RS Aggarwal Solutions for Class 12 Chapter 11: Applications of Derivatives

The various applications of derivatives in our daily lives are the concepts which are discussed under Chapter 11. To make students familiar with these concepts and help them score well in the exam, our set of faculty provides solutions, both chapter wise and exercise wise. Students can download the solutions while practising the chapter wise problems, based on their requirements. To build expertise in the concepts which are covered under this chapter, students can get RS Aggarwal Solutions for Class 12 Chapter 11 Applications of Derivatives PDF from the links provided here.

RS Aggarwal Solutions for Class 12 Chapter 11: Applications of Derivatives Download PDF

 

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Exercises of RS Aggarwal Solutions Class 12 Maths Chapter 11 – Applications of Derivatives

Exercise 11A Solutions

Exercise 11B Solutions

Exercise 11C Solutions

Exercise 11D Solutions

Exercise 11E Solutions

Exercise 11F Solutions

Exercise 11G Solutions

Exercise 11H Solutions

Access RS Aggarwal Solutions for Class 12 Chapter 11: Applications of Derivatives

Exercise 11A page: 461

1. The side of a square is increasing at the rate of 0.2 cm/s. Find the rate of increase of the perimeter of the square.

Solution:

Consider the side of a square as a

It is given that the rate of change of side is

da/ dt = 0.2 cm/s

We know that the perimeter of a square = 4a

So the rate of change of perimeter is

4 da/dt = 4 × 0.2

We get

dP/ dt = 0.8 cm/s

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

Solution:

Consider the radius of the circle as r

dr/dt = 0.7 cm/s

We know that circumference of the circle = 2πr 

So the rate of change of circumference = 2π dr/dt

By substituting the values we get

= 2 × 3.14 × 0.7

We get

dC/dt = 4.4 cm/s

3. The radius of a circle is increasing uniformly at the rate of 0.3 centimetre per second. At what rate is the area increasing when the radius is 10 cm? (Take π = 3.14.)

Solution:

Consider radius of the circle as r

dr/dt = 0.3 cm/s

We know that area of the circle = πr2

Here the rate of change of area = 2πr dr/dt

By substituting the values

= 2 × 3.14 × 10 × 0.3

We get

dA/dt = 18.84 cm2/s

4. The side of a square sheet of metal is increasing at 3 centimetres per minute. At what rate is the area increasing when the side is 10 cm long?

Solution:

Consider the side of square as a

So the rate of change of side is

da/dt = 3 cm/min

We know that area of the square = a2

Here the rate of change of area is

2a da/dt = 2 × 10 × 3

So we get

dA/dt = 60 cm2/ min

5. The radius of a circular soap bubble is increasing at the rate of 0.2 cm/s. Find the rate of increase of its surface area when the radius is 7 cm.

Solution:

We know that a soap bubble will be in sphere shape

Consider the radius of soap bubble as r

dr/dt = 0.2 cm/s

The surface area of soap bubble = 4 πr2

So the rate of change of surface area = 8πr dr/dt

By substituting the values

= 8 × 3.14 × 7 × 0.2

We get

dS/dt = 35.2 cm2/s

6. The radius of an air bubble is increasing at the rate of 0.5 centimetre per second. At what rate is the volume of the bubble increasing when the radius is 1 centimetre?

Solution:

We know that a soap bubble will be in sphere shape

Consider the radius of soap bubble as r

dr/dt = 0.5 cm/s

The volume of soap bubble = 4/3 πr3

Here the rate of change of volume = 4 πr2 dr/dt

By substituting the values

= 4 × 3.14 × 12 × 0.5

We get

dV/dt = 6.28 cm3/s

7. The volume of a spherical balloon is increasing at the rate of 25 cubic centimetres per second. Find the rate of change of its surface at the instant when its radius is 5 cm.

Solution:

Consider radius of balloon as r

Volume of spherical balloon is V

V = 4/3 πr3

Rate of change of volume of spherical balloon

dV/dt = 4πr2 dr/dt

By substituting the values

25 cm3/s = 4 × π × 52 × dr/dt

So we get

dr/dt = 1/4π

Here the surface area of spherical balloon = 4 πr2

So the rate of change of surface area = 8πr dr/dt

By substituting the values

= 8 × π × 5 × 1/4π

We get

dS/dt = 10 cm2/s

8. A balloon which always remains spherical is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon is increasing when the radius is 15 cm.

Solution:

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11A Image 1

9. The bottom of a rectangular swimming tank is 25 m by 40 m. Water is pumped into the tank at the rate of 500 cubic metres per minute. Find the rate at which the level of water in the tank is rising.

Solution:

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11A Image 2

10. A stone is dropped into a quiet lake and waves move in circles at a speed of 3.5 cm per second. At the instant when the radius of the circular wave is 7.5 cm, how fast is the enclosed area increasing? (Take π = 22/7).

Solution:

Consider the radius of circle as r

dr/dt = 3.5 cm/s

We know that area of circle = πr2

So the rate of change of area of circle = 2πr dr/dt

By substituting the values

= 2 × 3.14 × 7.5 × 3.5

On further calculation

= 165 cm2/s

11. A 2-m tall man walks at a uniform speed of 5 km per hour away from a 6-metre high lamp post. Find the rate at which the length of his shadow increases.

Solution:

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11A Image 3

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11A Image 4

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11A Image 5


Exercise 11B page: 469

Using differentials, find the approximate value of:

1. √37

Solution:

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11B Image 1

2. ∛29

Solution:

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11B Image 2

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11B Image 3

3. ∛27

Solution:

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11B Image 4

4. √0.24

Solution:

Consider y = √x

Here x = 0.25 and Δ x = -0.01

By differentiating w.r.t. x

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11B Image 5

5. √49.5

Solution:

Consider y = √x

Here x = 49 and Δ x = 0.5

By differentiating w.r.t. x

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11B Image 6

We know that

Δ y = f (x + Δ x) – f (x)

By substituting the values

0.0357 = √ (49 + 0.5) – √49

It can be written as

0.0357 = √49.5 – 7

We get

√49.5 = 7.0357

6. RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11B Image 7

Solution:

Consider
RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11B Image 8

Here x = 16 and Δ x = 1

By differentiating w.r.t. x

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11B Image 9

7. RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11B Image 10

Solution:

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11B Image 11

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11B Image 12

8. loge 10.02, given that loge 10 = 2.3026

Solution:

Consider y = loge x

Here x = 10 and Δ x = 0.02

By differentiating w.r.t. x

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11B Image 13

On further calculation

Δ y = 0.02/ 10 = 0.002

We know that

Δ y = f (x + Δ x) – f (x)

By substituting the values

0.002 = loge (10 + 0.02) – loge 10

It can be written as

0.002 = loge 10.02 – 2.3026

We get

loge 10.02 = 2.3046

9. log10 (4.04), it being given that log10 4 = 0.6021 and log10 e = 0.4343

Solution:

Consider y = log10 x

It can be written as

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11B Image 14

On further calculation

Δ y = 0.017372/ 4 = 0.004343

We know that

Δ y = f (x + Δ x) – f (x)

By substituting the values

0.004343 = loge (4 + 0.04) – loge 4

It can be written as

0.004343 = loge 4.04 – 0.6021

We get


Exercise 11C PAGE: 483

Verify Rolle’s Theorem for each of the following functions:

1. f (x) = x2 on [- 1, 1]

Solution:

We know that

(i) f (x) = x2 is a polynomial which is continuous for all x ϵ R

Hence, f (x) = x2 is continuous on [- 1, 1]

(ii) f’(x) = 2x exist in [- 1, 1]

Hence, f (x) = x2 is differentiable on (- 1, 1)

(iii) We know that

f (- 1) = (- 1)2 = 1

Similarly f (1) = 11 = 1

Here f (-1) = f(1)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (-1, 1) where f’(c) = 0

2c = 0 which gives c = 0

We know that value of c = 0 ϵ (-1, 1)

Therefore, Rolle’s Theorem is satisfied.

