Q3: The radius of a circle is increasing uniformly at the rate of 3 ~cm / s. Find the rate at which the area of the circle is increasing when the radius…

Let: Let r be the radius of the circle. Let A be the area of the circle. Given: The radius of the circle is increasing uniformly at the rate of 3 cm/s. So, dr/dt = 3 cm/s. To Find: The rate at which the area of the circle is increasing, dA/dt, when the radius is 10 cm (i.e., when r = 10 cm…

Q4: An edge of a variable cube is increasing at the rate of 3 ~cm / s. How fast is the volume of the cube increasing when the edge is 10 ~cm long?

Let: Let x be the length of an edge of the variable cube. Let V be the volume of the cube. Given: The edge of the cube is increasing at the rate of 3 cm/s. So, dx/dt = 3 cm/s. To Find: How fast the volume of the cube is increasing, dV/dt, when the edge is 10 cm long (i.e., when x = 10 cm )._x000D_…

Q6: The radius of a circle is increasing at the rate of 0.7 ~cm / s. What is the rate of increase of its circumference?

Given: The radius of a circle, r, is increasing at the rate of 0.7 cm/s. So, dr/dt = 0.7 cm/s. To Find: The rate of increase of its circumference, dC/dt. Formula: The circumference of a circle with radius r is given by C = 2π r. Solution: We have the circumference C = 2π r. To find the rate of…

Q7: The length x of a rectangle is decreasing at the rate of 5 ~cm / minute and the width y is increasing at the rate of 4 ~cm / minute. When x=8 ~cm and…

Given: For a rectangle, let the length be x and the width be y. The length x is decreasing at the rate of 5 cm/minute. So, dx/dt = -5 cm/min. The width y is increasing at the rate of 4 cm/minute. So, dy/dt = 4 cm/min. We need to find the rates of change when x = 8 cm and y = 6 cm. To Find: (a) The…

Q13: A balloon, which always remains spherical, has a variable diameter 3/2(2 x+1). Find the rate of change of its volume with respect to x.

Given: A spherical balloon has a variable diameter D = 3/2(2x+1). To Find: The rate of change of its volume with respect to x, which is dV/dx. Solution: First, we find the radius r of the balloon in terms of x. r = D/2 = 1/2 left[ 3/2(2x+1) right] = 3/4(2x+1) The volume V of a sphere with radius r…

Q14: Sand is pouring from a pipe at the rate of 12 ~cm^3 / s. The falling sand forms a cone on the ground in such a way that the height of the cone is…

Given: Sand is pouring from a pipe at a rate of 12 cm^3/s. This is the rate of change of the volume of the sand cone, so dV/dt = 12 cm^3/s. The falling sand forms a cone on the ground such that the height of the cone ( h ) is always one-sixth of the radius of the base ( r ). So, h = 1/6r, which…

Application of Derivatives Class 12 Mathematics NCERT Solutions

Q1EXERCISE 6.1

Find the rate of change of the area of a circle with respect to its radius $r$ when

(a)

r=3 \mathrm{~cm}

(b)

r=4 \mathrm{~cm}

Solution

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

A = \pi r^2

, where

r

is the radius.

To Find: The rate of change of the area of the circle with respect to its radius

r

, i.e.,

\frac{dA}{dr}

, when: (a)

r = 3 \text{ cm}

(b)

r = 4 \text{ cm}

Solution: We have the area of the circle,

A = \pi r^2

. To find the rate of change of area with respect to the radius, we differentiate

A

with respect to

r

.

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

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

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

r = 3

into the expression for

\frac{dA}{dr}

:

\frac{dA}{dr} = 2\pi (3) = 6\pi

The rate of change of the area is

6\pi \text{ cm}^2 / \text{cm}

.

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

r = 4

into the expression for

\frac{dA}{dr}

:

\frac{dA}{dr} = 2\pi (4) = 8\pi

The rate of change of the area is

8\pi \text{ cm}^2 / \text{cm}

.

Final Answer: The rate of change of the area of the circle with respect to its radius is: (a)

6\pi \text{ cm}^2 / \text{cm}

when

r = 3 \text{ cm}

. (b)

8\pi \text{ cm}^2 / \text{cm}

when

r = 4 \text{ cm}

.

Q2EXERCISE 6.1

The volume of a cube is increasing at the rate of $8 \mathrm{~cm}^{3} / \mathrm{s}$ . How fast is the surface area increasing when the length of an edge is $12 \mathrm{~cm}$ ?

Solution

Let: Let

x

be the length of an edge of the cube. Let

V

be the volume of the cube. Let

S

be the surface area of the cube.

Given: The volume of the cube is increasing at a rate of

8 \text{ cm}^3/\text{s}

. So,

\frac{dV}{dt} = 8 \text{ cm}^3/\text{s}

.

To Find: The rate at which the surface area is increasing,

\frac{dS}{dt}

, when the length of an edge is

12 \text{ cm}

(i.e., when

x = 12 \text{ cm}

).x000D

Formulas: Volume of a cube:

V = x^3

Surface area of a cube:

S = 6x^2

Solution: We need to find a relationship between

\frac{dS}{dt}

and

\frac{dV}{dt}

. We can do this by differentiating both formulas with respect to time

t

.

Differentiating the volume formula with respect to

t

:

\frac{dV}{dt} = \frac{d}{dt}(x^3) = 3x^2 \frac{dx}{dt}

We are given

\frac{dV}{dt} = 8

, so:

8 = 3x^2 \frac{dx}{dt}

This gives us an expression for

\frac{dx}{dt}

:

\frac{dx}{dt} = \frac{8}{3x^2} \quad ...(1)

Now, differentiating the surface area formula with respect to

t

:

\frac{dS}{dt} = \frac{d}{dt}(6x^2) = 12x \frac{dx}{dt} \quad ...(2)

Substitute the expression for

\frac{dx}{dt}

from equation (1) into equation (2):

\frac{dS}{dt} = 12x \left( \frac{8}{3x^2} \right) = \frac{96x}{3x^2} = \frac{32}{x}

Now we can find the rate of change of the surface area when

x = 12 \text{ cm}

:

\left. \frac{dS}{dt} \right|_{x=12} = \frac{32}{12} = \frac{8}{3}

Final Answer: The surface area of the cube is increasing at the rate of

\frac{8}{3} \text{ cm}^2/\text{s}

.

Q3EXERCISE 6.1

The radius of a circle is increasing uniformly at the rate of $3 \mathrm{~cm} / \mathrm{s}$ . Find the rate at which the area of the circle is increasing when the radius is $10 \mathrm{~cm}$ .

Solution

Let: Let

r

be the radius of the circle. Let

A

be the area of the circle.

Given: The radius of the circle is increasing uniformly at the rate of

3 \text{ cm/s}

. So,

\frac{dr}{dt} = 3 \text{ cm/s}

.

To Find: The rate at which the area of the circle is increasing,

\frac{dA}{dt}

, when the radius is

10 \text{ cm}

(i.e., when

r = 10 \text{ cm}

).x000D

Formula: Area of a circle:

A = \pi r^2

Solution: To find the rate of change of the area, we differentiate the area formula with respect to time

t

.

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

Using the chain rule:

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

Now, we substitute the given values into this equation. We have

r = 10 \text{ cm}

and

\frac{dr}{dt} = 3 \text{ cm/s}

.

\frac{dA}{dt} = 2\pi (10)(3)

\frac{dA}{dt} = 60\pi

The units for the rate of change of area are

\text{cm}^2/\text{s}

.

Final Answer: The area of the circle is increasing at the rate of

60\pi \text{ cm}^2/\text{s}

.

Q4EXERCISE 6.1

An edge of a variable cube is increasing at the rate of $3 \mathrm{~cm} / \mathrm{s}$ . How fast is the volume of the cube increasing when the edge is $10 \mathrm{~cm}$ long?

Solution

Let: Let

x

be the length of an edge of the variable cube. Let

V

be the volume of the cube.

Given: The edge of the cube is increasing at the rate of

3 \text{ cm/s}

. So,

\frac{dx}{dt} = 3 \text{ cm/s}

.

To Find: How fast the volume of the cube is increasing,

\frac{dV}{dt}

, when the edge is

10 \text{ cm}

long (i.e., when

x = 10 \text{ cm}

).x000D

Formula: Volume of a cube:

V = x^3

Solution: To find the rate of change of the volume, we differentiate the volume formula with respect to time

t

.

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

Using the chain rule:

\frac{dV}{dt} = 3x^2 \cdot \frac{dx}{dt}

Now, we substitute the given values into this equation. We have

x = 10 \text{ cm}

and

\frac{dx}{dt} = 3 \text{ cm/s}

.

\frac{dV}{dt} = 3(10)^2 (3)

\frac{dV}{dt} = 3(100)(3)

\frac{dV}{dt} = 900

The units for the rate of change of volume are

\text{cm}^3/\text{s}

.

Final Answer: The volume of the cube is increasing at the rate of

900 \text{ cm}^3/\text{s}

.

Q5EXERCISE 6.1

A stone is dropped into a quiet lake and waves move in circles at the speed of $5 \mathrm{~cm} / \mathrm{s}$ . At the instant when the radius of the circular wave is $8 \mathrm{~cm}$ , how fast is the enclosed area increasing?

Solution

Let: Let

r

be the radius of the circular wave. Let

A

be the area enclosed by the circular wave.

Given: A stone is dropped into a quiet lake and waves move in circles at the speed of

5 \text{ cm/s}

. This means the radius of the circular wave is increasing at this rate. So,

\frac{dr}{dt} = 5 \text{ cm/s}

.

To Find: How fast the enclosed area is increasing,

\frac{dA}{dt}

, at the instant when the radius of the circular wave is

8 \text{ cm}

(i.e., when

r = 8 \text{ cm}

).x000D

Formula: Area of a circle:

A = \pi r^2

Solution: To find the rate of change of the area, we differentiate the area formula with respect to time

t

.

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

Using the chain rule:

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

Now, we substitute the given values into this equation. We have

r = 8 \text{ cm}

and

\frac{dr}{dt} = 5 \text{ cm/s}

.

\frac{dA}{dt} = 2\pi (8)(5)

\frac{dA}{dt} = 80\pi

The units for the rate of change of area are

\text{cm}^2/\text{s}

.

Final Answer: The enclosed area is increasing at the rate of

80\pi \text{ cm}^2/\text{s}

.

Q6EXERCISE 6.1

The radius of a circle is increasing at the rate of $0.7 \mathrm{~cm} / \mathrm{s}$ . What is the rate of increase of its circumference?

Solution

Given: The radius of a circle,

r

, is increasing at the rate of

0.7 \text{ cm/s}

. So,

\frac{dr}{dt} = 0.7 \text{ cm/s}

.

To Find: The rate of increase of its circumference,

\frac{dC}{dt}

.

Formula: The circumference of a circle with radius

r

is given by

C = 2\pi r

.

Solution: We have the circumference

C = 2\pi r

. To find the rate of increase of the circumference, we differentiate

C

with respect to time

t

.

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

Using the chain rule, we get:

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

Now, we substitute the given value of

\frac{dr}{dt} = 0.7 \text{ cm/s}

.

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

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

Final Answer: The rate of increase of the circumference is

1.4\pi \text{ cm/s}

.

Q7EXERCISE 6.1

The length $x$ of a rectangle is decreasing at the rate of $5 \mathrm{~cm} /$ minute and the width $y$ is increasing at the rate of $4 \mathrm{~cm} /$ minute. When $x=8 \mathrm{~cm}$ and $y=6 \mathrm{~cm}$ , find the rates of change of (a) the perimeter, and (b) the area of the rectangle.

Solution

Given: For a rectangle, let the length be

x

and the width be

y

. The length

x

is decreasing at the rate of

5 \text{ cm/minute}

. So,

\frac{dx}{dt} = -5 \text{ cm/min}

. The width

y

is increasing at the rate of

4 \text{ cm/minute}

. So,

\frac{dy}{dt} = 4 \text{ cm/min}

. We need to find the rates of change when

x = 8 \text{ cm}

and

y = 6 \text{ cm}

.

To Find: (a) The rate of change of the perimeter,

\frac{dP}{dt}

. (b) The rate of change of the area,

\frac{dA}{dt}

.

Solution:

(a) Rate of change of the perimeter

Formula: The perimeter of a rectangle is given by

P = 2(x + y)

.

Calculation: Differentiating

P

with respect to time

t

:

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

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

Substituting the given values:

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

\frac{dP}{dt} = 2(-1) = -2

The negative sign indicates that the perimeter is decreasing.

Final Answer for (a): The perimeter of the rectangle is decreasing at the rate of

2 \text{ cm/min}

.

(b) Rate of change of the area

Formula: The area of a rectangle is given by

A = xy

.

Calculation: Differentiating

A

with respect to time

t

using the product rule:

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

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

Substituting the given values:

x = 8 \text{ cm}

,

y = 6 \text{ cm}

,

\frac{dx}{dt} = -5 \text{ cm/min}

, and

\frac{dy}{dt} = 4 \text{ cm/min}

.

\frac{dA}{dt} = (8)(4) + (6)(-5)

\frac{dA}{dt} = 32 - 30

\frac{dA}{dt} = 2

The positive sign indicates that the area is increasing.

Final Answer for (b): The area of the rectangle is increasing at the rate of

2 \text{ cm}^2\text{/min}

.

Q8EXERCISE 6.1

A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is $15 \mathrm{~cm}$ .

Solution

Given: A spherical balloon is being inflated by pumping in gas at a rate of

900 \text{ cm}^3\text{/s}

. This is the rate of change of volume, so

\frac{dV}{dt} = 900 \text{ cm}^3\text{/s}

. We need to find the rate at which the radius increases when the radius is

15 \text{ cm}

.

To Find: The rate at which the radius of the balloon increases,

\frac{dr}{dt}

, when

r = 15 \text{ cm}

.

Formula: The volume of a sphere with radius

r

is given by

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

.

Solution: We have the volume

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

. Differentiating both sides with respect to time

t

using the chain rule:

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

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

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

We need to find

\frac{dr}{dt}

. Rearranging the equation:

\frac{dr}{dt} = \frac{1}{4\pi r^2} \frac{dV}{dt}

Now, substitute the given values:

\frac{dV}{dt} = 900 \text{ cm}^3\text{/s}

and

r = 15 \text{ cm}

.

\frac{dr}{dt} = \frac{1}{4\pi (15)^2} (900)

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

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

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

Final Answer: The rate at which the radius of the balloon increases is

\frac{1}{\pi} \text{ cm/s}

.

Q9EXERCISE 6.1

A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is $10 \mathrm{~cm}$ .

Solution

Given: A spherical balloon has a variable radius

r

. We need to find the rate of change of volume with respect to the radius when the radius is

10 \text{ cm}

.

To Find: The rate at which the volume is increasing with the radius,

\frac{dV}{dr}

, when

r = 10 \text{ cm}

.

Formula: The volume of a sphere with radius

r

is given by

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

.

Solution: We have the volume

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

. We need to find the rate of change of volume with respect to the radius, so we differentiate

V

with respect to

r

.

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

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

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

Now, we evaluate this rate at

r = 10 \text{ cm}

.

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

= 4\pi (100)

= 400\pi

The units are cubic cm per cm.

Final Answer: The volume is increasing at the rate of

400\pi \text{ cm}^3/\text{cm}

with respect to the radius.

Q10EXERCISE 6.1

A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of $2 \mathrm{~cm} / \mathrm{s}$ . How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall ?

Solution

Given: A ladder of length 5 m is leaning against a wall. Let

x

be the distance of the foot of the ladder from the wall and

y

be the height of the top of the ladder on the wall. The length of the ladder is constant, so we can use the Pythagorean theorem:

x^2 + y^2 = 5^2

. The bottom of the ladder is pulled away from the wall at the rate of

2 \text{ cm/s}

. This means

\frac{dx}{dt} = 2 \text{ cm/s}

.

To Find: How fast the height on the wall is decreasing, i.e., find the value of

-\frac{dy}{dt}

, when the foot of the ladder is 4 m away from the wall (

x=4

m).

Let: Let's work with consistent units. We will convert all lengths to cm. Length of ladder =

5 \text{ m} = 500 \text{ cm}

. Distance from the wall,

x = 4 \text{ m} = 400 \text{ cm}

. Rate of change of

x

,

\frac{dx}{dt} = 2 \text{ cm/s}

.

Equation: The relationship between

x

and

y

is given by the Pythagorean theorem:

x^2 + y^2 = (500)^2

x^2 + y^2 = 250000

Solution: First, we find the height

y

when

x = 400 \text{ cm}

.

(400)^2 + y^2 = 250000

160000 + y^2 = 250000

y^2 = 250000 - 160000 = 90000

y = \sqrt{90000} = 300 \text{ cm}

(Since height must be positive).

Next, we differentiate the equation

x^2 + y^2 = 250000

with respect to time

t

.

\frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(250000)

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

We want to find

\frac{dy}{dt}

, so we rearrange the equation:

2y \frac{dy}{dt} = -2x \frac{dx}{dt}

\frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}

Now, substitute the known values:

x = 400 \text{ cm}

,

y = 300 \text{ cm}

, and

\frac{dx}{dt} = 2 \text{ cm/s}

.

\frac{dy}{dt} = -\frac{400}{300} (2)

\frac{dy}{dt} = -\frac{4}{3} (2)

\frac{dy}{dt} = -\frac{8}{3}

The negative sign indicates that the height

y

is decreasing. The rate of decrease is the positive value of this rate. Rate of decrease =

\frac{8}{3} \text{ cm/s}

.

Final Answer: The height on the wall is decreasing at a rate of

\frac{8}{3} \text{ cm/s}

.

Q11EXERCISE 6.1

A particle moves along the curve $6 y=x^{3}+2$ . Find the points on the curve at which the $y$ -coordinate is changing 8 times as fast as the $x$ -coordinate.

Solution

Given: A particle moves along the curve

6y = x^3 + 2

. The

y

-coordinate is changing 8 times as fast as the

x

-coordinate. This can be written as

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

.

To Find: The points

(x, y)

on the curve that satisfy the given condition.

Solution: The equation of the curve is

6y = x^3 + 2

.

Differentiating both sides with respect to time

t

, we get:

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

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

It is given that

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

. Substituting this into the differentiated equation:

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

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

Since the particle is moving, we can assume

\frac{dx}{dt} \neq 0

. Dividing both sides by

\frac{dx}{dt}

:

48 = 3x^2

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

x^2 = 16

x = \pm 4

Now we find the corresponding

y

-coordinates using the curve's equation

6y = x^3 + 2

.

Case 1: When

x = 4

6y = (4)^3 + 2

6y = 64 + 2

6y = 66

y = 11

So, one point is

(4, 11)

.

Case 2: When

x = -4

6y = (-4)^3 + 2

6y = -64 + 2

6y = -62

y = -\frac{62}{6} = -\frac{31}{3}

So, the other point is

(-4, -\frac{31}{3})

.

Final Answer: The required points on the curve are

(4, 11)

and

(-4, -\frac{31}{3})

.

Q12EXERCISE 6.1

The radius of an air bubble is increasing at the rate of $\frac{1}{2} \mathrm{~cm} / \mathrm{s}$ . At what rate is the volume of the bubble increasing when the radius is $1 \mathrm{~cm}$ ?

Solution

Given: The radius of a spherical air bubble is increasing at the rate of

\frac{1}{2} \text{ cm/s}

. This means

\frac{dr}{dt} = \frac{1}{2} \text{ cm/s}

.

To Find: The rate at which the volume of the bubble is increasing when the radius is

1 \text{ cm}

.

Formula: The volume

V

of a sphere with radius

r

is given by:

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

Solution: Differentiating the volume formula with respect to time

t

, we get:

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

Using the chain rule:

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

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

Now, we substitute the given values:

r = 1 \text{ cm}

and

\frac{dr}{dt} = \frac{1}{2} \text{ cm/s}

.

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

\frac{dV}{dt} = 4\pi \cdot 1 \cdot \frac{1}{2}

\frac{dV}{dt} = 2\pi

The unit for the rate of change of volume is

\text{cm}^3/\text{s}

.

Final Answer: The volume of the bubble is increasing at the rate of

2\pi \text{ cm}^3/\text{s}

.

Q13EXERCISE 6.1

A balloon, which always remains spherical, has a variable diameter $\frac{3}{2}(2 x+1)$ . Find the rate of change of its volume with respect to $x$ .

Solution

Given: A spherical balloon has a variable diameter

D = \frac{3}{2}(2x+1)

.

To Find: The rate of change of its volume with respect to

x

, which is

\frac{dV}{dx}

.

Solution: First, we find the radius

r

of the balloon in terms of

x

.

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

The volume

V

of a sphere with radius

r

is given by the formula:

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

Substitute the expression for

r

in terms of

x

into the volume formula:

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

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

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

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

Now, we need to find the rate of change of volume with respect to

x

, so we differentiate

V

with respect to

x

.

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

Using the chain rule:

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

\frac{dV}{dx} = \frac{27\pi}{16}(2x+1)^2 \cdot 2

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

Final Answer: The rate of change of the balloon's volume with respect to

x

is

\frac{27\pi}{8}(2x+1)^2

.

Q14EXERCISE 6.1

Sand is pouring from a pipe at the rate of $12 \mathrm{~cm}^{3} / \mathrm{s}$ . The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is $4 \mathrm{~cm}$ ?

Solution

Given: Sand is pouring from a pipe at a rate of

12 \text{ cm}^3/\text{s}

. This is the rate of change of the volume of the sand cone, so

\frac{dV}{dt} = 12 \text{ cm}^3/\text{s}

. The falling sand forms a cone on the ground such that the height of the cone (

h

) is always one-sixth of the radius of the base (

r

). So,

h = \frac{1}{6}r

, which implies

r = 6h

.

To Find: How fast the height of the sand cone is increasing (

rac{dh}{dt}

) when the height is

4 \text{ cm}

.

Formula: The volume

V

of a cone with radius

r

and height

h

is given by:

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

Solution: To find

\frac{dh}{dt}

, we should express the volume

V

in terms of

h

only. We use the given relation

r = 6h

. Substitute

r = 6h

into the volume formula:

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

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

V = 12\pi h^3

Now, differentiate both sides with respect to time

t

:

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

Using the chain rule:

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

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

We are given

\frac{dV}{dt} = 12 \text{ cm}^3/\text{s}

and we need to find

\frac{dh}{dt}

when

h = 4 \text{ cm}

. Substitute these values into the equation:

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

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

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

Now, solve for

\frac{dh}{dt}

:

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

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

The unit for the rate of change of height is cm/s.

Final Answer: The height of the sand cone is increasing at the rate of

\frac{1}{48\pi} \text{ cm/s}

.

Q15EXERCISE 6.1

The total cost $\mathrm{C}(x)$ in Rupees associated with the production of $x$ units of an item is given by $\mathrm{C}(x)=0.007 x^{3}-0.003 x^{2}+15 x+4000.$ Find the marginal cost when 17 units are produced.

Solution

Given: The total cost

\mathrm{C}(x)

in Rupees for producing

x

units of an item is given by the function:

\mathrm{C}(x) = 0.007x^3 - 0.003x^2 + 15x + 4000

To Find: The marginal cost when 17 units are produced (

x=17

).

Concept: The marginal cost (MC) is the rate of change of the total cost with respect to the number of units produced. It is found by differentiating the cost function

\mathrm{C}(x)

with respect to

x

.

\text{MC}(x) = \frac{dC}{dx}

Solution: First, we find the marginal cost function,

\text{MC}(x)

, by differentiating

\mathrm{C}(x)

:

\text{MC}(x) = \frac{d}{dx} (0.007x^3 - 0.003x^2 + 15x + 4000)

\text{MC}(x) = 0.007(3x^2) - 0.003(2x) + 15(1) + 0

\text{MC}(x) = 0.021x^2 - 0.006x + 15

Now, we need to find the marginal cost when

x=17

. We substitute

x=17

into the

\text{MC}(x)

function:

\text{MC}(17) = 0.021(17)^2 - 0.006(17) + 15

We know that

17^2 = 289

. So:

\text{MC}(17) = 0.021(289) - 0.006(17) + 15

Let's calculate the terms:

0.021 \times 289 = 6.069

0.006 \times 17 = 0.102

Now substitute these values back:

\text{MC}(17) = 6.069 - 0.102 + 15

\text{MC}(17) = 5.967 + 15

\text{MC}(17) = 20.967

The marginal cost is in Rupees.

Final Answer: The marginal cost when 17 units are produced is ₹ 20.967.

Q16EXERCISE 6.1

The total revenue in Rupees received from the sale of $x$ units of a product is given by $R(x)=13 x^{2}+26 x+15$ Find the marginal revenue when $x=7$ .

Solution

Given: The total revenue function is

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

, where

x

is the number of units sold.

To Find: The marginal revenue when

x = 7

.

Concept: Marginal revenue (MR) is the rate of change of the total revenue with respect to the number of units sold. It is calculated by finding the derivative of the total revenue function

R(x)

.

MR = \frac{dR}{dx}

Solution: First, we find the marginal revenue function by differentiating

R(x)

with respect to

x

.

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

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

MR(x) = 2 \cdot 13x + 26 + 0

MR(x) = 26x + 26

Now, we need to find the marginal revenue when

x = 7

. We substitute

x = 7

into the

MR(x)

function.

MR(7) = 26(7) + 26

MR(7) = 182 + 26

MR(7) = 208

Final Answer: The marginal revenue when

x=7

is Rs 208.

Q17EXERCISE 6.1

The rate of change of the area of a circle with respect to its radius $r$ at $r=6 \mathrm{~cm}$ is

(A)

10 \pi

(B)

12 \pi

(C)

8 \pi

(D)

11 \pi

Solution

Given: The radius of a circle is

r

. We need to find the rate of change of its area with respect to its radius at

r = 6 \text{ cm}

.

To Find: The value of

\frac{dA}{dr}

when

r=6 \text{ cm}

.

Formula: The area

A

of a circle with radius

r

is given by:

A = \pi r^2

Solution: We need to find the rate of change of the area with respect to the radius, which is the derivative of

A

with respect to

r

.

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

Using the power rule for differentiation, we get:

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

Now, we evaluate this rate of change at

r = 6 \text{ cm}

.

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

\left. \frac{dA}{dr} \right|_{r=6} = 12\pi

The rate of change of the area of the circle with respect to its radius at

r=6 \text{ cm}

is

12\pi \text{ cm}^2/\text{cm}

.

Comparing this result with the given options: (A)

10 \pi

(B)

12 \pi

(C)

8 \pi

(D)

11 \pi

The calculated value matches option (B).

Final Answer: The correct option is (B)

12 \pi

.

Q18EXERCISE 6.1

The total revenue in Rupees received from the sale of $x$ units of a product is given by $\mathrm{R}(x)=3 x^{2}+36 x+5$ . The marginal revenue, when $x=15$ is

(A)

116

(B)

96

(C)

90

(D)

126

Solution

Given: The total revenue function is

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

, where

x

is the number of units sold.

To Find: The marginal revenue when

x = 15

.

Concept: Marginal revenue (MR) is the instantaneous rate of change of total revenue with respect to the number of items sold. It is the derivative of the revenue function

R(x)

.

MR = \frac{dR}{dx}

Solution: First, we find the marginal revenue function,

MR(x)

, by differentiating

R(x)

with respect to

x

.

