RD Sharma Solutions For Class 12 Maths Exercise 15.2 Chapter 15 Mean Value Theorems

Exercise 15.2 of Chapter 15 consists of problems based on verification of Lagrange’s theorem. Students who are unable to solve exercise-wise problems as per RD Sharma textbook can make use of solutions designed by expert faculty at BYJU’S. These solutions are described in the best possible way to provide a clear idea about the concepts to students. The PDF of RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorem Exercise 15.2 is provided here. This PDF can be easily downloaded by the students. Some of the important topics of this exercise are listed below.

  • Lagrange’s mean value theorem
  • Geometrical interpretation of Lagrange’s mean value theorem
  • Verification of Lagrange’s mean value theorem
  • Proving inequalities by using Lagrange’s mean value theorem
  • Miscellaneous applications of Lagrange’s mean value theorem

RD Sharma Solutions For Class 12 Mean Value Theorems Exercise 15.2:Download PDF

RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems Exercise 15.2
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Exercise 15.1 Solutions

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Page No: 15.17

1. Verify Lagrange’s mean value theorem for the following functions on the indicated intervals. In each case find a point ‘c’ in the indicated interval as stated by the Lagrange’s mean value theorem:

(i) f (x) = x2 – 1 on [2, 3]

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 72

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 73

(ii) f (x) = x3 – 2x2 – x + 3 on [0, 1]

Solution:

Given f (x) = x3 – 2x2 – x + 3 on [0, 1]

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous in [0, 1] and differentiable in (0, 1). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

Therefore, there exist a point c ∈ (0, 1) such that:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 74

f (x) = x3 – 2x2 – x + 3

Differentiating with respect to x

f’(x) = 3x2 – 2(2x) – 1

= 3x2 – 4x – 1

For f’(c), put the value of x=c in f’(x)

f’(c)= 3c2 – 4c – 1

For f (1), put the value of x = 1 in f(x)

f (1)= (1)3 – 2(1)2 – (1) + 3

= 1 – 2 – 1 + 3

= 1

For f (0), put the value of x=0 in f(x)

f (0)= (0)3 – 2(0)2 – (0) + 3

= 0 – 0 – 0 + 3

= 3

∴ f’(c) = f(1) – f(0)

⇒ 3c2 – 4c – 1 = 1 – 3

⇒ 3c2 – 4c = 1 + 1 – 3

⇒ 3c2 – 4c = – 1

⇒ 3c2 – 4c + 1 = 0

⇒ 3c2 – 3c – c + 1 = 0

⇒ 3c(c – 1) – 1(c – 1) = 0

⇒ (3c – 1) (c – 1) = 0

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 75

(iii) f (x) = x (x – 1) on [1, 2]

Solution:

Given f (x) = x (x – 1) on [1, 2]

= x2 – x

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous in [1, 2] and differentiable in (1, 2). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

Therefore, there exist a point c ∈ (1, 2) such that:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 76

f (x) = x2 – x

Differentiating with respect to x

f’(x) = 2x – 1

For f’(c), put the value of x=c in f’(x):

f’(c)= 2c – 1

For f (2), put the value of x = 2 in f(x)

f (2) = (2)2 – 2

= 4 – 2

= 2

For f (1), put the value of x = 1 in f(x):

f (1)= (1)2 – 1

= 1 – 1

= 0

∴ f’(c) = f(2) – f(1)

⇒ 2c – 1 = 2 – 0

⇒ 2c = 2 + 1

⇒ 2c = 3

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 77

(iv) f (x) = x2 – 3x + 2 on [-1, 2]

Solution:

Given f (x) = x2 – 3x + 2 on [– 1, 2]

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous in [– 1, 2] and differentiable in (– 1, 2). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

Therefore, there exist a point c ∈ (-1, 2) such that:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 78

f (x) = x2 – 3x + 2

Differentiating with respect to x

f’(x) = 2x – 3

For f’(c), put the value of x = c in f’(x):

f’(c)= 2c – 3

For f (2), put the value of x = 2 in f(x)

f (2) = (2)2 – 3 (2) + 2

= 4 – 6 + 2

= 0

For f (– 1), put the value of x = – 1 in f(x):

f (– 1) = (– 1)2 – 3 (– 1) + 2

= 1 + 3 + 2

= 6

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 79

(v) f (x) = 2x2 – 3x + 1 on [1, 3]

Solution:

Given f (x) = 2x2 – 3x + 1 on [1, 3]

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous in [1, 3] and differentiable in (1, 3). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