2. f (x) = x2 – x – 12 in [-3, 4]

Solution:

We know that

(i) f (x) = x2 – x – 12 is a polynomial which is continuous for all x ϵ R

Hence, f (x) = x2 – x – 12 is continuous on [-3, 4]

(ii) f’(x) = 2x – 1 exist in [-3, 4]

Hence, f (x) = x2 – x – 12 is differentiable on (-3, 4)

(iii) We know that

f (- 3) = (- 3)2 – 3 – 12= 0

Similarly f (4) = 42 – 4 – 12 = 0

Here f (-3) = f(4)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (-3, 4) where f’(c) = 0

2c – 1 = 0 which gives c = ½

We know that value of c = 1/2 ϵ (-3, 4)

Therefore, Rolle’s Theorem is satisfied.

3. f (x) = x2 – 5x + 6 in [2, 3]

Solution:

We know that

(i) f (x) = x2 – 5x + 6 is a polynomial which is continuous for all x ϵ R

Hence, f (x) = x2 – 5x + 6 is continuous on [2, 3]

(ii) f’(x) = 2x – 5 exist in [2, 3]

Hence, f (x) = x2 – 5x + 6 is differentiable on (2, 3)

(iii) We know that

f (2) = (2)2 – 5(2) + 6 = 0

Similarly f (3) = 32 – 5(3) + 6 = 0

Here f (2) = f (3)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (2, 3) where f’(c) = 0

2c – 5 = 0 which gives c = 5/2

We know that value of c = 5/2 ϵ (2, 3)

Therefore, Rolle’s Theorem is satisfied.

4. f (x) = x2 – 3x – 18 in [-3, 6]

Solution:

We know that

(i) f (x) = x2 – 3x – 18 is a polynomial which is continuous for all x ϵ R

Hence, f (x) = x2 – 3x – 18 is continuous on [-3, 6]

(ii) f’(x) = 2x – 3 exist in [-3, 6]

Hence, f (x) = x2 – 3x – 18 is differentiable on (- 3, 6)

(iii) We know that

f (-3) = (-3)2 – 3(-3) – 18 = 0

Similarly f (6) = 62 – 5(6) – 18 = 0

Here f (-3) = f(6)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (- 3, 6) where f’(c) = 0

2c – 3 = 0 which gives c = 3/2

We know that value of c = 3/2 ϵ (-3, 6)

Therefore, Rolle’s Theorem is satisfied.

5. f (x) = x2 – 4x + 3 in [1, 3]

Solution:

We know that

(i) f (x) = x2 – 4x + 3 is a polynomial which is continuous for all x ϵ R

Hence, f (x) = x2 – 4x + 3 is continuous on [1, 3]

(ii) f’(x) = 2x – 4 exist in [1, 3]

Hence, f (x) = x2 – 4x + 3 is differentiable on (1, 3)

(iii) We know that

f (1) = (1)2 – 4(1) + 3 = 0

Similarly f (3) = 32 – 4(3) + 3 = 0

Here f (1) = f(3)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (1, 3) where f’(c) = 0

2c – 4 = 0 which gives c = 4/2 = 2

We know that value of c = 2 ϵ (1, 3)

Therefore, Rolle’s Theorem is satisfied.

6. f (x) = x (x – 4)2 in [0, 4]

Solution:

We know that

(i) f (x) = x (x – 4)2 is a polynomial which is continuous for all x ϵ R

Hence, f (x) = x (x – 4)2 is continuous on [0, 4]

(ii) f’(x) = (x – 4)2 + 2x (x – 4) exist in [0, 4]

Hence, f (x) = x (x – 4)2 is differentiable on (0, 4)

(iii) We know that

f (0) = 0(0 – 4)2 = 0

Similarly f (4) = 4 (4 – 4)2 = 0

Here f (0) = f(4)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (0, 4) where f’(c) = 0

(c – 4)2 + 2c (c – 4) = 0

It can be written as

(c – 4) (3c – 4) = 0

Here c = 4 or c = ¾

We know that value of c = 3/4 ϵ (0, 4)

Therefore, Rolle’s Theorem is satisfied.

7. f (x) = x3 – 7x2 + 16x – 12 in [2, 3]

Solution:

We know that

(i) f (x) = x3 – 7x2 + 16x – 12 is a polynomial which is continuous for all x ϵ R

Hence, f (x) = x3 – 7x2 + 16x – 12 is continuous on [2, 3]

(ii) f’(x) = 3x2 – 14x + 16 exist in [2, 3]

Hence, f (x) = x3 – 7x2 + 16x – 12 is differentiable on (2, 3)

(iii) We know that

f (2) = (2)3 – 7 (2)2 + 16(2) – 12 = 0

Similarly f (3) = 33 – 7(3)2 + 16 (3) – 12 = 0

Here f (2) = f(3)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (2, 3) where f’(c) = 0

3c2 – 14c + 16 = 0

It can be written as

(c – 2) (3c – 7) = 0 which gives c = 7/3

We know that value of c = 7/3 ϵ (2, 3)

Therefore, Rolle’s Theorem is satisfied.

8. f (x) = x3 + 3x2 – 24x – 80 in [- 4, 5]

Solution:

We know that

(i) f (x) = x3 + 3x2 – 24x – 80 is a polynomial which is continuous for all x ϵ R

Hence, f (x) = x3 + 3x2 – 24x – 80 is continuous on [- 4, 5]

(ii) f’(x) = 3x2 + 6x – 24 exist in [- 4, 5]

Hence, f (x) = x3 + 3x2 – 24x – 80 is differentiable on (- 4, 5)

(iii) We know that

f (- 4) = (- 4)3 + 3 (4)2 – 24(4) – 80 = 0

Similarly f (5) = 53 + 3(5)2 – 24 (5) – 80 = 0

Here f (- 4) = f (5)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (- 4, 5) where f’(c) = 0

3c2 + 6c – 24 = 0

Which gives c = – 4 or 2

We know that value of c = 2 ϵ (- 4, 5)

Therefore, Rolle’s Theorem is satisfied.

9. f (x) = (x – 1) (x – 2) (x – 3) in [1, 3]

Solution:

We know that

(i) f (x) = (x – 1) (x – 2) (x – 3) is a polynomial which is continuous for all x ϵ R

Hence, f (x) = (x – 1) (x – 2) (x – 3) is continuous on [1, 3]

(ii) f’(x) = (x – 2) (x – 3) + (x – 1) (x – 3) + (x – 1) (x – 2) exist in [1, 3]

Hence, f (x) = (x – 1) (x – 2) (x – 3) is differentiable on (1, 3)

(iii) We know that

f (1) = (1 – 1) (1 – 2) (1 – 3) = 0

Similarly f (3) = (3 – 1) (3 – 2) (3 – 3) = 0

Here f (1) = f(3)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (1, 3) where f’(c) = 0

So we get

(c – 2) (c – 3) + (c – 1) (c – 3) + (c – 1) (c – 2) = 0

On further calculation

(c – 3) (2c – 3) + (c – 1) (c – 2) = 0

By further simplification

(2c2 – 9c + 9) + (c2 – 3c + 2) = 0

We know that

3c2 – 12c + 11 = 0 where

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11C Image 1

Which gives c = 2.58 or 1.42

We know that value of c = 1.42 ϵ (1, 3) and c = 2.58 ϵ (1, 3)

Therefore, Rolle’s Theorem is satisfied.