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

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

MR(x) = 2 \cdot 3x + 36 + 0

MR(x) = 6x + 36

Now, we find the marginal revenue when

x = 15

by substituting

x=15

into the

MR(x)

function.

MR(15) = 6(15) + 36

MR(15) = 90 + 36

MR(15) = 126

The marginal revenue is Rs 126.

Comparing this result with the given options: (A) 116 (B) 96 (C) 90 (D) 126

The calculated value matches option (D).

Final Answer: The correct option is (D) 126.

Q1EXERCISE 6.2

Show that the function given by $f(x)=3 x+17$ is increasing on $\mathbf{R}$ .

Solution

Given: The function

f(x) = 3x + 17

.

To Prove: The function

f(x)

is increasing on

\mathbf{R}

.

Concept: A function

f(x)

is said to be increasing on an interval if its derivative

f'(x) > 0

for all

x

in that interval.

Proof: We have the function

f(x) = 3x + 17

. First, we find the derivative of the function with respect to

x

.

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

f'(x) = 3

Since

f'(x) = 3

, which is a positive constant, we have

f'(x) > 0

for all values of

x \in \mathbf{R}

.

Because the derivative of the function is positive for all real numbers, the function

f(x) = 3x + 17

is strictly increasing on

\mathbf{R}

. A strictly increasing function is also an increasing function.

Hence Proved.

Q2EXERCISE 6.2

Show that the function given by $f(x)=e^{2 x}$ is increasing on $\mathbf{R}$ .

Solution

Given: The function

f(x) = e^{2x}

.

To Prove: The function

f(x)

is increasing on

\mathbf{R}

.

Concept: A function is increasing on an interval if its first derivative is positive throughout that interval, i.e.,

f'(x) > 0

.

Proof: We are given the function

f(x) = e^{2x}

. First, we find the derivative of

f(x)

with respect to

x

using the chain rule.

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 determine the sign of

f'(x)

for all

x \in \mathbf{R}

. The exponential function

e^y

is always positive for any real number

y

. Therefore,

e^{2x} > 0

for all

x \in \mathbf{R}

. Since

e^{2x} > 0

, it follows that

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

is also positive for all

x \in \mathbf{R}

.

f'(x) > 0 \text{ for all } x \in \mathbf{R}

As the derivative

f'(x)

is positive for all real numbers, the function

f(x) = e^{2x}

is strictly increasing on

\mathbf{R}

.

Hence Proved.

Q3EXERCISE 6.2

Show that the function given by $f(x)=\sin x$ 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

Given: The function

f(x) = \sin x

.

To Show: (a)

f(x)

is increasing in

(0, \frac{\pi}{2})

(b)

f(x)

is decreasing in

(\frac{\pi}{2}, \pi)

(c)

f(x)

is neither increasing nor decreasing in

(0, \pi)

Solution: First, we find the derivative of the function

f(x)

.

f(x) = \sin x

f'(x) = \cos x

A function

f(x)

is increasing in an interval if

f'(x) > 0

and decreasing if

f'(x) < 0

in that interval.

(a) Interval $(0, \frac{\pi}{2})$ For any

x

in the interval

(0, \frac{\pi}{2})

, we are in the first quadrant. In the first quadrant,

\cos x

is positive. So,

f'(x) = \cos x > 0

for all

x \in (0, \frac{\pi}{2})

. Therefore, the function

f(x) = \sin x

is increasing in

(0, \frac{\pi}{2})

.

(b) Interval $(\frac{\pi}{2}, \pi)$ For any

x

in the interval

(\frac{\pi}{2}, \pi)

, we are in the second quadrant. In the second quadrant,

\cos x

is negative. So,

f'(x) = \cos x < 0

for all

x \in (\frac{\pi}{2}, \pi)

. Therefore, the function

f(x) = \sin x

is decreasing in

(\frac{\pi}{2}, \pi)

.

(c) Interval $(0, \pi)$ From parts (a) and (b), we see that in the interval

(0, \pi)

, the derivative

f'(x) = \cos x

is positive for

x \in (0, \frac{\pi}{2})

and negative for

x \in (\frac{\pi}{2}, \pi)

. Since the sign of

f'(x)

is not the same throughout the interval

(0, \pi)

, the function

f(x)

is neither increasing nor decreasing in

(0, \pi)

.

Final Answer: (a) The function is increasing in

(0, \frac{\pi}{2})

. (b) The function is decreasing in

(\frac{\pi}{2}, \pi)

. (c) The function is neither increasing nor decreasing in

(0, \pi)

.

Q4EXERCISE 6.2

Find the intervals in which the function $f$ given by $f(x)=2 x^{2}-3 x$ is

(a)

increasing

(b)

decreasing

Solution

Given: The function

f(x) = 2x^2 - 3x

.

To Find: The intervals in which the function is (a) increasing and (b) decreasing.

Solution: First, we find the derivative of the function

f(x)

.

f(x) = 2x^2 - 3x

f'(x) = 4x - 3

To find the critical points, we set

f'(x) = 0

.

4x - 3 = 0

4x = 3

x = \frac{3}{4}

The critical point

x = \frac{3}{4}

divides the real number line into two disjoint intervals:

(-\infty, \frac{3}{4})

and

(\frac{3}{4}, \infty)

.

We will now check the sign of

f'(x)

in each interval.

Interval 1: $(-\infty, \frac{3}{4})$ Let's take a test point in this interval, for example,

x=0

.

f'(0) = 4(0) - 3 = -3

Since

f'(x) < 0

in this interval, the function

f(x)

is decreasing in

(-\infty, \frac{3}{4})

.

Interval 2: $(\frac{3}{4}, \infty)$ Let's take a test point in this interval, for example,

x=1

.

f'(1) = 4(1) - 3 = 1

Since

f'(x) > 0

in this interval, the function

f(x)

is increasing in

(\frac{3}{4}, \infty)

.

Final Answer: (a) The function is increasing in the interval

(\frac{3}{4}, \infty)

. (b) The function is decreasing in the interval

(-\infty, \frac{3}{4})

.

Q5EXERCISE 6.2

Find the intervals in which the function $f$ given by $f(x)=2 x^{3}-3 x^{2}-36 x+7$ is

(a)

increasing

(b)

decreasing

Solution

Given: The function

f(x) = 2x^3 - 3x^2 - 36x + 7

.

To Find: The intervals in which the function is (a) increasing and (b) decreasing.

Solution: First, we find the derivative of the function

f(x)

.

f(x) = 2x^3 - 3x^2 - 36x + 7

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

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

To find the critical points, we set

f'(x) = 0

.

6(x^2 - x - 6) = 0

x^2 - x - 6 = 0

Factorizing the quadratic equation:

x^2 - 3x + 2x - 6 = 0

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

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

The critical points are

x = 3

and

x = -2

.

These points divide the real number line into three disjoint intervals:

(-\infty, -2)

,

(-2, 3)

, and

(3, \infty)

.

We will now check the sign of

f'(x)

in each interval.

Interval 1: $(-\infty, -2)$ Let's take a test point, for example,

x=-3

.

f'(-3) = 6((-3)^2 - (-3) - 6) = 6(9 + 3 - 6) = 6(6) = 36

Since

f'(x) > 0

, the function

f(x)

is increasing in

(-\infty, -2)

.

Interval 2: $(-2, 3)$ Let's take a test point, for example,

x=0

.

f'(0) = 6(0^2 - 0 - 6) = 6(-6) = -36

Since

f'(x) < 0

, the function

f(x)

is decreasing in

(-2, 3)

.

Interval 3: $(3, \infty)$ Let's take a test point, for example,

x=4

.

f'(4) = 6(4^2 - 4 - 6) = 6(16 - 10) = 6(6) = 36

Since

f'(x) > 0

, the function

f(x)

is increasing in

(3, \infty)

.

Final Answer: (a) The function is increasing in the intervals

(-\infty, -2) \cup (3, \infty)

. (b) The function is decreasing in the interval

(-2, 3)

.

Q6EXERCISE 6.2

Find the intervals in which the following functions are strictly increasing or decreasing:

(a)

x^{2}+2 x-5

(b)

10-6 x-2 x^{2}

(c)

-2 x^{3}-9 x^{2}-12 x+1

(d)

6-9 x-x^{2}

(e)

(x+1)^{3}(x-3)^{3}

Solution

To Find: The intervals in which the following functions are strictly increasing or decreasing.

(a) $f(x) = x^2 + 2x - 5$ Solution:

f'(x) = 2x + 2 = 2(x+1)

Setting

f'(x) = 0

gives

x = -1

. The intervals are

(-\infty, -1)

and

(-1, \infty)

. In

(-\infty, -1)

,

f'(x) < 0

. So,

f

is strictly decreasing in

(-\infty, -1)

. In

(-1, \infty)

,

f'(x) > 0

. So,

f

is strictly increasing in

(-1, \infty)

.

(b) $f(x) = 10 - 6x - 2x^2$ Solution:

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

Setting

f'(x) = 0

gives

3 + 2x = 0

, so

x = -\frac{3}{2}

. The intervals are

(-\infty, -\frac{3}{2})

and

(-\frac{3}{2}, \infty)

. In

(-\infty, -\frac{3}{2})

,

f'(x) > 0

. So,

f

is strictly increasing in

(-\infty, -\frac{3}{2})

. In

(-\frac{3}{2}, \infty)

,

f'(x) < 0

. So,

f

is strictly decreasing in

(-\frac{3}{2}, \infty)

.

(c) $f(x) = -2x^3 - 9x^2 - 12x + 1$ Solution:

f'(x) = -6x^2 - 18x - 12 = -6(x^2 + 3x + 2) = -6(x+1)(x+2)

Setting

f'(x) = 0

gives

x = -1

and

x = -2

. The intervals are

(-\infty, -2)

,

(-2, -1)

, and

(-1, \infty)

. In

(-\infty, -2)

,

f'(x) < 0

. So,

f

is strictly decreasing. In

(-2, -1)

,

f'(x) > 0

. So,

f

is strictly increasing. In

(-1, \infty)

,

f'(x) < 0

. So,

f

is strictly decreasing. Strictly increasing in

(-2, -1)

. Strictly decreasing in

(-\infty, -2) \cup (-1, \infty)

.

(d) $f(x) = 6 - 9x - x^2$ Solution:

f'(x) = -9 - 2x = -(9 + 2x)

Setting

f'(x) = 0

gives

9 + 2x = 0

, so

x = -\frac{9}{2}

. The intervals are

(-\infty, -\frac{9}{2})

and

(-\frac{9}{2}, \infty)

. In

(-\infty, -\frac{9}{2})

,

f'(x) > 0

. So,

f

is strictly increasing in

(-\infty, -\frac{9}{2})

. In

(-\frac{9}{2}, \infty)

,

f'(x) < 0

. So,

f

is strictly decreasing in

(-\frac{9}{2}, \infty)

.

(e) $f(x) = (x+1)^3(x-3)^3$ Solution:

f(x) = [(x+1)(x-3)]^3 = (x^2 - 2x - 3)^3

Using the chain rule:

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

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

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

The terms

(x-3)^2

and

(x+1)^2

are always non-negative. For strictly increasing/decreasing, we consider where

f'(x) \neq 0

. So, we must have

x \neq -1, 1, 3

. The sign of

f'(x)

is determined by the sign of

(x-1)

. For

f(x)

to be strictly increasing,

f'(x) > 0

. This means

6(x-3)^2(x+1)^2(x-1) > 0

, which simplifies to

x-1 > 0

and

x \neq 3

. So,

x > 1

and

x \neq 3

. The intervals are

(1, 3)

and

(3, \infty)

. So,

f

is strictly increasing in

(1, 3) \cup (3, \infty)

. For

f(x)

to be strictly decreasing,

f'(x) < 0

. This means

x-1 < 0

and

x \neq -1

. So,

x < 1

and

x \neq -1

. The intervals are

(-\infty, -1)

and

(-1, 1)

. So,

f

is strictly decreasing in

(-\infty, -1) \cup (-1, 1)

.

Q7EXERCISE 6.2

Show that $y=\log (1+x)-\frac{2 x}{2+x}, x>-1$ , is an increasing function of $x$ throughout its domain.

Solution

Given: The function

y = \log(1+x) - \frac{2x}{2+x}

, for

x > -1

.

To Show: The function is an increasing function of

x

throughout its domain.

Solution: The domain of the function is given as

x > -1

, which is

(-1, \infty)

. To show that the function is increasing, we need to show that its derivative is positive, i.e.,

\frac{dy}{dx} > 0

for all

x

in the domain.

First, we find the derivative of

y

with respect to

x

.

y = \log(1+x) - \frac{2x}{2+x}

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

Derivative of the first term:

\frac{d}{dx}(\log(1+x)) = \frac{1}{1+x}

Derivative of the second term using the quotient rule, where

u=2x

and

v=2+x

:

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

= \frac{(2+x)(2) - (2x)(1)}{(2+x)^2}

= \frac{4 + 2x - 2x}{(2+x)^2}

= \frac{4}{(2+x)^2}

Now, combine the derivatives:

\frac{dy}{dx} = \frac{1}{1+x} - \frac{4}{(2+x)^2}

To determine the sign of

\frac{dy}{dx}

, we combine the terms into a single fraction:

\frac{dy}{dx} = \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2}

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

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

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

Now we need to analyze the sign of

\frac{dy}{dx}

for

x > -1

.

The numerator is $x^2$ . For any real number $x$ , $x^2 \geq 0$ . It is zero only at $x=0$ .
The denominator contains $(2+x)^2$ . For any real number $x$ , $(2+x)^2 \geq 0$ . It is zero only at $x=-2$ , which is not in the domain $x > -1$ . So, $(2+x)^2 > 0$ for all $x > -1$ .
The denominator also contains $(1+x)$ . Since the domain is $x > -1$ , we have $1+x > 0$ .

So, for any

x

in the domain

(-1, \infty)

:

$x^2 \geq 0$
$1+x > 0$
$(2+x)^2 > 0$

Therefore, the derivative

\frac{dy}{dx} = \frac{x^2}{(1+x)(2+x)^2} \geq 0

for all

x

in the domain

(-1, \infty)

. The derivative is equal to zero only at

x=0

. Since the derivative is non-negative and is zero only at a single point, the function is increasing throughout its domain.

Hence, $y=\log (1+x)-\frac{2 x}{2+x}$ is an increasing function of $x$ for all $x > -1$ .

Q8EXERCISE 6.2

Find the values of $x$ for which $y=[x(x-2)]^{2}$ is an increasing function.

Solution

Given: The function is

y = [x(x-2)]^2

.

To Find: The values of

x

for which the function is increasing.

Solution: Let

f(x) = y = [x(x-2)]^2 = (x^2 - 2x)^2

. A function is increasing if its derivative is non-negative, i.e.,

f'(x) \geq 0

.

First, we find the derivative of

f(x)

using the chain rule:

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

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

f'(x) = 2(x^2 - 2x)(2x - 2)

f'(x) = 2 \cdot x(x-2) \cdot 2(x-1)

f'(x) = 4x(x-1)(x-2)

For the function to be increasing, we must have

f'(x) \geq 0

.

4x(x-1)(x-2) \geq 0

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

The critical points are

x=0

,

x=1

, and

x=2

. These points divide the number line into four intervals:

(-\infty, 0)

,

(0, 1)

,

(1, 2)

, and

(2, \infty)

.

We analyze the sign of

f'(x)

in each interval:

Interval $(-\infty, 0)$ : Let's take $x = -1$ . $f'(-1) = 4(-1)(-1-1)(-1-2) = 4(-1)(-2)(-3) = -24 < 0$ . The function is decreasing.
Interval $(0, 1)$ : Let's take $x = 0.5$ . $f'(0.5) = 4(0.5)(0.5-1)(0.5-2) = 4(0.5)(-0.5)(-1.5) = 1.5 > 0$ . The function is increasing.
Interval $(1, 2)$ : Let's take $x = 1.5$ . $f'(1.5) = 4(1.5)(1.5-1)(1.5-2) = 4(1.5)(0.5)(-0.5) = -1.5 < 0$ . The function is decreasing.
Interval $(2, \infty)$ : Let's take $x = 3$ . $f'(3) = 4(3)(3-1)(3-2) = 4(3)(2)(1) = 24 > 0$ . The function is increasing.

The function is increasing when

f'(x) \geq 0

. This occurs in the intervals

[0, 1]

and

[2, \infty)

.

Final Answer: The function is increasing for

x \in [0, 1] \cup [2, \infty)

.

Q9EXERCISE 6.2

Prove that $y=\frac{4 \sin \theta}{(2+\cos \theta)}-\theta$ is an increasing function of $\theta$ in $0, \frac{\pi}{2}$ .

Solution

To Prove: The function

y = \frac{4 \sin \theta}{(2+\cos \theta)} - \theta

is an increasing function of

\theta

in

[0, \frac{\pi}{2}]

.

Proof: Let the given function be

f(\theta) = \frac{4 \sin \theta}{2+\cos \theta} - \theta

. To prove that the function is increasing, we need to show that its derivative,

f'(\theta)

, is non-negative, i.e.,

f'(\theta) \geq 0

for

\theta \in [0, \frac{\pi}{2}]

.

First, we differentiate

f(\theta)

with respect to

\theta

. Using the quotient rule for the first term:

\frac{d}{d\theta} \left( \frac{4 \sin \theta}{2+\cos \theta} \right) = \frac{(2+\cos \theta) \frac{d}{d\theta}(4 \sin \theta) - (4 \sin \theta) \frac{d}{d\theta}(2+\cos \theta)}{(2+\cos \theta)^2}

= \frac{(2+\cos \theta)(4 \cos \theta) - (4 \sin \theta)(-\sin \theta)}{(2+\cos \theta)^2}

= \frac{8 \cos \theta + 4 \cos^2 \theta + 4 \sin^2 \theta}{(2+\cos \theta)^2}

Using the identity

\sin^2 \theta + \cos^2 \theta = 1

:

= \frac{8 \cos \theta + 4(\cos^2 \theta + \sin^2 \theta)}{(2+\cos \theta)^2} = \frac{8 \cos \theta + 4}{(2+\cos \theta)^2}

Now, we find the derivative of the entire function

f(\theta)

:

f'(\theta) = \frac{d}{d\theta} \left( \frac{4 \sin \theta}{2+\cos \theta} - \theta \right)

f'(\theta) = \frac{8 \cos \theta + 4}{(2+\cos \theta)^2} - 1

To analyze the sign of

f'(\theta)

, we combine the terms:

f'(\theta) = \frac{8 \cos \theta + 4 - (2+\cos \theta)^2}{(2+\cos \theta)^2}

f'(\theta) = \frac{8 \cos \theta + 4 - (4 + 4 \cos \theta + \cos^2 \theta)}{(2+\cos \theta)^2}

f'(\theta) = \frac{8 \cos \theta + 4 - 4 - 4 \cos \theta - \cos^2 \theta}{(2+\cos \theta)^2}

f'(\theta) = \frac{4 \cos \theta - \cos^2 \theta}{(2+\cos \theta)^2}

f'(\theta) = \frac{\cos \theta (4 - \cos \theta)}{(2+\cos \theta)^2}

Now, we check the sign of

f'(\theta)

in the interval

[0, \frac{\pi}{2}]

. For

\theta \in [0, \frac{\pi}{2}]

:

The denominator $(2+\cos \theta)^2$ is always positive, because $-1 \leq \cos \theta \leq 1$ , so $2+\cos \theta$ is always positive.
In the numerator, we analyze each factor:
- $\cos \theta$ : For $\theta \in [0, \frac{\pi}{2}]$ , we have $0 \leq \cos \theta \leq 1$ . So, $\cos \theta \geq 0$ .
- $(4 - \cos \theta)$ : Since the maximum value of $\cos \theta$ is 1, the minimum value of $(4 - \cos \theta)$ is $4-1=3$ . So, $(4 - \cos \theta)$ is always positive.

Since the numerator

\cos \theta (4 - \cos \theta)

is a product of a non-negative term and a positive term, it is non-negative (

\ geq 0

) for

\theta \in [0, \frac{\pi}{2}]

. The denominator is always positive. Therefore,

f'(\theta) = \frac{\text{non-negative}}{\text{positive}} \geq 0

for all

\theta \in [0, \frac{\pi}{2}]

.

Since

f'(\theta) \geq 0

in the interval

[0, \frac{\pi}{2}]

, the function

y

is increasing in this interval.

Hence Proved.

Q10EXERCISE 6.2

Prove that the logarithmic function is increasing on $(0, \infty)$ .

Solution

To Prove: The logarithmic function is increasing on

(0, \infty)

.

Proof: Let the logarithmic function be

f(x) = \log x

. In calculus, it is standard convention that

\log x

denotes the natural logarithm,

\ln x

, with base

e

. So, we consider the function

f(x) = \ln x

.

The domain of the logarithmic function

f(x) = \ln x

is

(0, \infty)

.

To prove that the function is increasing on its domain, we need to show that its derivative,

f'(x)

, is positive for all

x

in the interval

(0, \infty)

.

First, we find the derivative of

f(x)

:

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

f'(x) = \frac{1}{x}

Now, we analyze the sign of

f'(x)

for

x

in the interval

(0, \infty)

. For any value of

x

such that

x \in (0, \infty)

, we have

x > 0

.

Since

x

is positive, its reciprocal,

\frac{1}{x}

, is also positive. Therefore,

f'(x) = \frac{1}{x} > 0

for all

x \in (0, \infty)

.

Since the derivative of the function is strictly positive throughout its domain, the function is strictly increasing on

(0, \infty)

. A strictly increasing function is also an increasing function.

Hence Proved.

Q11EXERCISE 6.2

Prove that the function $f$ given by $f(x)=x^{2}-x+1$ is neither strictly increasing nor decreasing on $(-1,1)$ .

Solution

To Prove: The function

f

given by

f(x)=x^{2}-x+1

is neither strictly increasing nor decreasing on

(-1,1)

.

Proof: Let the given function be

f(x) = x^2 - x + 1

. To determine the intervals of increase or decrease, we first find the derivative of the function.

f'(x) = \frac{d}{dx}(x^2 - x + 1)

f'(x) = 2x - 1

Now, we find the critical points by setting the derivative to zero:

f'(x) = 0

2x - 1 = 0

x = \frac{1}{2}

The critical point

x = \frac{1}{2}

lies within the given interval

(-1, 1)

. This point divides the interval

(-1, 1)

into two sub-intervals:

(-1, \frac{1}{2})

and

(\frac{1}{2}, 1)

.

We will now analyze the sign of

f'(x)

in these sub-intervals.

Case 1: For the interval $(-1, \frac{1}{2})$ Let's choose a test point in this interval, for example,

x=0

.

f'(0) = 2(0) - 1 = -1

Since

f'(x) < 0

for all

x \in (-1, \frac{1}{2})

, the function

f(x)

is strictly decreasing on

(-1, \frac{1}{2})

.

Case 2: For the interval $(\frac{1}{2}, 1)$ Let's choose a test point in this interval, for example,

x=0.75

.

f'(0.75) = 2(0.75) - 1 = 1.5 - 1 = 0.5

Since

f'(x) > 0

for all

x \in (\frac{1}{2}, 1)

, the function

f(x)

is strictly increasing on

(\frac{1}{2}, 1)

.

Since the function is strictly decreasing on the interval

(-1, \frac{1}{2})

and strictly increasing on the interval

(\frac{1}{2}, 1)

, the function is not monotonic over the entire interval

(-1, 1)

.

Therefore, the function

f(x)=x^2-x+1

is neither strictly increasing nor strictly decreasing on

(-1, 1)

.

Hence Proved.

Q12EXERCISE 6.2

Which of the following functions are decreasing on $0, \frac{\pi}{2}$ ?

(A)

\cos x

(B)

\cos 2 x

(C)

\cos 3 x

(D)

\tan x

Solution

To Find: Which of the given functions are decreasing on the interval

(0, \frac{\pi}{2})

.

Solution: A function

f(x)

is decreasing on an interval if its derivative

f'(x) \leq 0

on that interval. We will check the derivative of each function.

The given interval is

I = (0, \frac{\pi}{2})

.

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

f'(x) = -\sin x

For

x \in (0, \frac{\pi}{2})

, we know that

\sin x > 0

. Therefore,

f'(x) = -\sin x < 0

. Since

f'(x) < 0

on

(0, \frac{\pi}{2})

, the function

f(x) = \cos x

is decreasing on this interval.

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

f'(x) = -2 \sin 2x

For

x \in (0, \frac{\pi}{2})

, the argument

2x

lies in the interval

(0, \pi)

. In the interval

(0, \pi)

,

\sin(2x) > 0

. Therefore,

f'(x) = -2 \sin 2x < 0

. Since

f'(x) < 0

on

(0, \frac{\pi}{2})

, the function

f(x) = \cos 2x

is decreasing on this interval.

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

f'(x) = -3 \sin 3x

For

x \in (0, \frac{\pi}{2})

, the argument

3x

lies in the interval

(0, \frac{3\pi}{2})

. The sign of

\sin(3x)

is positive for

3x \in (0, \pi)

and negative for

3x \in (\pi, \frac{3\pi}{2})

. This corresponds to

x \in (0, \frac{\pi}{3})

and

x \in (\frac{\pi}{3}, \frac{\pi}{2})

respectively.

For $x \in (0, \frac{\pi}{3})$ , $\sin(3x) > 0$ , so $f'(x) = -3\sin(3x) < 0$ (decreasing).
For $x \in (\frac{\pi}{3}, \frac{\pi}{2})$ , $\sin(3x) < 0$ , so $f'(x) = -3\sin(3x) > 0$ (increasing). Since the derivative changes sign in the interval $(0, \frac{\pi}{2})$ , the function $f(x) = \cos 3x$ is not decreasing on the entire interval.

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

f'(x) = \sec^2 x

For

x \in (0, \frac{\pi}{2})

,

\cos x \neq 0

, so

\sec x

is well-defined. The square of any non-zero real number is positive. Therefore,

f'(x) = \sec^2 x > 0

. Since

f'(x) > 0

on

(0, \frac{\pi}{2})

, the function

f(x) = \tan x

is increasing on this interval.

Based on the analysis, the functions that are decreasing on

(0, \frac{\pi}{2})

are

\cos x

and

\cos 2x

. The question is in a multiple-choice format, which typically expects a single best answer. However, both (A) and (B) are correct. In many textbooks, such questions might be intended as multiple-correct-answer types or there might be an error in the question framing. Assuming we must choose from the options, both (A) and (B) satisfy the condition.

Final Answer: The functions in options (A) and (B) are decreasing on

(0, \frac{\pi}{2})

. (A)

\cos x

(B)

\cos 2x

Q13EXERCISE 6.2

On which of the following intervals is the function $f$ given by $f(x)=x^{100}+\sin x-1$ decreasing ?

(A)

(0,1)

(B)

\frac{\pi}{2}, \pi

(C)

0, \frac{\pi}{2}

(D)

None of these

Solution

Given: The function

f(x) = x^{100} + \sin x - 1

.

To Find: The interval on which the function

f(x)

is decreasing.

Solution: A function

f(x)

is decreasing if its derivative

f'(x) < 0

. First, we find the derivative of the given function:

f(x) = x^{100} + \sin x - 1

f'(x) = \frac{d}{dx}(x^{100} + \sin x - 1)

f'(x) = 100x^{99} + \cos x

Now, we analyze the sign of

f'(x)

in each of the given intervals.