Therefore, there exist a point c ∈ (1, 3) such that:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 80

f (x) = 2x2 – 3x + 1

Differentiating with respect to x

f’(x) = 2(2x) – 3

= 4x – 3

For f’(c), put the value of x = c in f’(x):

f’(c)= 4c – 3

For f (3), put the value of x = 3 in f(x):

f (3) = 2 (3)2 – 3 (3) + 1

= 2 (9) – 9 + 1

= 18 – 8 = 10

For f (1), put the value of x = 1 in f(x):

f (1) = 2 (1)2 – 3 (1) + 1

= 2 (1) – 3 + 1

= 2 – 2 = 0

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 81

(vi) f (x) = x2 – 2x + 4 on [1, 5]

Solution:

Given f (x) = x2 – 2x + 4 on [1, 5]

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous in [1, 5] and differentiable in (1, 5). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

Therefore, there exist a point c ∈ (1, 5) such that:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 82

f (x) = x2 – 2x + 4

Differentiating with respect to x:

f’(x) = 2x – 2

For f’(c), put the value of x=c in f’(x):

f’(c)= 2c – 2

For f (5), put the value of x=5 in f(x):

f (5)= (5)2 – 2(5) + 4

= 25 – 10 + 4

= 19

For f (1), put the value of x = 1 in f(x)

f (1) = (1)2 – 2 (1) + 4

= 1 – 2 + 4

= 3

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 83

(vii) f (x) = 2x – x2 on [0, 1]

Solution:

Given f (x) = 2x – x2 on [0, 1]

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous in [0, 1] and differentiable in (0, 1). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

Therefore, there exist a point c ∈ (0, 1) such that:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 84

⇒ f’(c) = f(1) – f(0)

f (x) = 2x – x2

Differentiating with respect to x:

f’(x) = 2 – 2x

For f’(c), put the value of x = c in f’(x):

f’(c)= 2 – 2c

For f (1), put the value of x = 1 in f(x):

f (1)= 2(1) – (1)2

= 2 – 1

= 1

For f (0), put the value of x = 0 in f(x):

f (0) = 2(0) – (0)2

= 0 – 0

= 0

f’(c) = f(1) – f(0)

⇒ 2 – 2c = 1 – 0

⇒ – 2c = 1 – 2

⇒ – 2c = – 1

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 85

(viii) f (x) = (x – 1) (x – 2) (x – 3)

Solution:

Given f (x) = (x – 1) (x – 2) (x – 3) on [0, 4]

= (x2 – x – 2x + 3) (x – 3)

= (x2 – 3x + 3) (x – 3)

= x3 – 3x2 + 3x – 3x2 + 9x – 9

= x3 – 6x2 + 12x – 9 on [0, 4]

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous in [0, 4] and differentiable in (0, 4). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

Therefore, there exist a point c ∈ (0, 4) such that:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 86

f (x) = x3 – 6x2 + 12x – 9

Differentiating with respect to x:

f’(x) = 3x2 – 6(2x) + 12

= 3x2 – 12x + 12

For f’(c), put the value of x = c in f’(x):

f’(c)= 3c2 – 12c + 12

For f (4), put the value of x = 4 in f(x):

f (4)= (4)3 – 6(4)2 + 12 (4) – 9

= 64 – 96 + 48 – 9

= 7

For f (0), put the value of x = 0 in f(x):

f (0)= (0)3 – 6 (0)2 + 12 (0) – 9

= 0 – 0 + 0 – 9

= – 9

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 87

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 88

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 89

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 90

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 91

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 92

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 93

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 94

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 95

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

Solution:

Given f (x) = tan – 1 x on [0, 1]

Tan – 1 x has unique value for all x between 0 and 1.

∴ f (x) is continuous in [0, 1]

f (x) = tan – 1 x

Differentiating with respect to x:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 96

x2 always has value greater than 0.

⇒ 1 + x2 > 0

∴ f (x) is differentiable in (0, 1)

So both the necessary conditions of Lagrange’s mean value theorem is satisfied. Therefore, there exist a point c ∈ (0, 1) such that:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 97

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 98

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 99

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 100

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 101

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 102

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 103

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 104

(xii) f (x) = x (x + 4)2 on [0, 4]

Solution:

Given f (x) = x (x + 4)2 on [0, 4]

= x [(x)2 + 2 (4) (x) + (4)2]

= x (x2 + 8x + 16)

= x3 + 8x2 + 16x on [0, 4]

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous in [0, 4] and differentiable in (0, 4). So both the necessary conditions of Lagrange’s mean value theorem is satisfied. Therefore, there exist a point c ∈ (0, 4) such that:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 105

f (x) = x3 + 8x2 + 16x

Differentiating with respect to x:

f’(x) = 3x2 + 8(2x) + 16

= 3x2 + 16x + 16

For f’(c), put the value of x = c in f’(x):

f’(c)= 3c2 + 16c + 16

For f (4), put the value of x = 4 in f(x):

f (4)= (4)3 + 8(4)2 + 16(4)