10. f (x) = (x – 1) (x – 2)2 in [1, 2]

Solution:

We know that

(i) f (x) = (x – 1) (x – 2)2 is a polynomial which is continuous for all x ϵ R

Hence, f (x) = (x – 1) (x – 2)2 is continuous on [1, 2]

(ii) f’(x) = (x – 2)2 + 2 (x – 1) (x – 2) exist in [1, 2]

Hence, f (x) = (x – 1) (x – 2)2 is differentiable on (1, 2)

(iii) We know that

f (1) = (1 – 1) (1 – 2)2 = 0

Similarly f (2) = (2 – 1) (2 – 2)2 = 0

Here f (1) = f (2)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (1, 2) where f’(c) = 0

So we get

(c – 2)2 + 2 (c – 1) (c – 2) = 0

We know that

(3c – 4) (c – 2) = 0

Which gives c = 2 or 4/3

We know that value of c = 4/3 ϵ (1, 2)

Therefore, Rolle’s Theorem is satisfied.

11. f (x) = (x – 2)4 (x – 3)3 in [2, 3]

Solution:

We know that

(i) f (x) = (x – 2)4 (x – 3)3 is a polynomial which is continuous for all x ϵ R

Hence, f (x) = (x – 2)4 (x – 3)3 is continuous on [2, 3]

(ii) f’(x) = 4 (x – 2)3 (x – 3)3 + 3 (x – 2)4 (x – 3)2 exist in [2, 3]

Hence, f (x) = (x – 2)4 (x – 3)3 is differentiable on (2, 3)

(iii) We know that

f (2) = (2 – 2)4 (2 – 3)3 = 0

Similarly f (3) = (3 – 2)4 (3 – 3)3 = 0

Here f (2) = f (3)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (2, 3) where f’(c) = 0

So we get

4 (c – 2)3 (c – 3)3 + 3 (c – 2)4 (c – 3)2 = 0

We know that

(c – 2)3 (c – 3)2 (7c – 18) = 0

Which gives c = 2 or 3 or 18/7

We know that value of c = 18/7 ϵ (2, 3)

Therefore, Rolle’s Theorem is satisfied.

12. f (x) = √(1 – x2) in [- 1, 1]

Solution:

We know that

(i) f (x) = √(1 – x2) is a polynomial which is continuous for all x ϵ R

Hence, f (x) = √ (1 – x2) is continuous on [- 1, 1]

(ii)
RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11C Image 2exist in [- 1, 1]

Hence, f (x) = √ (1 – x2) is differentiable on (- 1, 1)

(iii) We know that

f (- 1) = √ (1 – (-1)2) = 0

Similarly f (1) = √ (1 – 12) = 0

Here f (-1) = f (1)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (-1, 1) where f’(c) = 0

So we get

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11C Image 3

Which gives c = 0

We know that value of c = 0 ϵ (-1, 1)

Therefore, Rolle’s Theorem is satisfied.

13. f (x) = cos x in [-π/2, π/2]

Solution:

We know that

(i) f (x) = cos x is a trigonometric function which is continuous

Hence, f (x) = cos x is continuous on [-π/2, π/2]

(ii) f’(x) = – sin x exist in [-π/2, π/2]

Hence, f (x) = cos x is differentiable on (-π/2, π/2)

(iii) We know that

f (-π/2) = cos (-π/2) = 0

Similarly f (π/2) = cos (π/2) = 0

Here f (-π/2) = f (π/2)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (-π/2, π/2) where f’(c) = 0

– sin c = 0

Which gives c = 0

We know that value of c = 0 ϵ (-π/2, π/2)

Therefore, Rolle’s Theorem is satisfied.

14. f (x) = cos 2x in [0, π]

Solution:

We know that

(i) f (x) = cos 2x is a trigonometric function which is continuous

Hence, f (x) = cos 2x is continuous on [0, π]

(ii) f’(x) = – 2sin 2x exist in [0, π]

Hence, f (x) = cos 2x is differentiable on (0, π)

(iii) We know that

f (0) = cos 2(0) = 0

Similarly f (π) = cos 2(π) = 0

Here f (0) = f (π)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (0, π) where f’(c) = 0

– 2sin 2c = 0

Which gives 2c = π

We know that value of c = π/2 ϵ (0, π)

Therefore, Rolle’s Theorem is satisfied.


Exercise 11d PAGE: 490

Verify Lagrange’s mean-value theorem for each of the following functions:

1. f (x) = x2 + 2x + 3 on [4, 6]

Solution:

We know that f (x) = x2 + 2x + 3 on [4, 6] is a polynomial function which is continuous.

Similarly f’(x) = 2x + 2 exists for all x in (4, 6) which is differential.

Hence, both the condition of Lagrange’s mean value theorem are satisfied

There exists c ϵ (4, 6) such that

f (a) = f (4) = 42 + 2 (4) + 3 = 27

f (b) = f (6) = 62 + 2 (6) + 3 = 51

It can be written as

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11D Image 1

By substituting the values

2c + 2 = (51 – 27)/ 2

On further calculation

2c + 2 = 12

We can write it as

2c = 10

By division

c = 5 ∈ (4, 6)

Therefore, Lagrange’s mean-value theorem is verified.

2. f (x) = x2 + x – 1 on [0, 4]

Solution:

We know that f (x) = x2 + x – 1 on [0, 4] is a polynomial function which is continuous.

Similarly f’(x) = 2x + 1 exists for all x in (0, 4) which is differential.

Hence, both the condition of Lagrange’s mean value theorem are satisfied

There exists c ϵ (0, 4) such that

f (a) = f (4) = 42 + 4 – 1 = 19

f (b) = f (0) = 02 + 0 – 1 = -1

It can be written as

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11D Image 2

By substituting the values

2c + 1 = 19 – (- 1)/ 4

On further calculation

2c + 1 = 20/4 = 5

We can write it as

2c = 4

By division

c = 2 ∈ (0, 4)

Therefore, Lagrange’s mean-value theorem is verified.

3. f (x) = 2x2 – 3x + 1 on [1, 3]

Solution:

We know that f (x) = 2x2 – 3x + 1 on [1, 3] is a polynomial function which is continuous.

Similarly f’(x) = 4x – 3 exists for all x in (1, 3) which is differential.

Hence, both the condition of Lagrange’s mean value theorem are satisfied

There exists c ϵ (1, 3) such that

f (a) = f (1) = 2(1)2 – 3(1) + 1 = 0

f (b) = f (3) = 2(3)2 – 3(3) + 1 = 10

It can be written as

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11D Image 3

By substituting the values

4c – 3 = 10 – 0/2

On further calculation

4c – 3 = 5

We can write it as

4c = 8

By division

c = 2 ∈ (1, 3)

Therefore, Lagrange’s mean-value theorem is verified.