(A) Interval $(0, 1)$ : For any

x \in (0, 1)

, we have

x > 0

, which implies

x^{99} > 0

, and thus

100x^{99} > 0

. Also, the interval

(0, 1)

in radians is a subset of

(0, \frac{\pi}{2})

, as

1 < \frac{\pi}{2} \approx 1.57

. For

x

in the first quadrant,

\cos x > 0

. Since both terms are positive, their sum is positive:

f'(x) = 100x^{99} + \cos x > 0

. Therefore,

f(x)

is increasing on

(0, 1)

.

(B) Interval $(\frac{\pi}{2}, \pi)$ : For any

x \in (\frac{\pi}{2}, \pi)

, we have

x > \frac{\pi}{2} > 1

. This implies

x^{99} > 1

, and thus

100x^{99} > 100

. For

x

in the second quadrant, we have

-1 < \cos x < 0

. The value of

f'(x)

is:

f'(x) = 100x^{99} + \cos x

Since

100x^{99} > 100

and

\cos x > -1

, we have:

f'(x) > 100 - 1 = 99

So,

f'(x) > 0

. Therefore,

f(x)

is increasing on

(\frac{\pi}{2}, \pi)

.

(C) Interval $(0, \frac{\pi}{2})$ : For any

x \in (0, \frac{\pi}{2})

, we have

x > 0

, which implies

100x^{99} > 0

. Also, for

x

in the first quadrant,

\cos x > 0

. So,

f'(x) = 100x^{99} + \cos x > 0

. Therefore,

f(x)

is increasing on

(0, \frac{\pi}{2})

.

Since

f(x)

is increasing on all the intervals given in options (A), (B), and (C), it is not decreasing on any of them.

Final Answer: The correct option is (D) None of these.

Q14EXERCISE 6.2

For what values of $a$ the function $f$ given by $f(x)=x^{2}+a x+1$ is increasing on $[1,2]$ ?

Solution

Given: The function

f(x) = x^2 + ax + 1

is increasing on the interval

[1, 2]

.

To Find: The values of

a

.

Solution: For a function to be increasing on an interval, its derivative must be non-negative in that interval. First, we find the derivative of the function

f(x)

:

f(x) = x^2 + ax + 1

f'(x) = \frac{d}{dx}(x^2 + ax + 1)

f'(x) = 2x + a

For

f(x)

to be increasing on

[1, 2]

, we must have

f'(x) \geq 0

for all

x \in [1, 2]

.

2x + a \geq 0 \quad \text{for all } x \in [1, 2]

Let's analyze the function

g(x) = 2x + a

. This is a linear function with a positive slope (2). This means that the function

g(x)

is itself an increasing function. Therefore, its minimum value on the closed interval

[1, 2]

will occur at the leftmost point, which is

x=1

.

If the condition

g(x) \geq 0

is satisfied at its minimum value in the interval, it will be satisfied for the entire interval. So, we need to ensure that

f'(1) \geq 0

.

Substitute

x=1

into the inequality for

f'(x)

:

2(1) + a \geq 0

2 + a \geq 0

a \geq -2

Thus, for all values of

a \geq -2

, the function

f(x)

is increasing on the interval

[1, 2]

.

Final Answer: The function is increasing on

[1, 2]

for

a \geq -2

.

Q15EXERCISE 6.2

Let I be any interval disjoint from $[-1,1]$ . Prove that the function $f$ given by $f(x)=x+\frac{1}{x}$ is increasing on I .

Solution

To Prove: The function

f

given by

f(x) = x + \frac{1}{x}

is increasing on any interval I disjoint from

[-1, 1]

.

Proof: Given: The function

f(x) = x + \frac{1}{x}

. The domain of the function is all real numbers except

x=0

. The interval I is any interval disjoint from

[-1, 1]

. This means that for any

x \in I

,

x

does not belong to the interval

[-1, 1]

.

An interval I disjoint from

[-1, 1]

implies that for any

x \in I

, we have either

x < -1

or

x > 1

. This can be written as

I \subseteq (-\infty, -1) \cup (1, \infty)

.

To prove that the function is increasing, we need to show that its derivative

f'(x) > 0

for all

x \in I

.

First, we find the derivative of

f(x)

:

f(x) = x + \frac{1}{x} = x + x^{-1}

f'(x) = \frac{d}{dx}(x + x^{-1})

f'(x) = 1 - 1 \cdot x^{-2} = 1 - \frac{1}{x^2}

Now we analyze the sign of

f'(x)

for

x \in I

. Since

x \in I

, we have

x < -1

or

x > 1

. In both cases, squaring the inequality gives:

x^2 > 1

Since

x^2

is positive, we can take the reciprocal and reverse the inequality sign:

\frac{1}{x^2} < 1

Now, subtract

\frac{1}{x^2}

from 1:

1 - \frac{1}{x^2} > 0

This means

f'(x) > 0

for all

x \in I

.

Since the derivative

f'(x)

is positive for all

x

in any interval I disjoint from

[-1, 1]

, the function

f(x)

is strictly increasing on I.

Hence Proved.

Q16EXERCISE 6.2

Prove that the function $f$ given by $f(x)=\log \sin x$ is increasing on $0, \frac{\pi}{2}$ and decreasing on $\frac{\pi}{2}, \pi$ .

Solution

To Prove: The function

f

given by

f(x) = \log \sin x

is increasing on

(0, \frac{\pi}{2})

and decreasing on

(\frac{\pi}{2}, \pi)

.

Proof: Given: The function

f(x) = \log \sin x

. The domain of this function requires

\sin x > 0

, which is true for

x \in (0, \pi)

. The given intervals are within this domain.

To determine where the function is increasing or decreasing, we need to find its derivative,

f'(x)

. Using the chain rule:

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 = \frac{\cos x}{\sin x} = \cot x

Now, we analyze the sign of

f'(x)

in the specified intervals.

Part 1: Increasing on $(0, \frac{\pi}{2})$ For any

x

in the interval

(0, \frac{\pi}{2})

,

x

lies in the first quadrant. In the first quadrant:

$\sin x > 0$
$\cos x > 0$ Therefore, $f'(x) = \cot x = \frac{\cos x}{\sin x}$ is positive. Since $f'(x) > 0$ for all $x \in (0, \frac{\pi}{2})$ , the function $f(x)$ is strictly increasing on $(0, \frac{\pi}{2})$ .

Part 2: Decreasing on $(\frac{\pi}{2}, \pi)$ For any

x

in the interval

(\frac{\pi}{2}, \pi)

,

x

lies in the second quadrant. In the second quadrant:

$\sin x > 0$
$\cos x < 0$ Therefore, $f'(x) = \cot x = \frac{\cos x}{\sin x}$ is negative (a negative number divided by a positive number). Since $f'(x) < 0$ for all $x \in (\frac{\pi}{2}, \pi)$ , the function $f(x)$ is strictly decreasing on $(\frac{\pi}{2}, \pi)$ .

Thus, we have proved that

f(x) = \log \sin x

is increasing on

(0, \frac{\pi}{2})

and decreasing on

(\frac{\pi}{2}, \pi)

.

Hence Proved.

Q17EXERCISE 6.2

Prove that the function $f$ given by $f(x)=\log |\cos x|$ is decreasing on $\left(0, \frac{\pi}{2}\right)$ and increasing on $\left(\frac{3 \pi}{2}, 2 \pi\right)$ .

Solution

To Prove: The function

f

given by

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)

.

Proof: Given: The function

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

. The domain of this function requires

|\cos x| > 0

, which means

\cos x \neq 0

. This is true for all

x

except

x = \frac{\pi}{2} + n\pi

for any integer

n

.

We need to analyze the function in two separate intervals.

Part 1: Interval $\left(0, \frac{\pi}{2}\right)$ For any

x

in the interval

\left(0, \frac{\pi}{2}\right)

,

x

lies in the first quadrant. In the first quadrant,

\cos x > 0

. Therefore,

|\cos x| = \cos x

. So, for

x \in \left(0, \frac{\pi}{2}\right)

, the function is

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

.

Now, we find the derivative

f'(x)

using the chain rule:

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) = -\frac{\sin x}{\cos x} = -\tan x

In the interval

\left(0, \frac{\pi}{2}\right)

,

\tan x > 0

. Therefore,

f'(x) = -\tan x < 0

. Since

f'(x) < 0

for all

x \in \left(0, \frac{\pi}{2}\right)

, the function

f(x)

is strictly decreasing on

\left(0, \frac{\pi}{2}\right)

.

Part 2: Interval $\left(\frac{3 \pi}{2}, 2 \pi\right)$ For any

x

in the interval

\left(\frac{3 \pi}{2}, 2 \pi\right)

,

x

lies in the fourth quadrant. In the fourth quadrant,

\cos x > 0

. Therefore,

|\cos x| = \cos x

. So, for

x \in \left(\frac{3 \pi}{2}, 2 \pi\right)

, the function is again

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

.

The derivative is the same as before:

f'(x) = -\tan x

In the interval

\left(\frac{3 \pi}{2}, 2 \pi\right)

(the fourth quadrant),

\tan x < 0

. Therefore,

f'(x) = -\tan x > 0

(since it is the negative of a negative number). Since

f'(x) > 0

for all

x \in \left(\frac{3 \pi}{2}, 2 \pi\right)

, the function

f(x)

is strictly increasing on

\left(\frac{3 \pi}{2}, 2 \pi\right)

.

Thus, we have proved that

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)

.

Hence Proved.

Q18EXERCISE 6.2

Prove that the function given by $f(x)=x^{3}-3 x^{2}+3 x-100$ is increasing in $\mathbf{R}$ .

Solution

To Prove: The function given by

f(x)=x^{3}-3 x^{2}+3 x-100

is increasing in

\mathbf{R}

.

Given: Function

f(x)=x^{3}-3 x^{2}+3 x-100

.

Proof: To determine if the function is increasing, we need to analyze its first derivative,

f'(x)

. A function is increasing on an interval if its derivative is non-negative, i.e.,

f'(x) \geq 0

, on that interval.

First, we find the derivative of

f(x)

with respect to

x

:

f'(x) = \frac{d}{dx}(x^{3}-3 x^{2}+3 x-100)

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

Now, we can simplify the expression for

f'(x)

by factoring out the common term:

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 is:

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

For any real number

x

, the term

(x-1)

is a real number. The square of any real number is always non-negative. Therefore,

(x-1)^2 \geq 0

for all

x \in \mathbf{R}

.

Since

(x-1)^2 \geq 0

, and it is multiplied by a positive constant 3, we have:

f'(x) = 3(x-1)^2 \geq 0 \text{ for all } x \in \mathbf{R}

The derivative

f'(x)

is equal to zero only at

x=1

, and is positive for all other real numbers. Since

f'(x) \geq 0

for all

x \in \mathbf{R}

, the function

f(x)

is an increasing function on the entire set of real numbers

\mathbf{R}

.

Hence Proved.

Q19EXERCISE 6.2

The interval in which $y=x^{2} e^{-x}$ is increasing is

(A)

(-\infty, \infty)

(B)

(-2,0)

(C)

(2, \infty)

(D)

(0,2)

Solution

Given: The function

y = x^2 e^{-x}

.

To Find: The interval in which the function is increasing.

Solution: A function is increasing in an interval where its first derivative is positive. First, we find the derivative of

y

with respect to

x

, using the product rule. Let

u(x) = x^2

and

v(x) = e^{-x}

. Then

u'(x) = 2x

and

v'(x) = -e^{-x}

.

\frac{dy}{dx} = u(x)v'(x) + v(x)u'(x)

\frac{dy}{dx} = x^2(-e^{-x}) + e^{-x}(2x)

\frac{dy}{dx} = -x^2e^{-x} + 2xe^{-x}

Factor out the common terms

xe^{-x}

:

\frac{dy}{dx} = xe^{-x}(2 - x)

For the function to be increasing, we must have

\frac{dy}{dx} > 0

.

xe^{-x}(2 - x) > 0

Since the exponential function

e^{-x}

is always positive for all real values of

x

, we can divide the inequality by

e^{-x}

without changing the direction of the inequality sign.

x(2 - x) > 0

To solve this inequality, we find the critical points by setting the expression to zero:

x = 0

and

2 - x = 0 \implies x = 2

.

These points divide the number line into three intervals:

(-\infty, 0)

,

(0, 2)

, and

(2, \infty)

. We test the sign of

x(2-x)

in each interval.

Interval $(-\infty, 0)$ : Let's pick $x = -1$ . $(-1)(2 - (-1)) = (-1)(3) = -3 < 0$ . The function is decreasing.
Interval $(0, 2)$ : Let's pick $x = 1$ . $(1)(2 - 1) = (1)(1) = 1 > 0$ . The function is increasing.
Interval $(2, \infty)$ : Let's pick $x = 3$ . $(3)(2 - 3) = (3)(-1) = -3 < 0$ . The function is decreasing.

So, the function

y = x^2 e^{-x}

is increasing in the interval

(0, 2)

.

Comparing this with the given options: (A)

(-\infty, \infty)

(B)

(-2,0)

(C)

(2, \infty)

(D)

(0,2)

Final Answer: The correct option is (D), as the interval in which

y=x^{2} e^{-x}

is increasing is

(0,2)

.

Q1EXERCISE 6.3

Find the maximum and minimum values, if any, of the following functions given by

(i)

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

(ii)

$f(x)=9 x^{2}+12 x+2$

(iii)

$f(x)=-(x-1)^{2}+10$

(iv)

$g(x)=x^{3}+1$

Solution

To Find: The maximum and minimum values, if any, of the given functions.

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

Solution: The term

(2x-1)^2

is the square of a real number. The minimum value of a squared term is 0. So,

(2x-1)^2 \geq 0

for all

x \in \mathbf{R}

. The minimum value of

(2x-1)^2

is 0, which occurs when

2x-1 = 0

, i.e.,

x = \frac{1}{2}

.

Therefore, the minimum value of

f(x)

is

0 + 3 = 3

.

As

x

approaches

\infty

or

-\infty

,

(2x-1)^2

approaches

\infty

. Thus,

f(x)

can be made arbitrarily large. So, the function has no maximum value.

Final Answer for (i): Minimum value is 3. No maximum value.

(ii) $f(x)=9 x^{2}+12 x+2$

Solution: We can find the minimum value by completing the square.

f(x) = 9x^2 + 12x + 2

f(x) = (3x)^2 + 2(3x)(2) + 2

To complete the square, we add and subtract

2^2=4

.

f(x) = [(3x)^2 + 2(3x)(2) + 2^2] - 4 + 2

f(x) = (3x+2)^2 - 2

The term

(3x+2)^2

is a square, so its minimum value is 0. This occurs when

3x+2 = 0

, i.e.,

x = -\frac{2}{3}

.

Therefore, the minimum value of

f(x)

is

0 - 2 = -2

.

Since

(3x+2)^2

can be arbitrarily large, the function has no maximum value.

Final Answer for (ii): Minimum value is -2. No maximum value.

(iii) $f(x)=-(x-1)^{2}+10$

Solution: The term

(x-1)^2

is always non-negative, i.e.,

(x-1)^2 \geq 0

. Multiplying by -1 reverses the inequality:

-(x-1)^2 \leq 0

. The maximum value of

-(x-1)^2

is 0, which occurs when

x-1 = 0

, i.e.,

x=1

.

Therefore, the maximum value of

f(x)

is

0 + 10 = 10

.

As

x

approaches

\infty

or

-\infty

,

(x-1)^2

approaches

\infty

, so

-(x-1)^2

approaches

-\infty

. Thus,

f(x)

can be made arbitrarily small (large negative). So, the function has no minimum value.

Final Answer for (iii): Maximum value is 10. No minimum value.

(iv) $g(x)=x^{3}+1$

Solution: The function

g(x) = x^3 + 1

is a cubic polynomial. The range of

x^3

is

(-\infty, \infty)

. As

x \to \infty

,

g(x) \to \infty

. As

x \to -\infty

,

g(x) \to -\infty

. Therefore, the function does not attain a maximum or a minimum value.

Alternatively, using calculus:

g'(x) = 3x^2

. Setting

g'(x) = 0

gives

x=0

.

g''(x) = 6x

. At

x=0

,

g''(0)=0

. The second derivative test fails. Since

g'(x) = 3x^2 \geq 0

for all

x

, the function is always increasing. An increasing function over

\mathbf{R}

does not have a global maximum or minimum.

Final Answer for (iv): Neither a maximum value nor a minimum value.

Q2EXERCISE 6.3

Find the maximum and minimum values, if any, of the following functions given by

(i)

$f(x)=|x+2|-1$

(ii)

$g(x)=-|x+1|+3$

(iii)

$h(x)=\sin (2 x)+5$

(iv)

$f(x)=|\sin 4 x+3|$

(v)

$h(x)=x+1, x \in(-1,1)$

Solution

To Find: The maximum and minimum values, if any, of the given functions.

(i) $f(x)=|x+2|-1$

Solution: The absolute value function

|x+2|

is always non-negative. Its minimum value is 0.

|x+2| \geq 0

for all

x \in \mathbf{R}

. The minimum value of

|x+2|

is 0, which occurs when

x+2=0

, i.e.,

x=-2

.

Therefore, the minimum value of

f(x)

is

0 - 1 = -1

. Since

|x+2|

can be arbitrarily large,

f(x)

has no maximum value.

Final Answer for (i): Minimum value is -1. No maximum value.

(ii) $g(x)=-|x+1|+3$

Solution: We know that

|x+1| \geq 0

for all

x \in \mathbf{R}

. Multiplying by -1 reverses the inequality:

-|x+1| \leq 0

. The maximum value of

-|x+1|

is 0, which occurs when

x+1=0

, i.e.,

x=-1

.

Therefore, the maximum value of

g(x)

is

0 + 3 = 3

. Since

-|x+1|

can be arbitrarily small (large negative),

g(x)

has no minimum value.

Final Answer for (ii): Maximum value is 3. No minimum value.

(iii) $h(x)=\sin (2 x)+5$

Solution: The range of the sine function is

[-1, 1]

. So,

-1 \leq \sin(2x) \leq 1

for all

x \in \mathbf{R}

. Adding 5 to all parts of the inequality:

-1 + 5 \leq \sin(2x) + 5 \leq 1 + 5

4 \leq h(x) \leq 6

Thus, the minimum value of

h(x)

is 4 and the maximum value is 6.

Final Answer for (iii): Minimum value is 4. Maximum value is 6.

(iv) $f(x)=|\sin 4 x+3|$

Solution: First, consider the expression inside the absolute value,

\sin(4x) + 3

. We know that

-1 \leq \sin(4x) \leq 1

. Adding 3 to all parts of the inequality:

-1 + 3 \leq \sin(4x) + 3 \leq 1 + 3

2 \leq \sin(4x) + 3 \leq 4

Since the expression

\sin(4x) + 3

is always positive (its values are between 2 and 4), the absolute value does not change its value. So,

f(x) = |\sin(4x) + 3| = \sin(4x) + 3

. Therefore,

2 \leq f(x) \leq 4

.

The minimum value of

f(x)

is 2 and the maximum value is 4.

Final Answer for (iv): Minimum value is 2. Maximum value is 4.

(v) $h(x)=x+1, x \in(-1,1)$

Solution: The function

h(x) = x+1

is a linear function with a positive slope, so it is strictly increasing. The domain is the open interval

(-1, 1)

.

As

x

approaches the lower bound of the interval,

-1

, the value of

h(x)

approaches

h(-1) = -1 + 1 = 0

. However, since

x > -1

,

h(x) > 0

. The value 0 is never attained.

As

x

approaches the upper bound of the interval,

1

, the value of

h(x)

approaches

h(1) = 1 + 1 = 2

. However, since

x < 1

,

h(x) < 2

. The value 2 is never attained.

The range of the function on the interval

(-1, 1)

is

(0, 2)

. Since the function never reaches the endpoints of its range, there is no minimum or maximum value.

Final Answer for (v): The function has neither a maximum value nor a minimum value.

Q3EXERCISE 6.3

Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:

(i)

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

(ii)

$g(x)=x^{3}-3 x$

(iii)

$h(x)=\sin x+\cos x, 0<x<\frac{\pi}{2}$

(iv)

$f(x)=\sin x-\cos x, 0<x<2 \pi$

(v)

$f(x)=x^{3}-6 x^{2}+9 x+15$

(vi)

$g(x)=\frac{x}{2}+\frac{2}{x}, x>0$

(vii)

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

(viii)

$f(x)=x \sqrt{1-x}, 0<x<1$

Solution

To Find: The local maxima and local minima, if any, and the corresponding local maximum and local minimum values.

We will use the second derivative test. For a function

f(x)

:

Find critical points by solving $f'(x)=0$ .
For a critical point $c$ $c$ , calculate $f''(c)$ $f^{''} (c)$ .
- If $f''(c) 0$ , $c$ is a point of local minimum.
- If $f''(c) = 0$ , the test is inconclusive.

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

f'(x) = 2x

. Setting

f'(x)=0

gives

x=0

.

f''(x) = 2

. Since

f''(0) = 2 > 0

,

x=0

is a point of local minimum. Local minimum value:

f(0) = 0^2 = 0

. There are no points of local maximum.

Final Answer for (i): Local minimum at

x=0

. Local minimum value is 0.

(ii) $g(x)=x^{3}-3 x$

g'(x) = 3x^2 - 3 = 3(x^2 - 1)

. Setting

g'(x)=0

gives

x^2=1

, so

x=1

and

x=-1

.

g''(x) = 6x

. At

x=1

:

g''(1) = 6(1) = 6 > 0

. So,

x=1

is a point of local minimum. Local minimum value:

g(1) = 1^3 - 3(1) = -2

. At

x=-1

:

g''(-1) = 6(-1) = -6 0

. So,

x=\frac{7\pi}{4}

is a point of local minimum. Local minimum value:

f(\frac{7\pi}{4}) = \sin(\frac{7\pi}{4}) - \cos(\frac{7\pi}{4}) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2}

.

Final Answer for (iv): Local maximum at

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

, value is

\sqrt{2}

. Local minimum at

x=\frac{7\pi}{4}

, value is

-\sqrt{2}

.

(v) $f(x)=x^{3}-6 x^{2}+9 x+15$

f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)

. Setting

f'(x)=0

gives

x=1

and

x=3

.

f''(x) = 6x - 12

. At

x=1

:

f''(1) = 6(1) - 12 = -6 0

. So,

x=3

is a point of local minimum. Local minimum value:

f(3) = 3^3 - 6(3)^2 + 9(3) + 15 = 27 - 54 + 27 + 15 = 15

.

Final Answer for (v): Local maximum at

x=1

, value is 19. Local minimum at

x=3

, value is 15.

(vi) $g(x)=\frac{x}{2}+\frac{2}{x}, x>0$

g'(x) = \frac{1}{2} - \frac{2}{x^2}

. Setting

g'(x)=0

gives

\frac{1}{2} = \frac{2}{x^2}

, so

x^2 = 4

. Since

x>0

, we have

x=2

.

g''(x) = -2(-2x^{-3}) = \frac{4}{x^3}

. At

x=2

:

g''(2) = \frac{4}{2^3} = \frac{4}{8} = \frac{1}{2} > 0

. So,

x=2

is a point of local minimum. Local minimum value:

g(2) = \frac{2}{2} + \frac{2}{2} = 1 + 1 = 2

.

Final Answer for (vi): Local minimum at

x=2

. Local minimum value is 2.

(vii) $g(x)=\frac{1}{x^{2}+2}$

g(x) = (x^2+2)^{-1}

.

g'(x) = -1(x^2+2)^{-2}(2x) = \frac{-2x}{(x^2+2)^2}

. Setting

g'(x)=0

gives

-2x=0

, so

x=0

. Using the quotient rule for

g''(x)

:

u=-2x, v=(x^2+2)^2

.

u'=-2, v'=2(x^2+2)(2x)=4x(x^2+2)

.

g''(x) = \frac{v u' - u v'}{v^2} = \frac{(x^2+2)^2(-2) - (-2x)(4x(x^2+2))}{((x^2+2)^2)^2} = \frac{-2(x^2+2) + 8x^2}{(x^2+2)^3} = \frac{6x^2-4}{(x^2+2)^3}

. At

x=0

:

g''(0) = \frac{6(0)^2-4}{(0^2+2)^3} = \frac{-4}{8} = -\frac{1}{2} 0

, so

f'(x) > 0

. For

x > \frac{2}{3}

,

2-3x < 0

, so

f'(x) < 0

. The sign of

f'(x)

changes from positive to negative at

x=\frac{2}{3}

. Thus, it is a point of local maximum. Local maximum value:

f(\frac{2}{3}) = \frac{2}{3}\sqrt{1-\frac{2}{3}} = \frac{2}{3}\sqrt{\frac{1}{3}} = \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{9}

.

Final Answer for (viii): Local maximum at

x=\frac{2}{3}

. Local maximum value is

\frac{2\sqrt{3}}{9}

.

Q4EXERCISE 6.3

Prove that the following functions do not have maxima or minima:

(i)

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

(ii)

$g(x)=\log x$

(iii)

$h(x)=x^{3}+x^{2}+x+1$

Solution

To prove that a function does not have maxima or minima, we can show that its first derivative is never zero, which means there are no critical points. A function that is always increasing or always decreasing does not have any local maximum or minimum values.

(i) $f(x) = e^x$

Given: The function is

f(x) = e^x

.

Solution:

Find the first derivative of the function: $f'(x) = \frac{d}{dx}(e^x) = e^x$
To find critical points, we set the first derivative to zero: $f'(x) = 0 \implies e^x = 0$
The exponential function $e^x$ is always positive for any real value of $x$ . Thus, $e^x$ can never be equal to zero.
Since there are no real values of $x$ for which $f'(x) = 0$ , there are no critical points.
Furthermore, $f'(x) = e^x > 0$ for all $x \in \mathbb{R}$ . This means the function $f(x)$ is strictly increasing on its entire domain.

Conclusion: A strictly monotonic function does not have a maximum or minimum value. Therefore,

f(x) = e^x

has no maxima or minima.

(ii) $g(x) = \log x$

Given: The function is

g(x) = \log x

. The domain of this function is

x > 0

, i.e.,

(0, \infty)

.

Solution:

Find the first derivative of the function: $g'(x) = \frac{d}{dx}(\log x) = \frac{1}{x}$
To find critical points, we set the first derivative to zero: $g'(x) = 0 \implies \frac{1}{x} = 0$
This equation has no solution for $x$ .
Since there are no values of $x$ for which $g'(x) = 0$ , there are no critical points in the domain of the function.
For all $x$ in the domain $(0, \infty)$ , $g'(x) = \frac{1}{x} > 0$ . This means the function $g(x)$ is strictly increasing on its domain.

Conclusion: A strictly monotonic function does not have a maximum or minimum value. Therefore,

g(x) = \log x

has no maxima or minima.

(iii) $h(x) = x^3 + x^2 + x + 1$

Given: The function is

h(x) = x^3 + x^2 + x + 1

.

Solution:

Find the first derivative of the function: $h'(x) = \frac{d}{dx}(x^3 + x^2 + x + 1) = 3x^2 + 2x + 1$
To find critical points, we set the first derivative to zero: $3x^2 + 2x + 1 = 0$
To check for real roots of this quadratic equation, we compute the discriminant, $D = b^2 - 4ac$ . Here, $a=3$ , $b=2$ , and $c=1$ . $D = (2)^2 - 4(3)(1) = 4 - 12 = -8$
Since the discriminant $D < 0$ , the quadratic equation $3x^2 + 2x + 1 = 0$ has no real roots. Therefore, there are no critical points.
Since the leading coefficient ( $a=3$ ) is positive and the discriminant is negative, the quadratic expression $3x^2 + 2x + 1$ is always positive for all real numbers $x$ . So, $h'(x) > 0$ for all $x \in \mathbb{R}$ .

