Chapter 6 Application Of Derivatives

Given:

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

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

We need to find the rate of change when $r = 5 \text{ cm}$.

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To Find:

The rate of change of the area of the circle with respect to its radius $r$, which is $\frac{dA}{dr}$, evaluated at $r = 5 \text{ cm}$.

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Solution:

The area of a circle is:

To find the rate of change of the area with respect to the radius, we differentiate equation (i) with respect to $r$:

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

We are asked to find the rate of change when the radius is $r = 5 \text{ cm}$. Substitute $r = 5$ into equation (ii):

The units of $\frac{dA}{dr}$ are $\frac{\text{Area Units}}{\text{Radius Units}}$, which is $\text{cm}^2/\text{cm}$.

The phrasing "per second with respect to its radius $r$" is interpreted as the rate of change of area with respect to radius $\frac{dA}{dr}$, as explicitly requested. The "per second" part does not affect the calculation of $\frac{dA}{dr}$.

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Final Answer:

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

Given:

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

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

Rate of increase of volume: $\frac{dV}{dt} = 9 \text{ cm}^3/\text{s}$.

Instant of interest: $s = 10 \text{ cm}$.

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To Find:

The rate of increase of the surface area, $\frac{dA}{dt}$, when $s = 10 \text{ cm}$.

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Solution:

The volume of the cube is $V = s^3$.

Differentiate the volume $V$ with respect to time $t$ using the chain rule:

Substitute the given rate $\frac{dV}{dt} = 9$ into the equation:

Solving for the rate of change of the edge length $\frac{ds}{dt}$:

The surface area of the cube is $A = 6s^2$.

Differentiate the surface area $A$ with respect to time $t$ using the chain rule:

Substitute the expression for $\frac{ds}{dt}$ from equation (i) into equation (ii):

Now, we evaluate $\frac{dA}{dt}$ at the instant when the edge length is $s = 10 \text{ cm}$:

The unit for the rate of change of surface area is $\text{cm}^2/\text{s}$.

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Final Answer:

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

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 $t$, so $\frac{dr}{dt} = 4 \text{ cm/s}$.

Instant of interest: $r = 10 \text{ cm}$.

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To Find:

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

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Solution:

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

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

Substitute the given values $r = 10 \text{ cm}$ and $\frac{dr}{dt} = 4 \text{ cm/s}$ into equation (i):

The rate of change of area is in $\text{cm}^2/\text{s}$.

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Final Answer:

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

Given:

Length: $x$, Width: $y$.

Rate of change of length (decreasing): $\frac{dx}{dt} = -3 \text{ cm/min}$.

Rate of change of width (increasing): $\frac{dy}{dt} = 2 \text{ cm/min}$.

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

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To Find:

(a) Rate of change of the perimeter $\left( \frac{dP}{dt} \right)$.

(b) Rate of change of the area $\left( \frac{dA}{dt} \right)$.

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Solution (a) Rate of change of Perimeter:

Let $P$ be the perimeter of the rectangle.

Differentiate $P$ with respect to time $t$:

Substitute the given rates $\frac{dx}{dt} = -3$ and $\frac{dy}{dt} = 2$ into equation (i):

Since $\frac{dP}{dt}$ is negative, the perimeter is decreasing.

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Solution (b) Rate of change of Area:

Let $A$ be the area of the rectangle.

Differentiate $A$ with respect to time $t$ using the Product Rule $\left( \frac{d}{dt}(uv) = u\frac{dv}{dt} + v\frac{du}{dt} \right)$:

Substitute the values $x = 10$, $y = 6$, $\frac{dx}{dt} = -3$, and $\frac{dy}{dt} = 2$ into equation (ii):

Since $\frac{dA}{dt}$ is positive, the area is increasing.

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Final Answer:

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

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

Given:

Total cost function: $C(x) = 0.005 x^3 – 0.02 x^2 + 30x + 5000$.

Marginal cost is $\frac{dC}{dx}$.

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To Find:

The marginal cost when $x = 3$ units are produced, i.e., $C'(3)$.

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Solution:

The marginal cost ($MC$) is the derivative of the total cost function $C(x)$ with respect to the number of units produced $x$:

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

Using the power rule, the marginal cost function is:

To find the marginal cost when $x = 3$, substitute $x = 3$ into equation (i):

The units for marginal cost are Rupees per unit.

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Final Answer:

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

Given:

Total revenue function: $R(x) = 3x^2 + 36x + 5$.

Marginal revenue is the rate of change of total revenue, $\frac{dR}{dx}$.

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To Find:

The marginal revenue when $x = 5$ units are sold, i.e., $R'(5)$.

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Solution:

The marginal revenue ($MR$) is the derivative of the total revenue function $R(x)$ with respect to the number of units sold $x$:

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

Using the power rule and sum rule, the marginal revenue function is:

To find the marginal revenue when $x = 5$, substitute $x = 5$ into equation (i):

The units for marginal revenue are Rupees per unit.

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Final Answer:

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

Given:

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

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To Find:

The rate of change of the area of the circle with respect to its radius $\frac{dA}{dr}$ when:

(a) $r = 3 \text{ cm}$

(b) $r = 4 \text{ cm}$

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Solution:

The area of a circle is:

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

Using the power rule $\left( \frac{d}{dr}(r^n) = nr^{n-1} \right)$:

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(a) Rate of change when $r = 3 \text{ cm}$:

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

The rate of change is $6\pi \text{ cm}^2/\text{cm}$.

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(b) Rate of change when $r = 4 \text{ cm}$:

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

The rate of change is $8\pi \text{ cm}^2/\text{cm}$.

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Final Answer:

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

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

Given:

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

Rate of increase of volume: $\frac{dV}{dt} = 8 \text{ cm}^3/\text{s}$.

Instant of interest: $s = 12 \text{ cm}$.

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To Find:

The rate of increase of the surface area, $\frac{dA}{dt}$, when $s = 12 \text{ cm}$.

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Solution:

The volume $V$ of a cube is $V = s^3$. Differentiate $V$ with respect to time $t$:

Substitute $\frac{dV}{dt} = 8$ and solve for $\frac{ds}{dt}$:

The surface area $A$ of a cube is $A = 6s^2$. Differentiate $A$ with respect to time $t$:

Substitute $\frac{ds}{dt}$ from equation (i) into equation (ii):

Evaluate $\frac{dA}{dt}$ at $s = 12 \text{ cm}$:

The rate of increase of the surface area is $\frac{8}{3} \text{ cm}^2/\text{s}$.

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Final Answer:

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

Given:

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

Rate of increase of radius: $\frac{dr}{dt} = 3 \text{ cm/s}$.

Instant of interest: $r = 10 \text{ cm}$.

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To Find:

The rate at which the area of the circle is increasing, $\frac{dA}{dt}$, when $r = 10 \text{ cm}$.

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Solution:

The area $A$ of a circle is $A = \pi r^2$.

Differentiate $A$ with respect to time $t$ using the chain rule:

Substitute the given values $r = 10 \text{ cm}$ and $\frac{dr}{dt} = 3 \text{ cm/s}$ into equation (i):

The rate of increase of the area is $60\pi \text{ cm}^2/\text{s}$.

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Final Answer:

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

Given:

Let $s$ be the length of an edge of the cube and $V$ be its volume.

Rate of increase of edge length: $\frac{ds}{dt} = 3 \text{ cm/s}$.

Instant of interest: $s = 10 \text{ cm}$.

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To Find:

The rate of increase of the volume, $\frac{dV}{dt}$, when $s = 10 \text{ cm}$.

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Solution:

The volume $V$ of a cube is $V = s^3$.

Differentiate $V$ with respect to time $t$ using the chain rule:

Substitute the given values $s = 10 \text{ cm}$ and $\frac{ds}{dt} = 3 \text{ cm/s}$ into equation (i):

The rate of increase of the volume is $900 \text{ cm}^3/\text{s}$.

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Final Answer:

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

Given:

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

Speed of waves (Rate of change of radius): $\frac{dr}{dt} = 5 \text{ cm/s}$.

Instant of interest: $r = 8 \text{ cm}$.

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To Find:

The rate at which the enclosed area is increasing, $\frac{dA}{dt}$, when $r = 8 \text{ cm}$.

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Solution:

The area $A$ of a circle is $A = \pi r^2$.

Differentiate $A$ with respect to time $t$ using the chain rule:

Substitute the given values $r = 8 \text{ cm}$ and $\frac{dr}{dt} = 5 \text{ cm/s}$ into equation (i):

The rate of increase of the area is $80\pi \text{ cm}^2/\text{s}$.

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Final Answer:

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

Given:

Let $r$ be the radius and $C$ be the circumference of the circle.

Rate of increase of radius: $\frac{dr}{dt} = 0.7 \text{ cm/s}$.

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To Find:

The rate of increase of the circumference, $\frac{dC}{dt}$.

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Solution:

The circumference $C$ of a circle is $C = 2\pi r$.

Differentiate $C$ with respect to time $t$ using the chain rule:

Substitute the given rate $\frac{dr}{dt} = 0.7 \text{ cm/s}$ into equation (i):

The rate of increase of the circumference is $1.4\pi \text{ cm/s}$.

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Final Answer:

The rate of increase of its circumference is $\mathbf{1.4\pi \text{ cm/s}}$.

Given:

Length: $x$, Width: $y$.

Rate of change of length (decreasing): $\frac{dx}{dt} = -5 \text{ cm/min}$.

Rate of change of width (increasing): $\frac{dy}{dt} = 4 \text{ cm/min}$.

Instant of interest: $x = 8 \text{ cm}$ and $y = 6 \text{ cm}$.

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To Find:

(a) Rate of change of the perimeter $\left( \frac{dP}{dt} \right)$.

(b) Rate of change of the area $\left( \frac{dA}{dt} \right)$.

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Solution (a) Rate of change of Perimeter:

The perimeter $P$ is $P = 2x + 2y$. Differentiate $P$ with respect to $t$:

Substitute $\frac{dx}{dt} = -5$ and $\frac{dy}{dt} = 4$ into equation (i):

The perimeter is decreasing at $2 \text{ cm/min}$.

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Solution (b) Rate of change of Area:

The area $A$ is $A = xy$. Differentiate $A$ with respect to $t$ using the Product Rule:

Substitute $x = 8$, $y = 6$, $\frac{dx}{dt} = -5$, and $\frac{dy}{dt} = 4$ into equation (ii):

The area is increasing at $2 \text{ cm}^2/\text{min}$.

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Final Answer:

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

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

Given:

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

Rate of change of volume: $\frac{dV}{dt} = 900 \text{ cm}^3/\text{s}$.

Instant of interest: $r = 15 \text{ cm}$.

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To Find:

The rate at which the radius increases, $\frac{dr}{dt}$, when $r = 15 \text{ cm}$.

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Solution:

The volume $V$ of a sphere is $V = \frac{4}{3}\pi r^3$.

Differentiate $V$ with respect to time $t$ using the chain rule:

Substitute the given values $\frac{dV}{dt} = 900$ and $r = 15$ into equation (i):

Solve for $\frac{dr}{dt}$:

The rate of increase of the radius is $\frac{1}{\pi} \text{ cm/s}$.

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Final Answer:

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

Given:

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

The volume of a sphere is $V = \frac{4}{3}\pi r^3$.

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To Find:

The rate of change of volume with respect to the radius $\frac{dV}{dr}$, evaluated at $r = 10 \text{ cm}$.

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Solution:

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

Using the power rule:

Evaluate $\frac{dV}{dr}$ at $r = 10 \text{ cm}$ by substituting $r = 10$ into equation (i):

The rate of change of volume with respect to radius is $400\pi \text{ cm}^3/\text{cm}$.

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Final Answer:

The rate at which its volume is increasing with the radius when the radius is $10 \text{ cm}$ is $\mathbf{400\pi \text{ cm}^3/\text{cm}}$.

Given:

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

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

Rate of pulling the bottom away: $\frac{dx}{dt} = 2 \text{ cm/s} = 0.02 \text{ m/s}$.

Instant of interest: $x = 4 \text{ m}$.

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To Find:

The rate at which the height on the wall is decreasing, $\frac{dy}{dt}$, when $x = 4 \text{ m}$.

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Solution:

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

First, find the value of $y$ when $x = 4 \text{ m}$:

Now, differentiate the Pythagorean equation with respect to time $t$ using the chain rule:

Substitute the values $x = 4$, $y = 3$, and $\frac{dx}{dt} = 0.02$ into equation (i):

Solve for $\frac{dy}{dt}$:

Convert the rate back to $\text{cm/s}$ ($\frac{dy}{dt} \times 100$):

The negative sign indicates that the height is decreasing.

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Final Answer:

The height on the wall is decreasing at the rate of $\mathbf{\frac{8}{3} \text{ cm/s}}$.

Given:

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

The rate of change of the $y$-coordinate with respect to time $t$ is $8$ times the rate of change of the $x$-coordinate with respect to time $t$.

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To Find:

The points $(x, y)$ on the curve that satisfy the given condition.

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Solution:

The equation of the curve is:

Differentiate the equation of the curve with respect to time $t$ using the chain rule:

Substitute the given condition from equation (i), $\frac{dy}{dt} = 8 \frac{dx}{dt}$, into equation (ii):

Rearrange the equation and factor out $\frac{dx}{dt}$:

Since the coordinates are "changing" (implying $\frac{dx}{dt} \neq 0$), we must have:

Now, find the corresponding $y$-coordinates using $6y = x^3 + 2$:

Case 1: When $x = 4$

The point is $\mathbf{(4, 11)}$.

Case 2: When $x = -4$

The point is $\mathbf{(-4, -\frac{31}{3})}$.

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Final Answer:

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

Given:

Let $r$ be the radius and $V$ be the volume of the spherical air bubble.

Rate of increase of radius: $\frac{dr}{dt} = \frac{1}{2} \text{ cm/s}$.

Instant of interest: $r = 1 \text{ cm}$.

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To Find:

The rate at which the volume of the bubble is increasing, $\frac{dV}{dt}$, when $r = 1 \text{ cm}$.

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Solution:

The volume $V$ of a sphere is $V = \frac{4}{3}\pi r^3$.

Differentiate $V$ with respect to time $t$ using the chain rule:

Substitute the given values $r = 1 \text{ cm}$ and $\frac{dr}{dt} = \frac{1}{2} \text{ cm/s}$ into equation (i):

The rate of increase of the volume is $2\pi \text{ cm}^3/\text{s}$.

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Final Answer:

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

Given:

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

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To Find:

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

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Solution:

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

The volume $V$ of a sphere is $V = \frac{4}{3}\pi r^3$.

Substitute the expression for $r$ into the volume formula:

Differentiate $V$ with respect to $x$ using the chain rule $\left( \frac{d}{dx}(u^n) = nu^{n-1} \frac{du}{dx} \right)$:

Simplify the fraction $\frac{54}{16}$:

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Final Answer:

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

Given:

Let $h$ be the height, $r$ the radius, and $V$ the volume of the cone.

Rate of pouring sand: $\frac{dV}{dt} = 12 \text{ cm}^3/\text{s}$.

Relationship between height and radius: $h = \frac{1}{6} r \implies r = 6h$.

Instant of interest: $h = 4 \text{ cm}$.

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To Find:

The rate at which the height is increasing, $\frac{dh}{dt}$, when $h = 4 \text{ cm}$.

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Solution:

The volume $V$ of a cone is $V = \frac{1}{3}\pi r^2 h$.

Substitute $r = 6h$ to express $V$ in terms of $h$ only:

Differentiate $V$ with respect to time $t$ using the chain rule:

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

Solve for $\frac{dh}{dt}$:

The rate of increase of the height is $\frac{1}{48\pi} \text{ cm/s}$.

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Final Answer:

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

Given:

Total cost function: $C(x) = 0.007x^3 – 0.003x^2 + 15x + 4000$.

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To Find:

The marginal cost $C'(x)$ when $x = 17$.

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Solution:

The marginal cost $C'(x)$ is the derivative of the total cost function $C(x)$ with respect to $x$:

Using the rules of differentiation:

Evaluate the marginal cost when $x = 17$:

The marginal cost is $\textsf{₹} \ 20.967$ per unit.

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Final Answer:

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

Given:

Total revenue function: $R(x) = 13x^2 + 26x + 15$.

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To Find:

The marginal revenue $R'(x)$ when $x = 7$.