4. f (x) = x3 + x2 – 6x on [- 1, 4]

Solution:

We know that f (x) = x3 + x2 – 6x on [-1, 4] is a polynomial function which is continuous.

Similarly f’(x) = 3x2 + 2x – 6 exists for all x in (-1, 4) which is differential.

Hence, both the condition of Lagrange’s mean value theorem are satisfied

There exists c ϵ (-1, 4) such that

f (a) = f (-1) = (-1)3 + (-1)2 – 6 (-1) = 6

f (b) = f (4) = 43 + 42 – 6(4) = 56

It can be written as

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11D Image 4

By substituting the values

3c2 + 2c – 6 = 56 – 6/5

On further calculation

3c2 + 2c – 6 = 10

We can write it as

3c2 + 2c – 16 = 0

c (3c + 8) – 2 (3c + 8) = 0

(c – 2) (3c + 8) = 0

where c = 2, -8/3 i.e. c ≠ -8/3

By division

c = 2 ∈ (-1, 4)

Therefore, Lagrange’s mean-value theorem is verified.

5. f (x) = (x – 4) (x – 6) (x – 8) on [4, 10]

Solution:

We know that f (x) = (x – 4) (x – 6) (x – 8) can be written as f (x) = x3 – 18x2 + 104x – 192 on [4, 10] is a polynomial function which is continuous.

Similarly f’(x) = 3x2 – 36x + 104 exists for all x in (4, 10) which is differential.

Hence, both the condition of Lagrange’s mean value theorem are satisfied

There exists c ϵ (4, 10) such that

f (a) = f (4) = (4 – 4) (4 – 6) (4 – 8) = 0

f (b) = f (10) = (10 – 4) (10 – 6) (10 – 8) = 48

It can be written as

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11D Image 5

By substituting the values

3c2 – 36c + 104 = 48 – 0/6

On further calculation

3c2 – 36c + 104 – 8 = 0

We can write it as

3c2 – 36c + 96 = 0

3 (c2 – 12c + 32) = 0

c2 – 12c + 32 = 0

So we get

c2 – 8c – 4c + 32 = 0

Taking c as common

c (c – 8) – 4 (c – 8) = 0

(c – 4) (c – 8) = 0

where c = 4, 8

By division

c = 8 ∈ (4, 10) and c = 4 ∉ (4, 10)

Therefore, Lagrange’s mean-value theorem is verified.

6. f (x) = ex on [0, 1]

Solution:

We know that f (x) = ex on [0, 1] is an exponential function which is continuous.

Similarly f’(x) = ex exists for all x in (0, 1) which is differential.

Hence, both the condition of Lagrange’s mean value theorem are satisfied

There exists c ϵ (0, 1) such that

f (a) = f (0) = e0 = 1

f (b) = f (1) = e1 = e

It can be written as

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11D Image 6

By substituting the values

ec = (e – 1)/ 1 – 0

So we get

ec = (e – 1)

Taking log on both sides

loge ec = loge (e – 1)

where c = loge (e – 1) ϵ (0, 1)

Therefore, Lagrange’s mean-value theorem is verified.

7. f (x) = x2/3 on [0, 1]

Solution:

We know that f (x) = x2/3 on [0, 1] is a polynomial function which is continuous.

Similarly f’(x) = 2/3 x -1/3 exists for all x in (0, 1) which is differential.

Hence, both the condition of Lagrange’s mean value theorem are satisfied

There exists c ϵ (0, 1) such that

f (a) = f (0) = 02/3 = 0

f (b) = f (1) = 12/3 = 1

It can be written as

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11D Image 7

By substituting the values

2/3 c -1/3 = 1 – 0/1 – 0

On further calculation

2/3 1/c 1/3 = 1

We can write it as

c 1/3 = 2/3

So we get

c = (2/3)3 = 8/27 ∈ (0, 1)

Therefore, Lagrange’s mean-value theorem is verified.

8. f (x) = log x on [1, e]

Solution:

We know that f (x) = log x on [1, e] is a logarithmic function which is continuous.

Similarly f’(x) = 1/x exists for all x in (1, e) which is differential.

Hence, both the condition of Lagrange’s mean value theorem are satisfied

There exists c ϵ (1, e) such that

f (a) = f (1) = log 1 = 0

f (b) = f (e) = log e = 1

It can be written as

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11D Image 8

By substituting the values

1/c = 1 – 0/e – 1

On further calculation

1/c = 1/e – 1

We can write it as

c = e – 1 ∈ (1, e)

Therefore, Lagrange’s mean-value theorem is verified.

9. f (x) = tan -1 x on [0, 1]

Solution:

We know that f (x) = tan -1 x on [0, 1] is a trigonometric function which is continuous.

Similarly f’(x) = 1/1 + x2 exists for all x in (0, 1) which is differential.

Hence, both the condition of Lagrange’s mean value theorem are satisfied

There exists c ϵ (0, 1) such that

f (a) = f (0) = tan -1 (0) = 0

f (b) = f (1) = tan -1 (1) = π/4

It can be written as

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11D Image 9

By substituting the values

1/1 + c2 = π/4 – 0/1

On further calculation

1/1 + c2 = π/4

We can write it as

1 + c2 = 4/ π

Where

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11D Image 10∈ (0, 1)

Therefore, Lagrange’s mean-value theorem is verified.


Exercise 11E page: 506

Find the maximum or minimum values, if any, without using derivatives, of the functions:

1. (5x – 1)2 + 4

Solution:

It is given that f (x) = (5x – 1)2 + 4 with minimum value 4 where 5x – 1 = 0

Here the maximum value cannot be found i.e. when x increases even f (x) increases

Therefore, f (x) does not have a maximum value.

2. – (x – 3)2 + 9

Solution:

It is given that f (x) = – (x – 3)2 + 9

Here (x – 3)2 ≥ 0

We can write it as

– (x – 3)2 ≤ 0

By adding 9 on both sides

– (x – 3)2 + 9 ≤ 0 + 9

Where f (x) ≤ 9

So the maximum value of f (x) = 9 which occurs at x = 3 which is the point of absolute maxima and minimum value does not exist.

3. – |x + 4| + 6

Solution:

We can write it as

|x + 4| ≥ 0

By taking – ve sign

– |x + 4| ≤ 0

By adding 6 on both sides

– (x + 4) + 6 ≤ 0 + 6

Where f (x) ≤ 6

Here the maximum value of f (x) = 6 which occurs at x = – 4 which is the point of absolute maxima and minimum value of f (x) does not exists.

4. sin 2x + 5

Solution:

For all x

– 1 ≤ sin 2x ≤ 1

By adding 5

5 – 1 ≤ sin 2x + 5 ≤ 5 + 1

On further calculation

4 ≤ f (x) ≤ 6

Here the minimum value of f (x) = 4 and minimum value of f (x) = 6.

5. |sin 4x + 3|

Solution:

For all x

– 1 ≤ sin 4x ≤ 1

By adding 3

3 – 1 ≤ sin 4x + 3 ≤ 1 + 3

On further calculation

2 ≤ sin 4x + 3 ≤ 4

We know that

|2| ≤ |sin 4x + 3| ≤ |4|

2 ≤ f (x) ≤ 4

Here the minimum value of f (x) = 2 and maximum value of f (x) = 4.