Conclusion: Since

h'(x)

is always positive, the function

h(x)

is strictly increasing on

\mathbb{R}

. A strictly increasing function does not have any maximum or minimum value. Therefore,

h(x) = x^3 + x^2 + x + 1

has no maxima or minima.

Q5EXERCISE 6.3

Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

(i)

$f(x)=x^{3}, x \in[-2,2]$

(ii)

$f(x)=\sin x+\cos x, x \in[0, \pi]$

(iii)

$f(x)=4 x-\frac{1}{2} x^{2}, x \in\left[-2, \frac{9}{2}\right]$

(iv)

$f(x)=(x-1)^{2}+3, x \in[-3,1]$

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. The largest value is the absolute maximum and the smallest value is the absolute minimum.

(i) $f(x) = x^3, x \in [-2, 2]$

Given:

f(x) = x^3

on the interval

[-2, 2]

.

Solution:

Find the derivative: $f'(x) = 3x^2$ .
Find critical points by setting $f'(x) = 0$ : $3x^2 = 0 \implies x=0$ . This point lies in the interval $[-2, 2]$ .
Evaluate $f(x)$ $f (x)$ at the critical point $x=0$ $x = 0$ and the endpoints $x=-2, x=2$ $x = - 2, x = 2$ :
- $f(0) = (0)^3 = 0$
- $f(-2) = (-2)^3 = -8$
- $f(2) = (2)^3 = 8$
The values are -8, 0, and 8. The largest is 8 and the smallest is -8.

Final Answer: The absolute maximum value is 8 and the absolute minimum value is -8.

(ii) $f(x) = \sin x + \cos x, x \in [0, \pi]$

Given:

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

on the interval

[0, \pi]

.

Solution:

Find the derivative: $f'(x) = \cos x - \sin x$ .
Find critical points by setting $f'(x) = 0$ : $\cos x - \sin x = 0 \implies \tan x = 1$ . The solution in $[0, \pi]$ is $x = \frac{\pi}{4}$ .
Evaluate $f(x)$ $f (x)$ at the critical point $x=\frac{\pi}{4}$ $x = \frac{π}{4}$ and the endpoints $x=0, x=\pi$ $x = 0, x = π$ :
- $f(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$
- $f(0) = \sin(0) + \cos(0) = 0 + 1 = 1$
- $f(\pi) = \sin(\pi) + \cos(\pi) = 0 + (-1) = -1$
The values are $\sqrt{2}$ , 1, and -1. Note that $\sqrt{2} \approx 1.414$ .

Final Answer: The absolute maximum value is

\sqrt{2}

and the absolute minimum value is -1.

(iii) $f(x) = 4x - \frac{1}{2}x^2, x \in \left[-2, \frac{9}{2}\right]$

Given:

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

on the interval

\left[-2, \frac{9}{2}\right]

.

Solution:

Find the derivative: $f'(x) = 4 - x$ .
Find critical points by setting $f'(x) = 0$ : $4 - x = 0 \implies x=4$ . This point lies in the interval $\left[-2, \frac{9}{2}\right]$ as $\frac{9}{2}=4.5$ .
Evaluate $f(x)$ $f (x)$ at the critical point $x=4$ $x = 4$ and the endpoints $x=-2, x=\frac{9}{2}$ $x = - 2, x = \frac{9}{2}$ :
- $f(4) = 4(4) - \frac{1}{2}(4)^2 = 16 - 8 = 8$
- $f(-2) = 4(-2) - \frac{1}{2}(-2)^2 = -8 - 2 = -10$
- $f(\frac{9}{2}) = 4(\frac{9}{2}) - \frac{1}{2}(\frac{9}{2})^2 = 18 - \frac{81}{8} = \frac{144-81}{8} = \frac{63}{8} = 7.875$
The values are 8, -10, and 7.875.

Final Answer: The absolute maximum value is 8 and the absolute minimum value is -10.

(iv) $f(x) = (x-1)^2 + 3, x \in [-3, 1]$

Given:

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

on the interval

[-3, 1]

.

Solution:

Find the derivative: $f'(x) = 2(x-1)$ .
Find critical points by setting $f'(x) = 0$ : $2(x-1) = 0 \implies x=1$ . This point is an endpoint of the interval.
We evaluate $f(x)$ $f (x)$ at the endpoints $x=-3$ $x = - 3$ and $x=1$ $x = 1$ :
- $f(-3) = (-3-1)^2 + 3 = (-4)^2 + 3 = 16 + 3 = 19$
- $f(1) = (1-1)^2 + 3 = 0^2 + 3 = 3$
The values are 19 and 3.

Final Answer: The absolute maximum value is 19 and the absolute minimum value is 3.

Q6EXERCISE 6.3

Find the maximum profit that a company can make, if the profit function is given by $p(x)=41-72 x-18 x^{2}$

Solution

Given: The profit function is

p(x) = 41 - 72x - 18x^2

.

To Find: The maximum profit.

Solution: To find the maximum profit, we need to find the maximum value of the function

p(x)

. We will use the first and second derivative tests.

Find the first derivative of the profit function with respect to $x$ : $p'(x) = \frac{d}{dx}(41 - 72x - 18x^2) = -72 - 36x$
Set the first derivative to zero to find the critical points: $p'(x) = 0$ $-72 - 36x = 0$ $-36x = 72$ $x = \frac{72}{-36} = -2$ So, the only critical point is $x = -2$ .
Find the second derivative to determine if this critical point corresponds to a maximum or a minimum: $p''(x) = \frac{d}{dx}(-72 - 36x) = -36$
Evaluate the second derivative at the critical point $x=-2$ : $p''(-2) = -36$ Since $p''(-2) < 0$ , the function $p(x)$ has a local maximum at $x = -2$ . As this is the only critical point for the quadratic function (which represents a downward-opening parabola), this local maximum is also the absolute maximum.
Calculate the maximum profit by substituting $x=-2$ into the profit function $p(x)$ : $p(-2) = 41 - 72(-2) - 18(-2)^2$ $p(-2) = 41 + 144 - 18(4)$ $p(-2) = 185 - 72$ $p(-2) = 113$

Final Answer: The maximum profit that the company can make is 113.

Q7EXERCISE 6.3

Find both the maximum value and the minimum value of $3 x^{4}-8 x^{3}+12 x^{2}-48 x+25$ on the interval $[0,3]$ .

Solution

Given: The function is

f(x) = 3x^4 - 8x^3 + 12x^2 - 48x + 25

on the interval

[0, 3]

.

To Find: The absolute maximum and absolute minimum values of the function on the given interval.

Method: To find the absolute extrema of a continuous function on a closed interval, we follow these steps:

Find the critical points of the function by finding where the first derivative is zero or undefined.
Evaluate the function at the critical points that lie within the interval.
Evaluate the function at the endpoints of the interval.
The largest of these values is the absolute maximum, and the smallest is the absolute minimum.

Solution:

Find the first derivative of $f(x)$ : $f'(x) = \frac{d}{dx}(3x^4 - 8x^3 + 12x^2 - 48x + 25)$ $f'(x) = 12x^3 - 24x^2 + 24x - 48$
Set $f'(x) = 0$ to find the critical points: $12x^3 - 24x^2 + 24x - 48 = 0$ 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: $x^2 + 2 = 0$ or $x - 2 = 0$ . The equation $x^2 = -2$ has no real solutions. The equation $x - 2 = 0$ gives $x = 2$ .
The only critical point is $x=2$ . This point lies within the given interval $[0, 3]$ .
Now, we evaluate the function $f(x)$ at the critical point $x=2$ and at the endpoints of the interval, $x=0$ and $x=3$ .
- At $x = 0$ : $f(0) = 3(0)^4 - 8(0)^3 + 12(0)^2 - 48(0) + 25 = 25$
- At $x = 2$ : $f(2) = 3(2)^4 - 8(2)^3 + 12(2)^2 - 48(2) + 25$ $f(2) = 3(16) - 8(8) + 12(4) - 96 + 25$ $f(2) = 48 - 64 + 48 - 96 + 25$ $f(2) = 121 - 160 = -39$
- At $x = 3$ : $f(3) = 3(3)^4 - 8(3)^3 + 12(3)^2 - 48(3) + 25$ $f(3) = 3(81) - 8(27) + 12(9) - 144 + 25$ $f(3) = 243 - 216 + 108 - 144 + 25$ $f(3) = (243 + 108 + 25) - (216 + 144) = 376 - 360 = 16$
The values of the function at these points are $25$ , $-39$ , and $16$ . The maximum value among these is 25. The minimum value among these is -39.

Final Answer: On the interval

[0, 3]

, the maximum value of the function is 25 and the minimum value is -39.

Q8EXERCISE 6.3

At what points in the interval $[0,2 \pi]$ , does the function $\sin 2 x$ attain its maximum value?

Solution

Given: The function is

f(x) = \sin(2x)

on the interval

[0, 2\pi]

.

To Find: The points in the interval

[0, 2\pi]

where the function attains its maximum value.

Method: We know that the maximum value of the sine function,

\sin(\theta)

, is 1. Therefore, the maximum value of

f(x) = \sin(2x)

is also 1. We need to find the values of

x

in the interval

[0, 2\pi]

for which

\sin(2x) = 1

.

Solution:

Set the function equal to its maximum value: $\sin(2x) = 1$
The general solution for $\sin(\theta) = 1$ is $\theta = 2n\pi + \frac{\pi}{2}$ , where $n$ is an integer. In our case, $\theta = 2x$ . So, we have: $2x = 2n\pi + \frac{\pi}{2}$ $x = n\pi + \frac{\pi}{4}$
Now, we need to find the values of $x$ that lie in the interval $[0, 2\pi]$ by substituting integer values for $n$ .
- For $n=0$ : $x = 0 \cdot \pi + \frac{\pi}{4} = \frac{\pi}{4}$ This value is in the interval $[0, 2\pi]$ .
- For $n=1$ : $x = 1 \cdot \pi + \frac{\pi}{4} = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$ This value is in the interval $[0, 2\pi]$ .
- For $n=2$ : $x = 2\pi + \frac{\pi}{4} = \frac{9π}{4}$ This value is greater than $2\pi$ , so it is outside the interval.
- For $n=-1$ : $x = -\pi + \frac{\pi}{4} = -\frac{3\pi}{4}$ This value is less than 0, so it is outside the interval.

Thus, the only values of

x

in the interval

[0, 2\pi]

for which

\sin(2x)

is maximum are

x = \frac{\pi}{4}

and

x = \frac{5\pi}{4}

.

Alternatively, using calculus:

Find the derivative: $f'(x) = 2\cos(2x)$ .
Set $f'(x)=0$ : $2\cos(2x)=0 \implies \cos(2x)=0$ .
Let $u=2x$ . Since $x \in [0, 2\pi]$ , $u \in [0, 4\pi]$ . The solutions for $\cos(u)=0$ in $[0, 4\pi]$ are $u = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$ .
So, $2x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$ . This gives $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$ . These are the critical points.
Find the second derivative: $f''(x) = -4\sin(2x)$ .
Evaluate $f''(x)$ $f^{''} (x)$ at the critical points:
- $f''(\frac{\pi}{4}) = -4\sin(\frac{\pi}{2}) = -4 < 0$ (Maximum)
- $f''(\frac{3\pi}{4}) = -4\sin(\frac{3\pi}{2}) = 4 > 0$ (Minimum)
- $f''(\frac{5\pi}{4}) = -4\sin(\frac{5\pi}{2}) = -4 < 0$ (Maximum)
- $f''(\frac{7\pi}{4}) = -4\sin(\frac{7\pi}{2}) = 4 > 0$ (Minimum)
The points of local maxima are $x=\frac{\pi}{4}$ and $x=\frac{5\pi}{4}$ .
The maximum value is $f(\frac{\pi}{4}) = \sin(\frac{\pi}{2}) = 1$ and $f(\frac{5\pi}{4}) = \sin(\frac{5\pi}{2}) = 1$ . Since the absolute maximum value of $\sin(2x)$ is 1, these points are the points of absolute maxima.

Final Answer: 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]

.

Q9EXERCISE 6.3

What is the maximum value of the function $\sin x+\cos x$ ?

Solution

Given: The function

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

.

To Find: The maximum value of the function.

Solution: Let the given function be

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

.

First, we find the derivative of the function with respect to

x

:

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

To find the critical points, we set

f'(x) = 0

:

\cos x - \sin x = 0

\cos x = \sin x

\tan x = 1

The principal solutions for

x

in the interval

[0, 2\pi]

are

x = \frac{\pi}{4}

and

x = \frac{5\pi}{4}

.

Now, we find the second derivative to apply the second derivative test:

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

We evaluate the second derivative at the critical points:

At $x = \frac{\pi}{4}$ : $f''\left(\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2}$ Since $f''(\frac{\pi}{4}) < 0$ , the function has a local maximum at $x = \frac{\pi}{4}$ . The maximum value is: $f\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$
At $x = \frac{5\pi}{4}$ : $f''\left(\frac{5\pi}{4}\right) = -\sin\left(\frac{5\pi}{4}\right) - \cos\left(\frac{5\pi}{4}\right) = -\left(-\frac{1}{\sqrt{2}}\right) - \left(-\frac{1}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$ Since $f''(\frac{5\pi}{4}) > 0$ , the function has a local minimum at $x = \frac{5\pi}{4}$ .

As the function

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

is periodic, its absolute maximum value is the same as its local maximum value.

Final Answer: The maximum value of the function

\sin x + \cos x

is

\sqrt{2}

.

Q10EXERCISE 6.3

Find the maximum value of $2 x^{3}-24 x+107$ in the interval $[1,3]$ . Find the maximum value of the same function in $[-3,-1]$ .

Solution

Given: The function

f(x) = 2x^3 - 24x + 107

.

To Find: The maximum value of

f(x)

in the interval

[1, 3]

and in the interval

[-3, -1]

.

Solution: Let the given function be

f(x) = 2x^3 - 24x + 107

.

First, we find the derivative of the function:

f'(x) = \frac{d}{dx}(2x^3 - 24x + 107) = 6x^2 - 24

To find the critical points, we set

f'(x) = 0

:

6x^2 - 24 = 0

6x^2 = 24

x^2 = 4

x = \pm 2

The critical points of the function are

x = 2

and

x = -2

.

Part 1: Maximum value in the interval $[1, 3]$

For a closed interval, the absolute maximum value occurs either at the critical points within the interval or at the endpoints of the interval. The critical point

x=2

lies in the interval

[1, 3]

. The endpoints are

x=1

and

x=3

. We evaluate the function at these points:

At $x=1$ (endpoint): $f(1) = 2(1)^3 - 24(1) + 107 = 2 - 24 + 107 = 85$ .
At $x=2$ (critical point): $f(2) = 2(2)^3 - 24(2) + 107 = 2(8) - 48 + 107 = 16 - 48 + 107 = 75$ .
At $x=3$ (endpoint): $f(3) = 2(3)^3 - 24(3) + 107 = 2(27) - 72 + 107 = 54 - 72 + 107 = 89$ .

Comparing the values {85, 75, 89}, the maximum value in the interval

[1, 3]

is 89.

Part 2: Maximum value in the interval $[-3, -1]$

The critical point

x=-2

lies in the interval

[-3, -1]

. The endpoints are

x=-3

and

x=-1

. We evaluate the function at these points:

At $x=-3$ (endpoint): $f(-3) = 2(-3)^3 - 24(-3) + 107 = 2(-27) + 72 + 107 = -54 + 72 + 107 = 125$ .
At $x=-2$ (critical point): $f(-2) = 2(-2)^3 - 24(-2) + 107 = 2(-8) + 48 + 107 = -16 + 48 + 107 = 139$ .
At $x=-1$ (endpoint): $f(-1) = 2(-1)^3 - 24(-1) + 107 = 2(-1) + 24 + 107 = -2 + 24 + 107 = 129$ .

Comparing the values {125, 139, 129}, the maximum value in the interval

[-3, -1]

is 139.

Final Answer: The maximum value of the function in the interval

[1, 3]

is 89. The maximum value of the function in the interval

[-3, -1]

is 139.

Q11EXERCISE 6.3

It is given that at $x=1$ , the function $x^{4}-62 x^{2}+a x+9$ attains its maximum value, on the interval $[0,2]$ . Find the value of $a$ .

Solution

Given: The function

f(x) = x^4 - 62x^2 + ax + 9

attains its maximum value at

x=1

on the interval

[0, 2]

.

To Find: The value of

a

.

Solution: Let the given function be

f(x) = x^4 - 62x^2 + ax + 9

.

The function is differentiable, and its maximum value on the closed interval

[0, 2]

is given to be at

x=1

. Since

x=1

is an interior point of the interval

[0, 2]

, it must be a critical point of the function

f(x)

.

A necessary condition for

x=1

to be a point of local maximum is that the first derivative of the function at that point is zero, i.e.,

f'(1) = 0

.

First, we find the derivative of the function

f(x)

:

f'(x) = \frac{d}{dx}(x^4 - 62x^2 + ax + 9)

f'(x) = 4x^3 - 124x + a

Now, we set

f'(1) = 0

:

f'(1) = 4(1)^3 - 124(1) + a = 0

4 - 124 + a = 0

-120 + a = 0

a = 120

We can verify if this gives a maximum at

x=1

. For

a=120

,

f'(x) = 4x^3 - 124x + 120

. The second derivative is

f''(x) = 12x^2 - 124

. At

x=1

,

f''(1) = 12(1)^2 - 124 = 12 - 124 = -112

. Since

f''(1) < 0

,

x=1

is a point of local maximum.

We also need to check the function values at the endpoints of the interval

[0, 2]

to confirm it is the absolute maximum. For

a=120

,

f(x) = x^4 - 62x^2 + 120x + 9

.

f(0) = 0^4 - 62(0)^2 + 120(0) + 9 = 9

.

f(1) = 1^4 - 62(1)^2 + 120(1) + 9 = 1 - 62 + 120 + 9 = 68

.

f(2) = 2^4 - 62(2)^2 + 120(2) + 9 = 16 - 62(4) + 240 + 9 = 16 - 248 + 240 + 9 = 17

.

Comparing the values {9, 68, 17}, the maximum value on

[0, 2]

is indeed 68, which occurs at

x=1

. Thus, the value

a=120

is correct.

Final Answer: The value of

a

is 120.

Q12EXERCISE 6.3

Find the maximum and minimum values of $x+\sin 2 x$ on $[0,2\pi]$ .

Solution

Given: The function

f(x) = x + \sin 2x

on the interval

[0, 2\pi]

.

To Find: The maximum and minimum values of the function on the given interval.

Solution: Let the given function be

f(x) = x + \sin 2x

. To find the maximum and minimum values in a closed interval, we evaluate the function at its critical points and at the endpoints of the interval.

First, find the derivative of the function:

f'(x) = \frac{d}{dx}(x + \sin 2x) = 1 + 2\cos 2x

To find the critical points, set

f'(x) = 0

:

1 + 2\cos 2x = 0

2\cos 2x = -1

\cos 2x = -\frac{1}{2}

We need to find solutions for

x

in the interval

[0, 2\pi]

. This means

2x

will be in the interval

[0, 4\pi]

. The angles in

[0, 4\pi]

for which cosine is

-\frac{1}{2}

are

2x = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}

.

Solving for

x

:

x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}

All these critical points lie within the interval

[0, 2\pi]

.

The points at which we need to evaluate the function are the endpoints

x=0, x=2\pi

and the critical points

x=\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}

.

At $x=0$ (endpoint): $f(0) = 0 + \sin(0) = 0$ .
At $x=\frac{\pi}{3}$ (critical point): $f(\frac{\pi}{3}) = \frac{\pi}{3} + \sin(2 \cdot \frac{\pi}{3}) = \frac{\pi}{3} + \sin(\frac{2\pi}{3}) = \frac{\pi}{3} + \frac{\sqrt{3}}{2} \approx 1.047 + 0.866 = 1.913$ .
At $x=\frac{2\pi}{3}$ (critical point): $f(\frac{2\pi}{3}) = \frac{2\pi}{3} + \sin(2 \cdot \frac{2\pi}{3}) = \frac{2\pi}{3} + \sin(\frac{4\pi}{3}) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \approx 2.094 - 0.866 = 1.228$ .
At $x=\frac{4\pi}{3}$ (critical point): $f(\frac{4\pi}{3}) = \frac{4\pi}{3} + \sin(2 \cdot \frac{4\pi}{3}) = \frac{4\pi}{3} + \sin(\frac{8\pi}{3}) = \frac{4\pi}{3} + \sin(2\pi + \frac{2\pi}{3}) = \frac{4\pi}{3} + \frac{\sqrt{3}}{2} \approx 4.188 + 0.866 = 5.054$ .
At $x=\frac{5\pi}{3}$ (critical point): $f(\frac{5\pi}{3}) = \frac{5\pi}{3} + \sin(2 \cdot \frac{5\pi}{3}) = \frac{5\pi}{3} + \sin(\frac{10\pi}{3}) = \frac{5\pi}{3} + \sin(4\pi - \frac{2\pi}{3}) = \frac{5\pi}{3} - \sin(\frac{2\pi}{3}) = \frac{5\pi}{3} - \frac{\sqrt{3}}{2} \approx 5.236 - 0.866 = 4.37$ .
At $x=2\pi$ (endpoint): $f(2\pi) = 2\pi + \sin(4\pi) = 2\pi + 0 = 2\pi \approx 6.283$ .

Let's compare the values: {0, 1.913, 1.228, 5.054, 4.37, 6.283}. The minimum value is 0. The maximum value is

2\pi

.

Final Answer: The maximum value of the function on

[0, 2\pi]

is

2\pi

. The minimum value of the function on

[0, 2\pi]

is 0.

Q13EXERCISE 6.3

Find two numbers whose sum is 24 and whose product is as large as possible.

Solution

Given: The sum of two numbers is 24.

To Find: Two numbers such that their product is as large as possible.

Solution: Let the two numbers be

x

and

y

.

According to the problem statement, we have:

x + y = 24

From this, we can express

y

in terms of

x

:

y = 24 - x

Let

P

be the product of the two numbers. We want to maximize

P

.

P = x \cdot y

Substitute the expression for

y

into the product equation:

P(x) = x(24 - x)

P(x) = 24x - x^2

To find the value of

x

for which the product

P

is maximum, we use the first and second derivative tests.

First, find the derivative of

P(x)

with respect to

x

:

P'(x) = \frac{d}{dx}(24x - x^2) = 24 - 2x

Set the first derivative to zero to find the critical points:

P'(x) = 0

24 - 2x = 0

2x = 24

x = 12

Now, find the second derivative to determine if this critical point corresponds to a maximum:

P''(x) = \frac{d}{dx}(24 - 2x) = -2

Since

P''(x) = -2

, which is a constant and is less than 0, the product

P(x)

has a maximum at the critical point

x=12

.

Now we find the value of the second number,

y

:

y = 24 - x

y = 24 - 12

y = 12

The two numbers are 12 and 12. Their product is

12 \times 12 = 144

, which is the maximum possible product.

Final Answer: The two numbers whose sum is 24 and whose product is as large as possible are 12 and 12.

Q14EXERCISE 6.3

Find two positive numbers $x$ and $y$ such that $x+y=60$ and $x y^{3}$ is maximum.

Solution

Given: Two positive numbers,

x

and

y

, such that

x + y = 60

.

To Find: The values of

x

and

y

for which the product

P = xy^3

is maximum.

Solution: Let the two positive numbers be

x

and

y

. We are given the constraint:

x + y = 60

From this, we can express

x

in terms of

y

:

x = 60 - y

Since

x

and

y

are positive numbers,

x > 0

and

y > 0

.

60 - y > 0 \implies y < 60

. So,

0 < y < 60

.

Let

P

be the product we want to maximize:

P = xy^3

Substituting

x = 60 - y

into the expression for

P

:

P(y) = (60 - y)y^3

P(y) = 60y^3 - y^4

To find the maximum value of

P

, we need to find the critical points by differentiating

P(y)

with respect to

y

and setting the derivative to zero.

P'(y) = \frac{d}{dy}(60y^3 - y^4)

P'(y) = 180y^2 - 4y^3

Now, set

P'(y) = 0

:

180y^2 - 4y^3 = 0

4y^2(45 - y) = 0

This gives two possible values for

y

:

y = 0

or

y = 45

. Since

y

must be a positive number,

y=0

is not a valid solution. The only critical point in the interval

(0, 60)

is

y = 45

.

To check if this point corresponds to a maximum, we use the second derivative test. We find the second derivative of

P(y)

:

P''(y) = \frac{d}{dy}(180y^2 - 4y^3)

P''(y) = 360y - 12y^2

Now, we evaluate

P''(y)

at the critical point

y = 45

:

P''(45) = 360(45) - 12(45)^2

P''(45) = 45(360 - 12 \times 45)

P''(45) = 45(360 - 540)

P''(45) = 45(-180) = -8100

Since

P''(45) < 0

, the function

P(y)

has a local maximum at

y = 45

. As this is the only critical point in the interval

(0, 60)

, it is the point of absolute maximum.

Now, we find the corresponding value of

x

:

x = 60 - y = 60 - 45 = 15

Final Answer: The two positive numbers are

x=15

and

y=45

.

Q15EXERCISE 6.3

Find two positive numbers $x$ and $y$ such that their sum is 35 and the product $x^{2} y^{5}$ is a maximum.

Solution

Given: Two positive numbers,

x

and

y

, such that their sum is 35.

To Find: The values of

x

and

y

for which the product

P = x^2 y^5

is a maximum.

Solution: Let the two positive numbers be

x

and

y

. We are given the constraint:

x + y = 35

From this, we can express

x

in terms of

y

:

x = 35 - y

Since

x

and

y

are positive numbers,

x > 0

and

y > 0

.

35 - y > 0 \implies y < 35

. So,

0 < y < 35

.

Let

P

be the product we want to maximize:

P = x^2 y^5

Substituting

x = 35 - y

into the expression for

P

:

P(y) = (35 - y)^2 y^5

To find the maximum value of

P

, we differentiate

P(y)

with respect to

y

using the product rule and set the derivative to zero. Let

u = (35-y)^2

and

v = y^5

. Then

u' = 2(35-y)(-1) = -2(35-y)

and

v' = 5y^4

.

P'(y) = u'v + uv'

P'(y) = -2(35-y)y^5 + (35-y)^2(5y^4)

Factor out the common terms

(35-y)y^4

:

P'(y) = (35-y)y^4 [-2y + 5(35-y)]

P'(y) = (35-y)y^4 [-2y + 175 - 5y]

P'(y) = (35-y)y^4 (175 - 7y)

P'(y) = 7(35-y)y^4 (25 - y)

Now, set

P'(y) = 0

:

7(35-y)y^4 (25 - y) = 0

This gives possible values for

y

:

y = 35

,

y = 0

, or

y = 25

. Since

0 < y < 35

, the only valid critical point is

y = 25

. The values

y=0

and

y=35

would make one of the numbers zero, which is not positive.

