RD Sharma Class 12 Ex 15.1 Solutions Chapter 15 Mean Value Theorems

Here we provide RD Sharma Class 12 Ex 15.1 Solutions Chapter 15 Mean Value Theorems for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 15.1 Solutions Chapter 15 Mean Value Theorems book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter16
Exercise15.1
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 15.1 Solutions Chapter 15 Mean Value Theorems

Question 1 (i). Verify Lagrange’s mean value theorem for the following

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

(i) f(x) = 3 + (x – 2)2/3 on [1, 3]

Solution:

Rolle’s theorem states that if a function f is continuous on the closed interval [a, b] and 

differentiable on the open interval (a, b) such that f(a) = f(b), then f′(x) = 0 for some x with a ≤ x ≤ b.

We have, f'(x) = \frac{d}{dx}(3+(x-2)^{\frac{2}{3}})

⇒ f'(x) = \frac{d(3)}{dx}+\frac{d((x-2)^{\frac{2}{3}})}{dx}

⇒  f'(x) = 0 + \frac{2}{3}(x-2)^{\frac{2}{3}-1}

⇒  f'(x) = \frac{2}{3(x-2)^{\frac{1}{3}}}

To analyse differentiability at x = 2:

⇒  f'(2) = [\frac{2}{3(2-2)}]^{\frac{1}{3}}

= 2 / 0 ≠ 0 

Since f'(2) ≠ 0, f(x) is not differentiable in the interval [1, 3].

Hence, the Rolle’s Theorem is not applicable for the given function f(x) in interval [1, 3].

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

Solution:

Rolle’s theorem requires a function to be continuous on the closed interval [a, b].

The continuity of f(x) needs to be analyzed at x = 1:

(RHL at x = 1) = limx⇢(1+h)[x]

= limh⇢0[1 + h]

= 1              ….. (1)

(LHL at x = 1) = limx⇢(1-h)[x]

= limh⇢0[1 – h]

= 0            …… (2)

Since f(x) is not continuous at x = 1 or the interval [-1, 1].

Hence, the Rolle’s Theorem is not applicable for the given function f(x) in interval [-1, 1].

(iii) f(x) = sin(1/x) for -1 ≤ x ≤ 1

Solution:

Rolle’s theorem requires a function to be continuous on the closed interval [a, b].

The continuity of f(x) needs to be analyzed at x = 0:

(RHL at x = 0) = limx⇢(0+h)sin(1/x)

= limh⇢0sin(1/h)

= k                  …..(1)

(LHL at x = 0) = limx⇢(0-h)sin(1/x)

= limh⇢0sin(1/0-h)

= limh⇢0sin(-1/h)

= -limh⇢0sin(1/h)

= −k               ….(2)

Since f(x) is not continuous at x = 1 or the interval [-1,1].

Hence, the Rolle’s Theorem is not applicable for the given function f(x) in interval [-1, 1].

(iv) f(x) = 2x2 – 5x + 3 on [1, 3]

Solution:

Clearly f(x) being a polynomial function shall be continuous on the interval.

We need to verify whether f(a) = f(b) for the applicability of theorem.

Here, f(1) = 2(1)2 – 5(1) + 3

⇒ f(1) = 2 – 5 + 3

⇒ f(1) = 0                             …… (1)

Now, f(3) = 2(3)2 – 5(3) + 3

⇒ f(3) = 2(9) – 15 + 3

⇒ f(3) = 18 – 12

⇒ f(3) = 6                              …… (2)

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

We observe that f(a) ≠ f(b)

Hence, the Rolle’s Theorem is not applicable for the given function f(x) in interval [1, 3].

(v) f (x) = x2/3 on [-1, 1]

Solution:

Rolle’s theorem states that if a function f is continuous on the closed interval [a, b] and

differentiable on the open interval (a, b) such that f(a) = f(b), then f′(x) = 0 for some x with a ≤ x ≤ b.

f'(x) = \frac{d(x^{\frac{2}{3}})}{dx}

⇒  f'(x) = \frac{2}{3x^{\frac{1}{3}}}

Clearly f'(x) is undefined at x = 0.

Hence, the Rolle’s Theorem is not applicable for the given function f(x) in interval [-1, 1].