Find the points of local maxima or local minima and the corresponding local maximum and minimum values of each of the following functions:

6. f (x) = (x – 3)4

Solution:

It is given that

f (x) = (x – 3)4

By differentiating w.r.t. x

f’(x) = 4 (x – 3)3

We know that

f’(x) = 0

By substituting the values

4 (x – 3)3 = 0

So we get

x – 3 = 0 where x = 3

x = 3 is point of local minima and local minimum value is f (3) = 0

7. f (x) = x2

Solution:

It is given that

f (x) = x2

By differentiating w.r.t. x

f’(x) = 2x and f”(x) = 2

We know that

f’(x) = 0

By substituting the values

2x = 0 where x = 0

f”(0) = 2 > 0

Hence, x = 0 is point of local minimum and f (0) = 0 is the local minimum value.

8. f (x) = 2x3 – 21x2 + 36x – 20

Solution:

It is given that

f (x) = 2x3 – 21x2 + 36x – 20

By differentiating w.r.t. x

f’(x) = 6x2 – 42x + 36

We know that

f’(x) = 0

By substituting the values

6x2 – 42x + 36 = 0

By taking 6 as common

6 (x2 – 7x + 6) = 0

It can be written as

6 (x2 – 6x – x + 6) = 0

By further simplification

6 [x (x – 6) – 1 (x – 6)] = 0

So we get

6 (x – 6) (x – 1) = 0

x = 1, 6

Similarly

f’’(x) = 2x – 42

x = 1 is a point of local maximum and

f (1) = 2(1)3 – 21(1)2 + 36(1) – 20 = -3 is the local maximum value

Similarly x = 6 is a point of local minimum and

f’’(6) = 30 is the local minimum value.

f (6) = 2(6)3 – 21(6)2 + 36(6) – 20 = – 128

9. f (x) = x3 – 6x2 + 9x + 15

Solution:

It is given that

f (x) = x3 – 6x2 + 9x + 15

By differentiating w.r.t. x

f’ (x) = 3x2 – 12x + 9

On further differentiation

f’’(x) = 6x – 12

We know that

f’(x) = 0

By substituting the values

3x2 – 12x + 9 = 0

By taking 3 as common

3 (x2 – 4x + 3) = 0

It can be written as

3 (x – 1) (x – 3) = 0 where x = 1, 3

Here x = 1 is the point of local maximum where f” (1) = – 6

f (1) = 13 – 6 (1)2 + 9(1) + 15 = 19 is the local maximum value

x = 3 is the point of local minimum and

f (3) = 33 – 6 (3)2 + 9(3) + 15 = 15 is the local minimum value

10. f (x) = x4 – 62x2 + 120x + 9

Solution:

It is given that

f (x) = x4 – 62x2 + 120x + 9

By differentiating w.r.t. x

f’(x) = 4x3 – 124x + 120

On further differentiation

f’’(x) = 12x2 – 124

We know that

f’(x) = 0

By substituting the values

4x3 – 124x + 120 = 0

By taking 4 as common

4 (x3 – 31x + 30) = 0

So we get

4 (x – 1) (x2 + x – 30) = 0

It can be written as

4 (x – 1) (x + 6) (x – 5) = 0

Where x = – 6, 1, 5

If x = – 6 is a point of local minimum

f (- 6) = (-6)4 – 62 (-6)2 + 120 (-6) + 9 = – 1647 is the local minimum value

If x = 1 is a point of local maximum

f (1) = (1)4 – 62 (1)2 + 120 (1) + 9 = 68 is the local maximum value

If x = 5 is a point of local minimum

f (5) = (5)4 – 62 (5)2 + 120 (5) + 9 = – 316

11. f (x) = – x3 + 12x2 – 5

Solution:

It is given that

f (x) = – x3 + 12x2 – 5

By differentiating w.r.t. x

f’(x) = – 3x2 + 24x

On further differentiation

f’’(x) = – 6x + 24 = -6 (x – 4)

We know that

f’(x) = 0

By substituting the values

– 3x2 + 24x = 0

By taking – 3x as common

– 3x (x – 8) = 0

Here x = 0, 8

By substituting the values

f’’(0) = -6 (0 – 4) = 24 > 0

f” (8) = – 6 (8 – 4) = – 24 < 0

If x = 0 is a point of local minimum

f (0) = – (0)3 + 12 (0) – 5 = – 5 is the local minimum value

If x = 8 is a point of local maximum

f (8) = – (8)3 + 12 (8)2 – 5 = 251 is the local maximum value


Exercise 11F page: 521

1. Find two positive numbers whose product is 49 and the sum is minimum.

Solution:

Consider the number as x and 49/x

We know that

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 1

Here x = 7 is a point of minimum and 7, 7 are the required numbers.

2. Find two positive numbers whose sum is 16 and the sum of whose squares is minimum.

Solution:

Consider the number as x and 16 – x

We know that

S = x2 + (16 – x) 2

It can be written as

S = x2 + 256 – 32x + x2 = 2x2 – 32x + 256

By differentiating w.r.t. x

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 2

x = 8 is the point of minimum and 8, 8 are the required numbers.

3. Divide 15 into two parts such that the square of one number multiplied with the cube of the other number is maximum.

Solution:

Consider the number as x and 15 – x

We know that

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 3

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 4

When x = 9 the product is maximum.

Hence, 9, 6 are the required numbers.

4. Divide 8 into two positive parts such that the sum of the square of one and the cube of the other is minimum.

Solution:

Consider the number as x and 8 – x

We know that

S = x2 + (8 – x) 3

By differentiating w.r.t. x

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 5

Hence, x = 6 is a point of minimum and 6, 2 are the required numbers.

5. Divide a into two parts such that the product of the pth power of one part and the qth power of the second part may be maximum.

Solution:

Consider the number as x and a – x

We know that

y = xp (a – x) q

By differentiating w.r.t. x

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 6

Here, y is maximum for x = ap/ p + q

Hence, x = ap/ p + q and y = aq/ p + q are the two positive numbers required.

6. The rate of working of an engine is given by

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 7

and v is the speed of the engine. Show that R is the least when v = 20.

Solution:

It is given that

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 8

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 9

Hence, v = 20 is a point of minimum.

7. Find the dimensions of the rectangle of area 96 cm2 whose perimeter is the least. Also, find the perimeter of the rectangle.

Solution:

Consider P as the fixed perimeter and x, y as the sides.

So area = xy = 96

It can be written as

y = 96/x

Similarly perimeter can be written as

P = 2 (x + y)

By substituting the value of y

P = 2 (x + 96/x)

By differentiating w.r.t. x

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 10

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 11

8. Prove that the largest rectangle with a given perimeter is a square.

Solution:

Consider a as the fixed perimeter.

Take a rectangle having x and y as sides, a as perimeter and A as the area of the rectangle

So we get

2x + 2y = a

It can be written as

A = xy

Substituting the value of y

A = x [(a – 2x)/ 2]

By differentiating w.r.t. x

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 12

We get

y = ½ (a – 2x)

It can be written as

y = ½ (a – a/2) = a/4

We know that x = y = a/4

Therefore, the rectangle is a square.

9. Given the perimeter of a rectangle, show that its diagonal is minimum when it is a square.

Solution:

Consider P as the fixed perimeter where x and y are the sides, P is the perimeter and D is the diagonal.

P = 2 (x + y)

It can be written as

y = ½ (P – 2x)

Here D2 = Z = x2 + y2

By further simplification

Z = x2 + ¼ (P – 2x) 2

By differentiating w.r.t. x

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 13

Where the diagonal is minimum at x = p/4

Similarly

y = ½ (p – 2x)

By substituting the values

y = ½ (p – p/2) = p/4

We know that x = y = p/4

Therefore, the rectangle is a square.