To check if this point corresponds to a maximum, we can use the first derivative test. We check the sign of

P'(y)

around

y=25

. For a value slightly less than 25 (e.g.,

y=24

in the interval

(0, 25)

):

P'(24) = 7(35-24)(24)^4(25-24) = 7(11)(24)^4(1) > 0

. The function is increasing. For a value slightly greater than 25 (e.g.,

y=26

in the interval

(25, 35)

):

P'(26) = 7(35-26)(26)^4(25-26) = 7(9)(26)^4(-1) < 0

. The function is decreasing. Since

P'(y)

changes sign from positive to negative at

y=25

, this point corresponds to a local maximum. As it is the only critical point in the interval, it is the point of absolute maximum.

Now, we find the corresponding value of

x

:

x = 35 - y = 35 - 25 = 10

Final Answer: The two positive numbers are

x=10

and

y=25

.

Q16EXERCISE 6.3

Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

Solution

Given: Two positive numbers whose sum is 16.

To Find: The two numbers such that the sum of their cubes is minimum.

Solution: Let the two positive numbers be

x

and

y

. We are given the constraint:

x + y = 16

From this, we can express

y

in terms of

x

:

y = 16 - x

Since

x

and

y

are positive numbers,

x > 0

and

y > 0

.

16 - x > 0 \implies x < 16

. So,

0 < x < 16

.

Let

S

be the sum of their cubes:

S = x^3 + y^3

Substituting

y = 16 - x

into the expression for

S

:

S(x) = x^3 + (16 - x)^3

To find the minimum value of

S

, we need to find the critical points by differentiating

S(x)

with respect to

x

and setting the derivative to zero.

S'(x) = \frac{d}{dx}[x^3 + (16 - x)^3]

S'(x) = 3x^2 + 3(16 - x)^2 \cdot (-1)

S'(x) = 3x^2 - 3(16 - x)^2

Now, set

S'(x) = 0

:

3x^2 - 3(16 - x)^2 = 0

x^2 = (16 - x)^2

Taking the square root of both sides:

x = \pm(16 - x)

Case 1:

x = 16 - x

2x = 16

x = 8

Case 2:

x = -(16 - x)

x = -16 + x

0 = -16

This is impossible. So, the only critical point is

x = 8

. This point lies in the interval

(0, 16)

.

To check if this point corresponds to a minimum, we use the second derivative test. We find the second derivative of

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 > 0

for all values of

x

, the function

S(x)

has a local minimum at

x=8

. As this is the only critical point, it is the point of absolute minimum.

Now, we find the corresponding value of

y

:

y = 16 - x = 16 - 8 = 8

Final Answer: The two positive numbers are 8 and 8.

Q17EXERCISE 6.3

A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible.

Solution

Given: A square piece of tin of side 18 cm. A box without a top is made by cutting a square from each corner and folding up the flaps.

To Find: The side of the square to be cut off so that the volume of the box is maximum.

Solution: Let the side of the square cut from each corner be

x

cm. After cutting the squares, the remaining sheet is folded to form an open box. The dimensions of the resulting box will be: Length,

l = 18 - 2x

cm Width,

w = 18 - 2x

cm Height,

h = x

cm

The volume

V

of the box is given by

V = l \times w \times h

.

V(x) = (18 - 2x)(18 - 2x)x

V(x) = (18 - 2x)^2 x

For the dimensions of the box to be positive, we must have

x > 0

and

18 - 2x > 0

.

18 - 2x > 0 \implies 18 > 2x \implies x < 9

. So, the domain of the function is

0 < x < 9

.

Now, we expand the expression for the volume:

V(x) = (324 - 72x + 4x^2)x

V(x) = 4x^3 - 72x^2 + 324x

To find the maximum volume, we differentiate

V(x)

with respect to

x

and set the derivative to zero.

V'(x) = \frac{d}{dx}(4x^3 - 72x^2 + 324x)

V'(x) = 12x^2 - 144x + 324

Set

V'(x) = 0

to find critical points:

12x^2 - 144x + 324 = 0

Divide the equation by 12:

x^2 - 12x + 27 = 0

Factor the quadratic equation:

(x - 3)(x - 9) = 0

This gives two possible values for

x

:

x = 3

or

x = 9

. Since the domain for

x

is

(0, 9)

, we discard

x=9

. If

x=9

, the length and width of the box would be zero, resulting in zero volume. Thus, the only critical point in the domain is

x = 3

.

To confirm that this point gives a maximum volume, we use the second derivative test.

V''(x) = \frac{d}{dx}(12x^2 - 144x + 324)

V''(x) = 24x - 144

Evaluate

V''(x)

at

x = 3

:

V''(3) = 24(3) - 144 = 72 - 144 = -72

Since

V''(3) < 0

, the volume is maximum at

x = 3

.

Final Answer: The side of the square to be cut off should be 3 cm for the volume of the box to be maximum.

Q18EXERCISE 6.3

A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum?

Solution

Given: A rectangular sheet of tin 45 cm by 24 cm. A box without a top is made by cutting a square from each corner and folding up the flaps.

To Find: The side of the square to be cut off so that the volume of the box is maximum.

Solution: Let the side of the square cut from each corner be

x

cm. The original dimensions of the sheet are length = 45 cm and width = 24 cm. After cutting the squares, the remaining sheet is folded to form an open box. The dimensions of the resulting box will be: Length,

l = 45 - 2x

cm Width,

w = 24 - 2x

cm Height,

h = x

cm

The volume

V

of the box is given by

V = l \times w \times h

.

V(x) = (45 - 2x)(24 - 2x)x

For the dimensions of the box to be positive, we must have

x > 0

,

45 - 2x > 0

, and

24 - 2x > 0

.

45 - 2x > 0 \implies 45 > 2x \implies x < 22.5

.

24 - 2x > 0 \implies 24 > 2x \implies x < 12

. For all dimensions to be positive, we must satisfy the strictest condition, which is

x < 12

. Also,

x > 0

. So, the domain of the function is

0 < x < 12

.

Now, we expand the expression for the volume:

V(x) = (1080 - 90x - 48x + 4x^2)x

V(x) = (4x^2 - 138x + 1080)x

V(x) = 4x^3 - 138x^2 + 1080x

To find the maximum volume, we differentiate

V(x)

with respect to

x

and set the derivative to zero.

V'(x) = \frac{d}{dx}(4x^3 - 138x^2 + 1080x)

V'(x) = 12x^2 - 276x + 1080

Set

V'(x) = 0

to find critical points:

12x^2 - 276x + 1080 = 0

Divide the equation by 12:

x^2 - 23x + 90 = 0

Factor the quadratic equation:

(x - 5)(x - 18) = 0

This gives two possible values for

x

:

x = 5

or

x = 18

. Since the domain for

x

is

(0, 12)

, we discard

x=18

. If

x=18

, the width of the box would be

24 - 2(18) = -12

, which is not possible. Thus, the only critical point in the domain is

x = 5

.

To confirm that this point gives a maximum volume, we use the second derivative test.

V''(x) = \frac{d}{dx}(12x^2 - 276x + 1080)

V''(x) = 24x - 276

Evaluate

V''(x)

at

x = 5

:

V''(5) = 24(5) - 276 = 120 - 276 = -156

Since

V''(5) < 0

, the volume is maximum at

x = 5

.

Final Answer: The side of the square to be cut off should be 5 cm for the volume of the box to be maximum.

Q19EXERCISE 6.3

Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Solution

To Prove: Of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Proof: Let a rectangle with length

l

and breadth

b

be inscribed in a circle of fixed radius

R

. The diagonal of the rectangle is equal to the diameter of the circle.

Using Pythagoras' theorem on a triangle formed by two sides and a diagonal of the rectangle:

l^2 + b^2 = (2R)^2 = 4R^2

l = \sqrt{4R^2 - b^2}

The area of the rectangle is

A = l \times b

. Substituting the expression for

l

:

A = b \sqrt{4R^2 - b^2}

To simplify differentiation, we can work with the square of the area,

Z = A^2

. Maximizing

A

is equivalent to maximizing

Z

since area is non-negative.

Z = A^2 = b^2 (4R^2 - b^2) = 4R^2b^2 - b^4

Now, we differentiate

Z

with respect to

b

:

\frac{dZ}{db} = \frac{d}{db}(4R^2b^2 - b^4) = 8R^2b - 4b^3

For maximum or minimum area, we set

\frac{dZ}{db} = 0

:

8R^2b - 4b^3 = 0

4b(2R^2 - b^2) = 0

Since the breadth

b

cannot be zero, we have:

2R^2 - b^2 = 0

b^2 = 2R^2 \implies b = R\sqrt{2}

Now, we find the corresponding length

l

:

l^2 = 4R^2 - b^2 = 4R^2 - 2R^2 = 2R^2

l = R\sqrt{2}

Since

l = b = R\sqrt{2}

, the rectangle is a square.

To confirm that this corresponds to a maximum area, we use the second derivative test. Differentiating

\frac{dZ}{db}

with respect to

b

:

\frac{d^2Z}{db^2} = \frac{d}{db}(8R^2b - 4b^3) = 8R^2 - 12b^2

Substitute

b^2 = 2R^2

into the second derivative:

\left. \frac{d^2Z}{db^2} \right|_{b^2=2R^2} = 8R^2 - 12(2R^2) = 8R^2 - 24R^2 = -16R^2

Since

R

is a radius,

R^2 > 0

, so

-16R^2 < 0

. This indicates that the area is maximum when

l = b

.

Hence Proved. The rectangle of maximum area inscribed in a circle is a square.

Q20EXERCISE 6.3

Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.

Solution

To Prove: The right circular cylinder of a given surface area and maximum volume is such that its height is equal to the diameter of the base.

Proof: Let

r

be the radius of the base and

h

be the height of a right circular cylinder. Let

S

be the given (fixed) total surface area and

V

be the volume.

Formulas: Total Surface Area,

S = 2\pi r^2 + 2\pi rh

(This is a constant). Volume,

V = \pi r^2 h

.

We need to maximize

V

for a given

S

. First, we express

V

as a function of a single variable, say

r

. From the surface area formula, we can express

h

in terms of

r

and

S

:

2\pi rh = S - 2\pi r^2

h = \frac{S - 2\pi r^2}{2\pi r} = \frac{S}{2\pi r} - r

Now, substitute this expression for

h

into the volume formula:

V(r) = \pi r^2 \left( \frac{S}{2\pi r} - r \right)

V(r) = \frac{\pi r^2 S}{2\pi r} - \pi r^3

V(r) = \frac{Sr}{2} - \pi r^3

To find the value of

r

for which

V

is maximum, we differentiate

V(r)

with respect to

r

and set the derivative to zero:

\frac{dV}{dr} = \frac{d}{dr}\left( \frac{Sr}{2} - \pi r^3 \right) = \frac{S}{2} - 3\pi r^2

Set

\frac{dV}{dr} = 0

:

\frac{S}{2} - 3\pi r^2 = 0

\frac{S}{2} = 3\pi r^2 \implies S = 6\pi r^2

To check if this corresponds to a maximum volume, we use the second derivative test:

\frac{d^2V}{dr^2} = \frac{d}{dr}\left( \frac{S}{2} - 3\pi r^2 \right) = -6\pi r

Since the radius

r

must be positive,

\frac{d^2V}{dr^2} = -6\pi r < 0

. This confirms that the volume is maximum at this value of

r

.

Now, we establish the relationship between

h

and

r

. We have two expressions for

S

:

$S = 2\pi r^2 + 2\pi rh$ (original formula)
$S = 6\pi r^2$ (condition for maximum volume)

Equating these two expressions:

2\pi r^2 + 2\pi rh = 6\pi r^2

2\pi rh = 6\pi r^2 - 2\pi r^2

2\pi rh = 4\pi r^2

Dividing both sides by

2\pi r

(since

r \neq 0

):

h = 2r

This shows that the height of the cylinder is equal to the diameter of its base (

2r

).

Hence Proved.

Q21EXERCISE 6.3

Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area?

Solution

Given: A closed right circular cylindrical can has a fixed volume

V = 100 \text{ cm}^3

.

To Find: The dimensions (radius

r

and height

h

) of the can that has the minimum surface area.

Let: Let

r

be the radius and

h

be the height of the cylinder.

Formulas: Volume,

V = \pi r^2 h = 100

. Surface Area,

S = 2\pi r^2 + 2\pi rh

.

Solution: We need to minimize

S

. First, we express

S

as a function of a single variable,

r

. From the volume formula:

h = \frac{100}{\pi r^2}

Substitute this expression for

h

into the surface area formula:

S(r) = 2\pi r^2 + 2\pi r \left( \frac{100}{\pi r^2} \right)

S(r) = 2\pi r^2 + \frac{200}{r}

To find the value of

r

that minimizes

S

, we differentiate

S(r)

with respect to

r

and set the derivative to zero:

\frac{dS}{dr} = \frac{d}{dr}\left( 2\pi r^2 + 200r^{-1} \right) = 4\pi r - 200r^{-2} = 4\pi r - \frac{200}{r^2}

Set

\frac{dS}{dr} = 0

:

4\pi r - \frac{200}{r^2} = 0

4\pi r = \frac{200}{r^2}

4\pi r^3 = 200

r^3 = \frac{200}{4\pi} = \frac{50}{\pi}

r = \left( \frac{50}{\pi} \right)^{1/3}

To confirm that this value of

r

corresponds to a minimum surface area, we use the second derivative test:

\frac{d^2S}{dr^2} = \frac{d}{dr}\left( 4\pi r - 200r^{-2} \right) = 4\pi - 200(-2)r^{-3} = 4\pi + \frac{400}{r^3}

Since

r > 0

,

r^3 > 0

, and thus

\frac{d^2S}{dr^2} = 4\pi + \frac{400}{r^3}

is always positive. This confirms that the surface area is minimum at this value of

r

.

Now, we find the dimensions of the can. The radius is:

r = \left( \frac{50}{\pi} \right)^{1/3} \text{ cm}

The height is:

h = \frac{100}{\pi r^2} = \frac{100}{\pi \left( \left( \frac{50}{\pi} \right)^{1/3} \right)^2} = \frac{100}{\pi \left( \frac{50}{\pi} \right)^{2/3}}

h = \frac{100}{\pi^{1 - 2/3} \cdot 50^{2/3}} = \frac{100}{\pi^{1/3} \cdot 50^{2/3}} = \frac{2 \times 50}{\pi^{1/3} \cdot 50^{2/3}} = \frac{2 \cdot 50^{1 - 2/3}}{\pi^{1/3}}

h = \frac{2 \cdot 50^{1/3}}{\pi^{1/3}} = 2 \left( \frac{50}{\pi} \right)^{1/3} \text{ cm}

Final Answer: The dimensions of the can with minimum surface area are: Radius

r = \left( \frac{50}{\pi} \right)^{1/3} \text{ cm}

. Height

h = 2 \left( \frac{50}{\pi} \right)^{1/3} \text{ cm}

.

Q22EXERCISE 6.3

A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

Solution

Given: A wire of total length 28 m is cut into two pieces. One piece is used to make a square, and the other to make a circle.

To Find: The lengths of the two pieces so that the combined area of the square and the circle is minimum.

Let: Let the length of the piece used for the square be

x

meters. Then the length of the piece used for the circle is

(28 - x)

meters. The domain for

x

is

0 \leq x \leq 28

.

Square: Perimeter of the square =

x

. Side of the square,

s = \frac{x}{4}

. Area of the square,

A_s = s^2 = \left( \frac{x}{4} \right)^2 = \frac{x^2}{16}

.

Circle: Circumference of the circle =

28 - x

. Let the radius of the circle be

r

. Then

2\pi r = 28 - x

. Radius,

r = \frac{28 - x}{2\pi}

. Area of the circle,

A_c = \pi r^2 = \pi \left( \frac{28 - x}{2\pi} \right)^2 = \frac{\pi (28 - x)^2}{4\pi^2} = \frac{(28 - x)^2}{4\pi}

.

Solution: The combined area,

A(x)

, is the sum of the areas of the square and the circle:

A(x) = A_s + A_c = \frac{x^2}{16} + \frac{(28 - x)^2}{4\pi}

To find the value of

x

that minimizes

A(x)

, we differentiate

A(x)

with respect to

x

:

\frac{dA}{dx} = \frac{d}{dx} \left( \frac{x^2}{16} + \frac{(28 - x)^2}{4\pi} \right)

\frac{dA}{dx} = \frac{2x}{16} + \frac{2(28 - x)(-1)}{4\pi} = \frac{x}{8} - \frac{28 - x}{2\pi}

Set

\frac{dA}{dx} = 0

to find critical points:

\frac{x}{8} - \frac{28 - x}{2\pi} = 0

\frac{x}{8} = \frac{28 - x}{2\pi}

2\pi x = 8(28 - x)

\pi x = 4(28 - x) = 112 - 4x

\pi x + 4x = 112

x(\pi + 4) = 112

x = \frac{112}{\pi + 4}

To confirm this gives a minimum, we find the second derivative:

\frac{d^2A}{dx^2} = \frac{d}{dx} \left( \frac{x}{8} - \frac{14}{\pi} + \frac{x}{2\pi} \right) = \frac{1}{8} + \frac{1}{2\pi}

Since

\frac{d^2A}{dx^2} > 0

, the area

A(x)

is minimum at

x = \frac{112}{\pi + 4}

.

Now we find the lengths of the two pieces. Length of the piece for the square:

x = \frac{112}{\pi + 4} \text{ m}

Length of the piece for the circle:

28 - x = 28 - \frac{112}{\pi + 4} = \frac{28(\pi + 4) - 112}{\pi + 4}

= \frac{28\pi + 112 - 112}{\pi + 4} = \frac{28\pi}{\pi + 4} \text{ m}

Final Answer: For the combined area to be minimum, the lengths of the two pieces should be: Length for the square =

\frac{112}{\pi + 4}

m. Length for the circle =

\frac{28\pi}{\pi + 4}

m.

Q23EXERCISE 6.3

Prove that the volume of the largest cone that can be inscribed in a sphere of radius $R$ is $\frac{8}{27}$ of the volume of the sphere.

Solution

To Prove: The volume of the largest cone that can be inscribed in a sphere of radius

R

is

\frac{8}{27}

of the volume of the sphere.

Proof: Let a cone be inscribed in a sphere of fixed radius

R

. Let the center of the sphere be

O

. Let the radius of the cone's base be

r

and its height be

h

.

Let the vertex of the cone be

A

and the center of its base be

C

. Let

V

be the volume of the cone. The height of the cone is

h = AC

. The line segment

AC

lies along a diameter of the sphere. The center of the sphere

O

lies on the height

AC

. The distance from the center of the sphere to the base of the cone is

OC

. We have

h = AO + OC = R + OC

. So,

OC = h - R

.

Consider the right-angled triangle

ODC

, where

D

is a point on the circumference of the cone's base.

OD

is the radius of the sphere, so

OD = R

.

CD

is the radius of the cone, so

CD = r

.

OC = h-R

.

By Pythagoras' theorem in

\triangle ODC

:

OD^2 = OC^2 + CD^2

R^2 = (h-R)^2 + r^2

R^2 = h^2 - 2hR + R^2 + r^2

0 = h^2 - 2hR + r^2

r^2 = 2hR - h^2

The volume of the cone is given by

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

. We substitute the expression for

r^2

to get

V

as a function of

h

:

V(h) = \frac{1}{3} \pi (2hR - h^2)h

V(h) = \frac{1}{3} \pi (2Rh^2 - h^3)

(Note: The height

h

of the cone can range from

0

to

2R

).

To find the height that maximizes the volume, we differentiate

V(h)

with respect to

h

and set the derivative to zero:

\frac{dV}{dh} = \frac{1}{3} \pi \frac{d}{dh}(2Rh^2 - h^3)

\frac{dV}{dh} = \frac{1}{3} \pi (4Rh - 3h^2)

Set

\frac{dV}{dh} = 0

:

\frac{1}{3} \pi (4Rh - 3h^2) = 0

h(4R - 3h) = 0

Since

h \neq 0

for a cone, we have:

4R - 3h = 0 \implies h = \frac{4R}{3}

To confirm this is a maximum, we use the second derivative test:

\frac{d^2V}{dh^2} = \frac{1}{3} \pi \frac{d}{dh}(4Rh - 3h^2) = \frac{1}{3} \pi (4R - 6h)

Substitute

h = \frac{4R}{3}

:

\left. \frac{d^2V}{dh^2} \right|_{h=4R/3} = \frac{1}{3} \pi \left(4R - 6\left(\frac{4R}{3}\right)\right) = \frac{1}{3} \pi (4R - 8R) = -\frac{4\pi R}{3}

Since

R > 0

, the second derivative is negative, which confirms the volume is maximum at

h = \frac{4R}{3}

.

Now, we calculate the maximum volume of the cone using

h = \frac{4R}{3}

: First, find the corresponding

r^2

:

r^2 = 2hR - h^2 = 2\left(\frac{4R}{3}\right)R - \left(\frac{4R}{3}\right)^2 = \frac{8R^2}{3} - \frac{16R^2}{9} = \frac{24R^2 - 16R^2}{9} = \frac{8R^2}{9}

Maximum volume of the cone,

V_{cone}

:

V_{cone} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \left(\frac{8R^2}{9}\right) \left(\frac{4R}{3}\right) = \frac{32\pi R^3}{81}

The volume of the sphere is

V_{sphere} = \frac{4}{3} \pi R^3

.

Now, we find the ratio of the cone's volume to the sphere's volume:

\frac{V_{cone}}{V_{sphere}} = \frac{\frac{32\pi R^3}{81}}{\frac{4}{3} \pi R^3} = \frac{32}{81} \times \frac{3}{4} = \frac{8}{27}

V_{cone} = \frac{8}{27} V_{sphere}

Hence Proved. The volume of the largest cone that can be inscribed in a sphere of radius

R

is

\frac{8}{27}

of the volume of the sphere.

Q24EXERCISE 6.3

Show that the right circular cone of least curved surface and given volume has an altitude equal to $\sqrt{2}$ time the radius of the base.

Solution

To Prove: For a right circular cone of least curved surface area and given volume, the altitude is equal to

\sqrt{2}

times the radius of the base.

Let:

$r$ be the radius of the base of the cone.
$h$ be the altitude (height) of the cone.
$l$ be the slant height of the cone.
$V$ be the volume of the cone.
$S$ be the curved surface area of the cone.

Given: The volume

V

of the cone is constant.

Formulas: Volume of a cone,

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

. Since

V

is constant, we can write

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

. Curved surface area of a cone,

S = \pi r l

. Slant height,

l = \sqrt{r^2 + h^2}

.

Solution: We need to minimize the curved surface area

S

. Substituting the expressions for

l

and

h

into the formula for

S

:

S = \pi r \sqrt{r^2 + h^2} = \pi r \sqrt{r^2 + \left(\frac{3V}{\pi r^2}\right)^2}

S = \pi r \sqrt{r^2 + \frac{9V^2}{\pi^2 r^4}} = \sqrt{(\pi r)^2 \left(r^2 + \frac{9V^2}{\pi^2 r^4}\right)} = \sqrt{\pi^2 r^4 + \frac{9V^2}{r^2}}

Minimizing

S

is equivalent to minimizing

S^2

. Let

Z = S^2

.

Z = \pi^2 r^4 + \frac{9V^2}{r^2}

To find the value of

r

for which

S

is minimum, we differentiate

Z

with respect to

r

and set the derivative to zero.

\frac{dZ}{dr} = \frac{d}{dr} \left(\pi^2 r^4 + 9V^2 r^{-2}\right)

\frac{dZ}{dr} = 4\pi^2 r^3 - 18V^2 r^{-3} = 4\pi^2 r^3 - \frac{18V^2}{r^3}

For a minimum, we set

\frac{dZ}{dr} = 0

:

4\pi^2 r^3 - \frac{18V^2}{r^3} = 0

4\pi^2 r^3 = \frac{18V^2}{r^3}

4\pi^2 r^6 = 18V^2

r^6 = \frac{18V^2}{4\pi^2} = \frac{9V^2}{2\pi^2}

Now, substitute the formula for volume

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

back into this equation:

r^6 = \frac{9}{2\pi^2} \left(\frac{1}{3}\pi r^2 h\right)^2

r^6 = \frac{9}{2\pi^2} \left(\frac{1}{9}\pi^2 r^4 h^2\right)

r^6 = \frac{1}{2} r^4 h^2

Since

r \neq 0

, we can divide both sides by

r^4

:

r^2 = \frac{1}{2} h^2

h^2 = 2r^2

h = \sqrt{2} r \quad (\text{since } h \text{ and } r \text{ are positive})

Now, we use the second derivative test to confirm that this condition corresponds to a minimum curved surface area.

\frac{d^2Z}{dr^2} = \frac{d}{dr} \left(4\pi^2 r^3 - \frac{18V^2}{r^3}\right) = 12\pi^2 r^2 + \frac{54V^2}{r^4}

Since

r > 0

and

V > 0

, both terms are positive. Thus,

\frac{d^2Z}{dr^2} > 0

. This confirms that

Z

(and therefore

S

) is minimum when

h = \sqrt{2} r

.

Hence Proved.

Q25EXERCISE 6.3

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is $\tan ^{-1} \sqrt{2}$ .

Solution

To Prove: The semi-vertical angle of the cone of the maximum volume and of given slant height is

\tan^{-1} \sqrt{2}

.

Let:

$r$ be the radius of the base of the cone.
$h$ be the altitude (height) of the cone.
$l$ be the slant height of the cone.
$V$ be the volume of the cone.
$\alpha$ be the semi-vertical angle of the cone.

Given: The slant height

l

of the cone is constant.

Formulas: Volume of a cone,

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

. In terms of the semi-vertical angle

\alpha

and slant height

l

, we have:

r = l \sin \alpha

h = l \cos \alpha

Solution: We need to maximize the volume

V

. Substituting the expressions for

r

and

h

into the volume formula:

V = \frac{1}{3}\pi (l \sin \alpha)^2 (l \cos \alpha)

V(\alpha) = \frac{1}{3}\pi l^3 \sin^2 \alpha \cos \alpha

Since

l

is a constant, to maximize

V

, we need to maximize the function

f(\alpha) = \sin^2 \alpha \cos \alpha

. To find the value of

\alpha

for which

V

is maximum, we differentiate

V

with respect to

\alpha

and set the derivative to zero.

\frac{dV}{d\alpha} = \frac{d}{d\alpha} \left( \frac{1}{3}\pi l^3 \sin^2 \alpha \cos \alpha \right)

\frac{dV}{d\alpha} = \frac{1}{3}\pi l^3 \left[ (2\sin\alpha\cos\alpha)(\cos\alpha) + (\sin^2\alpha)(-\sin\alpha) \right]

\frac{dV}{d\alpha} = \frac{1}{3}\pi l^3 \left[ 2\sin\alpha\cos^2\alpha - \sin^3\alpha \right]

\frac{dV}{d\alpha} = \frac{1}{3}\pi l^3 \sin\alpha (2\cos^2\alpha - \sin^2\alpha)

For a maximum, we set

\frac{dV}{d\alpha} = 0

:

\frac{1}{3}\pi l^3 \sin\alpha (2\cos^2\alpha - \sin^2\alpha) = 0

Since for a cone,

0 < \alpha < \frac{\pi}{2}

, we have

\sin\alpha \neq 0

. Therefore:

2\cos^2\alpha - \sin^2\alpha = 0

2\cos^2\alpha = \sin^2\alpha

\frac{\sin^2\alpha}{\cos^2\alpha} = 2

\tan^2\alpha = 2

\tan\alpha = \sqrt{2} \quad (\text{since } \alpha \text{ is acute})

\alpha = \tan^{-1} \sqrt{2}

Now, we use the second derivative test to confirm that this is a maximum. Let

K = \sin\alpha (2\cos^2\alpha - \sin^2\alpha) = 2\sin\alpha\cos^2\alpha - \sin^3\alpha

.

