin(\pi/4) + \cos(\pi/4) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

Since the function is periodic, the maximum value attained is $\sqrt{2}$.

---

Therefore, the maximum value of the function $\sin x + \cos x$ is $\sqrt{2}$.

Given:

The function $f(x) = 2x^3 - 24x + 107$.

Two intervals: $[1, 3]$ and $[-3, -1]$.

---

To Find:

The maximum value of the function $f(x)$ in the interval $[1, 3]$.

The maximum value of the function $f(x)$ in the interval $[-3, -1]$.

---

Solution:

The given function is $f(x) = 2x^3 - 24x + 107$.

To find the absolute maximum value on a closed interval, we first find the critical points by calculating the derivative and setting it to zero.

Step 1: Find the derivative $f'(x)$.

$f'(x) = \frac{d}{dx}(2x^3 - 24x + 107)$

$f'(x) = 6x^2 - 24$

Step 2: Find the critical points by setting $f'(x) = 0$.

$6x^2 - 24 = 0$

$6x^2 = 24$

$x^2 = \frac{24}{6}$

$x^2 = 4$

$x = 2$ or $x = -2$

The critical points are $x = 2$ and $x = -2$.

---

Analysis for the interval [1, 3]:

Step 3a: Identify critical points in the interval (1, 3).

The critical points are $x=2$ and $x=-2$. Only $x = 2$ lies within the interval $[1, 3]$.

Step 4a: Evaluate $f(x)$ at the critical point ($x=2$) and the endpoints ($x=1, x=3$).

At the endpoint $x = 1$:

$f(1) = 2(1)^3 - 24(1) + 107 = 2 - 24 + 107 = 85$

At the critical point $x = 2$:

$f(2) = 2(2)^3 - 24(2) + 107 = 2(8) - 48 + 107 = 16 - 48 + 107 = 75$

At the endpoint $x = 3$:

$f(3) = 2(3)^3 - 24(3) + 107 = 2(27) - 72 + 107 = 54 - 72 + 107 = 89$

Step 5a: Determine the maximum value in [1, 3].

Comparing the values $f(1)=85$, $f(2)=75$, and $f(3)=89$, the largest value is 89.

The maximum value of the function in the interval $[1, 3]$ is 89.

---

Analysis for the interval [-3, -1]:

Step 3b: Identify critical points in the interval (-3, -1).

The critical points are $x=2$ and $x=-2$. Only $x = -2$ lies within the interval $[-3, -1]$.

Step 4b: Evaluate $f(x)$ at the critical point ($x=-2$) and the endpoints ($x=-3, x=-1$).

At the endpoint $x = -3$:

$f(-3) = 2(-3)^3 - 24(-3) + 107 = 2(-27) + 72 + 107 = -54 + 72 + 107 = 125$

At the critical point $x = -2$:

$f(-2) = 2(-2)^3 - 24(-2) + 107 = 2(-8) + 48 + 107 = -16 + 48 + 107 = 139$

At the endpoint $x = -1$:

$f(-1) = 2(-1)^3 - 24(-1) + 107 = 2(-1) + 24 + 107 = -2 + 24 + 107 = 129$

Step 5b: Determine the maximum value in [-3, -1].

Comparing the values $f(-3)=125$, $f(-2)=139$, and $f(-1)=129$, the largest value is 139.

The maximum value of the function in the interval $[-3, -1]$ is 139.

---

Summary:

The maximum value of $f(x) = 2x^3 - 24x + 107$ in the interval $[1, 3]$ is 89 (at $x=3$).

The maximum value of the same function in the interval $[-3, -1]$ is 139 (at $x=-2$).

Given:

The function $f(x) = x^4 - 62x^2 + ax + 9$.

The interval $[0, 2]$.

The function $f(x)$ attains its maximum value on the interval $[0, 2]$ at the point $x = 1$.

---

To Find:

The value of $a$.

---

Solution:

The given function is $f(x) = x^4 - 62x^2 + ax + 9$.

We are told that the function attains its maximum value on the closed interval $[0, 2]$ at $x = 1$.

Since $x = 1$ is an interior point of the interval $[0, 2]$ (i.e., $0 < 1 < 2$), and the function $f(x)$ is differentiable (being a polynomial), the derivative of the function must be zero at this point of maximum value.