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Solution:

The marginal revenue $R'(x)$ is the derivative of the total revenue function $R(x)$ with respect to $x$:

Using the rules of differentiation:

Evaluate the marginal revenue when $x = 7$:

The marginal revenue is $\textsf{₹} \ 208$ per unit.

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Final Answer:

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

Given:

Area of a circle $A = \pi r^2$.

We need to find $\frac{dA}{dr}$ when $r = 6 \text{ cm}$.

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Solution:

Differentiate the area $A$ with respect to the radius $r$:

Substitute $r = 6 \text{ cm}$:

The rate of change is $12\pi \text{ cm}^2/\text{cm}$.

This matches option (B).

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Final Answer:

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

Given:

Total revenue function: $R(x) = 3x^2 + 36x + 5$.

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To Find:

The marginal revenue $R'(x)$ when $x = 15$.

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Solution:

The marginal revenue $R'(x)$ is the derivative of the total revenue function $R(x)$ with respect to $x$:

Evaluate the marginal revenue when $x = 15$:

The marginal revenue is $\textsf{₹} \ 126$ per unit. This matches option (D).

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Final Answer:

The correct option is (D) 126.

Given:

The function is $f(x) = 7x – 3$.

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

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To Show:

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

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Solution (Method 1: Using Differentiation)

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

Find the derivative of $f(x)$ with respect to $x$:

Since the derivative $f'(x) = 7$ is a constant and is always positive ($7 > 0$), it follows that $f'(x) > 0$ for all $x \in R$.

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

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Alternative Solution (Method 2: 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 $x_1, x_2 \in R$ be any two real numbers such that:

Multiply both sides by $7$ (a positive number):

Subtract $3$ from both sides:

By the definition of the function $f(x) = 7x - 3$, this means:

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

Given:

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

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

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To Show:

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

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Solution:

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

Find the derivative of $f(x)$ with respect to $x$:

To show that $f'(x) \geq 0$ for all $x \in R$, we can complete the square for the quadratic expression $3x^2 – 6x + 4$:

Since $(x - 1)^2$ is the square of a real number, we know that for all $x \in R$:

Multiplying by the positive constant $3$ and adding $1$ (from equation (i)):

Since $f'(x) \geq 1$ for all $x \in R$, it is clear that $f'(x) > 0$ for all $x \in R$.

Therefore, $f'(x) \geq 0$ for all $x \in R$.

Thus, the function $f(x) = x^3 – 3x^2 + 4x$ is increasing on $R$. (Note: Since $f'(x) > 0$ for all $x$, the function is actually strictly increasing on R.)

Given:

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

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To Prove:

The properties of the function $f(x) = \cos x$ regarding its monotonic behaviour on the given intervals.

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Solution:

First, find the derivative of the function $f(x)$ with respect to $x$:

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

For $x \in (0, \pi)$, $x$ lies in the first and second quadrants. In these quadrants, the sine function is positive:

From (i) and (ii), for $x \in (0, \pi)$:

Since $f'(x) < 0$ on $(0, \pi)$, the function $f(x) = \cos x$ is decreasing in $(0, \pi)$.

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

For $x \in (\pi, 2\pi)$, $x$ lies in the third and fourth quadrants. In these quadrants, the sine function is negative:

From (i) and (iii), for $x \in (\pi, 2\pi)$:

Since $f'(x) > 0$ on $(\pi, 2\pi)$, the function $f(x) = \cos x$ is increasing in $(\pi, 2\pi)$.

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

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

In $(0, \pi)$, $f'(x) < 0$, so $f(x)$ is decreasing.

In $(\pi, 2\pi)$, $f'(x) > 0$, so $f(x)$ is increasing.

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

Given:

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

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To Find:

The intervals in which $f(x)$ is (a) increasing and (b) decreasing.

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Solution:

First, find the derivative of $f(x)$ with respect to $x$:

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

The critical point $x=2$ divides the real number line into two open intervals: $(-\infty, 2)$ and $(2, \infty)$.

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Analysis of Intervals:

Interval $(-\infty, 2)$:

Choose a test value, e.g., $x=0$. $f'(0) = 2(0) - 4 = -4$.

Since $f'(x) < 0$ for all $x \in (-\infty, 2)$, the function is strictly decreasing on $(-\infty, 2)$.

Interval $(2, \infty)$:

Choose a test value, e.g., $x=3$. $f'(3) = 2(3) - 4 = 6 - 4 = 2$.

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

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(a) The function $f(x)$ is increasing where $f'(x) \geq 0$, which is for $x \geq 2$.

The interval is $\mathbf{[2, \infty)}$.

(b) The function $f(x)$ is decreasing where $f'(x) \leq 0$, which is for $x \leq 2$.

The interval is $\mathbf{(-\infty, 2]}$.

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Final Answer:

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

(a) increasing on $\mathbf{[2, \infty)}$

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

Given:

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

---

To Find:

The intervals in which $f(x)$ is (a) increasing and (b) decreasing.

---

Solution:

First, find the derivative of $f(x)$ with respect to $x$:

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

Divide by 12:

Factor the quadratic equation:

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

---

Analysis of Intervals:

The factored derivative is $f'(x) = 12(x - 3)(x + 2)$.

Interval	Test Value $x$	$12(x-3)(x+2)$	Sign of $f'(x)$	Monotonicity
$(-\infty, -2)$	$-3$	$12(-6)(-1) = 72$	$+$	Increasing
$(-2, 3)$	$0$	$12(-3)(2) = -72$	$-$	Decreasing
$(3, \infty)$	$4$	$12(1)(6) = 72$	$+$	Increasing

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

The increasing interval is $\mathbf{(-\infty, -2] \cup [3, \infty)}$.

(b) The function $f(x)$ is decreasing where $f'(x) \leq 0$. This occurs on $(-2, 3)$. Since the function is continuous, we include the endpoints.

The decreasing interval is $\mathbf{[-2, 3]}$.

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Final Answer:

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

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

(b) decreasing on $\mathbf{[-2, 3]}$

Given:

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

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

---

To Find:

The intervals in which $f(x)$ is (a) increasing and (b) decreasing.

---

Solution:

First, find the derivative of $f(x)$ with respect to $x$ using the chain rule:

Find the critical points by setting $f'(x) = 0$ in the domain $\left[ 0, \frac{\pi}{2} \right]$:

Since $x \in \left[ 0, \frac{\pi}{2} \right]$, we have $3x \in \left[ 0, \frac{3\pi}{2} \right]$.

In this range, $\cos(3x) = 0$ when $3x = \frac{\pi}{2}$ or $3x = \frac{3\pi}{2}$.

The critical point in the interior of the interval is $x = \frac{\pi}{6}$. This divides the domain into two intervals: $\left[ 0, \frac{\pi}{6} \right]$ and $\left[ \frac{\pi}{6}, \frac{\pi}{2} \right]$.

---

Analysis of Intervals:

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

For $x \in \left( 0, \frac{\pi}{6} \right)$, $3x \in \left( 0, \frac{\pi}{2} \right)$, which is the first quadrant. In the first quadrant, $\cos(3x) > 0$.

Thus, $f'(x) = 3\cos(3x) > 0$. The function is increasing on $\mathbf{\left[ 0, \frac{\pi}{6} \right]}$.

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

For $x \in \left( \frac{\pi}{6}, \frac{\pi}{2} \right)$, $3x \in \left( \frac{\pi}{2}, \frac{3\pi}{2} \right)$, which are the second and third quadrants. In these quadrants, $\cos(3x) < 0$.

Thus, $f'(x) = 3\cos(3x) < 0$. The function is decreasing on $\mathbf{\left[ \frac{\pi}{6}, \frac{\pi}{2} \right]}$.

---

Final Answer:

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

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

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

Given:

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

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

---

To Find:

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

---

Solution:

First, find the derivative of $f(x)$ with respect to $x$:

Find the critical points by setting $f'(x) = 0$ in the domain $[0, 2\pi]$:

In the interval $[0, 2\pi]$, $\tan x = 1$ occurs in the first and third quadrants:

The critical points divide the domain into three intervals: $\left[ 0, \frac{\pi}{4} \right]$, $\left[ \frac{\pi}{4}, \frac{5\pi}{4} \right]$, and $\left[ \frac{5\pi}{4}, 2\pi \right]$.

---

Analysis of Intervals (Sign of $f'(x) = \cos x - \sin x$):

Interval	Test Value $x$	$f'(x) = \cos x - \sin x$	Sign of $f'(x)$	Monotonicity
$\left( 0, \frac{\pi}{4} \right)$	$\frac{\pi}{6}$	$\cos\left(\frac{\pi}{6}\right) - \sin\left(\frac{\pi}{6}\right) $$ = \frac{\sqrt{3}}{2} - \frac{1}{2} = \frac{\sqrt{3}-1}{2}$	$+$	Increasing
$\left( \frac{\pi}{4}, \frac{5\pi}{4} \right)$	$\frac{\pi}{2}$	$\cos\left(\frac{\pi}{2}\right) - \sin\left(\frac{\pi}{2}\right) $$ = 0 - 1 = -1$	$-$	Decreasing
$\left( \frac{5\pi}{4}, 2\pi \right)$	$\frac{3\pi}{2}$	$\cos\left(\frac{3\pi}{2}\right) - \sin\left(\frac{3\pi}{2}\right) $$ = 0 - (-1) = 1$	$+$	Increasing

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The function is increasing where $f'(x) \geq 0$ and decreasing where $f'(x) \leq 0$.

Increasing intervals: $\mathbf{\left[ 0, \frac{\pi}{4} \right] \cup \left[ \frac{5\pi}{4}, 2\pi \right]}$

Decreasing interval: $\mathbf{\left[ \frac{\pi}{4}, \frac{5\pi}{4} \right]}$

---

Final Answer:

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

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

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

Given:

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

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

---

To Show:

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

---

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 $R$.

---

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 $R$.

---

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.

---

To Show:

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

---

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 $R$.

---

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 $R$.

---

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 $(0, \pi)$.

---

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 $(0, \pi)$

---

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 $x$:

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

$f'(x) = \cos x$

---

(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.

---

(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.

---

(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).

---

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 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.

---

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

---

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

---

(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).

---

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 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 $(-2, 3)$.

---

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 $(3, \infty)$.

---

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 $[-2, 3]$.

---

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:

Four functions and the open interval $\left( 0, \frac{\pi}{2} \right)$.

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To Find:

The function(s) that are 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 $(a, b)$ if its first derivative $f'(x) \leq 0$ for all $x \in (a, b)$. A function is strictly decreasing if $f'(x) < 0$ for all $x \in (a, b)$.

We analyze the derivative of each function on the interval $\left( 0, \frac{\pi}{2} \right)$.

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(A) $f(x) = \cos x$

For $x \in \left( 0, \frac{\pi}{2} \right)$, $\sin x > 0$.

Since $f'(x) < 0$, $\mathbf{f(x) = \cos x \text{ is strictly decreasing on } \left( 0, \frac{\pi}{2} \right)}$.

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(B) $f(x) = \cos 2x$

For $x \in \left( 0, \frac{\pi}{2} \right)$, the argument $2x$ lies in $(0, \pi)$.

In the interval $(0, \pi)$, $\sin 2x > 0$.

Since $f'(x) < 0$, $\mathbf{f(x) = \cos 2x \text{ is strictly decreasing on } \left( 0, \frac{\pi}{2} \right)}$.

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

For $x \in \left( 0, \frac{\pi}{2} \right)$, the argument $3x$ lies in $\left( 0, \frac{3\pi}{2} \right)$.

• For $x \in \left( 0, \frac{\pi}{3} \right)$, $3x \in (0, \pi)$, so $\sin 3x > 0$. $\implies f'(x) < 0$.
• For $x \in \left( \frac{\pi}{3}, \frac{\pi}{2} \right)$, $3x \in \left( \pi, \frac{3\pi}{2} \right)$, so $\sin 3x < 0$. $\implies f'(x) > 0$.

Since $f'(x)$ changes sign in $\left( 0, \frac{\pi}{2} \right)$, $\mathbf{f(x) = \cos 3x \text{ is neither increasing nor decreasing}}$ on the entire interval $\left( 0, \frac{\pi}{2} \right)$.

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(D) $f(x) = \tan x$

For $x \in \left( 0, \frac{\pi}{2} \right)$, $\cos x \neq 0$, so $\sec^2 x > 0$.

Since $f'(x) > 0$, $\mathbf{f(x) = \tan x \text{ is strictly increasing on } \left( 0, \frac{\pi}{2} \right)}$.

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Conclusion:

Both $\cos x$ and $\cos 2x$ are strictly decreasing (and hence decreasing) on the interval $\left( 0, \frac{\pi}{2} \right)$.

Since the question is in a single-choice format but mathematically, both options (A) and (B) are correct:

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Final Answer:

The functions that are decreasing on $\left( 0, \frac{\pi}{2} \right)$ are $\mathbf{\cos x}$ and $\mathbf{\cos 2x}$.

The correct options are (A) $\cos x$ and (B) $\cos 2x$.

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.

---

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 is $f(x) = x^2 + ax + 1$.

The condition is that $f(x)$ is strictly increasing on the interval $(1, 2)$.

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To Find:

The least value of the constant $a$.

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Solution:

For a function $f(x)$ to be strictly increasing on an open interval $(a, b)$, its first derivative $f'(x)$ must be strictly positive for all $x$ in that interval ($f'(x) > 0$).

First, find the derivative of $f(x)$ with respect to $x$:

Since $f(x)$ is strictly increasing on $(1, 2)$, we must have $f'(x) > 0$ for all $x \in (1, 2)$.

Using equation (i):

Rearrange the inequality to isolate $a$:

This inequality must hold for all $x$ in the interval $(1, 2)$.

Since $x \in (1, 2)$, the possible values of $x$ satisfy the condition:

Multiply the inequality by $2$:

Multiply the inequality by $-1$ (this reverses the inequality signs):

Which can be written as:

From the strict increasing condition (ii), $a$ must be greater than every value of $-2x$.

From inequality (iii), the values of $-2x$ are in the open interval $(-4, -2)$. The maximum value that $-2x$ approaches in this interval is $-2$ (as $x$ approaches $1$ from the right).

To ensure $a$ is strictly greater than all values of $-2x$, $a$ must be strictly greater than the supremum of the values of $-2x$, which is $-2$.

The set of all possible values for $a$ is the open interval $(-2, \infty)$. The least value in the set that satisfies the strict inequality $a > -2$ is approached at the boundary, $\mathbf{-2}$.

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Final Answer:

The least value of $a$ such that the function $f(x) = x^2 + ax + 1$ is strictly increasing on $(1, 2)$ is $\mathbf{-2}$.

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.

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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 $\cos x \neq 0$.

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To Prove:

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

(a) decreasing on $\left( 0, \frac{\pi}{2} \right)$

(b) increasing on $\left( \frac{3\pi}{2}, 2 \pi \right)$

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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:

$f(x) = \log |\cos x|$

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.

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(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.

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(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.

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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).

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To Prove:

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

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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.

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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).

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To Find:

The interval among the given options in 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 ($\frac{dy}{dx} \geq 0$). 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 $(2-x)$.

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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 $(-\infty, 0)$.

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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 $(0, 2)$.

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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 $(2, \infty)$.

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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.

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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 $x = 2$.

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To Find:

The slope of the tangent to the curve $y = x^3 – x$ at $x = 2$.

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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 $\frac{2}{3}$.

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To Find:

The point $(x, y)$ on the curve where the slope of the tangent is $\frac{2}{3}$.

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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 $m = 2$.

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To Find:

The equation of all lines having a slope of 2 and being tangent to the given curve.

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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.

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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.

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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:

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 $y \neq 0$.

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(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.

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(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.

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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.

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To Find:

The equation of the tangent line to the curve $y = \frac{x - 7}{(x - 2) (x - 3)}$ at its x-intercept.

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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 $Ax + By + C = 0$:

$x - 20y - 7 = 0$

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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.

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To Find:

The equations of the tangent line and the normal line to the curve at the point $(1, 1)$.

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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 $x + y - 2 = 0$.

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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 $x - y = 0$.

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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 $t = \frac{\pi}{2}$.

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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 $m = \lim\limits_{t \to \frac{\pi}{2}} -\frac{b}{a} \cot t = \lim\limits_{t \to \frac{\pi}{2}} -\frac{b}{a} \frac{\cos t}{\sin t}$

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.

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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 $y = 3x^4 – 4x$.