10. Show that a rectangle of maximum perimeter which can be inscribed in a circle of radius a is a square of side √2 a.

Solution:

Consider ABCD as a rectangle which is inscribed in a circle having O as centre and a as radius

Now join OC where ∠COX = θ

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 14

So the co-ordinate of C are (a cos θ, a sin θ)

We get OM = a cos θ, MC = a sin θ

BC = 2MC = 2a and CD = 2OM = 2a cos θ

It can be written as

P = 2a (cos θ + sin θ)

By differentiating w.r.t. x

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 15

Here, P is maximum at θ = π/4

BC = √2a = CD

11. The sum of the perimeters of a square and a circle is given. Show that the sum of their areas is least when the side of the square is equal to the diameter of the circle.

Solution:

Consider the side of square as x and radius of circle as r

It is given that

4x + 2 πr = k

We can write it as

x = (k – 2 πr)/ 4

We know that

A = x2 + πr2 = [(k – 2 πr)/ 4]2 + πr2

So we get

A = 1/16 (k2 – 4k πr + 4 π2 r2) + πr2

By differentiating w.r.t. r

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 16

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 17

Hence, A is least when side of square is equal to the diameter of the circle.

12. Show that the right triangle of maximum area that can be inscribed in a circle is an isosceles triangle.

Solution:

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 18

We know that

AC = 2r cos θ

AB = 2r sin θ

It can be written as

△ABC = ½ × 2r cos θ × 2r sin θ

So we get

△ABC = r2 sin 2θ

Here the area is maximum when

sin 2θ = 1

So we get

2θ = 90o

Where θ = 45o

Therefore, △ABC is an isosceles triangle.

13. Prove that the perimeter of a right-angled triangle of given hypotenuse is maximum when the triangle is isosceles.

Solution:

Consider ABC as a right triangle with h as hypotenuse.

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 19

Take BC = x as base and BA = a as altitude.

It can be written as

h2 = x2 + a2

By further simplification

a = (h2 – x2)

We know that

P = a + x + h

By substituting the value of a

P = (h2 – x2) + x + h

By differentiating w.r.t. x

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 20

So we get

x = (h2 – x2)

By squaring on both sides

x2 = h2 – x2

So we get

x2 = h2/2

where x = h/ 2

For the value x = h/ 2

We know that

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 21

Therefore, P is maximum when x = h/ 2 and a = h/ 2 where x = a.

14. The perimeter of a triangle is 8 cm. If one of the sides of the triangle be 3 cm, what will be the other two sides for maximum area of the triangle?

Solution:

Consider x and 5 – x as the sides

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 22

We know that

S = (x + 5 – x + 3)/ 2 = 8/2 = 4

Here the area of △ABC can be written as

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 23

On further simplification

A = 2 (4 – x) (x – 1)

By multiplication

A = 2 (4x – 4 – x2 + x)

We get

A = 2 (5x – 4 – x2)

Now we have to find the maximum value of 5x – 4 – x2

Consider y = – x2 + 5x – 4

By differentiating w.r.t. x

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 24

Here y will be maximum when x = 5/2

Therefore, the area will be maximum when x = 5/2 with sides 5/2, 5/2.

15. A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 metres. Find the dimensions of the window to admit maximum light through it.

Solution:

Consider x metres as the length and y metres as the breadth of the rectangle

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 25

Here the radius of semicircle is x/2 metres

Perimeter of the window can be written as

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 26

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 27

16. A square piece of tin of side 12 cm is to be made into a box without a lid by cutting a square from each corner and folding up the flaps to form the sides. What should be the side of the square to be cut off so that the volume of the box is maximum? Also, find this maximum volume.

Solution:

Consider x cm as the length cut off from each corner

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 28

So the length of box is (12 – 2x) cm and x cm as its breadth and height

Here volume can be written as

V = (12 – 2x) 2 × x

On further simplification

V = 4x3 – 48x2 + 144x

By differentiating w.r.t. x

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 29

We know that V is maximum when x = 2

Here V = {(12 – 4)2 × 2} = {64 × 2} = 128 cm3

17. An open box with a square base is to be made out of a given cardboard of area c2 (square) units. Show that the maximum volume of the box is c3/ 6√3 (cubic) units.

Solution:

Consider each side of base a and h as the height

So we get c2 = (a2 + 4ah)

It can be written as

h = (c2 – a2)/ 4a

Here volume can be written as

V = a2 × h

By substituting the value of h

V = (c2a – a3)/ 4

By differentiating w.r.t a

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11F Image 30

Which is lesser than 0

Hence, V is maximum when a2 = c2/3 and h = c/23


Exercise 11G page: 549

1. Show that the function f (x) = 5x – 2 is a strictly increasing function on R.

Solution:

It is given that

f (x) = 5x – 2

By differentiating w.r.t. x

f’(x) = 5 > 0 for all x > 0

Therefore, f (x) is strictly increasing function for all x > 0.

2. Show that the function f (x) = -2x + 7 is a strictly decreasing function on R.

Solution:

It is given that

f (x) = -2x + 7

By differentiating w.r.t. x

f’ (x) = – 2 < 0 for all value of x

Therefore, f (x) is strictly decreasing function on R.

3. Prove that f(x) = ax + b, where a and b are constants an a > 0, is a strictly increasing function on R.

Solution:

It is given that

f(x) = ax + b

By differentiating w.r.t. x

f’ (x) = a > 0 for all x

Therefore, f (x) is strictly increasing function on R.

4. Prove that the function f (x) = e2x is strictly increasing on R.

Solution:

It is given that

f (x) = e2x

By differentiating w.r.t. x

f’ (x) = 2 (e2x) > 0

Therefore, f (x) is strictly increasing function on R.

5. Show that the function f (x) = x2 is

(a) strictly increasing on [0, ∞[

(b) strictly decreasing on ]- ∞, 0[

(c) neither strictly increasing nor strictly decreasing on R

Solution:

(a) It is given that f (x) = x2

By differentiating w.r.t x

f’ (x) = 2x > 0 for x ∈ (0, ∞)

Here f’ (x) > 0 which means 2x > 0

So we get x > 0

(b) We know that

If f’ (x) < 0 we get x < 0

(c) It is given that f (x) = x2 is strictly increasing on [0, ∞[ and strictly decreasing on ]- ∞, 0[

So it is neither strictly increasing nor strictly decreasing on the whole real line.

6. Show that the function f (x) = |x| is

(a) strictly increasing on ]0, ∞[

(b) strictly decreasing on ]- ∞, 0[

Solution:

It is given that f (x) = |x|

We can write it as

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11G Image 1

So f’ (x) > 0 for x ∈ (0, ∞)

(a) Here f (x) is strictly increasing at x ∈ (0, ∞) and f’ (x) < 0 for all x ∈ (-∞, 0)

(b) Here f’ (x) is strictly decreasing at x ∈ (-∞, 0).

7. Prove that the function f (x) = loge x is strictly increasing on] 0, ∞ [.

Solution:

It is given that

f (x) = loge x

By differentiating w.r.t x

f’ (x) = 1/x

Here 1/x > 0 for x ∈ (0, ∞)

Similarly f’ (x) > 0 for x ∈ (0, ∞) where f’ (x) is strictly increasing on] 0, ∞ [.