First, find the derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx} (x^4 - 62x^2 + ax + 9)$

$f'(x) = 4x^3 - 62(2x) + a(1) + 0$

$f'(x) = 4x^3 - 124x + a$

Since the maximum occurs at $x = 1$, we must have $f'(1) = 0$.

Substitute $x = 1$ into the expression for $f'(x)$:

$f'(1) = 4(1)^3 - 124(1) + a$

$f'(1) = 4(1) - 124 + a$

$f'(1) = 4 - 124 + a$

$f'(1) = -120 + a$

Now, set $f'(1)$ equal to zero:

$-120 + a = 0$

Solving for $a$:

$a = 120$

---

Therefore, the value of $a$ is 120.

Given:

The function $f(x) = x + \sin(2x)$.

The interval $[0, 2\pi]$.

---

To Find:

The absolute maximum and absolute minimum values of the function $f(x)$ on the interval $[0, 2\pi]$.

---

Solution:

We are given the function $f(x) = x + \sin(2x)$ on the interval $[0, 2\pi]$.

To find the absolute maximum and minimum values, we first find the critical points by calculating the derivative $f'(x)$ and setting it to zero.

Step 1: Find the derivative $f'(x)$.

$f'(x) = \frac{d}{dx}(x + \sin(2x))$

$f'(x) = 1 + \cos(2x) \cdot 2$

$f'(x) = 1 + 2\cos(2x)$

Step 2: Find the critical points by setting $f'(x) = 0$.

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

$2\cos(2x) = -1$

$\cos(2x) = -\frac{1}{2}$

Let $\theta = 2x$. We need to find the values of $\theta$ such that $\cos(\theta) = -1/2$.

The cosine function is negative in the second and third quadrants. The reference angle for $\cos(\theta) = 1/2$ is $\pi/3$.

So, the solutions for $\theta$ in the interval $[0, 4\pi]$ (since $x \in [0, 2\pi] \implies 2x \in [0, 4\pi]$) are:

$\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$

$\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$

$\theta = \frac{2\pi}{3} + 2\pi = \frac{8\pi}{3}$

$\theta = \frac{4\pi}{3} + 2\pi = \frac{10\pi}{3}$

Now, we solve for $x$ using $\theta = 2x$:

$2x = \frac{2\pi}{3} \implies x = \frac{\pi}{3}$

$2x = \frac{4\pi}{3} \implies x = \frac{2\pi}{3}$

$2x = \frac{8\pi}{3} \implies x = \frac{4\pi}{3}$

$2x = \frac{10\pi}{3} \implies x = \frac{5\pi}{3}$

All these critical points, $x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$, lie within the interval $[0, 2\pi]$.

Step 3: Evaluate $f(x)$ at the critical points and the endpoints ($x=0, x=2\pi$).

At the endpoint $x = 0$:

$f(0) = 0 + \sin(2 \cdot 0) = 0 + \sin(0) = 0$

At the critical point $x = \frac{\pi}{3}$:

$f\left(\frac{\pi}{3}\right) = \frac{\pi}{3} + \sin\left(2 \cdot \frac{\pi}{3}\right) = \frac{\pi}{3} + \sin\left(\frac{2\pi}{3}\right) = \frac{\pi}{3} + \frac{\sqrt{3}}{2}$

At the critical point $x = \frac{2\pi}{3}$:

$f\left(\frac{2\pi}{3}\right) = \frac{2\pi}{3} + \sin\left(2 \cdot \frac{2\pi}{3}\right) = \frac{2\pi}{3} + \sin\left(\frac{4\pi}{3}\right) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}$

At the critical point $x = \frac{4\pi}{3}$:

$f\left(\frac{4\pi}{3}\right) = \frac{4\pi}{3} + \sin\left(2 \cdot \frac{4\pi}{3}\right) = \frac{4\pi}{3} + \sin\left(\frac{8\pi}{3}\right) = \frac{4\pi}{3} + \sin\left(\frac{2\pi}{3}\right) = \frac{4\pi}{3} + \frac{\sqrt{3}}{2}$

At the critical point $x = \frac{5\pi}{3}$:

$f\left(\frac{5\pi}{3}\right) = \frac{5\pi}{3} + \sin\left(2 \cdot \frac{5\pi}{3}\right) = \frac{5\pi}{3} + \sin\left(\frac{10\pi}{3}\right) = \frac{5\pi}{3} + \sin\left(\frac{4\pi}{3}\right) = \frac{5\pi}{3} - \frac{\sqrt{3}}{2}$

At the endpoint $x = 2\pi$:

$f(2\pi) = 2\pi + \sin(2 \cdot 2\pi) = 2\pi + \sin(4\pi) = 2\pi + 0 = 2\pi$

Step 4: Determine the absolute maximum and minimum values.