The point of tangency has $x$-coordinate $x = 4$.

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To Find:

The slope of the tangent to the curve at $x = 4$, which is $\left(\frac{dy}{dx}\right)_{x=4}$.

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Solution:

The slope of the tangent is given by the derivative $\frac{dy}{dx}$.

Differentiate $y$ with respect to $x$ using the difference rule and power rule:

Substitute $x = 4$ into equation (i) to find the slope at the required point:

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Final Answer:

The slope of the tangent to the curve at $x = 4$ is $\mathbf{764}$.

Given:

The equation of the curve is $y = \frac{x - 1}{x - 2}$, $x \neq 2$.

The point of tangency has $x$-coordinate $x = 10$.

---

To Find:

The slope of the tangent to the curve at $x = 10$, which is $\left(\frac{dy}{dx}\right)_{x=10}$.

---

Solution:

Differentiate $y$ with respect to $x$ using the Quotient Rule $\left( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \right)$:

Let $u = x - 1 \implies u' = 1$.

Let $v = x - 2 \implies v' = 1$.

Substitute $x = 10$ into equation (i) to find the slope at the required point:

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Final Answer:

The slope of the tangent to the curve at $x = 10$ is $\mathbf{-\frac{1}{64}}$.

Given:

The equation of the curve is $y = x^3 – x + 1$.

The point of tangency has $x$-coordinate $x = 2$.

---

To Find:

The slope of the tangent to the curve at $x = 2$, which is $\left(\frac{dy}{dx}\right)_{x=2}$.

---

Solution:

The slope of the tangent is given by the derivative $\frac{dy}{dx}$.

Differentiate $y$ with respect to $x$:

Substitute $x = 2$ into equation (i) to find the slope at the required point:

---

Final Answer:

The slope of the tangent to the curve at the point whose $x$-coordinate is $2$ is $\mathbf{11}$.

Given:

The equation of the curve is $y = x^3 – 3x + 2$.

The point of tangency has $x$-coordinate $x = 3$.

---

To Find:

The slope of the tangent to the curve at $x = 3$, which is $\left(\frac{dy}{dx}\right)_{x=3}$.

---

Solution:

The slope of the tangent is given by the derivative $\frac{dy}{dx}$.

Differentiate $y$ with respect to $x$:

Substitute $x = 3$ into equation (i) to find the slope at the required point:

---

Final Answer:

The slope of the tangent to the curve at the point whose $x$-coordinate is $3$ is $\mathbf{24}$.

Given:

The parametric equations are $x = a \cos^3 \theta$ and $y = a \sin^3 \theta$.

We need to find the slope of the normal at $\theta = \frac{\pi}{4}$.

---

To Find:

The slope of the normal, $m_{normal}$, at $\theta = \frac{\pi}{4}$.

---

Solution:

The slope of the tangent, $m_{tangent} = \frac{dy}{dx}$, is given by $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$.

Find $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$ using the chain rule:

Calculate the slope of the tangent $\frac{dy}{dx}$ (assuming $\sin \theta \neq 0$ and $\cos \theta \neq 0$):

Evaluate $m_{tangent}$ at $\theta = \frac{\pi}{4}$:

The slope of the normal, $m_{normal}$, is the negative reciprocal of the tangent slope $m_{tangent}$:

---

Final Answer:

The slope of the normal to the curve at $\theta = \frac{\pi}{4}$ is $\mathbf{1}$.

Given:

The parametric equations are $x = 1 - a \sin \theta$ and $y = b \cos^2 \theta$.

We need to find the slope of the normal at $\theta = \frac{\pi}{2}$.

---

To Find:

The slope of the normal, $m_{normal}$, at $\theta = \frac{\pi}{2}$.

---

Solution:

Find $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$:

Calculate the slope of the tangent $\frac{dy}{dx}$ (assuming $\cos \theta \neq 0$):

Evaluate $m_{tangent}$ at $\theta = \frac{\pi}{2}$:

The slope of the normal, $m_{normal}$, is the negative reciprocal of the tangent slope $m_{tangent}$:

Note: Since $\left(\frac{dx}{d\theta}\right)_{\theta=\frac{\pi}{2}} = -a \cos\left(\frac{\pi}{2}\right) = 0$ and $\left(\frac{dy}{d\theta}\right)_{\theta=\frac{\pi}{2}} = -2b \sin\left(\frac{\pi}{2}\right) \cos\left(\frac{\pi}{2}\right) = 0$, the initial application of the formula $\frac{dy}{dx}$ results in $\frac{0}{0}$. The calculation using the simplified expression (i) is valid as $\theta$ approaches $\frac{\pi}{2}$ (or by applying L'Hopital's rule, which yields the same result). The tangent is well-defined and has a finite slope.

---

Final Answer:

The slope of the normal to the curve at $\theta = \frac{\pi}{2}$ is $\mathbf{-\frac{a}{2b}}$.

Given:

The equation of the curve is $y = x^3 – 3x^2 – 9x + 7$.

The tangent is parallel to the $x$-axis.

---

To Find:

The points $(x, y)$ on the curve where the tangent is parallel to the $x$-axis.

---

Solution:

If the tangent is parallel to the $x$-axis, its slope is zero. The slope of the tangent is $\frac{dy}{dx}$.

Differentiate $y$ with respect to $x$:

Set the derivative equal to zero to find the $x$-coordinates of the required points:

Divide the equation by 3:

Factor the quadratic equation:

The possible $x$-coordinates are $x = 3$ and $x = -1$.

Find the corresponding $y$-coordinates using $y = x^3 – 3x^2 – 9x + 7$:

Case 1: When $x = 3$

The point is $\mathbf{(3, -20)}$.

Case 2: When $x = -1$

The point is $\mathbf{(-1, 12)}$.

---

Final Answer:

The points on the curve where the tangent is parallel to the $x$-axis are $\mathbf{(3, -20)}$ and $\mathbf{(-1, 12)}$.

Given:

The equation of the curve is $y = (x – 2)^2$.

The chord joins the points A(2, 0) and B(4, 4).

The tangent at the required point is parallel to the chord AB.

---

To Find:

The point on the curve where the tangent is parallel to the chord AB.

---

Solution:

The slope of the chord AB ($m_{chord}$) is:

The slope of the tangent ($m_{tangent}$) is the derivative $\frac{dy}{dx}$:

Since the tangent is parallel to the chord, their slopes must be equal:

Find the $y$-coordinate by substituting $x = 3$ into the curve equation $y = (x - 2)^2$:

---

Final Answer:

The point on the curve is $\mathbf{(3, 1)}$.

Given:

The equation of the curve is $y = x^3 – 11x + 5$.

The equation of the tangent line is $y = x – 11$.

---

To Find:

The point of tangency $(x, y)$.

---

Solution:

The slope of the tangent line $y = x - 11$ is $m = 1$.

The slope of the tangent to the curve is $\frac{dy}{dx}$:

Equate the slope of the curve's tangent to the slope of the given line:

Find the corresponding $y$-coordinates using the curve equation $y = x^3 – 11x + 5$:

Case 1: When $x = 2$: $y = (2)^3 - 11(2) + 5 = 8 - 22 + 5 = -9$. Point: $(2, -9)$.

Case 2: When $x = -2$: $y = (-2)^3 - 11(-2) + 5 = -8 + 22 + 5 = 19$. Point: $(-2, 19)$.

The point of tangency must satisfy the equation of the tangent line $y = x - 11$.

Check $(2, -9)$: $-9 = 2 - 11 \implies -9 = -9$. (True)

Check $(-2, 19)$: $19 = -2 - 11 \implies 19 = -13$. (False)

---

Final Answer:

The point on the curve at which the tangent is $y = x – 11$ is $\mathbf{(2, -9)}$.

Given:

The equation of the curve is $y = \frac{1}{x-1} = (x-1)^{-1}$, where $x \neq 1$.

The slope of the tangent lines is $m = -1$.

---

To Find:

The equations of all tangent lines with slope $-1$.

---

Solution:

The slope of the tangent is $\frac{dy}{dx}$:

Set $\frac{dy}{dx} = -1$ to find the $x$-coordinates of the points of tangency:

Case 1: $x-1 = 1 \implies x = 2$

Case 2: $x-1 = -1 \implies x = 0$}

Find the corresponding $y$-coordinates using $y = \frac{1}{x-1}$:

For $x = 2$: $y = \frac{1}{2-1} = 1$. Point: $(2, 1)$.

For $x = 0$: $y = \frac{1}{0-1} = -1$. Point: $(0, -1)$.

Find the equations of the tangent lines using $y - y_1 = m(x - x_1)$ with $m = -1$:

Tangent at $\mathbf{(2, 1)}$:

Tangent at $\mathbf{(0, -1)}$:

---

Final Answer:

The equations of the tangent lines are $\mathbf{x + y - 3 = 0}$ and $\mathbf{x + y + 1 = 0}$.

Given:

The equation of the curve is $y = \frac{1}{x - 3} = (x - 3)^{-1}$, where $x \neq 3$.

The required slope of the tangent lines is $m = 2$.

---

To Find:

The equations of all tangent lines to the curve with slope $2$.

---

Solution:

The slope of the tangent is given by the derivative $\frac{dy}{dx}$.

Differentiate $y$ with respect to $x$ using the chain rule:

Set the derivative equal to the given slope $m = 2$ to find the $x$-coordinates of the points of tangency:

The left-hand side of equation (ii), $(x - 3)^2$, is the square of a real number and must be greater than or equal to zero ($\geq 0$).

The right-hand side, $-\frac{1}{2}$, is a negative number ($< 0$).

Since a non-negative number cannot be equal to a negative number, equation (ii) has no real solution for $x$.

This means there are no points on the curve where the tangent line has a slope of 2.

---

Final Answer:

There are no lines having slope $2$ that are tangents to the curve $y = \frac{1}{x - 3}$.

Given:

The equation of the curve is $y = \frac{1}{x^2 - 2x + 3} = (x^2 - 2x + 3)^{-1}$.

The required slope of the tangent lines is $m = 0$.

---

To Find:

The equations of all tangent lines to the curve with slope $0$ (horizontal tangents).

---

Solution:

The slope of the tangent is $\frac{dy}{dx}$. Differentiate $y$ with respect to $x$ using the chain rule:

Set $\frac{dy}{dx} = 0$ to find the $x$-coordinates of the points of tangency:

The numerator must be zero (and the denominator non-zero):

The denominator at $x=1$ is $((1)^2 - 2(1) + 3)^2 = (2)^2 = 4 \neq 0$. So, $x=1$ is a valid point.

Find the corresponding $y$-coordinate using $y = \frac{1}{x^2 - 2x + 3}$:

The point of tangency is $\left(1, \frac{1}{2}\right)$.

Find the equation of the tangent line (with $m = 0$) using the point-slope form $y - y_1 = m(x - x_1)$:

---

Final Answer:

The equation of the line having slope $0$ which is tangent to the curve is $\mathbf{y = \frac{1}{2}}$.

Given:

The equation of the curve (ellipse) is $\frac{x^2}{9} + \frac{y^2}{16} = 1$.

---

To Find:

The points on the curve where the tangents are (i) parallel to the $x$-axis and (ii) parallel to the $y$-axis.

---

Solution:

First, find the slope of the tangent $\frac{dy}{dx}$ using implicit differentiation:

---

(i) Tangents parallel to $x$-axis (Horizontal Tangents):

This occurs when the slope $\frac{dy}{dx}$ is zero (numerator is zero, denominator is non-zero).

Substitute $x=0$ into the curve equation $\frac{x^2}{9} + \frac{y^2}{16} = 1$:

The points are $\mathbf{(0, 4)}$ and $\mathbf{(0, -4)}$. (Denominator $9y \neq 0$ is satisfied).

---

(ii) Tangents parallel to $y$-axis (Vertical Tangents):

This occurs when the slope $\frac{dy}{dx}$ is undefined (denominator is zero, numerator is non-zero).

Substitute $y=0$ into the curve equation $\frac{x^2}{9} + \frac{y^2}{16} = 1$:

The points are $\mathbf{(3, 0)}$ and $\mathbf{(-3, 0)}$. (Numerator $-16x \neq 0$ is satisfied).

---

Final Answer:

The points are:

(i) parallel to $x$-axis: $\mathbf{(0, 4)}$ and $\mathbf{(0, -4)}$.

(ii) parallel to $y$-axis: $\mathbf{(3, 0)}$ and $\mathbf{(-3, 0)}$.

The general slope of the tangent is $\frac{dy}{dx}$. The slope of the normal is $m_{normal} = -\frac{1}{m_{tangent}}$.

---

(i) $y = x^4 – 6x^3 + 13x^2 – 10x + 5$ at $(0, 5)$

Slope of Tangent: $\frac{dy}{dx} = 4x^3 - 18x^2 + 26x - 10$.

Slope of Normal: $m_{normal} = -\frac{1}{-10} = \frac{1}{10}$.

Equation of Tangent: $y - 5 = -10(x - 0) \implies \mathbf{10x + y - 5 = 0}$.

Equation of Normal: $y - 5 = \frac{1}{10}(x - 0) \implies 10y - 50 = x \implies \mathbf{x - 10y + 50 = 0}$.

---

(ii) $y = x^4 – 6x^3 + 13x^2 – 10x + 5$ at $(1, 3)$

Slope of Tangent: $\frac{dy}{dx} = 4x^3 - 18x^2 + 26x - 10$.

Slope of Normal: $m_{normal} = -\frac{1}{2}$.

Equation of Tangent: $y - 3 = 2(x - 1) \implies y - 3 = 2x - 2 \implies \mathbf{2x - y + 1 = 0}$.

Equation of Normal: $y - 3 = -\frac{1}{2}(x - 1) \implies 2y - 6 = -x + 1 \implies \mathbf{x + 2y - 7 = 0}$.

---

(iii) $y = x^3$ at $(1, 1)$

Slope of Tangent: $\frac{dy}{dx} = 3x^2$.

Slope of Normal: $m_{normal} = -\frac{1}{3}$.

Equation of Tangent: $y - 1 = 3(x - 1) \implies y - 1 = 3x - 3 \implies \mathbf{3x - y - 2 = 0}$.

Equation of Normal: $y - 1 = -\frac{1}{3}(x - 1) \implies 3y - 3 = -x + 1 \implies \mathbf{x + 3y - 4 = 0}$.

---

(iv) $y = x^2$ at $(0, 0)$

Slope of Tangent: $\frac{dy}{dx} = 2x$.

Slope of Normal: $m_{normal} = -\frac{1}{0}$ (Undefined).

Equation of Tangent: $y - 0 = 0(x - 0) \implies \mathbf{y = 0}$ (The $x$-axis).

Equation of Normal: Since the tangent is horizontal, the normal is vertical. $\mathbf{x = 0}$ (The $y$-axis).

---

(v) $x = \cos t, y = \sin t$ at $t = \frac{\pi}{4}$

Point: $x_1 = \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$, $y_1 = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$. Point: $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$.

Slope of Tangent: $\frac{dy}{dx} = \frac{\cos t}{-\sin t} = -\cot t$.

Slope of Normal: $m_{normal} = -\frac{1}{-1} = 1$.

Equation of Tangent: $y - \frac{\sqrt{2}}{2} = -1 \left( x - \frac{\sqrt{2}}{2} \right) \implies x + y = \sqrt{2} \implies \mathbf{x + y - \sqrt{2} = 0}$.

Equation of Normal: $y - \frac{\sqrt{2}}{2} = 1 \left( x - \frac{\sqrt{2}}{2} \right) \implies \mathbf{x - y = 0}$.

Given:

Curve: $y = x^2 – 2x + 7$.

General slope of tangent: $\frac{dy}{dx} = 2x - 2$.

---

(a) Parallel to the line $2x – y + 9 = 0$

Slope of the line $2x - y + 9 = 0$ is $m_{L1}$. In $y = 2x + 9$, $\mathbf{m_{L1} = 2}$.

Since the tangent is parallel, $m_{tangent} = 2$.

Point of Tangency ($x=2$): $y = (2)^2 - 2(2) + 7 = 7$. Point: $(2, 7)$.

Equation of Tangent: $y - 7 = 2(x - 2) \implies y - 7 = 2x - 4 \implies \mathbf{2x - y + 3 = 0}$.

---

(b) Perpendicular to the line $5y – 15x = 13$

Slope of the line $5y - 15x = 13$ is $m_{L2}$. In $y = 3x + \frac{13}{5}$, $\mathbf{m_{L2} = 3}$.