8. Prove that the function f(x) = loga x is strictly increasing on] 0, ∞ [when a > 1 and strictly decreasing on] 0, ∞ [ when 0 < a < 1.

Solution:

It is given that

f(x) = loga x

(a) By differentiating w.r.t x

f’ (x) = 1/x log a > 0 for x ∈ (0, ∞) and a > 1

Here f’ (x) > 0 for x ∈ (0, ∞) and a > 1

Therefore, f’ (x) is strictly increasing function.

(b) We know that

1/ x log a < 0 for x ∈ (0, ∞) where 0 < a < 1

So f’ (x) < 0

Therefore, f’ (x) is strictly decreasing on (0, ∞) and 0 < a < 1.

9. Prove that f (x) = 3x is strictly increasing on R.

Solution:

It is given that

f (x) = 3x

By differentiating w.r.t x

f’ (x) = 3x log 3 > 0 for all value of R

Therefore, f (x) is strictly increasing function on R.

10. Show that f (x) = x3 – 15x2 + 75 x – 50 is increasing on R.

Solution:

It is given that

f (x) = x3 – 15x2 + 75 x – 50

By differentiating w.r.t x

f’ (x) = 3x2 – 30x + 75

By taking 3 as common

f’ (x) = 3 (x2 – 10x + 25)

On further simplification

f’ (x) = 3 (x – 5)2

Here f’ (x) ≥ 0 for all x ≥ 0

Therefore, f (x) is increasing function on R.

11. Show that f (x) = [x – 1/x] is increasing for all x ∈ R, where x ≠ 0.

Solution:

It is given that

f (x) = [x – 1/x]

By differentiating w.r.t. x

f’ (x) = 1 + 1/x2

On further calculation

f’ (x) = (x2 + 1)/ x2

Here f’ (x) ≥ 0 for all x ≠ 0

Therefore, f(x) is increasing function for all x ∈ R, where x ≠ 0.

12. Show that f (x) = [3/x + 5] is decreasing for all x ∈ R, where x ≠ 0.

Solution:

It is given that

f (x) = [3/x + 5]

By differentiating w.r.t. x

f’ (x) = 3 (-1/x2) = -3/x2 < 0 for all value of R

Therefore, f (x) is decreasing for all x ∈ R, where x ≠ 0.

13. Show that f(x) = 1/ (1 + x2) is increasing for all x ≤ 0.

Solution:

It is given that

f(x) = 1/ (1 + x2)

By differentiating w.r.t. x

f’ (x) = – 2x/ (1 + x2) 2 ≤ 0

Therefore, f (x) is increasing function for all x ≤ 0.

14. Show that f (x) = [x3 + 1/x3] is decreasing on]-1, 1[.

Solution:

It is given that

f (x) = [x3 + 1/x3]

By differentiating w.r.t. x

f’ (x) = 3x2 – 3x -4

By taking 3 as common

f’ (x) = 3 [x2 – 1/x4]

Taking LCM

f’ (x) = 3 [(x6 – 1)/ x4] = 3 [(x2)3 – 1]/ x4

On further calculation

f’ (x) = [3 (x2 – 1) (x4 + x2 + 1)]/ x4

So we get

f’ (x) = [3 (x – 1) (x + 1) (x4 + x2 + 1)]/ x4 < 0 for x ∈ (- 1, 1).

Therefore, f (x) is decreasing function on]-1, 1[.

15. Show that f (x) = x/sinx is increasing on] 0, π/2[.

Solution:

It is given that

f (x) = x/sinx

By differentiating w.r.t. x

f’ (x) = (sin x. 1 – x. cos x)/ (sin x)2 = (sin x – x cos x)/ (sin x)2

We know that

f’ (x) > 0

By substituting the values

(sin x – x cos x)/ (sin x) 2 > 0

So we get

(sin x – x cos x) > 0

Here tan x > x which is true on] 0, π/2[.

16. Prove that the function f (x) = log (1 + x) – 2x/ (x + 2) is increasing for all x > -1.

Solution:

It is given that

f (x) = log (1 + x) – 2x/ (x + 2)

By differentiating w.r.t x

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11G Image 2

Therefore, f (x) is increasing function for all x > – 1.


Exercise 11H PAGE: 564

1. Find the slope of the tangent to the curve

(i) y = (x3 – x) at x = 2

(ii) y = (2x2 + 3 sin x) at x = 0

(iii) y = (sin 2x + cot x + 2)2 at x = π/2

Solution:

(i) y = (x3 – x) at x = 2

Consider y = (x3 – x) as the equation of curve

By differentiating both sides w.r.t x

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11H Image 1

(ii) y = (2x2 + 3 sin x) at x = 0

Consider y = (2x2 + 3 sin x) as the equation of curve

By differentiating both sides w.r.t x

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11H Image 2

(iii) y = (sin 2x + cot x + 2)2 at x = π/2

Consider y = (sin 2x + cot x + 2)2 as the equation of curve

By differentiating both sides w.r.t x

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11H Image 3

Find the equations of the tangent and the normal to the given curve at the indicated point:

2. y = x3 – 2x + 7 at (1, 6)

Solution:

Consider y = x3 – 2x + 7 as the equation of curve

By differentiating both sides w.r.t x

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11H Image 4

We know that

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11H Image 5

Where m = 1

So the required equation of the tangent at point (1, 6),

(y – y1) = m (x – x1)

By further simplification

(y – y1)/ (x – x1) = m

By substituting the values

(y – 6)/ (x – 1) = 1

By cross multiplication

y – 6 = x – 1

So we get

y – 6 + 1 = x

y – 5 = x

It can be written as

x – y + 5 = 0

Here the required equation of normal at point (1, 6)

(y – y1) = -1/m (x – x1)

By cross multiplication

(y – y1)/ (x – x1) = -1/m

Substituting the values

(y – 6)/ (x – 1) = -1/1

It can be written as

y – 6 = -x + 1

By further simplification

x + 2y – 6 – 1 = 0

x + y – 7 = 0

3. y2 = 4ax at (a/m2, 2a/m)

Solution:

Consider y2 = 4ax as the equation of curve

By differentiating both sides w.r.t x

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11H Image 6

By substituting the values we get

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11H Image 7

It can be written as

m (ym2 – 2am) = – m2x + a

By multiplying inside the bracket

ym3 – 2am2 = – m2x + a

We get

m2x +ym3 – 2am2 – a = 0

4. RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11H Image 8

Solution:

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11H Image 9

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11H Image 10

On further calculation

yb cos θ – b2 sin θ cos θ = ax sin θ – a2 sin θ cos θ

It can be written as

a2 sin θ cos θ – b2 sin θ cos θ = ax sin θ – by cos θ

By taking sin θ cos θ as common

sin θ cos θ (a2 – b2) = ax sin θ – by cos θ

It can be written as

ax sec θ – by cosec θ = a2 – b2

5. RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11H Image 11

Solution:

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11H Image 12

By substituting the values

y – b tan θ = – (a tan θ/ b sec θ) (x – a sec θ)

On further calculation

by sec θ – b2 tan θ sec θ = – ax tan θ + a2 sec θ tan θ

By taking sec θ tan θ as common

ax tan θ + by sec θ = sec θ tan θ (a2 + b2)

So we get

ax cos θ + by cot θ = a2 + b2

6. y = x3 at P (1, 1).