\frac{d^2V}{d\alpha^2} = \frac{1}{3}\pi l^3 \frac{dK}{d\alpha}

\frac{dK}{d\alpha} = (2\cos\alpha)(\cos^2\alpha) + (2\sin\alpha)(-2\cos\alpha\sin\alpha) - 3\sin^2\alpha\cos\alpha

= 2\cos^3\alpha - 4\sin^2\alpha\cos\alpha - 3\sin^2\alpha\cos\alpha = 2\cos^3\alpha - 7\sin^2\alpha\cos\alpha

= \cos\alpha (2\cos^2\alpha - 7\sin^2\alpha)

At the critical point where

\tan^2\alpha = 2

, we have

\sin^2\alpha = 2\cos^2\alpha

. Substituting this:

\frac{dK}{d\alpha} = \cos\alpha (2\cos^2\alpha - 7(2\cos^2\alpha)) = \cos\alpha (2\cos^2\alpha - 14\cos^2\alpha) = -12\cos^3\alpha

Since

0 < \alpha < \frac{\pi}{2}

,

\cos\alpha > 0

. Thus,

\frac{dK}{d\alpha} < 0

, which implies

\frac{d^2V}{d\alpha^2} < 0

. This confirms that the volume is maximum when

\alpha = \tan^{-1} \sqrt{2}

.

Hence Proved.

Q26EXERCISE 6.3

Show that semi-vertical angle of right circular cone of given surface area and maximum volume is $\sin ^{-1} \frac{1}{3}$ .

Solution

To Prove: The semi-vertical angle of a right circular cone of given surface area and maximum volume is

\sin^{-1} \frac{1}{3}

.

Let:

$r$ be the radius of the base of the cone.
$h$ be the altitude (height) of the cone.
$l$ be the slant height of the cone.
$S$ be the total surface area of the cone.
$V$ be the volume of the cone.
$\alpha$ be the semi-vertical angle of the cone.

Given: The total surface area

S

of the cone is constant.

Formulas: Total Surface Area,

S = \pi r^2 + \pi r l

. Since

S

is constant, we can write

l = \frac{S - \pi r^2}{\pi r}

. Volume of a cone,

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

. Slant height,

l^2 = r^2 + h^2

, so

h = \sqrt{l^2 - r^2}

.

Solution: We need to maximize the volume

V

. First, express

V

in terms of a single variable, say

r

.

h = \sqrt{\left(\frac{S - \pi r^2}{\pi r}\right)^2 - r^2} = \sqrt{\frac{(S - \pi r^2)^2 - (\pi r^2)^2}{\pi^2 r^2}}

h = \frac{1}{\pi r} \sqrt{S^2 - 2S\pi r^2 + \pi^2 r^4 - \pi^2 r^4} = \frac{1}{\pi r} \sqrt{S^2 - 2S\pi r^2}

Now, substitute this into the volume formula:

V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 \left( \frac{1}{\pi r} \sqrt{S^2 - 2S\pi r^2} \right) = \frac{r}{3} \sqrt{S^2 - 2S\pi r^2}

Maximizing

V

is equivalent to maximizing

V^2

. Let

Z = V^2

.

Z = \frac{r^2}{9} (S^2 - 2S\pi r^2) = \frac{S^2}{9} r^2 - \frac{2S\pi}{9} r^4

To find the value of

r

for which

V

is maximum, we differentiate

Z

with respect to

r

and set the derivative to zero.

\frac{dZ}{dr} = \frac{d}{dr} \left( \frac{S^2}{9} r^2 - \frac{2S\pi}{9} r^4 \right)

\frac{dZ}{dr} = \frac{2S^2}{9} r - \frac{8S\pi}{9} r^3

For a maximum, we set

\frac{dZ}{dr} = 0

:

\frac{2S^2}{9} r - \frac{8S\pi}{9} r^3 = 0

\frac{2Sr}{9} (S - 4\pi r^2) = 0

Since

S \neq 0

and

r \neq 0

, we have:

S - 4\pi r^2 = 0

S = 4\pi r^2

Now, we substitute this back into the formula for

S = \pi r^2 + \pi r l

:

4\pi r^2 = \pi r^2 + \pi r l

3\pi r^2 = \pi r l

3r = l \quad (\text{since } r \neq 0)

The semi-vertical angle

\alpha

is given by

\sin\alpha = \frac{r}{l}

. Substituting

l=3r

:

\sin\alpha = \frac{r}{3r} = \frac{1}{3}

\alpha = \sin^{-1} \frac{1}{3}

To confirm this is a maximum, we check the second derivative:

\frac{d^2Z}{dr^2} = \frac{d}{dr} \left( \frac{2S^2}{9} r - \frac{8S\pi}{9} r^3 \right) = \frac{2S^2}{9} - \frac{24S\pi}{9} r^2

Substitute

S = 4\pi r^2

into this expression:

\frac{d^2Z}{dr^2} = \frac{2(4\pi r^2)^2}{9} - \frac{24(4\pi r^2)\pi}{9} r^2 = \frac{32\pi^2 r^4}{9} - \frac{96\pi^2 r^4}{9} = -\frac{64\pi^2 r^4}{9}

Since

r > 0

,

\frac{d^2Z}{dr^2} < 0

. This confirms that the volume is maximum when

\alpha = \sin^{-1} \frac{1}{3}

.

Hence Proved.

Q27EXERCISE 6.3

The point on the curve $x^{2}=2 y$ which is nearest to the point $(0,5)$ is

(A)

(2 \sqrt{2}, 4)

(B)

(2 \sqrt{2}, 0)

(C)

(0,0)

(D)

(2,2)

Solution

Given: The curve is

x^2 = 2y

. The point is

P(0, 5)

.

To Find: The point on the curve

x^2 = 2y

which is nearest to the point

(0, 5)

.

Let: Let the required point on the curve be

Q(x, y)

. From the equation of the curve, we have

y = \frac{x^2}{2}

. So, the point

Q

can be written as

(x, \frac{x^2}{2})

.

Solution: The distance between the points

P(0, 5)

and

Q(x, \frac{x^2}{2})

is given by the distance formula:

D = \sqrt{(x - 0)^2 + \left(\frac{x^2}{2} - 5\right)^2}

D = \sqrt{x^2 + \frac{x^4}{4} - 5x^2 + 25}

D = \sqrt{\frac{x^4}{4} - 4x^2 + 25}

To find the nearest point, we need to minimize the distance

D

. Minimizing

D

is equivalent to minimizing

D^2

. Let

Z = D^2

.

Z(x) = \frac{x^4}{4} - 4x^2 + 25

To find the value of

x

for which

D

is minimum, we differentiate

Z

with respect to

x

and set the derivative to zero.

\frac{dZ}{dx} = \frac{d}{dx} \left( \frac{x^4}{4} - 4x^2 + 25 \right)

\frac{dZ}{dx} = \frac{4x^3}{4} - 8x = x^3 - 8x

For a minimum, we set

\frac{dZ}{dx} = 0

:

x^3 - 8x = 0

x(x^2 - 8) = 0

This gives three possible values for

x

:

x=0

,

x=\sqrt{8} = 2\sqrt{2}

, and

x = -\sqrt{8} = -2\sqrt{2}

.

Now, we use the second derivative test to determine which value of

x

corresponds to a minimum distance.

\frac{d^2Z}{dx^2} = \frac{d}{dx} (x^3 - 8x) = 3x^2 - 8

At $x=0$ : $\frac{d^2Z}{dx^2} = 3(0)^2 - 8 = -8 < 0$ . This corresponds to a local maximum distance.
At $x=2\sqrt{2}$ : $\frac{d^2Z}{dx^2} = 3(2\sqrt{2})^2 - 8 = 3(8) - 8 = 24 - 8 = 16 > 0$ . This corresponds to a local minimum distance.
At $x=-2\sqrt{2}$ : $\frac{d^2Z}{dx^2} = 3(-2\sqrt{2})^2 - 8 = 3(8) - 8 = 24 - 8 = 16 > 0$ . This also corresponds to a local minimum distance.

So, the distance is minimum when

x = 2\sqrt{2}

or

x = -2\sqrt{2}

. Let's find the corresponding

y

-coordinate for these values of

x

. When

x = \pm 2\sqrt{2}

,

y = \frac{x^2}{2} = \frac{(2\sqrt{2})^2}{2} = \frac{8}{2} = 4

. Thus, the points on the curve nearest to

(0, 5)

are

(2\sqrt{2}, 4)

and

(-2\sqrt{2}, 4)

.

From the given options,

(2\sqrt{2}, 4)

is present.

Final Answer: The correct option is (A)

(2 \sqrt{2}, 4)

.

Q28EXERCISE 6.3

For all real values of $x$ , the minimum value of $\frac{1-x+x^{2}}{1+x+x^{2}}$ is

(A)

0

(B)

1

(C)

3

(D)

\frac{1}{3}

Solution

Given: The function

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

for all real values of

x

.

To Find: The minimum value of the function.

Solution: Let

y = f(x) = \frac{1-x+x^2}{1+x+x^2}

. To find the minimum value, we will find the critical points by taking the derivative of

y

with respect to

x

and setting it to zero. Using the quotient rule,

\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}

: Here,

u = 1-x+x^2

and

v = 1+x+x^2

. So,

u' = -1+2x

and

v' = 1+2x

.

\frac{dy}{dx} = \frac{(-1+2x)(1+x+x^2) - (1-x+x^2)(1+2x)}{(1+x+x^2)^2}

Expanding the numerator: Numerator =

(-1-x-x^2 + 2x+2x^2+2x^3) - (1+2x-x-2x^2+x^2+2x^3)

=

(-1+x+x^2+2x^3) - (1+x-x^2+2x^3)

=

-1+x+x^2+2x^3 - 1-x+x^2-2x^3

=

-2 + 2x^2 = 2(x^2 - 1)

So,

\frac{dy}{dx} = \frac{2(x^2-1)}{(1+x+x^2)^2}

For critical points, we set

\frac{dy}{dx} = 0

:

\frac{2(x^2-1)}{(1+x+x^2)^2} = 0

2(x^2-1) = 0

x^2 = 1

x = 1 \text{ or } x = -1

Now we evaluate the function at these critical points:

Case 1:

x = 1

y = \frac{1-1+1^2}{1+1+1^2} = \frac{1}{3}

Case 2:

x = -1

y = \frac{1-(-1)+(-1)^2}{1+(-1)+(-1)^2} = \frac{1+1+1}{1-1+1} = \frac{3}{1} = 3

To determine which is the minimum and which is the maximum, we can analyze the sign of

\frac{dy}{dx}

. The denominator

(1+x+x^2)^2

is always positive. The sign of

\frac{dy}{dx}

is determined by the sign of

2(x^2-1)

.

For $x < -1$ , $x^2 > 1$ , so $\frac{dy}{dx} > 0$ (function is increasing).
For $-1 < x < 1$ , $x^2 < 1$ , so $\frac{dy}{dx} < 0$ (function is decreasing).
For $x > 1$ , $x^2 > 1$ , so $\frac{dy}{dx} > 0$ (function is increasing).

Since the function changes from increasing to decreasing at

x=-1

,

x=-1

is a point of local maximum. The maximum value is

y=3

. Since the function changes from decreasing to increasing at

x=1

,

x=1

is a point of local minimum. The minimum value is

y=\frac{1}{3}

.

We should also check the behavior as

x \to \pm \infty

:

\lim_{x \to \pm \infty} \frac{1-x+x^2}{1+x+x^2} = \lim_{x \to \pm \infty} \frac{x^2(1/x^2 - 1/x + 1)}{x^2(1/x^2 + 1/x + 1)} = \frac{1}{1} = 1

Comparing the values

y=3

(local max),

y=1/3

(local min), and the limit

y=1

, the absolute minimum value of the function is

\frac{1}{3}

.

Final Answer: The correct option is (D)

\frac{1}{3}

.

Q29EXERCISE 6.3

The maximum value of $[x(x-1)+1]^{\frac{1}{3}}, 0 \leq x \leq 1$ is

(A)

\frac{1}{3}^{\frac{1}{3}}

(B)

\frac{1}{2}

(C)

1

(D)

0

Solution

Given: The function is

f(x) = [x(x-1)+1]^{\frac{1}{3}}

for

0 \leq x \leq 1

.

To Find: The maximum value of the function in the given interval.

Solution: Let the given function be

f(x) = (x^2 - x + 1)^{\frac{1}{3}}

. To find the maximum value, we first need to find the critical points by finding the first derivative of the function and setting it to zero.

Differentiating

f(x)

with respect to

x

using the chain rule:

f'(x) = \frac{d}{dx} (x^2 - x + 1)^{\frac{1}{3}}

f'(x) = \frac{1}{3}(x^2 - x + 1)^{\frac{1}{3}-1} \cdot \frac{d}{dx}(x^2 - x + 1)

f'(x) = \frac{1}{3}(x^2 - x + 1)^{-\frac{2}{3}} \cdot (2x - 1)

f'(x) = \frac{2x - 1}{3(x^2 - x + 1)^{\frac{2}{3}}}

Now, set

f'(x) = 0

to find the critical points:

\frac{2x - 1}{3(x^2 - x + 1)^{\frac{2}{3}}} = 0

This implies

2x - 1 = 0

, which gives

x = \frac{1}{2}

.

The critical point

x = \frac{1}{2}

lies within the given interval

[0, 1]

.

To find the absolute maximum value, we evaluate the function at the critical point and the endpoints of the interval

[0, 1]

.

At $x = 0$ : $f(0) = [0(0-1)+1]^{\frac{1}{3}} = (1)^{\frac{1}{3}} = 1$ .
At $x = 1$ : $f(1) = [1(1-1)+1]^{\frac{1}{3}} = (1)^{\frac{1}{3}} = 1$ .
At $x = \frac{1}{2}$ : $f\left(\frac{1}{2}\right) = \left[\frac{1}{2}\left(\frac{1}{2}-1\right)+1\right]^{\frac{1}{3}} = \left[\frac{1}{2}\left(-\frac{1}{2}\right)+1\right]^{\frac{1}{3}} = \left[-\frac{1}{4}+1\right]^{\frac{1}{3}} = \left(\frac{3}{4}\right)^{\frac{1}{3}}$ .

Now, we compare the values:

1

,

1

, and

\left(\frac{3}{4}\right)^{\frac{1}{3}}

. Since

\frac{3}{4} < 1

, we have

\left(\frac{3}{4}\right)^{\frac{1}{3}} < (1)^{\frac{1}{3}} = 1

.

The maximum value among these is 1.

Final Answer: The maximum value of the function is 1. This corresponds to option (C).

Q1Miscellaneous Exercise on Chapter 6

Show that the function given by $f(x)=\frac{\log x}{x}$ has maximum at $x=e$ .

Solution

Given: The function

f(x) = \frac{\log x}{x}

.

To Prove: The function has a maximum at

x=e

.

Proof: The given function is

f(x) = \frac{\log x}{x}

. The domain of the function is

x > 0

.

To find the points of local maxima or minima, we use the second derivative test.

First, we find the first derivative of

f(x)

using the quotient rule,

\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}

:

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

f'(x) = \frac{\frac{1}{x} \cdot x - \log x \cdot 1}{x^2}

f'(x) = \frac{1 - \log x}{x^2}

To find the critical points, we set

f'(x) = 0

:

\frac{1 - \log x}{x^2} = 0

Since

x>0

,

x^2 \neq 0

. Thus, we must have:

1 - \log x = 0

\log x = 1

Taking the base of the logarithm as

e

, we get:

x = e^1 = e

So,

x=e

is the only critical point.

Now, we find the second derivative,

f''(x)

, to determine if this point is a maximum.

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

Using the quotient rule again:

f''(x) = \frac{\frac{d}{dx}(1 - \log x) \cdot x^2 - (1 - \log x) \cdot \frac{d}{dx}(x^2)}{(x^2)^2}

f''(x) = \frac{(-\frac{1}{x}) \cdot x^2 - (1 - \log x) \cdot (2x)}{x^4}

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

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

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

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

f''(x) = \frac{2 \log x - 3}{x^3}

Now, we evaluate the second derivative at the critical point

x=e

:

f''(e) = \frac{2 \log e - 3}{e^3}

Since

\log e = 1

:

f''(e) = \frac{2(1) - 3}{e^3} = \frac{-1}{e^3}

Since

e > 0

,

e^3 > 0

, which means

f''(e) = -\frac{1}{e^3} < 0

.

According to the second derivative test, if

f'(c) = 0

and

f''(c) < 0

, then the function

f(x)

has a local maximum at

x=c

. Here,

f'(e) = 0

and

f''(e) < 0

. Therefore, the function

f(x)

has a maximum value at

x=e

.

Hence Proved.

Q2Miscellaneous Exercise on Chapter 6

The two equal sides of an isosceles triangle with fixed base $b$ are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base ?

Solution

Given: An isosceles triangle with a fixed base

b

. Let the length of the two equal sides be

a

. The equal sides are decreasing at a rate of 3 cm per second, so

\frac{da}{dt} = -3 \text{ cm/s}

.

To Find: How fast the area of the triangle is decreasing when the two equal sides are equal to the base, i.e., find

\frac{dA}{dt}

when

a=b

.

Solution: Let the isosceles triangle be ABC with base BC =

b

and equal sides AB = AC =

a

. Let AD be the altitude from vertex A to the base BC. In an isosceles triangle, the altitude to the base bisects the base. So, BD = DC =

\frac{b}{2}

.

The altitude

h = AD

can be found using the Pythagorean theorem in

\triangle ADC

:

a^2 = h^2 + \left(\frac{b}{2}\right)^2

h^2 = a^2 - \frac{b^2}{4}

h = \sqrt{a^2 - \frac{b^2}{4}}

The area of the triangle,

A

, is given by:

A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} b h

Substituting the expression for

h

:

A = \frac{1}{2} b \sqrt{a^2 - \frac{b^2}{4}}

Now, we differentiate the area

A

with respect to time

t

. Note that the base

b

is fixed (constant).

\frac{dA}{dt} = \frac{d}{dt} \left( \frac{b}{2} \sqrt{a^2 - \frac{b^2}{4}} \right)

Using the chain rule:

\frac{dA}{dt} = \frac{b}{2} \cdot \frac{1}{2\sqrt{a^2 - \frac{b^2}{4}}} \cdot \frac{d}{dt}\left(a^2 - \frac{b^2}{4}\right)

\frac{dA}{dt} = \frac{b}{4\sqrt{a^2 - \frac{b^2}{4}}} \cdot \left(2a \frac{da}{dt} - 0\right)

\frac{dA}{dt} = \frac{2ab}{4\sqrt{a^2 - \frac{b^2}{4}}} \frac{da}{dt} = \frac{ab}{2\sqrt{a^2 - \frac{b^2}{4}}} \frac{da}{dt}

We are asked to find

\frac{dA}{dt}

when

a=b

. We are given

\frac{da}{dt} = -3 \text{ cm/s}

. Substituting these values into the expression for

\frac{dA}{dt}

:

\frac{dA}{dt} = \frac{b \cdot b}{2\sqrt{b^2 - \frac{b^2}{4}}} \cdot (-3)

\frac{dA}{dt} = \frac{-3b^2}{2\sqrt{\frac{3b^2}{4}}}

\frac{dA}{dt} = \frac{-3b^2}{2 \cdot \frac{\sqrt{3}b}{2}}

\frac{dA}{dt} = \frac{-3b^2}{\sqrt{3}b}

\frac{dA}{dt} = -\frac{3}{\sqrt{3}} b = -\sqrt{3} b

The negative sign indicates that the area is decreasing.

Final Answer: The area is decreasing at a rate of

\sqrt{3}b \text{ cm}^2/\text{s}

.

Q3Miscellaneous Exercise on Chapter 6

Find the intervals in which the function $f$ given by $f(x)=\frac{4 \sin x-2 x-x \cos x}{2+\cos x}$ is (i) increasing (ii) decreasing.

Solution

Given: The function

f(x)=\frac{4 \sin x-2 x-x \cos x}{2+\cos x}

.

To Find: The intervals in which the function is (i) increasing and (ii) decreasing.

Solution: The given function can be simplified as follows:

f(x) = \frac{4 \sin x - x(2 + \cos x)}{2 + \cos x}

f(x) = \frac{4 \sin x}{2 + \cos x} - \frac{x(2 + \cos x)}{2 + \cos x}

f(x) = \frac{4 \sin x}{2 + \cos x} - x

To find the intervals of increase or decrease, we need to find the sign of the first derivative,

f'(x)

. Differentiating

f(x)

with respect to

x

using the quotient rule for the first term:

f'(x) = \frac{d}{dx} \left( \frac{4 \sin x}{2 + \cos x} \right) - \frac{d}{dx}(x)

f'(x) = \frac{(2+\cos x) \frac{d}{dx}(4 \sin x) - (4 \sin x) \frac{d}{dx}(2+\cos x)}{(2+\cos x)^2} - 1

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

f'(x) = \frac{8\cos x + 4\cos^2 x + 4\sin^2 x}{(2+\cos x)^2} - 1

Using the identity

\sin^2 x + \cos^2 x = 1

:

f'(x) = \frac{8\cos x + 4(\cos^2 x + \sin^2 x)}{(2+\cos x)^2} - 1

f'(x) = \frac{8\cos x + 4}{(2+\cos x)^2} - 1

Now, combine the terms:

f'(x) = \frac{8\cos x + 4 - (2+\cos x)^2}{(2+\cos x)^2}

f'(x) = \frac{8\cos x + 4 - (4 + 4\cos x + \cos^2 x)}{(2+\cos x)^2}

f'(x) = \frac{8\cos x + 4 - 4 - 4\cos x - \cos^2 x}{(2+\cos x)^2}

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

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

To determine the sign of

f'(x)

, we analyze each factor:

The denominator $(2+\cos x)^2$ : Since $-1 \leq \cos x \leq 1$ , we have $1 \leq 2+\cos x \leq 3$ . Thus, $(2+\cos x)^2$ is always positive.
The factor $(4 - \cos x)$ : Since $-1 \leq \cos x \leq 1$ , we have $3 \leq 4-\cos x \leq 5$ . Thus, $(4 - \cos x)$ is always positive.

Therefore, the sign of

f'(x)

is the same as the sign of

\cos x

.

Although the domain is not specified, trigonometric function problems are often analyzed over an interval of length

2\pi

, such as

[0, 2\pi]

.

(i) Increasing function: The function

f(x)

is increasing when

f'(x) > 0

. This implies

\cos x > 0

. In the interval

[0, 2\pi]

,

\cos x > 0

when

x \in (0, \frac{\pi}{2}) \cup (\frac{3\pi}{2}, 2\pi)

. So,

f(x)

is increasing in the intervals

(0, \frac{\pi}{2})

and

(\frac{3\pi}{2}, 2\pi)

.

(ii) Decreasing function: The function

f(x)

is decreasing when

f'(x) < 0

. This implies

\cos x < 0

. In the interval

[0, 2\pi]

,

\cos x < 0

when

x \in (\frac{\pi}{2}, \frac{3\pi}{2})

. So,

f(x)

is decreasing in the interval

(\frac{\pi}{2}, \frac{3\pi}{2})

.

Final Answer:

(i)

The function

f(x)

is increasing in the intervals

(0, \frac{\pi}{2}) \cup (\frac{3\pi}{2}, 2\pi)

.

(ii)

The function

f(x)

is decreasing in the interval

(\frac{\pi}{2}, \frac{3\pi}{2})

.

Q4Miscellaneous Exercise on Chapter 6

Find the intervals in which the function $f$ given by $f(x)=x^{3}+\frac{1}{x^{3}}, x \neq 0$ is

(i)

increasing

(ii)

decreasing.

Solution

Given: The function

f(x) = x^3 + \frac{1}{x^3}

, for

x \neq 0

.

To Find: The intervals in which the function is (i) increasing and (ii) decreasing.

Solution: The given function is

f(x) = x^3 + x^{-3}

. To find the intervals of increase and decrease, we need to find the sign of the first derivative,

f'(x)

.

Differentiating

f(x)

with respect to

x

:

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

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

f'(x) = 3x^2 - \frac{3}{x^4}

f'(x) = 3 \left( x^2 - \frac{1}{x^4} \right)

f'(x) = 3 \left( \frac{x^6 - 1}{x^4} \right)

To find the critical points, we set

f'(x)=0

:

3 \left( \frac{x^6 - 1}{x^4} \right) = 0

This implies

x^6 - 1 = 0

, since

x \neq 0

.

x^6 = 1

This gives

x = 1

and

x = -1

.

The domain of the function is

(-\infty, 0) \cup (0, \infty)

. The critical points

x=-1

and

x=1

divide the domain into four intervals:

(-\infty, -1)

,

(-1, 0)

,

(0, 1)

, and

(1, \infty)

.

Now, we check the sign of

f'(x) = \frac{3(x^6 - 1)}{x^4}

in each interval. Since

x^4

is always positive for

x \neq 0

, the sign of

f'(x)

is determined by the sign of

(x^6 - 1)

.

Interval $(-\infty, -1)$ : Let's take a test point $x = -2$ . $f'(-2) = \frac{3((-2)^6 - 1)}{(-2)^4} = \frac{3(64 - 1)}{16} = \frac{3 \times 63}{16} > 0$ . So, $f(x)$ is increasing in $(-\infty, -1)$ .
Interval $(-1, 0)$ : Let's take a test point $x = -0.5$ . $x^6 = (-0.5)^6 = 0.015625 < 1$ . So, $x^6 - 1 < 0$ . $f'(-0.5) < 0$ . So, $f(x)$ is decreasing in $(-1, 0)$ .
Interval $(0, 1)$ : Let's take a test point $x = 0.5$ . $x^6 = (0.5)^6 = 0.015625 < 1$ . So, $x^6 - 1 < 0$ . $f'(0.5) < 0$ . So, $f(x)$ is decreasing in $(0, 1)$ .
Interval $(1, \infty)$ : Let's take a test point $x = 2$ . $f'(2) = \frac{3(2^6 - 1)}{2^4} = \frac{3(64 - 1)}{16} = \frac{3 \times 63}{16} > 0$ . So, $f(x)$ is increasing in $(1, \infty)$ .

(i) Increasing function: The function is increasing when

f'(x) > 0

. This occurs in the intervals

(-\infty, -1)

and

(1, \infty)

.

(ii) Decreasing function: The function is decreasing when

f'(x) < 0

. This occurs in the intervals

(-1, 0)

and

(0, 1)

. We can combine these as

(-1, 0) \cup (0, 1)

or

(-1, 1) - \{0\}

.

Final Answer:

(i)

The function

f(x)

is increasing in the intervals

(-\infty, -1)

and

(1, \infty)

.

(ii)

The function

f(x)

is decreasing in the intervals

(-1, 0)

and

(0, 1)

.

Q5Miscellaneous Exercise on Chapter 6

Find the maximum area of an isosceles triangle inscribed in the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ with its vertex at one end of the major axis.