Since the tangent is perpendicular, $m_{tangent} = -\frac{1}{m_{L2}} = -\frac{1}{3}$.

Point of Tangency ($x=\frac{5}{6}$): $y = \left(\frac{5}{6}\right)^2 - 2\left(\frac{5}{6}\right) + 7 = \frac{25}{36} - \frac{10}{6} + 7 = \frac{25 - 60 + 252}{36} = \frac{217}{36}$. Point: $\left(\frac{5}{6}, \frac{217}{36}\right)$.

Equation of Tangent: $y - \frac{217}{36} = -\frac{1}{3}\left(x - \frac{5}{6}\right)$. Multiply by $36$:

---

Final Answer:

(a) The equation of the tangent line is $\mathbf{2x - y + 3 = 0}$.

(b) The equation of the tangent line is $\mathbf{12x + 36y - 227 = 0}$.

Given:

The equation of the curve is $y = 7x^3 + 11$.

---

To Prove:

The tangents at $x = 2$ and $x = -2$ are parallel (i.e., their slopes are equal).

---

Proof:

The slope of the tangent is $\frac{dy}{dx}$:

Find the slope at $x = 2$ (let $m_1$):

Find the slope at $x = -2$ (let $m_2$):

Since $\mathbf{m_1 = m_2 = 84}$, the tangents to the curve at $x = 2$ and $x = -2$ are parallel.

Given:

Curve: $y = x^3$.

Condition: Slope of tangent ($\frac{dy}{dx}$) $=$ $y$-coordinate ($y$).

---

To Find:

The points $(x, y)$ on the curve satisfying the condition.

---

Solution:

The slope of the tangent is $\frac{dy}{dx}$:

Apply the given condition $\frac{dy}{dx} = y$:

The point $(x, y)$ must also be on the curve $y = x^3$:

Equate the two expressions for $y$ from (i) and (ii):

The $x$-coordinates are $x = 0$ and $x = 3$.

Find the corresponding $y$-coordinates using $y = x^3$:

For $x = 0$: $y = (0)^3 = 0$. Point: $\mathbf{(0, 0)}$.

For $x = 3$: $y = (3)^3 = 27$. Point: $\mathbf{(3, 27)}$.

---

Final Answer:

The points on the curve are $\mathbf{(0, 0)}$ and $\mathbf{(3, 27)}$.

Given:

Curve: $y = 4x^3 – 2x^5$.

The tangent at point $(x_0, y_0)$ passes through the origin $(0, 0)$.

---

To Find:

The points $(x_0, y_0)$ on the curve satisfying the condition.

---

Solution:

The slope of the tangent at $(x_0, y_0)$ is $m = \frac{dy}{dx}\Big|_{x=x_0}$.

The tangent equation at $(x_0, y_0)$ is $Y - y_0 = m(X - x_0)$.

Since the tangent passes through $(0, 0)$, substitute $X=0$ and $Y=0$:

The point $(x_0, y_0)$ is on the curve $y = 4x^3 - 2x^5$:

Equate (i) and (ii):

The $x$-coordinates are $x_0 = 0, x_0 = 1, x_0 = -1$.

Find $y_0$ using $y_0 = 4x_0^3 - 2x_0^5$:

For $x_0 = 0$: $y_0 = 0$. Point: $\mathbf{(0, 0)}$.

For $x_0 = 1$: $y_0 = 4(1)^3 - 2(1)^5 = 2$. Point: $\mathbf{(1, 2)}$.

For $x_0 = -1$: $y_0 = 4(-1)^3 - 2(-1)^5 = -4 - (-2) = -2$. Point: $\mathbf{(-1, -2)}$.

---

Final Answer:

The points are $\mathbf{(0, 0)}$, $\mathbf{(1, 2)}$, and $\mathbf{(-1, -2)}$.

Given:

Curve: $x^2 + y^2 – 2x – 3 = 0$.

The tangents are parallel to the $x$-axis ($\frac{dy}{dx} = 0$).

---

To Find:

The points on the curve where the tangents are horizontal.

---

Solution:

Differentiate implicitly with respect to $x$ to find $\frac{dy}{dx}$:

Set $\frac{dy}{dx} = 0$ (for tangent parallel to $x$-axis):

Substitute $x = 1$ into the curve equation to find $y$ (and check $y \neq 0$):

The points are $\mathbf{(1, 2)}$ and $\mathbf{(1, -2)}$.

---

Final Answer:

The points on the curve at which the tangents are parallel to the $x$-axis are $\mathbf{(1, 2)}$ and $\mathbf{(1, -2)}$.

Given:

Curve: $ay^2 = x^3$.

Point: $P(x_1, y_1) = (am^2, am^3)$.

---

To Find:

The equation of the normal at $P$.

---

Solution:

Differentiate implicitly to find the tangent slope $\frac{dy}{dx}$:

Evaluate the tangent slope $m_{tangent}$ at $P(am^2, am^3)$:

The slope of the normal, $m_{normal}$, is $-\frac{1}{m_{tangent}}$:

The equation of the normal (with point-slope form $Y - y_1 = m_{normal}(X - x_1)$):

Multiply by $3m$:

Factoring $am^2$:

---

Final Answer:

The equation of the normal is $\mathbf{2x + 3my - am^2(2 + 3m^2) = 0}$ (or $\mathbf{2x + 3my - 2am^2 - 3am^4 = 0}$).

Given:

Curve: $y = x^3 + 2x + 6$.

Normals are parallel to the line $x + 14y + 4 = 0$.

---

To Find:

The equations of the normal lines.

---

Solution:

First, find the slope of the given line $x + 14y + 4 = 0$. In slope-intercept form ($y = -\frac{1}{14}x - \frac{2}{7}$), the slope is $\mathbf{m_{line} = -\frac{1}{14}}$.

Since the normals are parallel to this line, their slope is $\mathbf{m_{normal} = -\frac{1}{14}}$.

Next, find the slope of the tangent to the curve, $m_{tangent} = \frac{dy}{dx}$:

The slope of the normal is also given by $m_{normal} = -\frac{1}{m_{tangent}}$. Equate the two expressions for $m_{normal}$:

Find the points of tangency using $y = x^3 + 2x + 6$:

For $x = 2$: $y = (2)^3 + 2(2) + 6 = 18$. Point: $\mathbf{(2, 18)}$.

For $x = -2$: $y = (-2)^3 + 2(-2) + 6 = -6$. Point: $\mathbf{(-2, -6)}$.

Find the equations of the normal lines using $y - y_1 = m_{normal}(x - x_1)$ with $m_{normal} = -\frac{1}{14}$:

Normal at $\mathbf{(2, 18)}$:

Normal at $\mathbf{(-2, -6)}$:

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Final Answer:

The equations of the normals are $\mathbf{x + 14y - 254 = 0}$ and $\mathbf{x + 14y + 86 = 0}$.

Given:

Curve: $y^2 = 4ax$.

Point: $P(x_1, y_1) = (at^2, 2at)$.

---

To Find:

The equation of the tangent and normal at $P$.

---

Solution:

Differentiate implicitly with respect to $x$ to find the tangent slope $\frac{dy}{dx}$:

Evaluate the tangent slope $m_{tangent}$ at $P(at^2, 2at)$:

The slope of the normal, $m_{normal}$, is $-\frac{1}{m_{tangent}}$:

---

Equation of Tangent: $Y - 2at = \frac{1}{t}(X - at^2)$. Multiply by $t$:

Equation of Normal: $Y - 2at = -t(X - at^2)$

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Final Answer:

Equation of the tangent is: $\mathbf{t y = x + a t^2}$.

Equation of the normal is: $\mathbf{t x + y = 2 a t + a t^3}$.

Given:

Curve $C_1$: $x = y^2$

Curve $C_2$: $xy = k$

---

To Prove:

$C_1$ and $C_2$ cut orthogonally (at right angles) if $8k^2 = 1$.

---

Proof:

The curves cut orthogonally if $m_1 \times m_2 = -1$ at the point of intersection $P(x_0, y_0)$.

Find the point of intersection $P$: Substitute $x = y^2$ into $xy = k$:

Find the slopes of the tangents $m_1$ and $m_2$ at $P(k^{2/3}, k^{1/3})$:

Slope $\mathbf{m_1}$ for $\mathbf{x = y^2}$: $1 = 2y \frac{dy}{dx} \implies m_1 = \frac{1}{2y}$

Slope $\mathbf{m_2}$ for $\mathbf{xy = k}$: $x \frac{dy}{dx} + y = 0 \implies m_2 = -\frac{y}{x}$

Apply the orthogonality condition $m_1 \times m_2 = -1$ (assuming $k \neq 0$):

Cube both sides:

Thus, the curves cut at right angles if $\mathbf{8k^2 = 1}$.

Given:

Curve (Hyperbola): $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

Point: $P(x_0, y_0)$.

---

To Find:

The equation of the tangent and normal at $P(x_0, y_0)$.

---

Solution:

Differentiate implicitly to find the tangent slope $\frac{dy}{dx}$:

The slope of the tangent at $(x_0, y_0)$ is $m_{tangent} = \frac{b^2 x_0}{a^2 y_0}$.

The slope of the normal is $m_{normal} = -\frac{1}{m_{tangent}} = -\frac{a^2 y_0}{b^2 x_0}$.

---

Equation of Tangent: $Y - y_0 = \frac{b^2 x_0}{a^2 y_0}(X - x_0)$. Multiply by $a^2 b^2$ and rearrange:

Since $\frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} = 1$, the equation is:

Equation of Normal: $Y - y_0 = -\frac{a^2 y_0}{b^2 x_0}(X - x_0)$. Multiply by $b^2 x_0$:

Alternatively (dividing by $x_0 y_0$):

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Final Answer:

Equation of the tangent: $\mathbf{\frac{x x_0}{a^2} - \frac{y y_0}{b^2} = 1}$.

Equation of the normal: $\mathbf{a^2 y_0 (x - x_0) + b^2 x_0 (y - y_0) = 0}$ or $\mathbf{\frac{a^2 x}{x_0} + \frac{b^2 y}{y_0} = a^2 + b^2}$.

Given:

Curve: $y = \sqrt{3x-2}$.

Tangent is parallel to line $4x - 2y + 5 = 0$.

---

To Find:

The equation of the tangent line.

---

Solution:

Slope of the line $4x - 2y + 5 = 0$ ($m_{line}$): $2y = 4x + 5 \implies y = 2x + \frac{5}{2}$. $\mathbf{m_{line} = 2}$.

Since the tangent is parallel, $m_{tangent} = 2$.

The slope of the tangent is $\frac{dy}{dx} = \frac{d}{dx}((3x-2)^{1/2})$:

Set $\frac{dy}{dx} = 2$ to find the $x$-coordinate of the point of tangency:

Square both sides: $9 = 16(3x-2) \implies 9 = 48x - 32$.

Find the $y$-coordinate: $y = \sqrt{3\left(\frac{41}{48}\right) - 2} = \sqrt{\frac{41}{16} - \frac{32}{16}} = \sqrt{\frac{9}{16}} = \frac{3}{4}$. Point: $\left(\frac{41}{48}, \frac{3}{4}\right)$.

Equation of Tangent: $y - \frac{3}{4} = 2\left(x - \frac{41}{48}\right) \implies y - \frac{3}{4} = 2x - \frac{41}{24}$.

Multiply by $24$ to clear fractions: $24y - 18 = 48x - 41$.

---

Final Answer:

The equation of the tangent line is $\mathbf{48x - 24y - 23 = 0}$.

Given:

Curve: $y = 2x^2 + 3 \sin x$.

We need the slope of the normal at $x = 0$.

---

Solution:

The slope of the tangent is $\frac{dy}{dx}$:

The slope of the tangent $m_{tangent}$ at $x = 0$ is:

The slope of the normal $m_{normal}$ is $-\frac{1}{m_{tangent}}$:

This matches option (D).

---

Final Answer:

The correct option is (D) $\mathbf{-\frac{1}{3}}$.

Given:

Curve: $y^2 = 4x$.

Tangent line: $y = x + 1$.

---

To Find:

The point of tangency $(x, y)$.

---

Solution:

The point of tangency must lie on both the curve and the line. Substitute $y = x + 1$ into $y^2 = 4x$:

Substitute $x=1$ into the line equation $y = x + 1$:

The point of tangency is $\mathbf{(1, 2)}$. This matches option (A).

---

Final Answer:

The correct option is (A) $\mathbf{(1, 2)}$.

To Find:

Approximate value of $\sqrt{36.6}$ using differentials.

---

Solution:

Let $f(x) = \sqrt{x}$. We want to approximate $f(x + \Delta x)$, where $x = 36$ and $\Delta x = 0.6$.

The approximation formula is $f(x + \Delta x) \approx f(x) + dy$, where $dy = f'(x) \Delta x$.

Find the derivative $f'(x)$:

Evaluate $f(x)$ and $f'(x)$ at $x = 36$:

Calculate the differential $dy$:

Approximate $f(36.6)$:

---

Final Answer:

The approximate value of $\sqrt{36.6}$ is $\mathbf{6.05}$.

To Find:

Approximate value of $25^{1/3}$ using differentials.

---

Solution:

Let $f(x) = x^{1/3}$. We want to approximate $f(x + \Delta x)$, where $x = 27$ and $\Delta x = 25 - 27 = -2$.

The approximation formula is $f(x + \Delta x) \approx f(x) + dy$, where $dy = f'(x) \Delta x$.

Find the derivative $f'(x)$:

Evaluate $f(x)$ and $f'(x)$ at $x = 27$:

Calculate the differential $dy$:

Approximate $f(25)$:

In decimal form, $\frac{79}{27} \approx 2.9259...$

---

Final Answer:

The approximate value of $25^{1/3}$ is $\mathbf{\frac{79}{27}}$ (or approximately $\mathbf{2.926}$).

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 $f(x + \Delta x)$, where $x = 3$ and $\Delta x = 0.02$.

The approximation formula is $f(x + \Delta x) \approx f(x) + dy$, where $dy = f'(x) \Delta x$.

Find the derivative $f'(x)$:

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

Calculate the differential $dy$:

Approximate $f(3.02)$:

---

Final Answer:

The approximate value of $f(3.02)$ is $\mathbf{45.46}$.

Given:

Volume of a cube: $V = x^3$.

Change in side length: $\Delta x = 2\% \text{ of } x = 0.02x$ meters.

---

To Find:

The approximate change in the volume, $dV$, using differentials.

---

Solution:

The approximate change in volume $\Delta V$ is given by the differential $dV$:

Find the derivative of $V$ with respect to $x$:

Substitute $\frac{dV}{dx} = 3x^2$ and $\Delta x = 0.02x$ into the differential formula:

The units are cubic meters ($\text{m}^3$).

---

Final Answer:

The approximate change in the volume is $\mathbf{0.06x^3 \text{ m}^3}$.

Given:

Radius: $r = 9 \text{ cm}$.

Error in radius (change in $r$): $\Delta r = 0.03 \text{ cm}$.

Volume of a sphere: $V = \frac{4}{3}\pi r^3$.

---

To Find:

The approximate error in calculating the volume, $dV$.

---

Solution:

The approximate error in volume $\Delta V$ is given by the differential $dV$:

Find the derivative of $V$ with respect to $r$:

Evaluate the derivative at $r = 9 \text{ cm}$:

Substitute the derivative and $\Delta r$ into the differential formula:

The approximate error in volume is $9.72\pi \text{ cm}^3$.

---

Final Answer:

The approximate error in calculating the volume is $\mathbf{9.72 \pi \text{ cm}^3}$.

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 $f(x) = x^2$, with domain $R$ (all real numbers).

---

To Find:

The maximum and minimum values of $f(x)$, if any.

---

Solution:

Minimum Value:

The square of any real number is always non-negative:

The equality $f(x) = 0$ holds when $x = 0$.

Since $f(x) \geq 0$ for all $x \in R$, the minimum value is $f(0) = 0$.

Maximum Value:

As $x \to \infty$ or $x \to -\infty$, $f(x) = x^2 \to \infty$.

Since the values of $f(x)$ can increase indefinitely, the function does not have an upper bound.

---

Conclusion:

The function $f(x) = x^2$ has a minimum value of $\mathbf{0}$ at $x = 0$.

The function has no maximum value.

Given:

The function is $f(x) = |x|$, with domain $R$.

---

To Find:

The maximum and minimum values of $f(x)$, if any.

