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

RD Sharma books are widely used by students who wish to excel in board exams as it provides a vast number of questions to solve. For those students here we have provided the answers for RD Sharma Solutions in simple PDF format, which can be downloaded easily from the below provided links. Solving these questions makes them perfect in all topics included in the syllabus. Knowing the concept is necessary but learning their applications also plays an important role to score full marks in Maths.

The 15th Chapter Mean Value Theorems of RD Sharma Solutions for Class 12 Maths is the best material. This chapter explains the geometrical and algebraic interpretation of some important named theorems related to mean values. Based on these theorems BYJU’S expert faculty has solved this chapter in a unique way. Some of the essential topics of the chapter are listed below.

  • Rolle’s theorem
  • Geometrical interpretation of Rolle’s theorem
  • Algebraic interpretation of Rolle’s theorem
  • Checking the applicability of Rolle’s theorem
  • Verification of Rolle’s theorem for a given function defined on a given interval
  • Miscellaneous applications of Rolle’s theorem
  • 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 Maths Chapter 15 Mean Value Theorems: Download PDF

RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 1
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 2
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 3
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 4
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 5
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 6
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 7
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 8
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 9
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 10
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 11
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 12
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 13
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 14
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 15
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 16
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 17
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 18
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 19
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 20
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 21
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 22
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 23
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 24
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 25
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 26
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 27
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 28
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 29
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 30
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 31
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 32
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 33
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 34
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 35
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 36
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 37
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 38
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 39
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 40
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 41
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 42
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 43
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 44
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 45
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 46
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 47
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 48
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 49
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 50
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 51
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 52
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 53
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 54
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 55
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 56
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 57
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 58
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 59
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 60
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 61
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 62
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 63
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 64
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 65
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 66
RD Sharma Class 12 Maths Solutions Chapter 15 Mean Value Theorems 67

Access Answers to RD Sharma Solutions For Class 12 Maths Chapter 15 – Mean Value Theorems

Exercise 15.1 Page No: 15.9

1. Discuss the applicability of Rolle’s Theorem for the following functions on the indicated intervals:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 1

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 2

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 3

(ii) f (x) = [x] for -1 < x ≤ 1, where [x] denotes the greatest integer not exceeding x

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 4

(iii) f(x) = sin 1/x for -1 ≤ x ≤ 1

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 5

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 6

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 7

(iv) f (x) = 2x2 – 5x + 3 on [1, 3]

Solution:

Given function is f (x) = 2x2 – 5x + 3 on [1, 3]

Since given function f is a polynomial. So, it is continuous and differentiable everywhere.

Now, we find the values of function at the extreme values.

⇒ f (1) = 2(1)2–5(1) + 3

⇒ f (1) = 2 – 5 + 3

⇒ f (1) = 0…… (1)

⇒ f (3) = 2(3)2–5(3) + 3

⇒ f (3) = 2(9)–15 + 3

⇒ f (3) = 18 – 12

⇒ f (3) = 6…… (2)

From (1) and (2), we can say that, f (1) ≠ f (3)

∴ Rolle’s Theorem is not applicable for the function f in interval [1, 3].

(v) f (x) = x2/3 on [-1, 1]

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 8

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 9

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 10

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 11

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

2. Verify the Rolle’s Theorem for each of the following functions on the indicated intervals:

(i) f (x) = x2 – 8x + 12 on [2, 6]

Solution:

Given function is f (x) = x2 – 8x + 12 on [2, 6]

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e., on R.

Let us find the values at extremes:

⇒ f (2) = 22 – 8(2) + 12

⇒ f (2) = 4 – 16 + 12

⇒ f (2) = 0

⇒ f (6) = 62 – 8(6) + 12

⇒ f (6) = 36 – 48 + 12

⇒ f (6) = 0

∴ f (2) = f(6), Rolle’s theorem applicable for function f on [2,6].