We need to compare the following values:

$0$

$\frac{\pi}{3} + \frac{\sqrt{3}}{2} \approx 1.047 + 0.866 = 1.913$

$\frac{2\pi}{3} - \frac{\sqrt{3}}{2} \approx 2.094 - 0.866 = 1.228$

$\frac{4\pi}{3} + \frac{\sqrt{3}}{2} \approx 4.189 + 0.866 = 5.055$

$\frac{5\pi}{3} - \frac{\sqrt{3}}{2} \approx 5.236 - 0.866 = 4.370$

$2\pi \approx 6.283$

Comparing these values, the smallest value is 0 and the largest value is $2\pi$.

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Therefore, on the interval $[0, 2\pi]$:

The absolute maximum value of the function is $2\pi$ (which occurs at $x=2\pi$).

The absolute minimum value of the function is 0 (which occurs at $x=0$).

Problem Statement:

Find two numbers such that their sum is 24 and their product is as large as possible.

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Solution:

Let the two numbers be $x$ and $y$.

According to the first condition, their sum is 24:

From equation (i), we can express $y$ in terms of $x$:

$y = 24 - x$

According to the second condition, their product, let's call it $P$, should be as large as possible:

$P = x \cdot y$

Substitute the expression for $y$ from equation (i) into the product equation to get the product $P$ as a function of $x$:

$P(x) = x(24 - x)$

$P(x) = 24x - x^2$

To find the value of $x$ that maximizes the product $P(x)$, we use calculus.

Step 1: Find the derivative of $P(x)$ with respect to $x$.

$P'(x) = \frac{d}{dx}(24x - x^2)$

$P'(x) = 24 - 2x$

Step 2: Find the critical points by setting $P'(x) = 0$.

$24 - 2x = 0$

$24 = 2x$

$x = \frac{24}{2}$

$x = 12$

Step 3: Use the second derivative test to confirm if this critical point corresponds to a maximum.

Find the second derivative $P''(x)$:

$P''(x) = \frac{d}{dx}(24 - 2x)$

$P''(x) = -2$

Since $P''(x) = -2$, which is negative, the function $P(x)$ has a local maximum at the critical point $x = 12$. As $P(x)$ is a downward-opening parabola, this local maximum is also the absolute maximum.

Step 4: Find the corresponding value of $y$.

Substitute $x = 12$ back into the equation $y = 24 - x$:

$y = 24 - 12$

$y = 12$

The two numbers are $x=12$ and $y=12$.

Their sum is $12 + 12 = 24$, and their product is $12 \times 12 = 144$, which is the maximum possible product.

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Therefore, 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 $y$.

Their sum is 60, i.e., $x + y = 60$.

The expression $P = xy^3$ needs to be maximized.

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To Find:

The values of the positive numbers $x$ and $y$ that satisfy the given conditions.

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Solution:

We are given the constraint:

Since $x$ and $y$ must be positive, we have $x > 0$ and $y > 0$.

From condition (i), we can express $x$ in terms of $y$:

$x = 60 - y$

Since $x > 0$, we have $60 - y > 0$, which implies $y < 60$.

So, the constraint on $y$ is $0 < y < 60$.

We want to maximize the product $P = xy^3$. Substitute $x = 60 - y$ into the expression for $P$ to get $P$ as a function of $y$:

$P(y) = (60 - y)y^3$

$P(y) = 60y^3 - y^4$

Now, we find the value of $y$ in the interval $(0, 60)$ that maximizes $P(y)$. We use calculus.

Step 1: Find the derivative of $P(y)$ with respect to $y$.

$P'(y) = \frac{d}{dy}(60y^3 - y^4)$

$P'(y) = 180y^2 - 4y^3$

Step 2: Find the critical points by setting $P'(y) = 0$.

$180y^2 - 4y^3 = 0$

Factor out $4y^2$:

$4y^2(45 - y) = 0$

This gives possible solutions $y = 0$ or $y = 45$.

