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

Rolle’s theorem, the geometrical theorem of Rolle’s theorem of given functions with illustrative examples are the main topics which are explained under exercise 15.1. The exercise-wise solutions are prepared by our experts with explanations for each step to make it easier for students while solving problems. It helps to check their knowledge about the concepts, which are discussed in this chapter. In order to master Maths, students can use RD Sharma Solutions for Class 12 Maths Chapter 15 Mean Value Theorems Exercise 15.1 as their reference materials. Some of the essential topics of this exercise are enlisted here.

  • 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

RD Sharma Solutions Class 12 Mean Value Theorems Exercise 15.1:Download PDF Here

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

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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:

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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

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Solution:

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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)

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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)

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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):

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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)

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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):

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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:

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(ii) f (x) = sin 2x on [0, π/2]

Solution:

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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:

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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,

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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:

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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.

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Solution:

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(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

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Solution:

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

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(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,

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

Solution:

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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.

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⇒ 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

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Solution:

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Solution:

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(xv) f (x) = 4sin x on [0, π]

Solution:

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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’

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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

⇒ 
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⇒ 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).

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