(vi) f(x) =\begin{cases}-4x+5 \ \ \ \ \ \ \ \ ,0\lex\le1\\2x-3 \ \ \ \ \ \ \ \ \ \ \ ,1<x\le2\end{cases}

Solution:

Rolle’s theorem requires a function to be continuous on the closed interval [a, b].

The continuity of f(x) needs to be analyzed at x = 1:

(LHL at x = 1) = lim(x->1-h) (-4x + 5)

= lim(h->0) [-4(1 – h) + 5] 

= -4 + 5

= 1

(RHL at x = 1) =  lim(x->1+h) (2x – 3)

= lim(h->0) [2(1 + h) – 3] 

= 2(0) – 3 

= -1

We observe that LHL ≠ RHL.

Clearly f(x) is not continuous at x = 1.

Hence, the Rolle’s Theorem is not applicable for the f(x) function in interval [-1, 1].

Question 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.

We only need to check whether f(a) = f(b) or not.

Thus, f(2) = 22 – 8(2) + 12

⇒ f(2) = 4 – 16 + 12

⇒ f(2) = 0

Now, f(6) = 62 – 8(6) + 12

⇒ f(6) = 36 – 48 + 12

⇒ f(6) = 0

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

f(x) = x2 – 8x + 12 

f'(x) = 2x – 8

Then f'(c) = 0

2c – 8 = 0

c = 4 ∈ (2, 6)

Hence, the Rolle’s theorem is verified 

(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.

We only need to check whether f(a) = f(b) or not.

So f (1) = 12 – 4(1) + 3

⇒ f (1) = 1 – 4 + 3

⇒ f (1) = 0

Now, 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].

⇒ 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 ∈ (1, 3)

Hence, 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.

We only need to check whether f(a) = f(b) or not.

So f (1) = (1 – 1) (1 – 2)2

⇒ f (1) = 0(1)2

⇒ f (1) = 0

Now, f (2) = (2 – 1)(2 – 2)2

⇒ f (2) = 02

⇒ f (2) = 0

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

f(x) = (x – 1)(x – 2)2

f'(x) = (x – 1) × 2(x – 2) + (x – 2)2

f'(x) = (x – 2) (3x – 4) 

Then f'(c) = 0

(c – 2)(3c – 4) = 0

c = 2

or 

c = 4/3 ∈ (1, 2) 

Hence, Rolle’s Theorem is verified.

(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.

We only need to check whether f(a) = f(b) or not.

So f (0) = 0 (0 – 1)2

⇒ f (0) = 0

Now f (1) = 1 (1 – 1)2

⇒ f (1) = 02

⇒ f (1) = 0

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

f (x) = x (x – 1)2

f'(x) = (x – 1)2+ x × 2(x – 1)

= (x – 1)(x – 1 + 2x)

= (x – 1)(3x – 1)

Then f'(c) = 0

(c – 1)(3c – 1) = 0

c = 1 or 1/3 ∈ (0, 1) 

Hence, Rolle’s Theorem is verified.

(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.

We only need to check whether f(a) = f(b) or not.

So f(–1) = (( – 1)2 – 1)( – 1 – 2)

⇒ f ( – 1) = (1 – 1)( – 3)

⇒ f ( – 1) = (0)( – 3)

⇒ f ( – 1) = 0

Now, 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].

⇒ f’(x) = 3x2 – 4x – 1

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

Clearly, f'(c) = 0

3c2 – 4c – 1 = 0

c = (2 ±√7)/ 3 

c = 1/3(2 + √7) or 1/3(2 – √7) ∈ (-1, 2)

Hence, 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.

We only need to check whether f(a) = f(b) or not.

So f (0) = 0(0 – 4)2

⇒ f (0) = 0

Now, f (4) = 4(4 – 4)2

⇒ f (4) = 4(0)2

⇒ f (4) = 0

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

 f(x) = x(x – 4)2

 f'(x) = x × 2(x – 4) + (x – 4)2

= 2x2 – 8x + x2 + 16 – 8x

= 2x2 – 16x + 16

Then 

 f'(c) = 2c2 – 16c + 16

= (c – 4)(3c – 4)

c = 4 or 4/3 ∈ (0, 4)

Hence, Rolle’s Theorem is verified.

(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.

We only need to check whether f(a) = f(b) or not.