---

Solution:

Minimum Value:

The absolute value of any real number is always non-negative:

The equality $f(x) = 0$ holds when $x = 0$.

Since $f(x) \geq 0$ for all $x \in R$, the minimum value is $f(0) = 0$.

Maximum Value:

As $x \to \infty$ or $x \to -\infty$, $f(x) = |x| \to \infty$.

Since the values of $f(x)$ can increase indefinitely, the function does not have an upper bound.

---

Conclusion:

The function $f(x) = |x|$ has a minimum value of $\mathbf{0}$ at $x = 0$.

The function has no maximum value.

Given:

The function is $f(x) = x$, with domain the open interval $(0, 1)$.

---

To Find:

The maximum and minimum values of $f(x)$ on $(0, 1)$, if any.

---

Solution:

The domain of the function is the open interval $(0, 1)$, which means $0 < x < 1$.

Minimum Value:

The value of $f(x) = x$ approaches $0$ as $x$ approaches $0$ from the right. Since $x$ cannot reach $0$ (the interval is open), the function gets arbitrarily close to $0$ but never reaches a smallest value in the interval.

For any value $c \in (0, 1)$, we can find a value $x = c/2$ such that $f(x) < c$.

Therefore, the function has no minimum value in $(0, 1)$.

Maximum Value:

The value of $f(x) = x$ approaches $1$ as $x$ approaches $1$ from the left. Since $x$ cannot reach $1$ (the interval is open), the function gets arbitrarily close to $1$ but never reaches a largest value in the interval.

For any value $c \in (0, 1)$, we can find a value $x = (c+1)/2$ such that $f(x) > c$.

Therefore, the function has no maximum value in $(0, 1)$.

---

Conclusion:

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

---

Solution:

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

Set $f'(x) = 0$:

Use the Second Derivative Test to classify the critical points. Find $f''(x)$:

At $x = 1$: $f''(1) = 6(1) = 6$. Since $f''(1) > 0$, there is a local minimum at $x=1$.

At $x = -1$: $f''(-1) = 6(-1) = -6$. Since $f''(-1) < 0$, there is a local maximum at $x=-1$.

Find the local maximum and minimum values:

---

Conclusion:

The function has a local maximum at $\mathbf{x = -1}$ and a local minimum at $\mathbf{x = 1}$.

The points are $\mathbf{(-1, 5)}$ (local maximum) and $\mathbf{(1, 1)}$ (local minimum).

Given:

The function is $f(x) = 2x^3 – 6x^2 + 6x + 5$.

---

To Find:

All points of local maxima and local minima of $f(x)$.

---

Solution:

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

Set $f'(x) = 0$:

The only critical point is $x = 1$.

Use the First Derivative Test since the Second Derivative Test will fail ($f''(1) = 12(1) - 12 = 0$).

Examine the sign of $f'(x) = 6(x - 1)^2$ around $x = 1$:

For $x < 1$: $f'(x) = 6(\text{negative})^2 > 0$.

For $x > 1$: $f'(x) = 6(\text{positive})^2 > 0$.

Since the sign of $f'(x)$ does not change as $x$ passes through $x=1$, the function has neither a local maximum nor a local minimum at $x=1$.

---

Conclusion:

The function $f(x) = 2x^3 - 6x^2 + 6x + 5$ has no points of local maxima or local minima.

Given:

The function is $f(x) = 3 + |x|$, with domain $R$.

---

To Find:

The local minimum value of $f(x)$.

---

Solution:

We use the fundamental property of the absolute value function:

Adding 3 to the inequality:

The smallest value the function can attain is 3. This value is attained when $|x| = 0$, which is at $x=0$.

Since $f(x) \geq f(0)$ for all $x \in R$, $f(x)$ has an absolute minimum at $x=0$. An absolute minimum is also a local minimum.

---

Final Answer:

The local minimum value of the function $f(x) = 3 + |x|$ is $\mathbf{3}$, 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 $f(x)$.

---

Solution:

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

Set $f'(x) = 0$:

The critical points are $x = 0$, $x = -2$, and $x = 1$.

Use the Second Derivative Test. Find $f''(x)$:

At $x = 0$: $f''(0) = -24$. Since $f''(0) < 0$, $\mathbf{f(x)}$ has a local maximum at $x=0$.

At $x = 1$: $f''(1) = 36 + 24 - 24 = 36$. Since $f''(1) > 0$, $\mathbf{f(x)}$ has a local minimum at $x=1$.

At $x = -2$: $f''(-2) = 36(4) + 24(-2) - 24 = 144 - 48 - 24 = 72$. Since $f''(-2) > 0$, $\mathbf{f(x)}$ has a local minimum at $x=-2$.

Calculate the local maximum and minimum values:

---

Final Answer:

Local maximum value is $\mathbf{12}$ (at $x=0$).

Local minimum values are $\mathbf{7}$ (at $x=1$) and $\mathbf{-20}$ (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 $f(x)$.

---

Solution:

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

Set $f'(x) = 0 \implies 6(x - 1)^2 = 0 \implies x = 1$.

The only critical point is $x = 1$.

Use the First Derivative Test on $f'(x) = 6(x - 1)^2$ around $x = 1$:

For $x < 1$: $f'(x) > 0$.

For $x > 1$: $f'(x) > 0$.

Since the sign of $f'(x)$ does not change as $x$ passes through $x=1$ (it is positive on both sides), there is neither a local maximum nor a local minimum at $x=1$.

---

Conclusion:

The function $f(x) = 2x^3 – 6x^2 + 6x + 5$ has no points of local maxima or local minima.

Given:

Two positive numbers $x$ and $y$ such that $x + y = 15$.

---

To Find:

The values of $x$ and $y$ that minimize $S = x^2 + y^2$.

---

Solution:

From the constraint $y = 15 - x$. Substitute into $S$ to get $S(x)$:

The domain is $0 < x < 15$ (since $x>0$ and $y=15-x>0$).

Find the critical point by setting $S'(x) = 0$:

Use the Second Derivative Test to confirm a minimum:

Since $S''(7.5) = 4 > 0$, $S(x)$ has a minimum at $x = 7.5$.

Find the corresponding value of $y$:

---

Final Answer:

The two positive numbers are $\mathbf{7.5}$ and $\mathbf{7.5}$.

Given:

Parabola: $y = x^2$. Point: $P(0, c)$, with $\frac{1}{2} \leq c \leq 5$.

---

To Find:

The shortest distance $D$ between $P$ and the parabola.

---

Solution:

Let $Q(x, x^2)$ be a point on the parabola. Minimize the square of the distance, $Z = D^2$.

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

Since $c \geq \frac{1}{2}$, $x^2 = c - \frac{1}{2}$ has real solutions $x = \pm \sqrt{c - \frac{1}{2}}$.

The minimum distance occurs at $x = \pm \sqrt{c - \frac{1}{2}}$ (or at $x=0$ if $c=1/2$).

Calculate $Z_{min}$ by substituting $x^2 = c - \frac{1}{2}$ into $Z(x)$:

The shortest distance $D_{min} = \sqrt{Z_{min}}$:

---

Final Answer:

The shortest distance is $\mathbf{\sqrt{c - \frac{1}{4}}}$.

Given:

Vertical poles $AP=16 \text{ m}$ and $BQ=22 \text{ m}$. Bases $AB=20 \text{ m}$.

R is a point on AB. Minimize $S = RP^2 + RQ^2$.

---

To Find:

The distance AR $= x$.

---

Solution:

Let $A$ be at $(0, 0)$. Then $B$ is at $(20, 0)$, $P$ is at $(0, 16)$, and $Q$ is at $(20, 22)$.

Let $R$ be at $(x, 0)$, where $0 \le x \le 20$.

The sum of the squares of the distances is $S(x)$:

Find the critical point by setting $S'(x) = 0$:

The Second Derivative Test confirms a minimum: $S''(x) = 4 > 0$.

The minimum occurs at $x=10$. The distance of R from A is $AR = x = 10 \text{ m}$.

---

Final Answer:

The distance of a point R on AB from the point A such that $RP^2 + RQ^2$ is minimum is $\mathbf{10 \text{ m}}$.

Given:

A trapezium with three equal sides of $10 \text{ cm}$.

---

To Find:

The maximum area of the trapezium.

---

Solution:

Let the non-parallel sides and the shorter parallel side be $10 \text{ cm}$. Let the longer base be $b_1$ and the shorter parallel side be $b_2 = 10$. Let $h$ be the height.

Let $x$ be the length of the horizontal projection of the non-parallel sides onto the base $b_1$. Then $b_1 = 10 + 2x$.

By the Pythagorean theorem: $h^2 + x^2 = 10^2 \implies h = \sqrt{100 - x^2}$.

The area $A$ of the trapezium is $A = \frac{1}{2} (b_1 + b_2) h$.

To maximize $A(x)$, we find the critical points by setting $A'(x) = 0$ (or maximize $A(x)^2$).

Using the product rule:

Set $A'(x) = 0$:

Since $x$ must be a length, $x > 0$. Thus, $\mathbf{x = 5}$.

Substitute $x=5$ into the area formula to find the maximum area:

---

Final Answer:

The maximum area of the trapezium is $\mathbf{75\sqrt{3} \text{ cm}^2}$.

Given:

Cone of fixed radius $R$ and height $H$.

Inscribed cylinder of radius $r$ and height $h$.

---

To Prove:

Curved Surface Area (CSA) is maximum when $r = \frac{R}{2}$.

---

Proof:

The CSA of the cylinder is $S = 2 \pi r h$.

From similar triangles (cross-section), the relation between $r$ and $h$ is:

Substitute $h$ into the CSA formula to get $S(r)$:

Differentiate $S(r)$ with respect to $r$:

Set $\frac{dS}{dr} = 0$ for critical points (assuming $H \neq 0$):

Verify this is a maximum using the Second Derivative Test:

Since $R > 0$ and $H > 0$, $\frac{d^2S}{dr^2} < 0$, which confirms that the CSA 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:

Function: $f(x) = 2x^3 – 15x^2 + 36x + 1$. Interval: $[1, 5]$.

---

To Find:

The absolute maximum and minimum values of $f(x)$ on $[1, 5]$.

---

Solution:

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

Set $f'(x) = 0$: $6(x^2 - 5x + 6) = 0 \implies 6(x - 2)(x - 3) = 0$.

Critical points are $x = 2$ and $x = 3$. Both are in $[1, 5]$.

Evaluate $f(x)$ at the endpoints ($x=1, x=5$) and critical points ($x=2, x=3$):

---

Final Answer:

Absolute maximum value is $\mathbf{56}$ (at $x=5$).

Absolute minimum value is $\mathbf{24}$ (at $x=1$).

Given:

Function: $f(x) = 12x^{4/3} - 6x^{1/3}$. Interval: $[-1, 1]$.

---

To Find:

The absolute maximum and minimum values of $f(x)$ on $[-1, 1]$.

---

Solution:

Find the critical points by setting $f'(x) = 0$ or where $f'(x)$ is undefined.

Set $f'(x) = 0$: $16x - 2 = 0 \implies x = \frac{1}{8}$. (Critical point, in $[-1, 1]$)

$f'(x)$ is undefined at $x = 0$. (Critical point, in $[-1, 1]$)

Evaluate $f(x)$ at the endpoints ($x=-1, x=1$) and critical points ($x=0, x=1/8$):

---

Final Answer:

Absolute maximum value is $\mathbf{18}$ (at $x=-1$).

Absolute minimum value is $\mathbf{-\frac{9}{4}}$ (at $x=\frac{1}{8}$).

Given:

Helicopter path: $y = x^2 + 7$. Soldier: $S(3, 7)$.

---

To Find:

The nearest distance $D_{min}$.

---

Solution:

Let $P(x, x^2 + 7)$ be a point on the path. Minimize the square of the distance, $Z = D^2$.

Find the critical point by setting $Z'(x) = 0$:

By inspection, $x = 1$ is a root ($2(1) + 1 - 3 = 0$). Factoring gives $(x - 1)(2x^2 + 2x + 3) = 0$.

The quadratic factor $2x^2 + 2x + 3$ has a negative discriminant ($\Delta = -20$), so $x = 1$ is the only real critical point.

The Second Derivative Test confirms a minimum: $Z''(x) = 12x^2 + 2$. $Z''(1) = 14 > 0$.

The minimum value of $Z$ is at $x=1$:

The nearest distance $D_{min} = \sqrt{Z_{min}}$:

---

Final Answer:

The nearest distance is $\mathbf{\sqrt{5}}$ units.

(i) $f(x) = (2x – 1)^2 + 3$

We analyze the squared term $(2x - 1)^2$. Since the square of any real number is non-negative:

$(2x - 1)^2 \ge 0 \quad \text{for all } x \in \mathbb{R}$

Adding 3 to both sides to match the function $f(x)$:

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

$f(x) \ge 3$

Minimum Value: The minimum value is attained when the squared term is zero:
$2x - 1 = 0 \implies x = \frac{1}{2}$.
Thus, the minimum value is $f(\frac{1}{2}) = 3$.

Maximum Value: As $x$ tends to $\infty$ or $-\infty$, the term $(2x - 1)^2$ increases without bound. Therefore, the function has no maximum value.

Result: Minimum value = 3. No maximum value.

---

(ii) $f(x) = 9x^2 + 12x + 2$

We can rewrite this quadratic equation by completing the square to analyze it more easily.

$f(x) = (3x)^2 + 2(3x)(2) + 2$

To complete the square for $(3x+2)^2$, we need to add and subtract 4:

$f(x) = (9x^2 + 12x + 4) - 4 + 2$

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

Since $(3x + 2)^2 \ge 0$ for all real $x$:

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

$f(x) \ge -2$

Minimum Value: This occurs when $3x + 2 = 0 \implies x = -\frac{2}{3}$. The value is $-2$.

Maximum Value: As $x$ increases or decreases indefinitely, the function value increases towards infinity. There is no maximum value.

Result: Minimum value = -2. No maximum value.

---

(iii) $f(x) = – (x – 1)^2 + 10$

We know that $(x - 1)^2 \ge 0$ for all real $x$.

Multiplying by $-1$ reverses the inequality:

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

Adding 10 to both sides:

$-(x - 1)^2 + 10 \le 10$

$f(x) \le 10$

Maximum Value: The function reaches its highest point when the squared term is zero ($x=1$). The maximum value is 10.

Minimum Value: As $x$ moves away from 1, the term $-(x-1)^2$ becomes a large negative number. Thus, the function decreases without bound and has no minimum value.

Result: Maximum value = 10. No minimum value.

---

(iv) $g(x) = x^3 + 1$

This is a cubic polynomial.

As $x \to \infty$, $x^3 + 1 \to \infty$.

As $x \to -\infty$, $x^3 + 1 \to -\infty$.

Since the function range extends from negative infinity to positive infinity, it does not possess any global extreme points (absolute maximum or minimum).

Result: Neither a maximum value nor a minimum value exists.

(i) $f(x) = |x + 2| – 1$

The absolute value function $|x+2|$ is defined such that $|x+2| \ge 0$ for all $x \in \mathbb{R}$.

Subtracting 1 from both sides:

$|x + 2| - 1 \ge -1$

Minimum Value: The lowest value is $-1$, which occurs when $x+2=0$ (i.e., $x=-2$).

Maximum Value: Since $|x+2|$ can be infinitely large, there is no maximum value.

Result: Minimum value = -1. No maximum value.

---

(ii) $g(x) = – |x + 1| + 3$

We know that $|x + 1| \ge 0$.

Multiplying by negative one reverses the inequality: $-|x + 1| \le 0$.

Adding 3 to both sides:

$-|x + 1| + 3 \le 3$

Maximum Value: The highest value is 3, occurring when $x = -1$.

Minimum Value: As $x$ tends to infinity, the function tends to $-\infty$. There is no minimum value.

Result: Maximum value = 3. No minimum value.

---

(iii) $h(x) = \sin(2x) + 5$

The range of the sine function $\sin(\theta)$ is always $[-1, 1]$. Therefore:

$-1 \le \sin(2x) \le 1$

Adding 5 to the entire inequality:

$-1 + 5 \le \sin(2x) + 5 \le 1 + 5$

$4 \le h(x) \le 6$

Result: Minimum value = 4. Maximum value = 6.