Solution

Given: An isosceles triangle inscribed in the ellipse

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

, with its vertex at one end of the major axis.

To Find: The maximum area of the triangle.

Solution: Let the equation of the ellipse be

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

. The major axis is along the x-axis, with vertices at

(a, 0)

and

(-a, 0)

.

Let the vertex of the isosceles triangle be

A(a, 0)

. Since the triangle is isosceles, its base must be a vertical chord of the ellipse. Let the other two vertices be

P(x, y)

and

Q(x, -y)

, where

-a < x < a

.

The base of the triangle is

PQ = 2y

. The height of the triangle is the perpendicular distance from vertex

A(a, 0)

to the base

PQ

(the line

X=x

). Height

h = a - x

.

The area of

\triangle APQ

, denoted by

A_{area}

, is:

A_{area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2y) \times (a - x) = y(a - x)

Since the point

P(x, y)

lies on the ellipse, we have:

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

y^2 = b^2 \left(1 - \frac{x^2}{a^2}\right) = \frac{b^2}{a^2}(a^2 - x^2)

y = \frac{b}{a} \sqrt{a^2 - x^2} \quad (\text{since y > 0})

Substituting this into the area formula:

A_{area}(x) = \frac{b}{a} \sqrt{a^2 - x^2} (a - x)

To find the maximum area, we can maximize the square of the area,

Z = (A_{area})^2

, to simplify calculations.

Z(x) = \frac{b^2}{a^2} (a^2 - x^2) (a - x)^2

Z(x) = \frac{b^2}{a^2} (a - x)(a + x) (a - x)^2 = \frac{b^2}{a^2} (a - x)^3 (a + x)

Differentiating

Z

with respect to

x

using the product rule:

\frac{dZ}{dx} = \frac{b^2}{a^2} \left[ 3(a - x)^2(-1)(a + x) + (a - x)^3(1) \right]

\frac{dZ}{dx} = \frac{b^2}{a^2} (a - x)^2 \left[ -3(a + x) + (a - x) \right]

\frac{dZ}{dx} = \frac{b^2}{a^2} (a - x)^2 [-3a - 3x + a - x]

\frac{dZ}{dx} = \frac{b^2}{a^2} (a - x)^2 (-2a - 4x) = -\frac{2b^2}{a^2} (a - x)^2 (a + 2x)

For maximum or minimum area, set

\frac{dZ}{dx} = 0

:

- \frac{2b^2}{a^2} (a - x)^2 (a + 2x) = 0

This gives

x = a

or

x = -\frac{a}{2}

. If

x = a

, the height is 0, so the area is 0, which cannot be the maximum. So, we consider

x = -\frac{a}{2}

.

To confirm this is a maximum, we use the second derivative test:

\frac{d^2Z}{dx^2} = -\frac{2b^2}{a^2} \frac{d}{dx} \left[ (a^2 - 2ax + x^2)(a + 2x) \right]

\frac{d^2Z}{dx^2} = -\frac{2b^2}{a^2} \left[ 2(a-x)(-1)(a+2x) + (a-x)^2(2) \right]

At

x = -\frac{a}{2}

, the term

(a+2x)

is zero. So we evaluate at this point:

\left. \frac{d^2Z}{dx^2} \right|_{x=-a/2} = -\frac{2b^2}{a^2} \left[ -2(a - (-\frac{a}{2}))(a + 2(-\frac{a}{2})) + 2(a - (-\frac{a}{2}))^2(2) \right]

\left. \frac{d^2Z}{dx^2} \right|_{x=-a/2} = -\frac{2b^2}{a^2} \left[ -2(\frac{3a}{2})(0) + 2(\frac{3a}{2})^2 \right] = -\frac{2b^2}{a^2} \left[ 2 \times \frac{9a^2}{4} \right] = -\frac{2b^2}{a^2} \frac{9a^2}{2} = -9b^2

Since

\frac{d^2Z}{dx^2} < 0

at

x = -\frac{a}{2}

, the area is maximum at this point.

Now, we calculate the maximum area: When

x = -\frac{a}{2}

:

y = \frac{b}{a} \sqrt{a^2 - x^2} = \frac{b}{a} \sqrt{a^2 - \left(-\frac{a}{2}\right)^2} = \frac{b}{a} \sqrt{a^2 - \frac{a^2}{4}} = \frac{b}{a} \sqrt{\frac{3a^2}{4}} = \frac{b}{a} \frac{a\sqrt{3}}{2} = \frac{b\sqrt{3}}{2}

Base of the triangle

= 2y = b\sqrt{3}

. Height of the triangle

= a - x = a - \left(-\frac{a}{2}\right) = \frac{3a}{2}

. Maximum Area

= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (b\sqrt{3}) \times \left(\frac{3a}{2}\right) = \frac{3\sqrt{3}}{4}ab

.

Final Answer: The maximum area of the isosceles triangle is

\frac{3\sqrt{3}}{4}ab

square units.

Q6Miscellaneous Exercise on Chapter 6

A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is $8 \mathrm{~m}^{3}$ . If building of tank costs Rs 70 per sq metres for the base and Rs 45 per square metre for sides. What is the cost of least expensive tank?

Solution

Given: A tank with a rectangular base and rectangular sides, open at the top. Depth

h = 2

m. Volume

V = 8

m³. Cost of base = Rs 70 per m². Cost of sides = Rs 45 per m².

To Find: The cost of the least expensive tank.

Solution: Let the length and breadth of the rectangular base be

l

and

b

respectively.

Volume of the tank:

V = l \times b \times h

8 = l \times b \times 2

lb = 4 \quad \implies \quad b = \frac{4}{l} \quad ...(1)

Surface area of the tank: The tank is open at the top. Area of the base:

A_{base} = lb

. Area of the four sides:

A_{sides} = 2(lh + bh) = 2(l+b)h

. Since

h=2

m,

A_{sides} = 2(l+b)(2) = 4(l+b)

.

Cost of construction: Let

C

be the total cost. Cost = (Cost of base) + (Cost of sides)

C = (A_{base} \times 70) + (A_{sides} \times 45)

C = (lb \times 70) + (4(l+b) \times 45)

Using

lb=4

from equation (1):

C = (4 \times 70) + 180(l+b)

C = 280 + 180(l+b)

Now, express the cost

C

as a function of a single variable, say

l

, using

b = \frac{4}{l}

:

C(l) = 280 + 180\left(l + \frac{4}{l}\right)

To find the minimum cost, we need to find the derivative of

C(l)

with respect to

l

and set it to zero.

\frac{dC}{dl} = \frac{d}{dl} \left[ 280 + 180\left(l + \frac{4}{l}\right) \right]

\frac{dC}{dl} = 180\left(1 - \frac{4}{l^2}\right)

Set

\frac{dC}{dl} = 0

:

180\left(1 - \frac{4}{l^2}\right) = 0

1 - \frac{4}{l^2} = 0

l^2 = 4

l = 2 \quad (\text{since length cannot be negative})

Now, we use the second derivative test to confirm this is a minimum.

\frac{d^2C}{dl^2} = \frac{d}{dl} \left[ 180\left(1 - 4l^{-2}\right) \right]

\frac{d^2C}{dl^2} = 180(0 - (-2)4l^{-3}) = 180\left(\frac{8}{l^3}\right) = \frac{1440}{l^3}

At

l=2

:

\left. \frac{d^2C}{dl^2} \right|_{l=2} = \frac{1440}{2^3} = \frac{1440}{8} = 180 > 0

Since the second derivative is positive, the cost is minimum at

l=2

m.

Dimensions for the least expensive tank: Length

l = 2

m. Breadth

b = \frac{4}{l} = \frac{4}{2} = 2

m. Depth

h = 2

m (given). The tank is a cube with an open top.

Cost of the least expensive tank: Substitute

l=2

into the cost function:

C(2) = 280 + 180\left(2 + \frac{4}{2}\right)

C(2) = 280 + 180(2 + 2)

C(2) = 280 + 180(4)

C(2) = 280 + 720 = 1000

Final Answer: The cost of the least expensive tank is Rs 1000.

Q7Miscellaneous Exercise on Chapter 6

The sum of the perimeter of a circle and square is $k$ , where $k$ is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

Solution

Given: The sum of the perimeter of a circle and a square is a constant,

k

.

To Prove: The sum of their areas is least when the side of the square is double the radius of the circle.

Proof: Let

r

be the radius of the circle and

x

be the side length of the square.

Perimeters: Perimeter of the circle =

2\pi r

. Perimeter of the square =

4x

.

According to the given condition:

2\pi r + 4x = k \quad ...(1)

From this, we can express

x

in terms of

r

:

4x = k - 2\pi r

x = \frac{k - 2\pi r}{4} \quad ...(2)

Areas: Area of the circle,

A_c = \pi r^2

. Area of the square,

A_s = x^2

. Let

S

be the sum of their areas:

S = A_c + A_s = \pi r^2 + x^2

Substitute the expression for

x

from (2) into the sum of areas equation to get

S

as a function of

r

:

S(r) = \pi r^2 + \left(\frac{k - 2\pi r}{4}\right)^2

S(r) = \pi r^2 + \frac{1}{16}(k^2 - 4k\pi r + 4\pi^2 r^2)

To find the value of

r

for which

S

is minimum, we find the derivative of

S(r)

with respect to

r

and set it to zero.

\frac{dS}{dr} = \frac{d}{dr} \left[ \pi r^2 + \frac{1}{16}(k^2 - 4k\pi r + 4\pi^2 r^2) \right]

\frac{dS}{dr} = 2\pi r + \frac{1}{16}(-4k\pi + 8\pi^2 r)

\frac{dS}{dr} = 2\pi r - \frac{k\pi}{4} + \frac{\pi^2 r}{2}

Set

\frac{dS}{dr} = 0

:

2\pi r - \frac{k\pi}{4} + \frac{\pi^2 r}{2} = 0

r \left( 2\pi + \frac{\pi^2}{2} \right) = \frac{k\pi}{4}

r \pi \left( 2 + \frac{\pi}{2} \right) = \frac{k\pi}{4}

r \left( \frac{4+\pi}{2} \right) = \frac{k}{4}

r = \frac{k}{4} \times \frac{2}{4+\pi} = \frac{k}{2(4+\pi)}

Now, we use the second derivative test to confirm this corresponds to a minimum.

\frac{d^2S}{dr^2} = \frac{d}{dr} \left[ 2\pi r - \frac{k\pi}{4} + \frac{\pi^2 r}{2} \right]

\frac{d^2S}{dr^2} = 2\pi + \frac{\pi^2}{2}

Since

2\pi + \frac{\pi^2}{2} > 0

, the sum of the areas

S

is minimum for this value of

r

.

Now, we find the relationship between the side of the square

x

and the radius

r

when the sum of areas is least. Substitute the value of

k

from equation (1) into the expression for

r

:

r = \frac{2\pi r + 4x}{2(4+\pi)}

2r(4+\pi) = 2\pi r + 4x

8r + 2\pi r = 2\pi r + 4x

8r = 4x

x = 2r

This shows that the sum of the areas is least when the side of the square (

x

) is double the radius of the circle (

r

).

Hence Proved.

Q8Miscellaneous Exercise on Chapter 6

A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m . Find the dimensions of the window to admit maximum light through the whole opening.

Solution

Given: A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m.

To Find: The dimensions of the window to admit maximum light through the whole opening.

Solution: Let the length of the rectangular part be

2x

and the breadth be

y

. The radius of the semicircular opening will then be

x

.

Dimensions of the window:

Rectangular part: length = $2x$ , breadth = $y$ .
Semicircular part: radius = $x$ .

Perimeter of the window: The perimeter consists of the two vertical sides of the rectangle, the base of the rectangle, and the circumference of the semicircle. Perimeter

P = (y) + (2x) + (y) + (\text{arc length of semicircle})

P = 2x + 2y + \pi x

We are given that the total perimeter is 10 m.

2x + 2y + \pi x = 10

x(2 + \pi) + 2y = 10

2y = 10 - x(2 + \pi)

y = 5 - \frac{x(2 + \pi)}{2} \quad ...(1)

Area of the window: To admit maximum light, the area of the window must be maximum. Area

A = (\text{Area of rectangle}) + (\text{Area of semicircle})

A = (2x)(y) + \frac{1}{2}\pi x^2

Substitute the expression for

y

from (1) into the area equation to get

A

as a function of

x

:

A(x) = 2x \left( 5 - \frac{x(2 + \pi)}{2} \right) + \frac{1}{2}\pi x^2

A(x) = 10x - x^2(2 + \pi) + \frac{1}{2}\pi x^2

A(x) = 10x - 2x^2 - \pi x^2 + \frac{1}{2}\pi x^2

A(x) = 10x - 2x^2 - \frac{1}{2}\pi x^2

A(x) = 10x - x^2\left(2 + \frac{\pi}{2}\right)

To find the value of

x

for which

A

is maximum, we find the derivative of

A(x)

and set it to zero.

\frac{dA}{dx} = \frac{d}{dx} \left[ 10x - x^2\left(2 + \frac{\pi}{2}\right) \right]

\frac{dA}{dx} = 10 - 2x\left(2 + \frac{\pi}{2}\right)

Set

\frac{dA}{dx} = 0

:

10 - 2x\left(2 + \frac{\pi}{2}\right) = 0

10 = 2x\left(\frac{4 + \pi}{2}\right)

10 = x(4 + \pi)

x = \frac{10}{4 + \pi}

Now, we use the second derivative test to confirm this is a maximum.

\frac{d^2A}{dx^2} = \frac{d}{dx} \left[ 10 - 2x\left(2 + \frac{\pi}{2}\right) \right]

\frac{d^2A}{dx^2} = -2\left(2 + \frac{\pi}{2}\right) = -(4 + \pi)

Since

(4 + \pi) > 0

,

\frac{d^2A}{dx^2} < 0

. Thus, the area is maximum at

x = \frac{10}{4 + \pi}

.

Dimensions for maximum light: Radius of semicircle:

x = \frac{10}{4 + \pi}

m. Length of rectangle:

2x = \frac{20}{4 + \pi}

m. Breadth of rectangle:

y = 5 - \frac{x(2 + \pi)}{2}

y = 5 - \frac{1}{2} \left( \frac{10}{4 + \pi} \right) (2 + \pi)

y = 5 - \frac{5(2 + \pi)}{4 + \pi}

y = \frac{5(4 + \pi) - 5(2 + \pi)}{4 + \pi}

y = \frac{20 + 5\pi - 10 - 5\pi}{4 + \pi}

y = \frac{10}{4 + \pi}

So, the required dimensions are: Length of the rectangular part =

\frac{20}{4 + \pi}

m. Breadth of the rectangular part =

\frac{10}{4 + \pi}

m. The radius of the semicircular opening is also

\frac{10}{4 + \pi}

m.

Final Answer: For maximum light, the dimensions of the window are: Length of the rectangle =

\frac{20}{4 + \pi}

m. Breadth of the rectangle =

\frac{10}{4 + \pi}

m.

Q9Miscellaneous Exercise on Chapter 6

A point on the hypotenuse of a triangle is at distance $a$ and $b$ from the sides of the triangle. Show that the minimum length of the hypotenuse is $\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}$ .

Solution

Given: A point on the hypotenuse of a right-angled triangle is at distances

a

and

b

from the other two sides.

To Show: The minimum length of the hypotenuse is

\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}

.

Proof: Let the right-angled triangle be

ABC

, with the right angle at

B

. Let the sides containing the right angle be along the coordinate axes. So,

B = (0,0)

,

A = (0, y_0)

and

C = (x_0, 0)

for some

x_0, y_0 > 0

.

The equation of the hypotenuse

AC

is

\frac{x}{x_0} + \frac{y}{y_0} = 1

. The length of the hypotenuse,

L

, is the distance between

A

and

C

.

L = \sqrt{(x_0-0)^2 + (0-y_0)^2} = \sqrt{x_0^2 + y_0^2}

Let

P

be a point on the hypotenuse

AC

. The coordinates of

P

are

(a, b)

because its distance from the side

BC

(y-axis) is

a

and from the side

AB

(x-axis) is

b

.

Since

P(a, b)

lies on the line

AC

, it must satisfy the line equation:

\frac{a}{x_0} + \frac{b}{y_0} = 1 \quad ...(1)

We need to minimize

L = \sqrt{x_0^2 + y_0^2}

. This is equivalent to minimizing

L^2 = x_0^2 + y_0^2

. Let

S = L^2 = x_0^2 + y_0^2

.

From (1), we can express one variable in terms of the other. Let's express

y_0

in terms of

x_0

:

\frac{b}{y_0} = 1 - \frac{a}{x_0} = \frac{x_0 - a}{x_0}

y_0 = \frac{bx_0}{x_0 - a}

Now substitute this into the expression for

S

:

S(x_0) = x_0^2 + \left(\frac{bx_0}{x_0 - a}\right)^2 = x_0^2 + \frac{b^2 x_0^2}{(x_0 - a)^2}

To find the minimum value of

S

, we differentiate with respect to

x_0

and set the derivative to zero.

\frac{dS}{dx_0} = 2x_0 + b^2 \left[ \frac{2x_0(x_0 - a)^2 - x_0^2 \cdot 2(x_0 - a)}{(x_0 - a)^4} \right]

\frac{dS}{dx_0} = 2x_0 + 2b^2 x_0 \left[ \frac{(x_0 - a) - x_0}{(x_0 - a)^3} \right]

\frac{dS}{dx_0} = 2x_0 + 2b^2 x_0 \left[ \frac{-a}{(x_0 - a)^3} \right] = 2x_0 - \frac{2ab^2 x_0}{(x_0 - a)^3}

Set

\frac{dS}{dx_0} = 0

:

2x_0 \left( 1 - \frac{ab^2}{(x_0 - a)^3} \right) = 0

Since

x_0 \neq 0

, we have:

1 = \frac{ab^2}{(x_0 - a)^3}

(x_0 - a)^3 = ab^2

x_0 - a = (ab^2)^{\frac{1}{3}} = a^{\frac{1}{3}}b^{\frac{2}{3}}

x_0 = a + a^{\frac{1}{3}}b^{\frac{2}{3}} = a^{\frac{1}{3}}(a^{\frac{2}{3}} + b^{\frac{2}{3}})

Now find the corresponding value of

y_0

from

y_0 = \frac{bx_0}{x_0 - a}

:

y_0 = \frac{b \cdot a^{\frac{1}{3}}(a^{\frac{2}{3}} + b^{\frac{2}{3}})}{a^{\frac{1}{3}}b^{\frac{2}{3}}} = b^{1 - \frac{2}{3}} (a^{\frac{2}{3}} + b^{\frac{2}{3}}) = b^{\frac{1}{3}}(a^{\frac{2}{3}} + b^{\frac{2}{3}})

Now calculate the minimum value of

L^2 = S = x_0^2 + y_0^2

:

L^2 = \left[ a^{\frac{1}{3}}(a^{\frac{2}{3}} + b^{\frac{2}{3}}) \right]^2 + \left[ b^{\frac{1}{3}}(a^{\frac{2}{3}} + b^{\frac{2}{3}}) \right]^2

L^2 = a^{\frac{2}{3}}(a^{\frac{2}{3}} + b^{\frac{2}{3}})^2 + b^{\frac{2}{3}}(a^{\frac{2}{3}} + b^{\frac{2}{3}})^2

L^2 = (a^{\frac{2}{3}} + b^{\frac{2}{3}})(a^{\frac{2}{3}} + b^{\frac{2}{3}})^2 = (a^{\frac{2}{3}} + b^{\frac{2}{3}})^3

Taking the square root to find the length

L

:

L = \sqrt{(a^{\frac{2}{3}} + b^{\frac{2}{3}})^3} = \left(a^{\frac{2}{3}} + b^{\frac{2}{3}}\right)^{\frac{3}{2}}

The second derivative of

S

is positive for

x_0 > a

, which confirms this is a minimum.

Alternative Method using Trigonometry: Let the hypotenuse make an angle

\theta

with the side of length

x_0

. Then

a = L_1 \sin \theta

and

b = L_2 \cos \theta

, where

L_1

and

L_2

are segments of the hypotenuse divided by point

P

. The length of hypotenuse

L = L_1 + L_2

. From similar triangles, we have

a = (L-L_2)\sin\theta

and

b = L_2 \cos\theta

. This can be simplified. A better way: The vertices are

(0,0)

,

(L\cos\theta, 0)

and

(0, L\sin\theta)

. The point is

(a,b)

. The hypotenuse line is

x/(L\cos\theta) + y/(L\sin\theta) = 1

. Since

(a,b)

is on it,

a/(L\cos\theta) + b/(L\sin\theta) = 1

. So

L = a/\cos\theta + b/\sin\theta = a\sec\theta + b\csc\theta

.

dL/d\theta = a\sec\theta\tan\theta - b\csc\theta\cot\theta = 0

.

a\frac{\sin\theta}{\cos^2\theta} = b\frac{\cos\theta}{\sin^2\theta} \implies a\sin^3\theta = b\cos^3\theta \implies \tan^3\theta = b/a \implies \tan\theta = (b/a)^{1/3}

. From this,

\sin\theta = b^{1/3}/(a^{2/3}+b^{2/3})^{1/2}

and

\cos\theta = a^{1/3}/(a^{2/3}+b^{2/3})^{1/2}

.

L = a/\cos\theta + b/\sin\theta = a\frac{(a^{2/3}+b^{2/3})^{1/2}}{a^{1/3}} + b\frac{(a^{2/3}+b^{2/3})^{1/2}}{b^{1/3}}

.

L = (a^{2/3}+b^{2/3})^{1/2} [a^{2/3} + b^{2/3}] = (a^{2/3}+b^{2/3})^{3/2}

.

Final Answer: The minimum length of the hypotenuse is

\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}

. Hence shown.

Q10Miscellaneous Exercise on Chapter 6

Find the points at which the function $f$ given by $f(x)=(x-2)^{4}(x+1)^{3}$ has

(i)

local maxima

(ii)

local minima

(iii)

point of inflexion

Solution

Given: The function is

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

.

To Find:

(i)

Points of local maxima

(ii)

Points of local minima

(iii)

Points of inflexion

Solution: We have the function

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

.

First, we find the derivative of

f(x)

with respect to

x

using the product rule,

(uv)' = u'v + uv'

. Let

u(x) = (x-2)^4

and

v(x) = (x+1)^3

. Then

u'(x) = 4(x-2)^3

and

v'(x) = 3(x+1)^2

.

f'(x) = \frac{d}{dx}[(x-2)^4(x+1)^3]

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

Factor out common terms

(x-2)^3

and

(x+1)^2

:

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

f'(x) = (x-2)^3(x+1)^2 [4x + 4 + 3x - 6]

f'(x) = (x-2)^3(x+1)^2 (7x - 2)

For critical points, we set

f'(x) = 0

.

(x-2)^3(x+1)^2 (7x - 2) = 0

This gives us the critical points

x=2

,

x=-1

, and

x = \frac{2}{7}

.

Now, we use the first derivative test to analyze the sign of

f'(x)

in the intervals around these critical points. The term

(x+1)^2

is always non-negative.

Case 1: Around

x = -1

For

x < -1

(e.g.,

x=-2

),

f'(x) = (-4)^3(-1)^2(-16) > 0

. For

-1 < x < \frac{2}{7}

(e.g.,

x=0

),

f'(x) = (-2)^3(1)^2(-2) = 16 > 0

. Since the sign of

f'(x)

does not change as

x

passes through

-1

,

x=-1

is neither a point of local maximum nor a point of local minimum. It is a point of inflexion.

Case 2: Around

x = \frac{2}{7}

For

-1 < x < \frac{2}{7}

(e.g.,

x=0

),

f'(x) > 0

. For

\frac{2}{7} < x < 2

(e.g.,

x=1

),

f'(x) = (-1)^3(2)^2(5) = -20 < 0

. Since

f'(x)

changes sign from positive to negative as

x

passes through

\frac{2}{7}

, the function has a local maximum at

x = \frac{2}{7}

.

Case 3: Around

x = 2

For

\frac{2}{7} < x < 2

(e.g.,

x=1

),

f'(x) < 0

. For

x > 2

(e.g.,

x=3

),

f'(x) = (1)^3(4)^2(19) > 0

. Since

f'(x)

changes sign from negative to positive as

x

passes through

2

, the function has a local minimum at

x = 2

.

Final Answer:

(i)

The function has a local maximum at the point

x = \frac{2}{7}

.

(ii)

The function has a local minimum at the point

x = 2

.

(iii)

The function has a point of inflexion at the point

x = -1

.

Q11Miscellaneous Exercise on Chapter 6

Find the absolute maximum and minimum values of the function $f$ given by $f(x)=\cos ^{2} x+\sin x, x \in[0, \pi]$

Solution

Given: The function

f(x) = \cos^2 x + \sin x

for

x \in [0, \pi]

.

To Find: The absolute maximum and minimum values of the function

f(x)

.

Solution: We have the function

f(x) = \cos^2 x + \sin x

. To find the absolute maximum and minimum values on a closed interval, we find the critical points by differentiating the function and setting the derivative to zero. Then we evaluate the function at these critical points and at the endpoints of the interval.

First, find the derivative of

f(x)

:

f'(x) = \frac{d}{dx}(\cos^2 x + \sin x)

f'(x) = 2\cos x (-\sin x) + \cos x

f'(x) = -2\sin x \cos x + \cos x

f'(x) = \cos x (1 - 2\sin x)

Now, set

f'(x) = 0

to find the critical points in the interval

[0, \pi]

.

\cos x (1 - 2\sin x) = 0

This implies either

\cos x = 0

or

1 - 2\sin x = 0

.

Case 1:

\cos x = 0

In the interval

[0, \pi]

,

\cos x = 0

when

x = \frac{\pi}{2}

.

Case 2:

1 - 2\sin x = 0

This means

\sin x = \frac{1}{2}

. In the interval

[0, \pi]

,

\sin x = \frac{1}{2}

when

x = \frac{\pi}{6}

and

x = \frac{5\pi}{6}

.

So, the critical points in

[0, \pi]

are

x = \frac{\pi}{6}

,

x = \frac{\pi}{2}

, and

x = \frac{5\pi}{6}

.

The points at which we need to evaluate

f(x)

are the critical points and the endpoints of the interval, which are

0, \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \pi

.

At $x=0$ : $f(0) = \cos^2(0) + \sin(0) = 1^2 + 0 = 1$ .
At $x = \frac{\pi}{6}$ : $f(\frac{\pi}{6}) = \cos^2(\frac{\pi}{6}) + \sin(\frac{\pi}{6}) = (\frac{\sqrt{3}}{2})^2 + \frac{1}{2} = \frac{3}{4} + \frac{1}{2} = \frac{5}{4}$ .
At $x = \frac{\pi}{2}$ : $f(\frac{\pi}{2}) = \cos^2(\frac{\pi}{2}) + \sin(\frac{\pi}{2}) = 0^2 + 1 = 1$ .
At $x = \frac{5\pi}{6}$ : $f(\frac{5\pi}{6}) = \cos^2(\frac{5\pi}{6}) + \sin(\frac{5\pi}{6}) = (-\frac{\sqrt{3}}{2})^2 + \frac{1}{2} = \frac{3}{4} + \frac{1}{2} = \frac{5}{4}$ .
At $x=\pi$ : $f(\pi) = \cos^2(\pi) + \sin(\pi) = (-1)^2 + 0 = 1$ .