Now we have to find the derivative of f(x)

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 13

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 14

(ii) f(x) = x2 – 4x + 3 on [1, 3]

Solution:

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

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e., on R. Let us find the values at extremes:

⇒ f (1) = 12 – 4(1) + 3

⇒ f (1) = 1 – 4 + 3

⇒ f (1) = 0

⇒ f (3) = 32 – 4(3) + 3

⇒ f (3) = 9 – 12 + 3

⇒ f (3) = 0

∴ f (1) = f(3), Rolle’s theorem applicable for function ‘f’ on [1,3].

Let’s find the derivative of f(x)

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

⇒ f’(x) = 2x – 4

We have f’(c) = 0, c ϵ (1, 3), from the definition of Rolle’s Theorem.

⇒ f’(c) = 0

⇒ 2c – 4 = 0

⇒ 2c = 4

⇒ c = 4/2

⇒ C = 2 ϵ (1, 3)

∴ Rolle’s Theorem is verified.

(iii) f (x) = (x – 1) (x – 2)2 on [1, 2]

Solution:

Given function is f (x) = (x – 1) (x – 2)2 on [1, 2]

Since, given function f is a polynomial it is continuous and differentiable everywhere that is on R.

Let us find the values at extremes:

⇒ f (1) = (1 – 1) (1 – 2)2

⇒ f (1) = 0(1)2

⇒ f (1) = 0

⇒ f (2) = (2 – 1)(2 – 2)2

⇒ f (2) = 02

⇒ f (2) = 0

∴ f (1) = f (2), Rolle’s Theorem applicable for function ‘f’ on [1, 2].

Let’s find the derivative of f(x)

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 16

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 17

(iv) f (x) = x (x – 1)2 on [0, 1]

Solution:

Given function is f (x) = x(x – 1)2 on [0, 1]

Since, given function f is a polynomial it is continuous and differentiable everywhere that is, on R.

Let us find the values at extremes

⇒ f (0) = 0 (0 – 1)2

⇒ f (0) = 0

⇒ f (1) = 1 (1 – 1)2

⇒ f (1) = 02

⇒ f (1) = 0

∴ f (0) = f (1), Rolle’s theorem applicable for function ‘f’ on [0,1].

Let’s find the derivative of f(x)

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 18

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 19

(v) f (x) = (x2 – 1) (x – 2) on [-1, 2]

Solution:

Given function is f (x) = (x2 – 1) (x – 2) on [– 1, 2]

Since, given function f is a polynomial it is continuous and differentiable everywhere that is on R.

Let us find the values at extremes:

⇒ f ( – 1) = (( – 1)2 – 1)( – 1 – 2)

⇒ f ( – 1) = (1 – 1)( – 3)

⇒ f ( – 1) = (0)( – 3)

⇒ f ( – 1) = 0

⇒ f (2) = (22 – 1)(2 – 2)

⇒ f (2) = (4 – 1)(0)

⇒ f (2) = 0

∴ f (– 1) = f (2), Rolle’s theorem applicable for function f on [ – 1,2].

Let’s find the derivative of f(x)

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 20

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 21

(vi) f (x) = x (x – 4)2 on [0, 4]

Solution:

Given function is f (x) = x (x – 4)2 on [0, 4]

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e., on R.

Let us find the values at extremes:

⇒ f (0) = 0(0 – 4)2

⇒ f (0) = 0

⇒ f (4) = 4(4 – 4)2

⇒ f (4) = 4(0)2

⇒ f (4) = 0

∴ f (0) = f (4), Rolle’s theorem applicable for function ‘f’ on [0,4].

Let’s find the derivative of f(x):

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 22

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 23

(vii) f (x) = x (x – 2)2 on [0, 2]

Solution:

Given function is f (x) = x (x – 2)2 on [0, 2]

Since, given function f is a polynomial it is continuous and differentiable everywhere that is on R.