---

(iv) $f(x) = |\sin 4x + 3|$

Start with the range of sine:

$-1 \le \sin(4x) \le 1$

Add 3 to the inequality:

$2 \le \sin(4x) + 3 \le 4$

Since the term $(\sin 4x + 3)$ is always positive (between 2 and 4), the absolute value brackets do not change the value. Thus, $f(x) \in [2, 4]$.

Result: Minimum value = 2. Maximum value = 4.

---

(v) $h(x) = x + 1, \, x \in (-1, 1)$

This function is defined on an open interval $(-1, 1)$. This means $x$ can get infinitely close to -1 and 1 but can never equal -1 or 1.

Since $h(x)$ is strictly increasing ($h'(x)=1 > 0$):

• As $x \to 1$, $h(x) \to 2$. However, $h(x)$ never actually reaches 2.
• As $x \to -1$, $h(x) \to 0$. However, $h(x)$ never actually reaches 0.

Because the boundary points are not included in the domain, the function never attains a maximum or minimum.

Result: Neither a maximum value nor a minimum value exists.

Method: We will use the Second Derivative Test. If $f'(c) = 0$:
1. If $f''(c) < 0$, $x=c$ is a local maximum.
2. If $f''(c) > 0$, $x=c$ is a local minimum.

---

(i) $f(x) = x^2$

$f'(x) = 2x$. Setting $f'(x) = 0 \implies x = 0$.

$f''(x) = 2$. Since $f''(0) = 2 > 0$, $x=0$ is a local minimum.

Local Min Value: $f(0) = 0$. No Local Max.

---

(ii) $g(x) = x^3 - 3x$

$g'(x) = 3x^2 - 3$. Setting $g'(x) = 0$:

$3(x^2 - 1) = 0 \implies x = \pm 1$.

$g''(x) = 6x$.

• At $x = 1$: $g''(1) = 6 > 0$ (Min). Value: $1^3 - 3(1) = -2$.
• At $x = -1$: $g''(-1) = -6 < 0$ (Max). Value: $(-1)^3 - 3(-1) = 2$.

Local Min at $x=1$ (Value: -2). Local Max at $x=-1$ (Value: 2).

---

(iii) $h(x) = \sin x + \cos x, \quad 0 < x < \frac{\pi}{2}$

$h'(x) = \cos x - \sin x$. Set $h'(x) = 0 \implies \sin x = \cos x \implies \tan x = 1$.

In the interval $(0, \pi/2)$, $x = \frac{\pi}{4}$.

$h''(x) = -\sin x - \cos x$.

$h''(\pi/4) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2} < 0$ (Max).

Value: $\sin(\pi/4) + \cos(\pi/4) = \sqrt{2}$.

Local Max at $x = \pi/4$ (Value: $\sqrt{2}$). No Local Min.

---

(iv) $f(x) = \sin x - \cos x, \quad 0 < x < 2\pi$

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

Set $f'(x) = 0 \implies \sin x = -\cos x \implies \tan x = -1$.

In $(0, 2\pi)$, $x = \frac{3\pi}{4}$ and $x = \frac{7\pi}{4}$.

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

• At $x = \frac{3\pi}{4}$: $f''(x) = -\frac{1}{\sqrt{2}} + (-\frac{1}{\sqrt{2}}) = -\sqrt{2} < 0$ (Max).
  Value: $\frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}}) = \sqrt{2}$.
• At $x = \frac{7\pi}{4}$: $f''(x) = -(-\frac{1}{\sqrt{2}}) + \frac{1}{\sqrt{2}} = \sqrt{2} > 0$ (Min).
  Value: $-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2}$.

Local Max at $x=3\pi/4$ (Value: $\sqrt{2}$). Local Min at $x=7\pi/4$ (Value: $-\sqrt{2}$).

---

(v) $f(x) = x^3 - 6x^2 + 9x + 15$

$f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)$.

Critical points: $x = 1, x = 3$.

$f''(x) = 6x - 12$.

• At $x=1$: $f''(1) = 6 - 12 = -6 < 0$ (Max). Value: $1 - 6 + 9 + 15 = 19$.
• At $x=3$: $f''(3) = 18 - 12 = 6 > 0$ (Min). Value: $27 - 54 + 27 + 15 = 15$.

Local Max at $x=1$ (Value: 19). Local Min at $x=3$ (Value: 15).

---

(vi) $g(x) = \frac{x}{2} + \frac{2}{x}, \quad x > 0$

$g'(x) = \frac{1}{2} - \frac{2}{x^2}$. Set to 0: $\frac{1}{2} = \frac{2}{x^2} \implies x^2 = 4$.

Since $x > 0$, critical point is $x = 2$.

$g''(x) = \frac{4}{x^3}$. At $x=2$, $g''(2) = 4/8 > 0$ (Min).

Value: $1 + 1 = 2$.

Local Min at $x=2$ (Value: 2). No Local Max.

---

(vii) $g(x) = \frac{1}{x^2+2} = (x^2+2)^{-1}$

$g'(x) = -1(x^2+2)^{-2} \cdot 2x = \frac{-2x}{(x^2+2)^2}$.

Set to 0: $-2x = 0 \implies x = 0$.

Using the first derivative test: For $x < 0$, $g'(x) > 0$. For $x > 0$, $g'(x) < 0$. The slope goes from positive to negative.

Thus, $x=0$ is a local maximum.

Value: $1/(0+2) = 1/2$.

Local Max at $x=0$ (Value: 1/2). No Local Min.

---

(viii) $f(x) = x\sqrt{1-x}, \quad 0 < x < 1$

Using Product Rule: $f'(x) = 1\cdot\sqrt{1-x} + x\left(\frac{-1}{2\sqrt{1-x}}\right)$

$f'(x) = \frac{2(1-x) - x}{2\sqrt{1-x}} = \frac{2 - 3x}{2\sqrt{1-x}}$

Set $f'(x) = 0 \implies 2 - 3x = 0 \implies x = \frac{2}{3}$.

First derivative test:

• When $x < 2/3$, numerator is positive ($f' > 0$).
• When $x > 2/3$, numerator is negative ($f' < 0$).

Change from increasing to decreasing indicates a maximum.

Value: $\frac{2}{3}\sqrt{1 - \frac{2}{3}} = \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{9}$.

Local Max at $x=2/3$ (Value: $\frac{2\sqrt{3}}{9}$). No Local Min.

To prove a differentiable function has no local extrema, we can show that its derivative is never zero.

---

(i) $f(x) = e^x$

$f'(x) = e^x$.

The exponential function $e^x$ is strictly positive for all real numbers ($e^x > 0$). It is never zero.

Since $f'(x) \ne 0$ for any $x$, there are no critical points. The function is strictly increasing everywhere.

Therefore, $f(x)$ has no maxima or minima.

---

(ii) $g(x) = \log x$

Defined for $x > 0$.

$g'(x) = \frac{1}{x}$.

For any $x > 0$, $\frac{1}{x} > 0$. The derivative is never zero.

Since the function is strictly increasing on its domain, it does not turn around.

Therefore, $g(x)$ has no maxima or minima.

---

(iii) $h(x) = x^3 + x^2 + x + 1$

$h'(x) = 3x^2 + 2x + 1$.

We check if $h'(x) = 0$ has real solutions by calculating the discriminant ($\Delta = b^2 - 4ac$) of the quadratic:

$\Delta = (2)^2 - 4(3)(1) = 4 - 12 = -8$.

Since $\Delta < 0$, the quadratic equation $3x^2 + 2x + 1 = 0$ has no real roots. Furthermore, since the leading coefficient (3) is positive, $h'(x)$ is always positive ($h'(x) > 0$).

The function is strictly increasing.

Therefore, $h(x)$ has no maxima or minima.

Method: Evaluate the function at all Critical Points inside the interval AND at the Endpoints of the interval.

---

(i) $f(x) = x^3, \quad x \in [-2, 2]$

1. Derivative: $f'(x) = 3x^2$.

2. Critical Point: $3x^2 = 0 \implies x = 0$. ($0 \in [-2, 2]$)

3. Evaluate:

• At endpoint $x = -2$: $f(-2) = (-2)^3 = -8$
• At critical point $x = 0$: $f(0) = 0$
• At endpoint $x = 2$: $f(2) = 2^3 = 8$

Absolute Maximum = 8. Absolute Minimum = -8.

---

(ii) $f(x) = \sin x + \cos x, \quad x \in [0, \pi]$

1. Derivative: $f'(x) = \cos x - \sin x$.

2. Critical Point: $\cos x = \sin x \implies \tan x = 1$. In $[0, \pi]$, $x = \pi/4$.

3. Evaluate:

• At endpoint $x = 0$: $f(0) = 0 + 1 = 1$
• At critical point $x = \pi/4$: $f(\pi/4) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \approx 1.414$
• At endpoint $x = \pi$: $f(\pi) = 0 - 1 = -1$

Absolute Maximum = $\sqrt{2}$. Absolute Minimum = -1.

---

(iii) $f(x) = 4x - \frac{1}{2}x^2, \quad x \in [-2, \frac{9}{2}]$

1. Derivative: $f'(x) = 4 - x$.

2. Critical Point: $4 - x = 0 \implies x = 4$. ($4 \in [-2, 4.5]$)

3. Evaluate:

• At endpoint $x = -2$: $f(-2) = 4(-2) - \frac{1}{2}(4) = -8 - 2 = -10$
• At critical point $x = 4$: $f(4) = 16 - \frac{1}{2}(16) = 16 - 8 = 8$
• At endpoint $x = 9/2$: $f(4.5) = 4(4.5) - \frac{1}{2}(20.25) = 18 - 10.125 = 7.875$

Absolute Maximum = 8. Absolute Minimum = -10.

---

(iv) $f(x) = (x - 1)^2 + 3, \quad x \in [-3, 1]$

1. Derivative: $f'(x) = 2(x - 1)$.

2. Critical Point: $2(x - 1) = 0 \implies x = 1$. This is also an endpoint.

3. Evaluate:

• At endpoint $x = -3$: $f(-3) = (-3 - 1)^2 + 3 = 16 + 3 = 19$
• At endpoint/critical point $x = 1$: $f(1) = (1 - 1)^2 + 3 = 3$

Absolute Maximum = 19. Absolute Minimum = 3.

Given:

The profit function is given by:

$p(x) = 41 - 72x - 18x^2$

---

To Find:

The maximum profit the company can make.

---

Solution:

To find the maximum profit, we need to determine the maximum value of the function $p(x)$. We will use the second derivative test.

Step 1: Find the first derivative $p'(x)$.

Differentiating $p(x)$ with respect to $x$:

$p'(x) = \frac{d}{dx} (41 - 72x - 18x^2)$

$p'(x) = 0 - 72 - 18(2x)$

$p'(x) = -72 - 36x$

Step 2: Find the critical points.

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

$-72 - 36x = 0$

$-36x = 72$

$x = \frac{72}{-36}$

$x = -2$

Step 3: Use the Second Derivative Test.

Find the second derivative $p''(x)$:

$p''(x) = \frac{d}{dx} (-72 - 36x)$

$p''(x) = -36$

Since $p''(x) = -36 < 0$, the second derivative is negative. This indicates that the critical point $x = -2$ is a point of local maximum.

Step 4: Calculate the Maximum Profit.

Substitute $x = -2$ back into the original profit function $p(x)$:

$p(-2) = 41 - 72(-2) - 18(-2)^2$

$p(-2) = 41 + 144 - 18(4)$

$p(-2) = 41 + 144 - 72$

$p(-2) = 185 - 72$

$p(-2) = 113$

---

Result: The maximum profit the company can make is 113 units.

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 $f(x)$.

---

Solution:

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

We can rewrite this expression by multiplying and dividing by $\sqrt{1^2 + 1^2} = \sqrt{2}$.

$f(x) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x \right)$

We know that $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$.

So, $f(x) = \sqrt{2} \left( \sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4} \right)$

Using the compound angle formula $\sin(A + B) = \sin A \cos B + \cos A \sin B$:

$f(x) = \sqrt{2} \sin \left( x + \frac{\pi}{4} \right)$

We know that the maximum value of the sine function $\sin \theta$ is 1.

Therefore, the maximum value of $\sin \left( x + \frac{\pi}{4} \right)$ is 1.

Maximum value of $f(x) = \sqrt{2} \times 1 = \sqrt{2}$.

---

Alternate Method (Using Calculus):

$f'(x) = \cos x - \sin x$. For critical points, set $f'(x) = 0 \implies \tan x = 1 \implies x = \pi/4$.

$f''(x) = -\sin x - \cos x$. At $x = \pi/4$, $f''(\pi/4) = -\sqrt{2} < 0$ (Maxima).

Max Value $f(\pi/4) = \sin(\pi/4) + \cos(\pi/4) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$.

Result: The maximum value is $\sqrt{2}$.

Given: function $f(x) = 2x^3 - 24x + 107$.

To Find: Absolute maximum in intervals $[1, 3]$ and $[-3, -1]$.

---

Solution:

First, find the derivative to locate critical points.

$f'(x) = 6x^2 - 24 = 6(x^2 - 4)$

Set $f'(x) = 0 \implies x^2 = 4 \implies x = 2, -2$.

Case 1: Interval [1, 3]

The critical point $x = 2$ lies in $[1, 3]$. We evaluate $f(x)$ at the endpoints and the critical point.

• $f(1) = 2(1)^3 - 24(1) + 107 = 2 - 24 + 107 = 85$
• $f(2) = 2(2)^3 - 24(2) + 107 = 16 - 48 + 107 = 75$
• $f(3) = 2(3)^3 - 24(3) + 107 = 54 - 72 + 107 = 89$

Maximum value in $[1, 3]$ is 89.

Case 2: Interval [-3, -1]

The critical point $x = -2$ lies in $[-3, -1]$. We evaluate $f(x)$ at the endpoints and the critical point.

• $f(-3) = 2(-3)^3 - 24(-3) + 107 = -54 + 72 + 107 = 125$
• $f(-2) = 2(-2)^3 - 24(-2) + 107 = -16 + 48 + 107 = 139$
• $f(-1) = 2(-1)^3 - 24(-1) + 107 = -2 + 24 + 107 = 129$

Maximum value in $[-3, -1]$ is 139.

---

Result: Max value in [1, 3] is 89. Max value in [-3, -1] is 139.

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: $f(x) = x + \sin 2x$ on $[0, 2\pi]$.

---

Solution:

1. Find Derivative: $f'(x) = 1 + 2\cos 2x$.

2. Find Critical Points: Set $f'(x) = 0$.

$1 + 2\cos 2x = 0 \implies \cos 2x = -\frac{1}{2}$.

Since $x \in [0, 2\pi]$, the angle $2x \in [0, 4\pi]$. The reference angle for $\cos \theta = 1/2$ is $\pi/3$. Cosine is negative in Quadrants II and III.

Values for $2x$: $\frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}$.

Values for $x$: $\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$.

3. Evaluate $f(x)$ at endpoints and critical points:

• $f(0) = 0 + \sin 0 = 0$
• $f(\pi/3) = \pi/3 + \sin(2\pi/3) = \pi/3 + \sqrt{3}/2 \approx 1.91$
• $f(2\pi/3) = 2\pi/3 + \sin(4\pi/3) = 2\pi/3 - \sqrt{3}/2 \approx 1.23$
• $f(4\pi/3) = 4\pi/3 + \sin(8\pi/3) = 4\pi/3 + \sqrt{3}/2 \approx 5.05$
• $f(5\pi/3) = 5\pi/3 + \sin(10\pi/3) = 5\pi/3 - \sqrt{3}/2 \approx 4.37$
• $f(2\pi) = 2\pi + \sin(4\pi) = 2\pi \approx 6.28$

Comparing the values:

Minimum Value = 0 (at $x=0$).

Maximum Value = $2\pi$ (at $x=2\pi$).

---

Result: Maximum value = $2\pi$. Minimum value = 0.

Problem Statement:

Find two numbers such that their sum is 24 and their product is as large as possible.

---

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.

---

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.

---

To Find:

The values of the positive numbers $x$ and $y$ that satisfy the given conditions.

---

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.

---

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.

---

To Find:

The values of the positive numbers $x$ and $y$ that satisfy the given conditions.

---

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.

---

Therefore, the two positive numbers are 10 and 25.

Problem Statement:

Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

---

Solution:

Let the two positive numbers be $x$ and $y$.

According to the first condition, their sum is 16:

Since $x$ and $y$ are positive, we have $x > 0$ and $y > 0$. This implies $0 < x < 16$ and $0 < y < 16$.