So f (0) = 0(0 – 2)2

⇒ f (0) = 0

Now, 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].

c = 12/6 or 4/6

c = 2 or 2/3

So, c = 2/3 since c ∈ (0, 2)

Hence, 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. 

We only need to check whether f(a) = f(b) or not.

So f(–3) = (–3)2 + 5(–3) + 6

⇒ f(–3) = 9 – 15 + 6

⇒ f(–3) = 0

Now, f(–2) = (–2)2 + 5(–2) + 6

⇒ f(–2) = 4 – 10 + 6

⇒ f(–2) = 0

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

f(x) = x2 + 5x + 6

f'(x) = 2x + 5

Then

f(c) = 0

2c + 5 = 0

c = -5/2 ∈ (-3, -2)

Hence, Rolle’s Theorem is verified.

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

(i) f(x) = cos2(x – π/4) on [0, π/2]

Solution:

Clearly the cosine function is continuous and differentiable everywhere.

Hence, the given function f(x) is continuous in [0, π/2] and differentiable in [0, π/2]

So,

f(0) = cos2(0 – π/4) = 0

f(π/2) = cos2(π/2 – π/4) = 0

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

Now, f'(x) = \frac{d(cos2(x-\frac{π}{4}))}{dx}

= -2sin(2x – π/2)

If f'(c) = 0

⇒  -2sin(2c – π/2) = 0

⇒ c = π/4 ∈ [0, π/2]

Clearly c belongs to the given interval.

Hence, Rolle’s Theorem is verified.

(ii) f(x) = sin 2x on [0, π/2]

Solution:

Since sin 2x is everywhere continuous and differentiable.

So, sin2x is continuous on (0, π/2) and differentiable on (0, π/2).

f(0) = sin 0 = 0

f(π/2) = sin π/2 = 0

f(π/2) = f(0) = 0

Thus, f(x) satisfies all the conditions of Rolle’s theorem.

Now, we have to show that there exists c in (0, π/2) such that f'(c) = 0.

We have, f'(x) = 0

⇒ cos2x = 0

⇒ x = π/4

So, f'(c) = 2cos2c

2cos2c = 0

Thus, c = π/4 in (0, π/2).

​Hence, Rolle’s theorem is verified.

(iii) f(x) = cos2x on [-π/4, π/4]

Solution:

Since cos 2x is everywhere continuous and differentiable.

So, cos2x is continuous on [-π/4, π/4] and differentiable on [-π/4, π/4].

cos(-π/4) = cos2(-π/4) = cos(-π/2) = 0 

cos(π/4) = cos2(π/4) = cos(π/2) = 0 

Also, cos(-π/4) = cos(π/4)  

Thus, f(x) satisfies all the conditions of Rolle’s theorem.

Now, we have to show that there exists c in (0, π/2) such that f'(c) = 0.

We have f'(x) = 0

 ⇒ sin2c = 0

⇒ 2c = 0

So, c = 0 as c ∈ (-π/4, π/4)

Hence, Rolle’s Theorem is verified for the given function f(x).

(iv) esin x on [0, π]

Solution:

The given function is a composition of exponential and trigonometric functions. Hence it is differential and continuous everywhere.

So, the given function f(x) is continuous on [0, π] and differentiable on (0, π).

f(0) = esin 0 = 0

f(π) = eπ sin π = 0

f(0) = f(π)

Now, we have to show that there exists c in (0, π) such that f'(c) = 0.

So f'(x) = ex (sin x + cos x)

⇒ ex (sin x + cos x) = 0

⇒ sin x + cos x = 0

Dividing both sides by cos x, we have

tan x = -1

⇒  x = π – π/4 = 3π/4

Since c = 3π/4 in (0, π) such that f'(c) = 0.

​Hence, Rolle’s theorem is verified for the given function f(x).

(v) f(x) = ex cos x on [−π/2, π/2]

Solution:

The given function is a composition of exponential and trigonometric functions. Hence it is differential and continuous everywhere.

So, the given function f(x) is continuous on [-π/2, π/2] and differentiable on (-π/2, π/2).

f(−π/2) = eπ/2 cos(−π/2) = 0

f(π/2) = eπ/2 cos(π/2) = 0

f(−π/2) = f(π/2)

Now, we have to show that there exists c in (-π/2, π/2) such that f'(c) = 0.

 f'(x) = ex (cos x – sin x)

⇒ ex (cos x – sin x) = 0

⇒ sin x – cos x = 0

Dividing both sides by cos x, we have

 tan x = 1

 ⇒ x = π/4

Since c = π/4 in (-π/2, π/2) such that f’(c) = 0.