Let us find the values at extremes:

⇒ f (0) = 0(0 – 2)2

⇒ f (0) = 0

⇒ f (2) = 2(2 – 2)2

⇒ f (2) = 2(0)2

⇒ f (2) = 0

f (0) = f(2), Rolle’s theorem applicable for function f on [0,2].

Let’s find the derivative of f(x)

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 24

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 25

(viii) f (x) = x2 + 5x + 6 on [-3, -2]

Solution:

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

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e., on R. Let us find the values at extremes:

⇒ f ( – 3) = ( – 3)2 + 5( – 3) + 6

⇒ f ( – 3) = 9 – 15 + 6

⇒ f ( – 3) = 0

⇒ f ( – 2) = ( – 2)2 + 5( – 2) + 6

⇒ f ( – 2) = 4 – 10 + 6

⇒ f ( – 2) = 0

∴ f (– 3) = f( – 2), Rolle’s theorem applicable for function f on [ – 3, – 2].

Let’s find the derivative of f(x):

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 26

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 27

3. Verify the Rolle’s Theorem for each of the following functions on the indicated intervals:

(i) f (x) = cos 2 (x – π/4) on [0, π/2]

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 28

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 29

(ii) f (x) = sin 2x on [0, π/2]

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 30

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 31

(iii) f (x) = cos 2x on [-π/4, π/4]

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 32

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 33

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 34

(iv) f (x) = ex sin x on [0, π]

Solution:

Given function is f (x) = ex sin x on [0, π]

We know that exponential and sine functions are continuous and differentiable on R. Let’s find the values of the function at an extreme,

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 35

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 36

(v) f (x) = ex cos x on [-π/2, π/2]

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 37

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 38

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 39

(vi) f (x) = cos 2x on [0, π]

Solution:

Given function is f (x) = cos 2x on [0, π]

We know that cosine function is continuous and differentiable on R. Let’s find the values of function at extreme,

⇒ f (0) = cos2(0)

⇒ f (0) = cos(0)

⇒ f (0) = 1

⇒ f (π) = cos2(π)

⇒ f (π) = cos(2 π)

⇒ f (π) = 1

We have f (0) = f (π), so there exist a c belongs to (0, π) such that f’(c) = 0.

Let’s find the derivative of f(x)

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 40

Hence Rolle’s Theorem is verified.

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 41

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 42

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 43

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 44

(viii) f (x) = sin 3x on [0, π]

Solution:

Given function is f (x) = sin3x on [0, π]

We know that sine function is continuous and differentiable on R. Let’s find the values of function at extreme,

⇒ f (0) = sin3(0)

⇒ f (0) = sin0

⇒ f (0) = 0

⇒ f (π) = sin3(π)

⇒ f (π) = sin(3 π)

⇒ f (π) = 0

We have f (0) = f (π), so there exist a c belongs to (0, π) such that f’(c) = 0.

Let’s find the derivative of f(x)

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 45

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 46

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 47

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 48

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 49

(x) f (x) = log (x2 + 2) – log 3 on [-1, 1]

Solution:

Given function is f (x) = log(x2 + 2) – log3 on [– 1, 1]

We know that logarithmic function is continuous and differentiable in its own domain. We check the values of the function at the extreme,

⇒ f (– 1) = log((– 1)2 + 2) – log 3

⇒ f (– 1) = log (1 + 2) – log 3

⇒ f (– 1) = log 3 – log 3

⇒ f ( – 1) = 0

⇒ f (1) = log (12 + 2) – log 3

⇒ f (1) = log (1 + 2) – log 3

⇒ f (1) = log 3 – log 3

⇒ f (1) = 0

We have got f (– 1) = f (1). So, there exists a c such that c ϵ (– 1, 1) such that f’(c) = 0.

Let’s find the derivative of the function f,

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 50

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 51

(xi) f (x) = sin x + cos x on [0, π/2]

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 52

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 53

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

Solution:

Given function is f (x) = 2sinx + sin2x on [0, π]

We know that sine function continuous and differentiable over R.