Since we require $y > 0$, we only consider the critical point $y = 45$. This point lies within the interval $(0, 60)$.

Step 3: Use the second derivative test to confirm if $y = 45$ corresponds to a maximum.

Find the second derivative $P''(y)$:

$P''(y) = \frac{d}{dy}(180y^2 - 4y^3)$

$P''(y) = 360y - 12y^2$

Evaluate $P''(y)$ at the critical point $y = 45$:

$P''(45) = 360(45) - 12(45)^2$

$P''(45) = 12(45)(30 - 45)$

$P''(45) = 12(45)(-15)$

Since $P''(45)$ is negative, the function $P(y)$ has a local maximum at $y = 45$. As this is the only critical point in the interval $(0, 60)$, this corresponds to the absolute maximum.

Step 4: Find the corresponding value of $x$.

Substitute $y = 45$ back into the equation $x = 60 - y$:

$x = 60 - 45$

$x = 15$

The two positive numbers are $x = 15$ and $y = 45$. They satisfy $x+y = 15+45 = 60$ and are both positive.

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Therefore, the two positive numbers are 15 and 45.

Given:

Two positive numbers, $x$ and $y$.

Their sum is 35, i.e., $x + y = 35$.

The product $P = x^2 y^5$ needs to be maximized.

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To Find:

The values of the positive numbers $x$ and $y$ that satisfy the given conditions.

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Solution:

We are given the constraint:

Since $x$ and $y$ must be positive numbers, we have $x > 0$ and $y > 0$.

From equation (i), we can express $x$ in terms of $y$:

$x = 35 - y$

Since $x > 0$, we must have $35 - y > 0$, which implies $y < 35$.

So, the possible values for $y$ are in the interval $(0, 35)$.

We want to maximize the product $P = x^2 y^5$. Substitute $x = 35 - y$ into the expression for $P$ to get $P$ as a function of $y$:

$P(y) = (35 - y)^2 y^5$

To find the value of $y$ in the interval $(0, 35)$ that maximizes $P(y)$, we use calculus.

Step 1: Find the derivative of $P(y)$ with respect to $y$.

Using the product rule, let $u(y) = (35 - y)^2$ and $v(y) = y^5$.

$u'(y) = 2(35 - y)(-1) = -2(35 - y)$

$v'(y) = 5y^4$

$P'(y) = u'(y)v(y) + u(y)v'(y)$

$P'(y) = -2(35 - y) y^5 + (35 - y)^2 (5y^4)$

Factor out common terms $(35 - y)$ and $y^4$:

$P'(y) = y^4 (35 - y) [-2y + 5(35 - y)]$

$P'(y) = y^4 (35 - y) [-2y + 175 - 5y]$

$P'(y) = y^4 (35 - y) (175 - 7y)$

Step 2: Find the critical points by setting $P'(y) = 0$.

$y^4 (35 - y) (175 - 7y) = 0$

The possible solutions are:

$y^4 = 0 \implies y = 0$

$35 - y = 0 \implies y = 35$

$175 - 7y = 0 \implies 7y = 175 \implies y = \frac{175}{7} = 25$

We are looking for critical points within the interval $(0, 35)$. The only critical point in this interval is $y = 25$.

Step 3: Use the first derivative test to confirm if $y = 25$ corresponds to a maximum.

$P'(y) = y^4 (35 - y) (175 - 7y) = 7y^4 (35 - y)(25 - y)$.

For $y \in (0, 35)$, the term $7y^4$ is always positive.

If $0 < y < 25$: $(35 - y)$ is positive and $(25 - y)$ is positive. So, $P'(y) = (+)(+)(+) > 0$. The function is increasing.

If $25 < y < 35$: $(35 - y)$ is positive and $(25 - y)$ is negative. So, $P'(y) = (+)(+)(-) < 0$. The function is decreasing.

Since $P'(y)$ changes sign from positive to negative at $y = 25$, the function $P(y)$ has a local maximum at $y = 25$. As this is the only critical point in the interval $(0, 35)$, it corresponds to the absolute maximum.

Step 4: Find the corresponding value of $x$.

Substitute $y = 25$ back into the equation $x = 35 - y$:

$x = 35 - 25$

$x = 10$

The two positive numbers are $x = 10$ and $y = 25$. They satisfy $x+y = 10+25 = 35$ and are both positive.

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Therefore, the two positive numbers are 10 and 25.