We want to minimize the sum of their cubes. Let $S$ be the sum of the cubes:

$S = x^3 + y^3$

To minimize $S$, we first express it as a function of a single variable.

From equation (i), we can write $y = 16 - x$.

Substitute this into the expression for $S$:

$S(x) = x^3 + (16 - x)^3$

Now, we find the value of $x$ in the interval $(0, 16)$ that minimizes $S(x)$. We use calculus.

Step 1: Find the derivative of $S(x)$ with respect to $x$.

$S'(x) = \frac{d}{dx} [x^3 + (16 - x)^3]$

$S'(x) = 3x^2 + 3(16 - x)^2 \cdot \frac{d}{dx}(16 - x)$

$S'(x) = 3x^2 + 3(16 - x)^2 \cdot (-1)$

$S'(x) = 3x^2 - 3(16 - x)^2$

Step 2: Find the critical points by setting $S'(x) = 0$.

$3x^2 - 3(16 - x)^2 = 0$

Divide by 3:

$x^2 - (16 - x)^2 = 0$

$x^2 = (16 - x)^2$

Taking the square root of both sides gives two possibilities:

1) $x = 16 - x \implies 2x = 16 \implies x = 8$

2) $x = -(16 - x) \implies x = -16 + x \implies 0 = -16$, which is impossible.

So, the only critical point is $x = 8$. This critical point lies within the interval $(0, 16)$.

Step 3: Use the second derivative test to confirm if $x = 8$ corresponds to a minimum.

Find the second derivative $S''(x)$:

$S''(x) = \frac{d}{dx} [3x^2 - 3(16 - x)^2]$

$S''(x) = 6x - 3 \cdot 2(16 - x) \cdot (-1)$

$S''(x) = 6x + 6(16 - x)$

$S''(x) = 6x + 96 - 6x$

$S''(x) = 96$

Since $S''(x) = 96$, which is positive for all $x$ (including $x=8$), the function $S(x)$ has a local minimum at $x = 8$. Since it is the only critical point in the interval, this local minimum is the absolute minimum.

Step 4: Find the corresponding value of $y$.

Substitute $x = 8$ back into the equation $y = 16 - x$:

$y = 16 - 8$

$y = 8$

The two positive numbers are $x = 8$ and $y = 8$. They satisfy $x+y = 8+8=16$ and are both positive.

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Therefore, the two positive numbers whose sum is 16 and the sum of whose cubes is minimum are 8 and 8.

Problem Statement:

A square piece of tin of side 18 cm is to be made into an open-top box by cutting identical squares from each corner and folding up the flaps. Find the side length of the square to be cut off so that the volume of the box is maximum.

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Solution:

Let the side length of the square piece of tin be $L = 18$ cm.

Let the side length of the square cut from each corner be $x$ cm.

When the squares are cut from the corners and the flaps are folded up:

The length of the base of the box will be $L - 2x = 18 - 2x$ cm.

The width of the base of the box will be $L - 2x = 18 - 2x$ cm.

The height of the box will be $x$ cm.

The volume $V$ of the box is given by Length $\times$ Width $\times$ Height:

$V = (18 - 2x)(18 - 2x)(x)$

$V(x) = x(18 - 2x)^2$

For the dimensions of the box to be physically meaningful and positive:

Height $x > 0$.

Length $18 - 2x > 0 \implies 18 > 2x \implies 9 > x$.

Width $18 - 2x > 0 \implies 9 > x$.

So, the side length $x$ must be in the interval $(0, 9)$.

We need to find the value of $x$ in $(0, 9)$ that maximizes the volume $V(x)$.

Step 1: Find the derivative of $V(x)$ with respect to $x$.

We can expand $V(x)$ first or use the product and chain rules.

$V(x) = x(18 - 2x)^2$

Using product rule ($u=x, v=(18-2x)^2 \implies u'=1, v'=2(18-2x)(-2)=-4(18-2x)$):

$V'(x) = (1)(18 - 2x)^2 + x [-4(18 - 2x)]$

$V'(x) = (18 - 2x)^2 - 4x(18 - 2x)$

Factor out the common term $(18 - 2x)$:

$V'(x) = (18 - 2x)[(18 - 2x) - 4x]$

$V'(x) = (18 - 2x)(18 - 6x)$

$V'(x) = 2(9 - x) \cdot 6(3 - x)$

$V'(x) = 12(9 - x)(3 - x)$

Step 2: Find the critical points by setting $V'(x) = 0$.

$12(9 - x)(3 - x) = 0$

This implies $9 - x = 0$ or $3 - x = 0$.

The critical points are $x = 9$ and $x = 3$.

Step 3: Consider the valid interval for $x$.

We are considering the interval $(0, 9)$.

The critical point $x = 9$ is not within the open interval $(0, 9)$.

The critical point $x = 3$ is within the interval $(0, 9)$.

Step 4: Use the second derivative test to confirm if $x = 3$ corresponds to a maximum.

Find the second derivative $V''(x)$:

$V'(x) = 12(27 - 12x + x^2) = 324 - 144x + 12x^2$

$V''(x) = \frac{d}{dx}(324 - 144x + 12x^2)$

$V''(x) = -144 + 24x$

Evaluate $V''(x)$ at the critical point $x = 3$:

$V''(3) = -144 + 24(3)$

$V''(3) = -144 + 72$

$V''(3) = -72$

Since $V''(3) = -72 < 0$, the volume $V(x)$ has a local maximum at $x = 3$. As this is the only critical point in the interval $(0, 9)$, this corresponds to the absolute maximum volume.

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Therefore, the side of the square to be cut off so that the volume of the box is maximum possible should be 3 cm.

Let $x$ be the side of the square cut off.

Dimensions of the box:

• Length = $45 - 2x$
• Breadth = $24 - 2x$
• Height = $x$

Volume $V(x) = x(45 - 2x)(24 - 2x)$.

Differentiate:

$V'(x) = 12x^2 - 276x + 1080$.

Set $V'(x) = 0$ and divide by 12:

$x^2 - 23x + 90 = 0$

$(x - 18)(x - 5) = 0$

Possible values: $x = 18$ or $x = 5$.

However, the width of the sheet is 24 cm. If we cut 18 cm from each side ($2x = 36$), it exceeds the width. Thus, $x = 18$ is impossible.

So, $x = 5$.

Calculating $V''(x) = 24x - 276$. At $x=5$, $V''(5) = 120 - 276 < 0$ (Maximum).

Result: The side of the square to be cut off is 5 cm.

Let the circle have a fixed diameter $D$ (or radius $R$). Let the sides of the rectangle be $x$ and $y$.

By Pythagoras theorem: $x^2 + y^2 = D^2 = (2R)^2$.

Area $A = xy$. To make differentiation easier, we maximize $A^2$ (let $Z = A^2$).

$Z = x^2 y^2 = x^2 (4R^2 - x^2) = 4R^2 x^2 - x^4$.

Differentiate $Z$ with respect to $x$:

$Z' = 8R^2 x - 4x^3$.

Set $Z' = 0$:

$4x(2R^2 - x^2) = 0$.

Since side $x \ne 0$, we have $x^2 = 2R^2 \implies x = \sqrt{2}R$.

Find $y$ using (1): $y = \sqrt{4R^2 - 2R^2} = \sqrt{2R^2} = \sqrt{2}R$.

Since $x = y = \sqrt{2}R$, the rectangle is a square.

$Z'' = 8R^2 - 12x^2$. At $x^2 = 2R^2$, $Z'' = 8R^2 - 24R^2 < 0$ (Maximum).

Result: Since x = y, the rectangle with maximum area is a square.

Let $r$ be the radius and $h$ be the height.

Given: Surface Area $S$ is constant.

From (1), $h = \frac{S - 2\pi r^2}{2\pi r} = \frac{S}{2\pi r} - r$.

Maximize Volume: $V = \pi r^2 h$.

Substitute $h$:

$V(r) = \pi r^2 \left( \frac{S}{2\pi r} - r \right) = \frac{Sr}{2} - \pi r^3$.

Differentiate:

$V'(r) = \frac{S}{2} - 3\pi r^2$.

Set $V'(r) = 0 \implies \frac{S}{2} = 3\pi r^2 \implies S = 6\pi r^2$.

Check second derivative: $V''(r) = -6\pi r < 0$ (Maximum).

Now, substitute $S = 6\pi r^2$ back into the expression for $h$:

$h = \frac{6\pi r^2 - 2\pi r^2}{2\pi r} = \frac{4\pi r^2}{2\pi r} = 2r$.

Since $2r$ is the diameter of the base:

Result: The height is equal to the diameter of the base.

Let $r$ be the radius and $h$ be the height of the cylinder.

Given: Volume $V = \pi r^2 h = 100$ cm$^3$.

To Minimize: Surface Area $S = 2\pi r^2 + 2\pi rh$.

Substitute $h$ from (1):

$S(r) = 2\pi r^2 + 2\pi r \left( \frac{100}{\pi r^2} \right) = 2\pi r^2 + \frac{200}{r}$.

Differentiate $S(r)$ with respect to $r$:

$S'(r) = 4\pi r - \frac{200}{r^2}$.

Set $S'(r) = 0$ for critical points:

$4\pi r = \frac{200}{r^2} \implies r^3 = \frac{50}{\pi} \implies r = \sqrt[3]{\frac{50}{\pi}}$.

Check second derivative:

$S''(r) = 4\pi + \frac{400}{r^3}$. Since $r>0$, $S''(r) > 0$ (Minimum).

Find height $h$:

$h = \frac{100}{\pi r^2} = \frac{100}{\pi (\frac{50}{\pi})^{2/3}} = 2 \left( \frac{50}{\pi} \right)^{1/3} = 2r$.

Result: Dimensions are Radius $r = \sqrt[3]{\frac{50}{\pi}}$ cm and Height $h = 2\sqrt[3]{\frac{50}{\pi}}$ cm.

Let $x$ be the length of the wire for the square. Then $(28 - x)$ is for the circle.

Square: Perimeter $= x \implies$ Side $= x/4 \implies$ Area $A_s = \frac{x^2}{16}$.

Circle: Circumference $= 28 - x \implies 2\pi r = 28 - x \implies r = \frac{28 - x}{2\pi}$.

Area $A_c = \pi r^2 = \pi \left( \frac{28 - x}{2\pi} \right)^2 = \frac{(28 - x)^2}{4\pi}$.

Total Area: $A(x) = \frac{x^2}{16} + \frac{(28 - x)^2}{4\pi}$.

Differentiate:

$A'(x) = \frac{2x}{16} + \frac{2(28 - x)(-1)}{4\pi} = \frac{x}{8} - \frac{28 - x}{2\pi}$.

Set $A'(x) = 0$:

$\frac{x}{8} = \frac{28 - x}{2\pi} \implies \pi x = 4(28 - x) \implies \pi x = 112 - 4x$.

$x(\pi + 4) = 112 \implies x = \frac{112}{\pi + 4}$.

Length for Circle $= 28 - x = 28 - \frac{112}{\pi + 4} = \frac{28\pi}{\pi + 4}$.

Check $A''(x) = \frac{1}{8} + \frac{1}{2\pi} > 0$ (Minimum).

Result: Length for square = $\frac{112}{\pi + 4}$ m, Length for circle = $\frac{28\pi}{\pi + 4}$ m.

Let $R$ be the sphere radius, $r$ be the cone base radius, and $h$ be the cone height.

Let $x$ be the distance from the sphere's center to the cone's base. Then $h = R + x$.

Radius of cone base: $r^2 = R^2 - x^2$.

Volume of Cone $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (R^2 - x^2)(R + x)$.

To maximize $V$, differentiate with respect to $x$:

$V'(x) = \frac{\pi}{3} [ (R^2 - x^2)(1) + (R + x)(-2x) ] $$ = \frac{\pi}{3} (R + x) [ R - x - 2x ] = \frac{\pi}{3} (R + x)(R - 3x)$.

Set $V'(x) = 0 \implies x = R/3$ (since $x \ne -R$).

At $x = R/3$, $h = R + R/3 = 4R/3$ and $r^2 = R^2 - R^2/9 = 8R^2/9$.

Max Volume $V = \frac{1}{3}\pi \left( \frac{8R^2}{9} \right) \left( \frac{4R}{3} \right) = \frac{32\pi R^3}{81}$.

Volume of Sphere $V_s = \frac{4}{3}\pi R^3$.

Ratio $\frac{V}{V_s} = \frac{32\pi R^3 / 81}{4\pi R^3 / 3} = \frac{32}{81} \times \frac{3}{4} = \frac{8}{27}$.

Result: $V_{cone} = \frac{8}{27} V_{sphere}$.

Given: Volume $V = \frac{1}{3}\pi r^2 h$ (Constant) $\implies h = \frac{3V}{\pi r^2}$.

To Minimize: Curved Surface Area $S = \pi r l = \pi r \sqrt{h^2 + r^2}$.

Minimize $S^2$ to simplify: $Z = S^2 = \pi^2 r^2 (h^2 + r^2) = \pi^2 r^2 h^2 + \pi^2 r^4$.

Substitute $h$: $Z = \pi^2 r^2 \left( \frac{9V^2}{\pi^2 r^4} \right) + \pi^2 r^4 = \frac{9V^2}{r^2} + \pi^2 r^4$.

Differentiate $Z$ with respect to $r$:

$Z'(r) = -\frac{18V^2}{r^3} + 4\pi^2 r^3$.

Set $Z'(r) = 0 \implies 4\pi^2 r^3 = \frac{18V^2}{r^3} \implies 2\pi^2 r^6 = 9V^2$.

Substitute $V = \frac{1}{3}\pi r^2 h$ into $9V^2$:

$9 \left( \frac{1}{9} \pi^2 r^4 h^2 \right) = \pi^2 r^4 h^2$.

So, $2\pi^2 r^6 = \pi^2 r^4 h^2$. Divide by $\pi^2 r^4$:

$2r^2 = h^2 \implies h = \sqrt{2}r$.

Result: The altitude is $\sqrt{2}$ times the radius.

Let slant height $l$ be constant. $r^2 + h^2 = l^2 \implies r^2 = l^2 - h^2$.

To Maximize: Volume $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (l^2 - h^2)h = \frac{1}{3}\pi (l^2 h - h^3)$.

Differentiate with respect to $h$:

$V'(h) = \frac{\pi}{3} (l^2 - 3h^2)$.

Set $V'(h) = 0 \implies l^2 = 3h^2 \implies h = \frac{l}{\sqrt{3}}$.

Radius $r = \sqrt{l^2 - h^2} = \sqrt{l^2 - l^2/3} = \sqrt{\frac{2l^2}{3}} = l\sqrt{\frac{2}{3}}$.

Semi-vertical angle $\alpha$: $\tan \alpha = \frac{r}{h}$.

$\tan \alpha = \frac{l\sqrt{2/3}}{l/\sqrt{3}} = \sqrt{2}$.

$\alpha = \tan^{-1} \sqrt{2}$.

Result: Semi-vertical angle is $\tan^{-1} \sqrt{2}$.

Given: Surface Area $S = \pi r^2 + \pi r l$ (Constant). So $l = \frac{S - \pi r^2}{\pi r}$.

To Maximize: Volume $V = \frac{1}{3}\pi r^2 h$. Maximize $V^2$ for simplicity.

$V^2 = \frac{1}{9}\pi^2 r^4 (l^2 - r^2) = \frac{1}{9}\pi^2 r^4 \left[ \left( \frac{S - \pi r^2}{\pi r} \right)^2 - r^2 \right]$.

$V^2 = \frac{1}{9}\pi^2 r^4 \left[ \frac{S^2 - 2S\pi r^2 + \pi^2 r^4 - \pi^2 r^4}{\pi^2 r^2} \right] = \frac{1}{9} r^2 (S^2 - 2S\pi r^2) = \frac{S^2 r^2}{9} - \frac{2S\pi r^4}{9}$.

Differentiate $V^2$ with respect to $r$:

$\frac{d(V^2)}{dr} = \frac{2S^2 r}{9} - \frac{8S\pi r^3}{9}$.

Set derivative to 0: $2S^2 r = 8S\pi r^3 \implies S = 4\pi r^2$.

Substitute $S = 4\pi r^2$ into $l = \frac{S - \pi r^2}{\pi r}$:

$l = \frac{4\pi r^2 - \pi r^2}{\pi r} = \frac{3\pi r^2}{\pi r} = 3r$.