Solution:

Consider y = x3 as the equation of curve

By differentiating both sides w.r.t. x

dy/dx = 3x2

By substituting the values

(dy/dx)(1, 1) = 3 (1)2 = 3

Here the required equation of tangent at point (1, 1)

y – y1 = m (x – x1)

We can write it as

(y – y1)/ (x – x1) = m

By substituting the values

(y – 1)/ (x – 1) = 3

By cross multiplication

y – 1 = 3x – 3

So we get

3x – y – 2 = 0

Here the required equation of normal at point (1, 1)

y – y1 = – 1/m (x – x1)

We can write it as

(y – y1)/ (x – x1) = – 1/m

By substituting the values

(y – 1)/ (x – 1) = – 1/3

By cross multiplication

3y – 3 = – x + 1

So we get

x + 3y – 3 – 1 = 0

x + 3y – 4 = 0

7. y2 = 4ax at (at2, 2at)

Solution:

Consider y2 = 4ax as the equation of curve

By differentiating both sides w.r.t. x

2y dy/dx = 4a

We can write it as

dy/dx = 2a/ y

By substituting the values

(dy/dx)(at2, 2at) = 2a/2at = 1/t

Here the equation of tangent at point (at2, 2at)

y – 2at = 1/t (x – at2)

On further calculation

yt – 2at2 = x – at2

So we get

x – ty + at2 = 0

Here the equation of normal at point (at2, 2at)

y – 2at = – t (x – at2)

By further calculation

y – 2at = – tx + at3

We can write it as

y + tx = at3 + 2at

So we get

tx + y = at3 + 2at

8. y = cot2 x – 2 cot x + 2 at x = π/4

Solution:

Consider y = cot2 x – 2 cot x + 2 as the equation of curve

Take the value of x = π/4 so we get y = 1

By differentiating both sides w.r.t. x

dy/dx = 2 cot x (- cosec2 x) – 2 (- cosec2 x)

By further simplification

dy/dx = – 2 cot x cosec2 x + 2 cosec2 x

We get

(dy/dx)(x = π/4) = 0

Here the equation of tangent at point (π/4, 1)

y – 1 = 0 (x – π/4)

So we get

y – 1 = 0

Here y = 1

Here the equation of normal at point (π/4, 1)

y – π/4 = 1/0 (x – π/4)

We get

0 = x – π/4

Here x = π/4

9. 16x2 + 9y2 = 144 at (2, y1), where y1 > 0

Solution:

Consider 16x2 + 9y2 = 144 as the equation of curve

Taking x1 = 2 we get y1 = 45/3

By differentiating both sides w.r.t. x

32x + 18y dy/dx = 0

We can write it as

18y dy/dx = – 32x

So we get

dy/dx = -16x/ 9y

Here

(dy/dx)(2, 45/3) = (- 16×2)/ (9 × 45/3) = – 8/ 35

Here the equation of tangent at point (2, 45/3)

y – 45/3 = – 8/ 35 (x – 2)

On further calculation

35y – 20 = -8x + 16

We get

8x + 35y – 36 = 0

Here the equation of the normal at point (2, 45/3)

y – 45/3 = 35/8 (x – 2)

It can be written as

8y – 325/3 = 35x – 65

On further calculation

24y – 325 = 95x – 185

So we get

95x – 24y + 145 = 0

10. y = x4 – 6x3 + 13x2 – 10x + 5 at the point where x = 1

Solution:

It is given that

y = x4 – 6x3 + 13x2 – 10x + 5

By substituting the value of x = 1

y = (1)4 – 6(1)3 + 13(1)2 – 10(1) + 5

On further calculation

y = 1 – 6 + 13 – 10 + 5 = 3

Point of contact is (1, 3)

Consider y = x4 – 6x3 + 13x2 – 10x + 5 as the equation of curve

By differentiating w.r.t. x

dy/dx = 4x3 – 18x2 + 26x – 10

So we get

(dy/dx)(1, 3) = 4(1)3 – 18(1)2 + 26(1) – 10 = 2

Here the required equation of tangent is

y – y1 = m (x – x1)

By substituting the values

y – 3 = 2x – 2

On further calculation

2x – y – 2 + 3 = 0

So we get

2x – y + 1 = 0

Here the required equation of normal i

y – y1 = -1/m (x – x1)

By substituting the values

y – 3 = -1/2 (x – 1)

On further calculation

2y – 6 = – x + 1

So we get

2y – 6 + x – 1 = 0

x + 2y – 7 = 0

11. Find the equation of the tangent to the curve √x + √y = a at (a2/4, a2/4).

Solution:

Consider √x + √y = a as the equation of curve

By differentiating both sides w.r.t x

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11H Image 13

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11H Image 14

12. Show that the equation of the tangent to the hyperbola x2/a2 – y2/b2 = 1 at (x1, y1) is xx1/a2 – yy1/b2 = 1.

Solution:

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11H Image 15

RS Aggarwal Solutions for Class 12 Chapter 11 Ex 11H Image 16

13. Find the equation of the tangent to the curve y = (sec4 x – tan4 x) at x = π/3.

Solution:

It is given that y = (sec4 x – tan4 x)

By substituting the value of x = π/3

y = (sec4 (π/3) – tan4 (π/3)) = 24 – (√3)4 = 7

By differentiating both sides w.r.t x

dy/dx = 4 sec3 x sec x tan x – 4 tan3 x sec2 x

It can be written as

dy/dx = 4 sec2 x tan x (sec2 x – tan2 x) = 4 sec2 x tan x

So (dy/dx)x= π/3 = 4 sec2 (π/3) tan (π/3) = 16 √3

Here the equation of tangent at point (π/3, 7)

y – 7 = 16 √3 (x – π/3)

On further calculation

y – 7 = 16 √3x – 16 √3 π/3

By taking LCM as 3

y – 7 = (48√3x – 16√3 π)/ 3

We get

3y – 21 = 48√3x – 16√3 π

We can write it as

48√3x – 3y – 16√3 π + 21 = 0

Taking negative sign as common

3y – 48√3x + 16√3 π – 21 = 0

14. Find the equation of the normal to the curve y = (sin 2x + cot x + 2) 2 at x = π/2.

Solution:

Consider y = (sin 2x + cot x + 2) 2 as the equation of the curve

By substituting the value of x = π/2 we get y = 4

On differentiation of both sides w.r.t. x

dy/dx = 2 (sin 2x + cot x + 2) (2 cos 2x – cosec2 x)

We get

(dy/dx)x= π/2 = – 12

Here the equation of normal at point (π/2, 4)

y – 4 = 1/12 (x – π/2)

On further calculation

12y – 48 = x – π/2

We get

24y – 96 = 2x – π

It can be written as

24y – 96 – 2x + π = 0

24y – 2x + π – 96 = 0

15. Show that the tangents to the curve y = 2x3 – 4 at the points x = 2 and x = – 2 are parallel.

Solution:

Consider y = 2x3 – 4 as the equation of curve

By differentiating both sides w.r.t. x

dy/dx = 6x2

We get

(dy/dx)x=2 = 6 (2)2 = 24

When m1 = 24

dy/dx = 6x2

(dy/dx)x=2 = 6 (2)2 = 24

Similarly when m2 = 24

dy/dx = 6x2

(dy/dx)x=-2 = 6 (-2)2 = 24

Hence the tangents to the curve at the points x = 2 and x = -2 are parallel i.e. m1 = m2.

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