= 64 + 128 + 64

= 256

For f (0), put the value of x = 0 in f(x):

f (0)= (0)3 + 8(0)2 + 16(0)

= 0 + 0 + 0

= 0

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 106

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 107

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 108

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 109

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 110

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 111

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 112

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 113

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 114

(xiv) f (x) = x2 + x – 1 on [0, 4]

Solution:

Given f (x) = x2 + x – 1 on [0, 4]

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous in [0, 4] and differentiable in (0, 4). So both the necessary conditions of Lagrange’s mean value theorem is satisfied. Therefore, there exist a point c ∈ (0, 4) such that:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 115

f (x) = x2 + x – 1

Differentiating with respect to x:

f’(x) = 2x + 1

For f’(c), put the value of x = c in f’(x):

f’(c) = 2c + 1

For f (4), put the value of x = 4 in f(x):

f (4)= (4)2 + 4 – 1

= 16 + 4 – 1

= 19

For f (0), put the value of x = 0 in f(x):

f (0) = (0)2 + 0 – 1

= 0 + 0 – 1

= – 1

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 116

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 117

(xv) f (x) = sin x – sin 2x – x on [0, π]

Solution:

Given f (x) = sin x – sin 2x – x on [0, π]

Sin x and cos x functions are continuous everywhere on (−∞, ∞) and differentiable for all arguments. So both the necessary conditions of Lagrange’s mean value theorem is satisfied. Therefore, there exist a point c ∈ (0, π) such that:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 118

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 119

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 120

(xvi) f (x) = x3 – 5x2 – 3x on [1, 3]

Solution:

Given f (x) = x3 – 5x2 – 3x on [1, 3]

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous in [1, 3] and differentiable in (1, 3). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

Therefore, there exist a point c ∈ (1, 3) such that:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 121

f (x) = x3 – 5x2 – 3x

Differentiating with respect to x:

f’(x) = 3x2 – 5(2x) – 3

= 3x2 – 10x – 3

For f’(c), put the value of x=c in f’(x):

f’(c)= 3c2 – 10c – 3

For f (3), put the value of x = 3 in f(x):

f (3)= (3)3 – 5(3)2 – 3(3)

= 27 – 45 – 9

= – 27

For f (1), put the value of x = 1 in f(x):

f (1)= (1)3 – 5 (1)2 – 3 (1)

= 1 – 5 – 3

= – 7

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 122

⇒ 3c2 – 10c – 3 = – 10

⇒ 3c2 – 10c – 3 + 10 = 0

⇒ 3c2 – 10c + 7 = 0

⇒ 3c2 – 7c – 3c + 7 = 0

⇒ c (3c – 7) – 1(3c – 7) = 0

⇒ (3c – 7) (c – 1) = 0

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 123

2. Discuss the applicability of Lagrange’s mean value theorem for the function f(x) = |x| on [– 1, 1].

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 124

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 125

3. Show that the Lagrange’s mean value theorem is not applicable to the function f(x) = 1/x on [–1, 1].

Solution:

Given
RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 126

Here, x ≠ 0

⇒ f (x) exists for all values of x except 0

⇒ f (x) is discontinuous at x=0

∴ f (x) is not continuous in [– 1, 1]

Hence the Lagrange’s mean value theorem is not applicable to the function f (x) = 1/x on [-1, 1]

4. Verify the hypothesis and conclusion of Lagrange’s mean value theorem for the function

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 127

Solution:

Given

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 128

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 129

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 130

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 131

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 132

5. Find a point on the parabola y = (x – 4)2, where the tangent is parallel to the chord joining (4, 0) and (5, 1).

Solution:

Given f(x) = (x – 4)2 on [4, 5]

This interval [a, b] is obtained by x – coordinates of the points of the chord.

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous in [4, 5] and differentiable in (4, 5). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

Therefore, there exist a point c ∈ (4, 5) such that:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 133

⇒ f’(x) = 2 (x – 4) (1)

⇒ f’(x) = 2 (x – 4)

For f’(c), put the value of x=c in f’(x):

f’(c) = 2 (c – 4)

For f (5), put the value of x=5 in f(x):

f (5) = (5 – 4)2

= (1)2

= 1

For f (4), put the value of x=4 in f(x):

f (4) = (4 – 4)2

= (0)2

= 0

f’(c) = f(5) – f(4)

⇒ 2(c – 4) = 1 – 0

⇒ 2c – 8 = 1

⇒ 2c = 1 + 8

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 134

We know that, the value of c obtained in Lagrange’s Mean value Theorem is nothing but the value of x – coordinate of the point of the contact of the tangent to the curve which is parallel to the chord joining the points (4, 0) and (5, 1).

Now, put this value of x in f(x) to obtain y:

y = (x – 4)2

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 135

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