Let’s check the values of function f at the extremes

⇒ f (0) = 2sin(0) + sin2(0)

⇒ f (0) = 2(0) + 0

⇒ f (0) = 0

⇒ f (π) = 2sin(π) + sin2(π)

⇒ f (π) = 2(0) + 0

⇒ f (π) = 0

We have f (0) = f (π), so there exist a c belongs to (0, π) such that f’(c) = 0.

Let’s find the derivative of function f.

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 54

⇒ f’(x) = 2cosx + 2cos2x

⇒ f’(x) = 2cosx + 2(2cos2x – 1)

⇒ f’(x) = 4 cos2x + 2 cos x – 2

We have f’(c) = 0,

⇒ 4cos2c + 2 cos c – 2 = 0

⇒ 2cos2c + cos c – 1 = 0

⇒ 2cos2c + 2 cos c – cos c – 1 = 0

⇒ 2 cos c (cos c + 1) – 1 (cos c + 1) = 0

⇒ (2cos c – 1) (cos c + 1) = 0

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 55

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 56

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 57

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 58

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 59

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 60

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 61

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 62

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 63

(xv) f (x) = 4sin x on [0, π]

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 64

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 65

(xvi) f (x) = x2 – 5x + 4 on [0, π/6]

Solution:

Given function is f (x) = x2 – 5x + 4 on [1, 4]

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e., on R.

Let us find the values at extremes

⇒ f (1) = 12 – 5(1) + 4

⇒ f (1) = 1 – 5 + 4

⇒ f (1) = 0

⇒ f (4) = 42 – 5(4) + 4

⇒ f (4) = 16 – 20 + 4

⇒ f (4) = 0

We have f (1) = f (4). So, there exists a c ϵ (1, 4) such that f’(c) = 0.

Let’s find the derivative of f(x):

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 66

(xvii) f (x) = sin4 x + cos4 x on [0, π/2]

Solution:

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 67

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 68

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 69

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

Solution:

Given function is f (x) = sin x – sin2x on [0, π]

We know that sine function is continuous and differentiable over R.

Now we have to check the values of the function ‘f’ at the extremes.

⇒ f (0) = sin (0)–sin 2(0)

⇒ f (0) = 0 – sin (0)

⇒ f (0) = 0

⇒ f (π) = sin(π) – sin2(π)

⇒ f (π) = 0 – sin(2π)

⇒ f (π) = 0

We have f (0) = f (π). So, there exists a c ϵ (0, π) such that f’(c) = 0.

Now we have to find the derivative of the function ‘f’

RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 70

4. Using Rolle’s Theorem, find points on the curve y = 16 – x2, x ∈ [-1, 1], where tangent is parallel to x – axis.

Solution:

Given function is y = 16 – x2, x ϵ [– 1, 1]

We know that polynomial function is continuous and differentiable over R.

Let us check the values of ‘y’ at extremes

⇒ y (– 1) = 16 – (– 1)2

⇒ y (– 1) = 16 – 1

⇒ y (– 1) = 15

⇒ y (1) = 16 – (1)2

⇒ y (1) = 16 – 1

⇒ y (1) = 15

We have y (– 1) = y (1). So, there exists a c ϵ (– 1, 1) such that f’(c) = 0.

We know that for a curve g, the value of the slope of the tangent at a point r is given by g’(r).

Now we have to find the derivative of curve y


RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Image 71

⇒ y’ = – 2x

We have y’(c) = 0

⇒ – 2c = 0

⇒ c = 0 ϵ (– 1, 1)

Value of y at x = 1 is

⇒ y = 16 – 02

⇒ y = 16

∴ The point at which the curve y has a tangent parallel to x – axis (since the slope of x – axis is 0) is (0, 16).

Exercise 15.2 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

Also, Access Exercises of RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems

Exercise 15.1 Solutions

Exercise 15.2 Solutions

Leave a Comment

Your email address will not be published. Required fields are marked *

BOOK

Free Class