Semi-vertical angle $\alpha$: $\sin \alpha = \frac{r}{l} = \frac{r}{3r} = \frac{1}{3}$.

Result: $\alpha = \sin^{-1}(1/3)$.

Let point be $(x, y)$. Minimize distance squared $D^2 = x^2 + (y - 5)^2$.

Substitute $x^2 = 2y$: $Z(y) = 2y + (y - 5)^2 = 2y + y^2 - 10y + 25 = y^2 - 8y + 25$.

$Z'(y) = 2y - 8$. Set to 0: $y = 4$.

If $y = 4$, then $x^2 = 2(4) = 8 \implies x = \pm 2\sqrt{2}$.

Point is $(2\sqrt{2}, 4)$.

Correct Option: (A)

Let $y = \frac{x^2 - x + 1}{x^2 + x + 1}$.

$y' = \frac{(2x-1)(x^2+x+1) - (x^2-x+1)(2x+1)}{(x^2+x+1)^2} = \frac{2x^2 - 2}{(x^2+x+1)^2}$.

Set $y' = 0 \implies x^2 = 1 \implies x = \pm 1$.

$y(1) = 1/3$ (Minimum).

$y(-1) = 3$ (Maximum).

Correct Option: (D)

Let $f(x) = (x^2 - x + 1)^{1/3}$. Maximize $g(x) = x^2 - x + 1$ on $[0, 1]$.

Critical point: $g'(x) = 2x - 1 = 0 \implies x = 1/2$.

$g(0) = 1$.

$g(1) = 1$.

$g(1/2) = 3/4$.

Max value of $g(x)$ is 1. Thus max value of $f(x) = 1^{1/3} = 1$.

Correct Option: (C)

Let $x(t)$ denote the distance of the car from point P at time $t$, and $v(t)$ denote its velocity.

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Given:

The distance function is:

Simplifying the expression:

The car starts at P ($t=0$) and stops at Q.

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To Find:

1. Time taken to reach Q.
2. Distance between P and Q.

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Solution:

The velocity of the car is the rate of change of distance with respect to time.

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

The car stops at point Q, which means its velocity becomes zero ($v=0$).

Substitute $v=0$ in equation (ii):

$4t - t^2 = 0$

$t(4 - t) = 0$

This gives two values for time:

$t = 0$ (Start time at P)

$t = 4$ (Stop time at Q)

Thus, the time taken to reach Q is 4 seconds.

To find the distance between P and Q, we calculate the position $x$ at $t=4$.

Substituting $t=4$ in equation (i):

$x = (4)^2 \left( 2-\frac{4}{3} \right)$

$x = 16 \left( \frac{6-4}{3} \right)$

$x = 16 \left( \frac{2}{3} \right)$

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The time taken to reach Q is 4 seconds and the distance between P and Q is $\frac{32}{3}$ m.

Let $r$ be the radius of the water surface, $h$ be the height (depth) of the water, and $V$ be the volume of the water at any instant $t$.

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Given:

Rate of water pouring in, $\frac{dV}{dt} = 5 \text{ m}^3/\text{h}$.

Semi-vertical angle $\alpha = \tan^{-1}(0.5)$.

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To Find:

Rate of rise of water level, $\frac{dh}{dt}$, when $h = 4$ m.

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Solution:

From the given semi-vertical angle:

From the geometry of the cone:

The volume of water in the cone is given by:

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

Substituting the value of $r$ from (i):

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

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

Differentiating both sides with respect to time $t$:

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

Substitute $\frac{dV}{dt} = 5$ and $h = 4$ into equation (iii):

$5 = \frac{\pi}{4} (4)^2 \frac{dh}{dt}$

$5 = \frac{\pi}{4} (16) \frac{dh}{dt}$

$5 = 4\pi \frac{dh}{dt}$

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The rate at which the water level is rising is $\frac{5}{4\pi}$ m/h.

Let $AB$ be the lamp post and $CD$ be the man. Let $x$ be the distance of the man from the lamp post and $s$ be the length of his shadow.

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Given:

Height of lamp post, $H = 6$ m.

Height of man, $h = 2$ m.

Speed of walking, $\frac{dx}{dt} = 5$ km/h.

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To Find:

Rate of increase of shadow length, $\frac{ds}{dt}$.

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Solution:

From the figure, $\Delta ABE$ and $\Delta CDE$ are similar triangles.

$3 = \frac{x}{s} + 1$

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

Differentiating with respect to time $t$:

Substitute $\frac{dx}{dt} = 5$ km/h:

$5 = 2 \frac{ds}{dt}$

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The length of the shadow increases at the rate of 2.5 km/h.

Given: Curve $x^2 = 4y$ and a point $(1, 2)$ through which the normal passes.

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To Find: Equation of the normal.

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Solution:

Let the normal be drawn at the point $(x_1, y_1)$ on the curve.

Differentiating the curve equation with respect to $x$:

$2x = 4\frac{dy}{dx} \implies \frac{dy}{dx} = \frac{x}{2}$

The slope of the tangent at $(x_1, y_1)$ is $m_T = \frac{x_1}{2}$.

The slope of the normal at $(x_1, y_1)$ is:

The equation of the normal at $(x_1, y_1)$ is:

$y - y_1 = -\frac{2}{x_1}(x - x_1)$

Since this normal passes through $(1, 2)$, substitute $x=1, y=2$:

From (i), substitute $y_1 = \frac{x_1^2}{4}$ into (iii):

$2 - \frac{x_1^2}{4} = -\frac{2}{x_1} + 2$

$-\frac{x_1^2}{4} = -\frac{2}{x_1}$

$x_1^3 = 8 \implies x_1 = 2$

Now find $y_1$:

$y_1 = \frac{2^2}{4} = 1$

The point of contact is $(2, 1)$. The slope of the normal is $m_N = -\frac{2}{2} = -1$.

Equation of the normal:

$y - 1 = -1(x - 2)$

$y - 1 = -x + 2$

$x + y - 3 = 0$

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The required equation of the normal is $x + y - 3 = 0$.

Given: Curve $y = \cos(x+y)$ and parallel line $x + 2y = 0$.

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To Find: Equation of tangents parallel to the given line.

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Solution:

Slope of the line $x + 2y = 0$ is $m = -\frac{1}{2}$.

Since tangents are parallel, their slope must be:

Differentiating the curve $y = \cos(x+y)$ w.r.t $x$:

$\frac{dy}{dx} = -\sin(x+y) \left( 1 + \frac{dy}{dx} \right)$

Substitute $\frac{dy}{dx} = -\frac{1}{2}$:

$-\frac{1}{2} = -\sin(x+y) \left( 1 - \frac{1}{2} \right)$

$-\frac{1}{2} = -\sin(x+y) \left( \frac{1}{2} \right)$

Since $y = \cos(x+y)$, and $\sin^2\theta + \cos^2\theta = 1$:

$1^2 + \cos^2(x+y) = 1 \implies \cos(x+y) = 0 \implies y = 0$.

From (ii), if $y=0$, then $\sin(x) = 1$.

For $-2\pi \le x \le 2\pi$, possible values for $x$ where $\sin x = 1$ are:

$x = \frac{\pi}{2}$ and $x = -\frac{3\pi}{2}$.

So the points of contact are $P(\frac{\pi}{2}, 0)$ and $Q(-\frac{3\pi}{2}, 0)$.

Tangent at $P(\frac{\pi}{2}, 0)$:

$y - 0 = -\frac{1}{2}\left(x - \frac{\pi}{2}\right)$

$2y = -x + \frac{\pi}{2} \implies 2x + 4y - \pi = 0$

Tangent at $Q(-\frac{3\pi}{2}, 0)$:

$y - 0 = -\frac{1}{2}\left(x + \frac{3\pi}{2}\right)$

$2y = -x - \frac{3\pi}{2} \implies 2x + 4y + 3\pi = 0$

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The equations of the tangents are $2x + 4y - \pi = 0$ and $2x + 4y + 3\pi = 0$.

Solution:

We find the derivative $f'(x)$:

$f'(x) = \frac{3}{10}(4x^3) - \frac{4}{5}(3x^2) - 3(2x) + \frac{36}{5}$

$f'(x) = \frac{6}{5}x^3 - \frac{12}{5}x^2 - 6x + \frac{36}{5}$

Factor out $\frac{6}{5}$:

$f'(x) = \frac{6}{5} (x^3 - 2x^2 - 5x + 6)$

By inspection, $x=1$ is a root ($1 - 2 - 5 + 6 = 0$). Using polynomial division or synthetic division, we can factorize the cubic polynomial:

$x^3 - 2x^2 - 5x + 6 = (x-1)(x^2 - x - 6) = (x-1)(x-3)(x+2)$

The critical points are $x = -2, 1, 3$. We test the intervals.

Interval	Sign of $f'(x)$	Nature of function
$(-\infty, -2)$	$(-)(-)(-) < 0$	Decreasing
$(-2, 1)$	$(-)(-)(+) > 0$	Increasing
$(1, 3)$	$(+)(-)(+) < 0$	Decreasing
$(3, \infty)$	$(+)(+)(+) > 0$	Increasing

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(a) Increasing: $(-2, 1) \cup (3, \infty)$

(b) Decreasing: $(-\infty, -2) \cup (1, 3)$

To Prove: $f(x)$ is increasing in $\left( 0 , \frac{\pi}{4} \right)$.

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Proof:

Let $u = \sin x + \cos x$. Then $f(x) = \tan^{-1} u$.

$f'(x) = \frac{1}{1+u^2} \cdot \frac{du}{dx}$

Expanding the denominator:

$1 + \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + 1 + \sin 2x = 2 + \sin 2x$.

For the interval $x \in \left( 0 , \frac{\pi}{4} \right)$:

1. The denominator $2 + \sin 2x$ is always positive because $\sin 2x \ge -1$, so $2 + \sin 2x \ge 1$.
2. For the numerator: In $(0, \pi/4)$, $\cos x > \sin x$. Thus, $\cos x - \sin x > 0$.

Since both numerator and denominator are positive, $f'(x) > 0$ for all $x \in \left( 0 , \frac{\pi}{4} \right)$.

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Therefore, $f(x)$ is an increasing function in the given interval.

Given: Rate of change of radius $\frac{dr}{dt} = 0.05$ cm/s.

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To Find: Rate of change of area $\frac{dA}{dt}$ when $r = 3.2$ cm.

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Solution:

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

Differentiating with respect to time $t$:

Substitute $r = 3.2$ and $\frac{dr}{dt} = 0.05$:

$\frac{dA}{dt} = 2\pi (3.2)(0.05)$

$\frac{dA}{dt} = 2\pi (0.16)$

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The rate at which the area is increasing is $0.32\pi$ cm$^2$/s.

Let $x$ be the side of the square cut from each corner.

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Given: Sheet dimensions $8 \text{ m} \times 3 \text{ m}$.

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To Find: Maximum volume of the box.

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Solution:

After cutting squares of side $x$, the dimensions of the box will be:

Length $L = 8 - 2x$

Width $W = 3 - 2x$

Height $H = x$

Constraint: $3 - 2x > 0 \implies x < 1.5$. So $x \in (0, 1.5)$.

Volume $V(x) = (8 - 2x)(3 - 2x)x$

$V(x) = (24 - 16x - 6x + 4x^2)x$

Differentiating with respect to $x$:

$V'(x) = 12x^2 - 44x + 24$

For critical points, set $V'(x) = 0$:

$12x^2 - 44x + 24 = 0$

Divide by 4:

$3x^2 - 11x + 6 = 0$

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

Roots are $x = 3$ and $x = \frac{2}{3}$.

Since $x < 1.5$, we reject $x=3$. Thus, $x = \frac{2}{3}$.

To confirm maximum, find $V''(x) = 24x - 44$.

At $x = \frac{2}{3}$: $V''(\frac{2}{3}) = 24(\frac{2}{3}) - 44 = 16 - 44 < 0$. Maxima confirmed.

Max Volume $V(\frac{2}{3})$:

$V = \left(8 - 2(\frac{2}{3})\right)\left(3 - 2(\frac{2}{3})\right)\left(\frac{2}{3}\right)$

$V = \left(\frac{20}{3}\right) \left(\frac{5}{3}\right) \left(\frac{2}{3}\right)$

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The volume of the largest such box is $\frac{200}{27}$ m$^3$.

Let $x$ be the number of items.

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Given:

Selling Price per unit, $SP = \textsf{₹} \left(5 - \frac{x}{100}\right)$.

Total Cost Price, $C(x) = \textsf{₹} \left(\frac{x}{5} + 500\right)$.

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To Find: $x$ for maximum profit.

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Solution:

Total Revenue $R(x) = x \times SP = x\left(5 - \frac{x}{100}\right) = 5x - \frac{x^2}{100}$.

Profit function $P(x) = R(x) - C(x)$.

$P(x) = \left(5x - \frac{x^2}{100}\right) - \left(\frac{x}{5} + 500\right)$

$P(x) = 5x - \frac{x}{5} - \frac{x^2}{100} - 500$

Differentiating with respect to $x$:

$P'(x) = \frac{24}{5} - \frac{2x}{100} = \frac{24}{5} - \frac{x}{50}$

Set $P'(x) = 0$ for maxima:

$\frac{24}{5} = \frac{x}{50}$

$x = \frac{24 \times 50}{5} = 240$.

Second derivative test:

$P''(x) = -\frac{1}{50} < 0$. Maxima confirmed.

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The manufacturer should sell 240 items to earn maximum profit.

The approximate value of a function $f(x + \Delta x)$ is given by $f(x) + \Delta y$, where $\Delta y \approx f'(x) \Delta x$.

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(a) $\left( \frac{17}{81} \right)^{\frac{1}{4}}$

Consider the function $y = x^{\frac{1}{4}}$. Let $x = \frac{16}{81}$ and $\Delta x = \frac{1}{81}$.

Then $x + \Delta x = \frac{17}{81}$.

Now, $y = x^{\frac{1}{4}}$. Differentiating with respect to $x$:

At $x = \frac{16}{81}$:

$y = \left(\frac{16}{81}\right)^{\frac{1}{4}} = \frac{2}{3}$

And the derivative value is:

$\frac{dy}{dx} = \frac{1}{4\left(\frac{16}{81}\right)^{\frac{3}{4}}} = \frac{1}{4\left[\left(\frac{16}{81}\right)^{\frac{1}{4}}\right]^3} = \frac{1}{4\left(\frac{2}{3}\right)^3} = \frac{1}{4\left(\frac{8}{27}\right)} = \frac{27}{32}$.

Now, $\Delta y \approx \frac{dy}{dx} \cdot \Delta x$:

$\Delta y = \frac{27}{32} \times \frac{1}{81} = \frac{1}{32 \times 3} = \frac{1}{96} = 0.010$ (approx).

Approximate value = $y + \Delta y$

$= \frac{2}{3} + \frac{1}{96} = \frac{64+1}{96} = \frac{65}{96}$.

So, $\left( \frac{17}{81} \right)^{\frac{1}{4}} \approx 0.677$.

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(b) $(33)^{-\frac{1}{5}}$

Consider the function $y = x^{-\frac{1}{5}}$. Let $x = 32$ and $\Delta x = 1$.

Then $y = (32)^{-\frac{1}{5}} = \frac{1}{2} = 0.5$.

Differentiating with respect to $x$:

At $x = 32$:

$\frac{dy}{dx} = -\frac{1}{5(32)^{\frac{6}{5}}} = -\frac{1}{5(2)^6} = -\frac{1}{5(64)} = -\frac{1}{320} = -0.0031$.

Now, $\Delta y \approx \frac{dy}{dx} \cdot \Delta x$:

$\Delta y = -\frac{1}{320} \times 1 = -0.0031$.

Approximate value = $y + \Delta y = 0.5 - 0.0031 = 0.4969$.

Given: Function $f(x) = \frac{\log x}{x}$ for $x > 0$.

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To Prove: $f(x)$ has a maximum at $x = e$.

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Proof:

Differentiating $f(x)$ with respect to $x$ using the quotient rule:

For critical points, put $f'(x) = 0$:

$1 - \log x = 0 \implies \log x = 1 \implies x = e$.

Now, we check the second derivative to confirm maxima.

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

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

At $x = e$:

$f''(e) = \frac{2(1) - 3}{e^3} = \frac{-1}{e^3} < 0$.

Since the second derivative is negative at $x = e$, the function attains a maximum value at $x = e$.