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

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

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Exercise 15.1 Page No: 15.9

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:

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C:\Users\tnluser\Downloads\CodeCogsEqn - 2020-01-10T123317.045.gif

Solution:

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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:

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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)

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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)

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)

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f’(x) = 3x2 – 4x – 1

We have f’(c) = 0 c ∈ (-1, 2), from the definition of Rolle’s Theorem

f'(c) = 0

3c2 – 4c – 1 = 0

c = 4 ± √[(-4)2 – (4 x 3 x -1)]/ (2 x 3) [Using the Quadratic Formula]

c = 4 ± √[16 + 12]/ 6

c = (4 ± √28)/ 6

c = (4 ± 2√7)/ 6

c = (2 ±√7)/ 3 = 1.5 ± √7/3

c = 1.5 + √7/3 or 1.5 – √7/3

So,

c = 1.5 – √7/3 since c ∈ (-1, 2)

∴ Rolle’s Theorem is verified.

(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)

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

c = 12/6 or 4/6

c = 2 or 2/3

So,

c = 2/3 since c ∈ (0, 2)

∴ Rolle’s Theorem is verified.

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

(ii) f (x) = sin 2x on [0, π/2]

Solution:

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(iii) f (x) = cos 2x on [-π/4, π/4]

Solution:

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sin 2c = 0

⇒ 2c = 0

So,

c = 0 as c ∈ (-π/4, π/4)

∴ Rolle’s Theorem is verified.

(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(
https://gradeup-question-images.grdp.co/liveData/PROJ24056/1544073308542287.png)

⇒ 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)

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sin 2c = 0

So, 2c = 0 or π

c = 0 or π/2

But,

c = π/2 as c ∈ (0, π)

Hence, Rolle’s Theorem is verified.

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Solution:

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

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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)

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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:

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.

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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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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

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Solution:

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(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’

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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).

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:

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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

⇒ 2c = 1

⇒ c = ½ ∈ (-1, 2)

Hence, Lagrange’s mean value theorem is verified.

(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 + 2) (x – 3)

= (x2 – 3x + 2) (x – 3)

= x3 – 3x2 + 2x – 3x2 + 9x – 6

= x3 – 6x2 + 11x – 6 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 + 11x – 6

Differentiating with respect to x:

f’(x) = 3x2 – 6(2x) + 11

= 3x2 – 12x + 11

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

f’(c) = 3c2 – 12c + 11

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

f (4) = (4)3 – 6(4)2 + 11 (4) – 6

= 64 – 96 + 44 – 6

= 6

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

f (0) = (0)3 – 6 (0)2 + 11 (0) – 6

= 0 – 0 + 0 – 6

= – 6

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

3c2 – 12c + 11 = [6 – (-6)]/ 4

3c2 – 12c + 11 = 12/4

3c2 – 12c + 11 = 3

3c2 – 12c + 8 = 0

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

RD Sharma Class 12 Solutions Maths Chapter 15 Mean Value Theorems

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

Frequently Asked Questions on RD Sharma Solutions for Class 12 Maths Chapter 15

Q1

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Yes, students can download the PDF of RD Sharma Solutions for Class 12 Maths Chapter 15 for free from BYJU’S website. The solutions created by the experts help students to improve their logical and analytical thinking skills which are important from the exam point of view. The solutions are designed based on the latest CBSE syllabus and guidelines. All the answers are strictly in accordance with the textbook prescribed by the CBSE board. The chapter-wise and exercise-wise PDF links are provided to help students boost their exam preparation.
Q2

Is it necessary to learn all the questions provided in RD Sharma Solutions for Class 12 Maths Chapter 15?

Yes, to attempt all types of questions which appear in the exams it is necessary to solve all the questions provided in RD Sharma Solutions for Class 12 Maths Chapter 15. These solutions are formulated by experts at BYJU’S having excellent knowledge of Maths. These solutions are provided in student-friendly language to help students in solving difficult problems in an easier way.
Q3

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RD Sharma Solutions have been prescribed for years as a complete source of information to CBSE students, to develop their analytical skills. The RD Sharma Solutions for Class 12 Maths Chapter 15 explain the steps with precision, without missing out on essential aspects of solving a question. They have proven to be vital for learning the syllabus and developing the confidence that is required to face their exams.

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