Comparing the values obtained:

1, \frac{5}{4}, 1, \frac{5}{4}, 1

. Note that

\frac{5}{4} = 1.25

.

The maximum value among these is

\frac{5}{4}

. The minimum value among these is

1

.

Final Answer: The absolute maximum value of the function is

\frac{5}{4}

and the absolute minimum value is

1

.

Q12Miscellaneous Exercise on Chapter 6

Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius $r$ is $\frac{4 r}{3}$ .

Solution

To Prove: The altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius

r

is

\frac{4r}{3}

.

Proof: Let a right circular cone of height

h

and base radius

x

be inscribed in a sphere of radius

r

. Let the center of the sphere be

O

. Let the vertex of the cone be

V

and the center of its base be

C

. Let

A

be a point on the circumference of the base of the cone.

In the right-angled triangle

\triangle OCA

, we have:

OA^2 = OC^2 + AC^2

Here,

OA = r

(radius of the sphere) and

AC = x

(radius of the cone's base). The height of the cone is

h = VC = VO + OC = r + OC

. Thus,

OC = h - r

. Note that the cone's vertex is assumed to be above the sphere's center. If it is below,

h 0

, which means

V''\left(\frac{4r}{3}\right) < 0

. By the second derivative test, the volume of the cone is maximum when its altitude is

h = \frac{4r}{3}

.

Hence Proved.

Q13Miscellaneous Exercise on Chapter 6

Let $f$ be a function defined on $[a, b]$ such that $f^{\prime}(x)>0$ , for all $x \in(a, b)$ . Then prove that $f$ is an increasing function on $(a, b)$ .

Solution

To Prove: If

f

is a function defined on

[a, b]

such that

f'(x) > 0

for all

x \in (a, b)

, then

f

is an increasing function on

(a, b)

.

Proof: By definition, a function

f

is increasing on an interval

(a, b)

if for any two points

x_1, x_2 \in (a, b)

with

x_1 < x_2

, we have

f(x_1) < f(x_2)

.

Let

x_1

and

x_2

be any two distinct points in the interval

(a, b)

such that

x_1 < x_2

. Consider the function

f

on the closed interval

[x_1, x_2]

.

We are given that

f'(x)

exists for all

x \in (a, b)

. This implies that

f

is differentiable on

(a, b)

. Since differentiability implies continuity,

f

is also continuous on

(a, b)

.

As

[x_1, x_2]

is a subinterval of

(a, b)

, we have:

$f$ is continuous on the closed interval $[x_1, x_2]$ .
$f$ is differentiable on the open interval $(x_1, x_2)$ .

Thus, the conditions for the Mean Value Theorem (MVT) are satisfied for the function

f

on the interval

[x_1, x_2]

. By the Mean Value Theorem, there exists at least one point

c

in the open interval

(x_1, x_2)

such that:

f'(c) = \frac{f(x_2) - f(x_1)}{x_2 - x_1}

We are given that

f'(x) > 0

for all

x \in (a, b)

. Since

c \in (x_1, x_2)

and

(x_1, x_2) \subset (a, b)

, we must have

f'(c) > 0

.

Therefore, from the MVT equation, we have:

\frac{f(x_2) - f(x_1)}{x_2 - x_1} > 0

We chose

x_1

and

x_2

such that

x_1 < x_2

, which implies that the denominator

(x_2 - x_1)

is positive. For the fraction to be positive, the numerator must also be positive.

f(x_2) - f(x_1) > 0

f(x_2) > f(x_1)

or equivalently,

f(x_1) < f(x_2)

.

Since for any arbitrary points

x_1, x_2 \in (a, b)

with

x_1 < x_2

, we have shown that

f(x_1) < f(x_2)

, it follows by definition that

f

is an increasing function on

(a, b)

.

Hence Proved.

Q14Miscellaneous Exercise on Chapter 6

Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is $\frac{2 \mathrm{R}}{\sqrt{3}}$ . Also find the maximum volume.

Solution

To Prove: The height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is

\frac{2R}{\sqrt{3}}

. Also find the maximum volume.

Proof: Let a cylinder of height

h

and radius

r

be inscribed in a sphere of radius

R

. Let the center of the sphere be the origin

(0,0,0)

. The axis of the cylinder is along the z-axis. The top and bottom circular faces of the cylinder are at

z=h/2

and

z=-h/2

. The radius of the cylinder is

r

.

Consider a point on the rim of the cylinder's top face. Its coordinates would be

(x, y, h/2)

such that

x^2+y^2=r^2

. This point also lies on the sphere. The equation of the sphere is

x^2+y^2+z^2=R^2

. Substituting the point's coordinates into the sphere's equation:

r^2 + \left(\frac{h}{2}\right)^2 = R^2

r^2 = R^2 - \frac{h^2}{4}

The volume

V

of the cylinder is given by

V = \pi r^2 h

. Substitute the expression for

r^2

into the volume formula to express

V

as a function of

h

:

V(h) = \pi \left(R^2 - \frac{h^2}{4}\right) h

V(h) = \pi \left(R^2 h - \frac{h^3}{4}\right)

For maximum volume, we find the derivative of

V

with respect to

h

and set it to zero.

V'(h) = \frac{d}{dh} \left[ \pi \left(R^2 h - \frac{h^3}{4}\right) \right]

V'(h) = \pi \left(R^2 - \frac{3h^2}{4}\right)

Set

V'(h) = 0

to find critical points:

\pi \left(R^2 - \frac{3h^2}{4}\right) = 0

R^2 = \frac{3h^2}{4}

h^2 = \frac{4R^2}{3}

Since height

h

must be positive, we take the positive square root:

h = \frac{2R}{\sqrt{3}}

To confirm that this value of

h

gives a maximum volume, we use the second derivative test. Find

V''(h)

:

V''(h) = \frac{d}{dh} \left[ \pi \left(R^2 - \frac{3h^2}{4}\right) \right]

V''(h) = \pi \left(0 - \frac{6h}{4}\right) = -\frac{3\pi h}{2}

Since

h

is a height,

h > 0

. Therefore,

V''(h) = -\frac{3\pi h}{2} < 0

. This confirms that the volume is maximum at

h = \frac{2R}{\sqrt{3}}

.

Thus, the height of the cylinder of maximum volume is

\frac{2R}{\sqrt{3}}

.

Now, we find the maximum volume. Substitute

h = \frac{2R}{\sqrt{3}}

back into the volume equation

V(h) = \pi \left(R^2 h - \frac{h^3}{4}\right)

. Alternatively, we can use

V = \pi r^2 h

and find

r^2

for this

h

. From

h^2 = \frac{4R^2}{3}

, we have

r^2 = R^2 - \frac{h^2}{4} = R^2 - \frac{1}{4}\left(\frac{4R^2}{3}\right) = R^2 - \frac{R^2}{3} = \frac{2R^2}{3}

.

Maximum Volume,

V_{max} = \pi r^2 h

V_{max} = \pi \left(\frac{2R^2}{3}\right) \left(\frac{2R}{\sqrt{3}}\right)

V_{max} = \frac{4\pi R^3}{3\sqrt{3}}

Final Answer: The height of the cylinder of maximum volume is

\frac{2R}{\sqrt{3}}

. The maximum volume of the cylinder is

\frac{4\pi R^3}{3\sqrt{3}}

.

Q15Miscellaneous Exercise on Chapter 6

Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height $h$ and semi vertical angle $\alpha$ is one-third that of the cone and the greatest volume of cylinder is $\frac{4}{27} \pi h^{3} \tan ^{2} \alpha$ .

Solution

To Prove: The height of the cylinder of greatest volume which can be inscribed in a right circular cone of height

h

and semi-vertical angle

\alpha

is one-third that of the cone, i.e.,

\frac{h}{3}

. The greatest volume of the cylinder is

\frac{4}{27} \pi h^{3} \tan ^{2} \alpha

.

Proof: Let a right circular cone have height

h

and semi-vertical angle

\alpha

. Let a cylinder of radius

r

and height

H

be inscribed in the cone.

Let's denote the cone's vertex as A, the center of its base as O, and a point on the circumference of the base as B. So,

AO = h

and

\angle OAB = \alpha

. The radius of the cone's base is

OB = R = h \tan \alpha

.

Let the inscribed cylinder have its base on the base of the cone. Let the height of the cylinder be

H

and its radius be

r

. Let the top rim of the cylinder intersect the slant height AB at a point P. Let the center of the top face of the cylinder be O'.

From the cross-section, we can see two similar right-angled triangles,

\triangle APO'

and

\triangle ABO

. Here,

AO' = AO - O'O = h - H

and

O'P = r

.

By similarity of triangles:

\frac{AO'}{AO} = \frac{O'P}{OB}

\frac{h-H}{h} = \frac{r}{R}

Substituting

R = h \tan \alpha

:

\frac{h-H}{h} = \frac{r}{h \tan \alpha}

r = (h-H) \tan \alpha \quad ...(1)

The volume of the cylinder,

V

, is given by:

V = \pi r^2 H

Substituting the value of

r

from equation (1):

V(H) = \pi [(h-H) \tan \alpha]^2 H

V(H) = \pi \tan^2 \alpha \cdot (h-H)^2 H

V(H) = \pi \tan^2 \alpha \cdot (h^2 - 2hH + H^2) H

V(H) = \pi \tan^2 \alpha \cdot (h^2H - 2hH^2 + H^3)

To find the maximum volume, we need to find the derivative of

V

with respect to

H

and set it to zero.

\frac{dV}{dH} = \pi \tan^2 \alpha \cdot \frac{d}{dH}(h^2H - 2hH^2 + H^3)

\frac{dV}{dH} = \pi \tan^2 \alpha \cdot (h^2 - 4hH + 3H^2)

Set

\frac{dV}{dH} = 0

:

\pi \tan^2 \alpha \cdot (h^2 - 4hH + 3H^2) = 0

Since

\pi \tan^2 \alpha \neq 0

, we have:

3H^2 - 4hH + h^2 = 0

Factorizing the quadratic equation:

3H^2 - 3hH - hH + h^2 = 0

3H(H - h) - h(H - h) = 0

(3H - h)(H - h) = 0

This gives two possible values for

H

:

H = h

or

H = \frac{h}{3}

. If

H=h

, the volume of the cylinder is 0, which cannot be the maximum. So, we consider

H = \frac{h}{3}

.

Now, we use the second derivative test to confirm if the volume is maximum at this point.

\frac{d^2V}{dH^2} = \pi \tan^2 \alpha \cdot \frac{d}{dH}(h^2 - 4hH + 3H^2)

\frac{d^2V}{dH^2} = \pi \tan^2 \alpha \cdot (-4h + 6H)

Evaluate at

H = \frac{h}{3}

:

\left. \frac{d^2V}{dH^2} \right|_{H=h/3} = \pi \tan^2 \alpha \cdot (-4h + 6(\frac{h}{3}))

= \pi \tan^2 \alpha \cdot (-4h + 2h)

= -2h \pi \tan^2 \alpha

Since

h > 0

and

\tan^2 \alpha > 0

, the second derivative is negative. Thus, the volume is maximum when

H = \frac{h}{3}

.

This proves that the height of the cylinder of greatest volume is one-third that of the cone.

Now, we find the greatest volume by substituting

H = \frac{h}{3}

into the volume formula:

V_{max} = \pi \tan^2 \alpha \cdot (h - \frac{h}{3})^2 \cdot (\frac{h}{3})

V_{max} = \pi \tan^2 \alpha \cdot (\frac{2h}{3})^2 \cdot (\frac{h}{3})

V_{max} = \pi \tan^2 \alpha \cdot (\frac{4h^2}{9}) \cdot (\frac{h}{3})

V_{max} = \frac{4}{27} \pi h^3 \tan^2 \alpha

This is the greatest volume of the cylinder.

Hence Proved.

Q16Miscellaneous Exercise on Chapter 6

A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic metre per hour. Then the depth of the wheat is increasing at the rate of

(A)

1 m/h

(B)

0.1 m/h

(C)

1.1 m/h

(D)

0.5 m/h

Solution

Given: A cylindrical tank with radius

r = 10

m. The rate at which wheat is being filled, which is the rate of change of volume,

\frac{dV}{dt} = 314 \text{ m}^3/\text{h}

.

To Find: The rate at which the depth of the wheat is increasing, i.e.,

\frac{dh}{dt}

.

Formula: The volume

V

of a cylinder with radius

r

and height (or depth)

h

is given by:

V = \pi r^2 h

Solution: Since the radius of the tank is constant, we can differentiate the volume formula with respect to time

t

to relate the rates of change.

\frac{d}{dt}(V) = \frac{d}{dt}(\pi r^2 h)

\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}

(Since

r

is a constant)

Now, we substitute the given values into this equation:

r = 10

m

\frac{dV}{dt} = 314 \text{ m}^3/\text{h}

314 = \pi (10)^2 \frac{dh}{dt}

314 = 100\pi \frac{dh}{dt}

Now, we solve for

\frac{dh}{dt}

:

\frac{dh}{dt} = \frac{314}{100\pi}

The value 314 suggests using the approximation

\pi \approx 3.14

.

\frac{dh}{dt} = \frac{314}{100 \times 3.14}

\frac{dh}{dt} = \frac{314}{314}

\frac{dh}{dt} = 1 \text{ m/h}

Thus, the depth of the wheat is increasing at the rate of 1 m/h. This corresponds to option (A).

Final Answer: The correct option is (A) 1 m/h.

Application of DerivativesClass 12 Mathematics NCERT Solutions

Find the rate of change of the area of a circle with respect to its radius rrr when

Solution

The volume of a cube is increasing at the rate of 8 cm3/s8 \mathrm{~cm}^{3} / \mathrm{s}8 cm3/s. How fast is the surface area increasing when the length of an edge is 12 cm12 \mathrm{~cm}12 cm ?

Solution

The radius of a circle is increasing uniformly at the rate of 3 cm/s3 \mathrm{~cm} / \mathrm{s}3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm10 \mathrm{~cm}10 cm .

Solution

An edge of a variable cube is increasing at the rate of 3 cm/s3 \mathrm{~cm} / \mathrm{s}3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm10 \mathrm{~cm}10 cm long?

Solution

A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s5 \mathrm{~cm} / \mathrm{s}5 cm/s. At the instant when the radius of the circular wave is 8 cm8 \mathrm{~cm}8 cm , how fast is the enclosed area increasing?

Solution

The radius of a circle is increasing at the rate of 0.7 cm/s0.7 \mathrm{~cm} / \mathrm{s}0.7 cm/s. What is the rate of increase of its circumference?

Solution

Solution

A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm15 \mathrm{~cm}15 cm .

Solution

A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm10 \mathrm{~cm}10 cm .

Solution

A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s2 \mathrm{~cm} / \mathrm{s}2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall ?

Solution

A particle moves along the curve 6y=x3+26 y=x^{3}+26y=x3+2. Find the points on the curve at which the yyy-coordinate is changing 8 times as fast as the xxx-coordinate.

Solution

The radius of an air bubble is increasing at the rate of 12 cm/s\frac{1}{2} \mathrm{~cm} / \mathrm{s}21​ cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm1 \mathrm{~cm}1 cm ?

Solution

A balloon, which always remains spherical, has a variable diameter 32(2x+1)\frac{3}{2}(2 x+1)23​(2x+1). Find the rate of change of its volume with respect to xxx.

Solution

Solution

The total cost C(x)\mathrm{C}(x)C(x) in Rupees associated with the production of xxx units of an item is given by C(x)=0.007x3−0.003x2+15x+4000.\mathrm{C}(x)=0.007 x^{3}-0.003 x^{2}+15 x+4000.C(x)=0.007x3−0.003x2+15x+4000. Find the marginal cost when 17 units are produced.

Solution

The total revenue in Rupees received from the sale of xxx units of a product is given by R(x)=13x2+26x+15R(x)=13 x^{2}+26 x+15R(x)=13x2+26x+15 Find the marginal revenue when x=7x=7x=7.

Solution

The rate of change of the area of a circle with respect to its radius rrr at r=6 cmr=6 \mathrm{~cm}r=6 cm is

Solution

The total revenue in Rupees received from the sale of xxx units of a product is given by R(x)=3x2+36x+5\mathrm{R}(x)=3 x^{2}+36 x+5R(x)=3x2+36x+5. The marginal revenue, when x=15x=15x=15 is

Solution

Show that the function given by f(x)=3x+17f(x)=3 x+17f(x)=3x+17 is increasing on R\mathbf{R}R.

Solution

Show that the function given by f(x)=e2xf(x)=e^{2 x}f(x)=e2x is increasing on R\mathbf{R}R.

Solution

Show that the function given by f(x)=sin⁡xf(x)=\sin xf(x)=sinx is

Solution

Find the intervals in which the function fff given by f(x)=2x2−3xf(x)=2 x^{2}-3 xf(x)=2x2−3x is

Solution

Find the intervals in which the function fff given by f(x)=2x3−3x2−36x+7f(x)=2 x^{3}-3 x^{2}-36 x+7f(x)=2x3−3x2−36x+7 is

Solution

Find the intervals in which the following functions are strictly increasing or decreasing:

Solution

Show that y=log⁡(1+x)−2x2+x,x>−1y=\log (1+x)-\frac{2 x}{2+x}, x>-1y=log(1+x)−2+x2x​,x>−1, is an increasing function of xxx throughout its domain.

Solution

Find the values of xxx for which y=[x(x−2)]2y=[x(x-2)]^{2}y=[x(x−2)]2 is an increasing function.

Solution

Prove that y=4sin⁡θ(2+cos⁡θ)−θy=\frac{4 \sin \theta}{(2+\cos \theta)}-\thetay=(2+cosθ)4sinθ​−θ is an increasing function of θ\thetaθ in 0,π20, \frac{\pi}{2}0,2π​.

Solution

Prove that the logarithmic function is increasing on (0,∞)(0, \infty)(0,∞).

Solution

Prove that the function fff given by f(x)=x2−x+1f(x)=x^{2}-x+1f(x)=x2−x+1 is neither strictly increasing nor decreasing on (−1,1)(-1,1)(−1,1).

Solution

Which of the following functions are decreasing on 0,π20, \frac{\pi}{2}0,2π​?

Solution

On which of the following intervals is the function fff given by f(x)=x100+sin⁡x−1f(x)=x^{100}+\sin x-1f(x)=x100+sinx−1 decreasing ?

Solution

For what values of aaa the function fff given by f(x)=x2+ax+1f(x)=x^{2}+a x+1f(x)=x2+ax+1 is increasing on [1,2][1,2][1,2] ?

Solution

Let I be any interval disjoint from [−1,1][-1,1][−1,1]. Prove that the function fff given by f(x)=x+1xf(x)=x+\frac{1}{x}f(x)=x+x1​ is increasing on I .

Solution

Prove that the function fff given by f(x)=log⁡sin⁡xf(x)=\log \sin xf(x)=logsinx is increasing on 0,π20, \frac{\pi}{2}0,2π​ and decreasing on π2,π\frac{\pi}{2}, \pi2π​,π.

Solution

Prove that the function fff given by f(x)=log⁡∣cos⁡x∣f(x)=\log |\cos x|f(x)=log∣cosx∣ is decreasing on (0,π2)\left(0, \frac{\pi}{2}\right)(0,2π​) and increasing on (3π2,2π)\left(\frac{3 \pi}{2}, 2 \pi\right)(23π​,2π).

Solution

Prove that the function given by f(x)=x3−3x2+3x−100f(x)=x^{3}-3 x^{2}+3 x-100f(x)=x3−3x2+3x−100 is increasing in R\mathbf{R}R.

Solution

The interval in which y=x2e−xy=x^{2} e^{-x}y=x2e−x is increasing is

Solution

Find the maximum and minimum values, if any, of the following functions given by

f(x)=(2x−1)2+3f(x)=(2 x-1)^{2}+3f(x)=(2x−1)2+3

f(x)=9x2+12x+2f(x)=9 x^{2}+12 x+2f(x)=9x2+12x+2

f(x)=−(x−1)2+10f(x)=-(x-1)^{2}+10f(x)=−(x−1)2+10

g(x)=x3+1g(x)=x^{3}+1g(x)=x3+1

Solution

Find the maximum and minimum values, if any, of the following functions given by

Find the rate of change of the area of a circle with respect to its radius $r$ when

The volume of a cube is increasing at the rate of $8 \mathrm{~cm}^{3} / \mathrm{s}$ . How fast is the surface area increasing when the length of an edge is $12 \mathrm{~cm}$ ?

The radius of a circle is increasing uniformly at the rate of $3 \mathrm{~cm} / \mathrm{s}$ . Find the rate at which the area of the circle is increasing when the radius is $10 \mathrm{~cm}$ .

An edge of a variable cube is increasing at the rate of $3 \mathrm{~cm} / \mathrm{s}$ . How fast is the volume of the cube increasing when the edge is $10 \mathrm{~cm}$ long?

A stone is dropped into a quiet lake and waves move in circles at the speed of $5 \mathrm{~cm} / \mathrm{s}$ . At the instant when the radius of the circular wave is $8 \mathrm{~cm}$ , how fast is the enclosed area increasing?

The radius of a circle is increasing at the rate of $0.7 \mathrm{~cm} / \mathrm{s}$ . What is the rate of increase of its circumference?

A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is $15 \mathrm{~cm}$ .

A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is $10 \mathrm{~cm}$ .

A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of $2 \mathrm{~cm} / \mathrm{s}$ . How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall ?

A particle moves along the curve $6 y=x^{3}+2$ . Find the points on the curve at which the $y$ -coordinate is changing 8 times as fast as the $x$ -coordinate.

The radius of an air bubble is increasing at the rate of $\frac{1}{2} \mathrm{~cm} / \mathrm{s}$ . At what rate is the volume of the bubble increasing when the radius is $1 \mathrm{~cm}$ ?

A balloon, which always remains spherical, has a variable diameter $\frac{3}{2}(2 x+1)$ . Find the rate of change of its volume with respect to $x$ .

The total cost $\mathrm{C}(x)$ in Rupees associated with the production of $x$ units of an item is given by $\mathrm{C}(x)=0.007 x^{3}-0.003 x^{2}+15 x+4000.$ Find the marginal cost when 17 units are produced.

The total revenue in Rupees received from the sale of $x$ units of a product is given by $R(x)=13 x^{2}+26 x+15$ Find the marginal revenue when $x=7$ .

The rate of change of the area of a circle with respect to its radius $r$ at $r=6 \mathrm{~cm}$ is

The total revenue in Rupees received from the sale of $x$ units of a product is given by $\mathrm{R}(x)=3 x^{2}+36 x+5$ . The marginal revenue, when $x=15$ is

Show that the function given by $f(x)=3 x+17$ is increasing on $\mathbf{R}$ .

Show that the function given by $f(x)=e^{2 x}$ is increasing on $\mathbf{R}$ .

Show that the function given by $f(x)=\sin x$ is

Find the intervals in which the function $f$ given by $f(x)=2 x^{2}-3 x$ is

Find the intervals in which the function $f$ given by $f(x)=2 x^{3}-3 x^{2}-36 x+7$ is

Show that $y=\log (1+x)-\frac{2 x}{2+x}, x>-1$ , is an increasing function of $x$ throughout its domain.

Find the values of $x$ for which $y=[x(x-2)]^{2}$ is an increasing function.

Prove that $y=\frac{4 \sin \theta}{(2+\cos \theta)}-\theta$ is an increasing function of $\theta$ in $0, \frac{\pi}{2}$ .

Prove that the logarithmic function is increasing on $(0, \infty)$ .

Prove that the function $f$ given by $f(x)=x^{2}-x+1$ is neither strictly increasing nor decreasing on $(-1,1)$ .

Which of the following functions are decreasing on $0, \frac{\pi}{2}$ ?

On which of the following intervals is the function $f$ given by $f(x)=x^{100}+\sin x-1$ decreasing ?

For what values of $a$ the function $f$ given by $f(x)=x^{2}+a x+1$ is increasing on $[1,2]$ ?

Let I be any interval disjoint from $[-1,1]$ . Prove that the function $f$ given by $f(x)=x+\frac{1}{x}$ is increasing on I .

Prove that the function $f$ given by $f(x)=\log \sin x$ is increasing on $0, \frac{\pi}{2}$ and decreasing on $\frac{\pi}{2}, \pi$ .

Prove that the function $f$ given by $f(x)=\log |\cos x|$ is decreasing on $\left(0, \frac{\pi}{2}\right)$ and increasing on $\left(\frac{3 \pi}{2}, 2 \pi\right)$ .

Prove that the function given by $f(x)=x^{3}-3 x^{2}+3 x-100$ is increasing in $\mathbf{R}$ .

The interval in which $y=x^{2} e^{-x}$ is increasing is

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

$f(x)=9 x^{2}+12 x+2$

$f(x)=-(x-1)^{2}+10$

$g(x)=x^{3}+1$

$f(x)=|x+2|-1$

$g(x)=-|x+1|+3$

$h(x)=\sin (2 x)+5$

$f(x)=|\sin 4 x+3|$

$h(x)=x+1, x \in(-1,1)$

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

$g(x)=x^{3}-3 x$

$h(x)=\sin x+\cos x, 0<x<\frac{\pi}{2}$

$f(x)=\sin x-\cos x, 0<x<2 \pi$

$f(x)=x^{3}-6 x^{2}+9 x+15$

$g(x)=\frac{x}{2}+\frac{2}{x}, x>0$

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

$f(x)=x \sqrt{1-x}, 0<x<1$

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

$g(x)=\log x$

$h(x)=x^{3}+x^{2}+x+1$

$f(x)=x^{3}, x \in[-2,2]$

$f(x)=\sin x+\cos x, x \in[0, \pi]$

$f(x)=4 x-\frac{1}{2} x^{2}, x \in\left[-2, \frac{9}{2}\right]$

$f(x)=(x-1)^{2}+3, x \in[-3,1]$

Find the maximum profit that a company can make, if the profit function is given by $p(x)=41-72 x-18 x^{2}$

Find both the maximum value and the minimum value of $3 x^{4}-8 x^{3}+12 x^{2}-48 x+25$ on the interval $[0,3]$ .

At what points in the interval $[0,2 \pi]$ , does the function $\sin 2 x$ attain its maximum value?

What is the maximum value of the function $\sin x+\cos x$ ?

Find the maximum value of $2 x^{3}-24 x+107$ in the interval $[1,3]$ . Find the maximum value of the same function in $[-3,-1]$ .

It is given that at $x=1$ , the function $x^{4}-62 x^{2}+a x+9$ attains its maximum value, on the interval $[0,2]$ . Find the value of $a$ .

Find the maximum and minimum values of $x+\sin 2 x$ on $[0,2\pi]$ .

Find two positive numbers $x$ and $y$ such that $x+y=60$ and $x y^{3}$ is maximum.

Find two positive numbers $x$ and $y$ such that their sum is 35 and the product $x^{2} y^{5}$ is a maximum.

Prove that the volume of the largest cone that can be inscribed in a sphere of radius $R$ is $\frac{8}{27}$ of the volume of the sphere.

Show that the right circular cone of least curved surface and given volume has an altitude equal to $\sqrt{2}$ time the radius of the base.

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is $\tan ^{-1} \sqrt{2}$ .

Show that semi-vertical angle of right circular cone of given surface area and maximum volume is $\sin ^{-1} \frac{1}{3}$ .