Hence, Rolle’s theorem is verified for the given function f(x).

(vi) f (x) = cos 2x on [0, π]

Solution:

Since cos 2x is everywhere continuous and differentiable.

So, cos2x is continuous on [0, π] and differentiable on (0, π).

f(0) = cos(0) = 1

f(π) = cos(2π) = 1

f(0) = f(π)

Thus, f(x) satisfies all the conditions of Rolle’s theorem.

There must be a c belongs to (0, π) such that f’(c) = 0.

⇒ 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 for the given function f(x).

(vii) f(x) = sinx/eon 0 ≤ x ≤ π

Solution:

The given function is a composition of exponential and trigonometric functions. 

Hence it is differential and continuous everywhere.

So, sinx/ex is continuous on [0, π] and differentiable on (0, π).

f(0) = sin(0)/e0 = 0

f(π) = sin(π)/eπ = 1

f(0) = f(π)

Now, we have to show that there exists c in (0, π/2) such that f'(c) = 0.

 f'(x) = ex (cos x – sin x)

⇒ ex (cos x – sin x) = 0

⇒ sin x – cos x = 0

⇒ tan x = 1

⇒  x = π/4

Since c = π/4 in (-π/4, π/4) such that f’(c) = 0.

​Hence, Rolle’s theorem is verified for the given function f(x).

(viii) f(x) = sin 3x on [0, π]

Solution:

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

Since sine function is continuous and differentiable on R. 

We need to check whether f(a) = f(b) or not.

So f (0) = sin 3(0)

⇒ f (0) = sin0

⇒ f (0) = 0

Now, f (π) = sin 3π

⇒ 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.

⇒ 3cos(3x) = 0

⇒ cos 3x = 0

⇒ 3x = π / 2

⇒ x = π / 6 ∈ [0, π]

​Hence, Rolle’s theorem is verified for the given function f(x).

(ix) f(x) = {e^{1 - x}}^2       on [−1, 1]

Solution:

We know that the exponential function is continuous and differentiable everywhere.

So, {e^{1 - x}}^2      is continuous on [-1, 1] and differentiable on (-1, 1).

f(-1) = e1-1 = 1

f(1) = e1-1 = 1

Also, f(-1) = f(1) = 1

Thus, Rolle’s theorem is applicable. So there must exist c ∈  [−1, 1] such that f'(c) = 0.

 f'(x) = -2x [e^{1 - x^2}]

⇒ -2x e^{1 – x^2}  = 0

⇒ x = 0 ∈  [−1, 1]

Hence, Rolle’s theorem is verified.

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

Solution:

Clearly the logarithmic function is continuous and differentiable in its own domain. 

We need to verify whether f(a) = f(b) or not.

So f (– 1) = log((– 1)2 + 2) – log 3

⇒ f (– 1) = log (1 + 2) – log 3

⇒ f (– 1) = log 3 – log 3

⇒ f ( – 1) = 0

Now, 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.

Now, f'(x) = \frac{2x}{x^2+2}

f'(c) = 0

⇒ c = 0 ϵ (– 1, 1)

​Hence, Rolle’s theorem is verified.

(xi) f(x) = sin x + cos x on [0, π/2]

Solution:

Since sin and cos functions are continuous and differentiable everywhere, 

f(x) is continuous on [0, π/2] and differentiable on (0, π/2).

So, f(0) = sin 0 + cos 0 = 1

f(π/2) = sin π/2 + cos π/2 = 1

Also, f(0) = f(π/2) = 1.

Hence, Rolle’s theorem is applicable. So there must exist c ∈  [−1, 1] such that f'(c) = 0.

Now, f'(x) = cos x – sin x

⇒ cos x – sin x = 0

⇒ tan x = 1

⇒ x = π/4

Thus, c = π/4 on (0, π/2).

​Hence, Rolle’s theorem is verified for the given function f(x).

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

Solution:

Since sin function continuous and differentiable over R, f(x) is continuous on [0, π] and differentiable on (0, π).

We need to verify whether f(a) = f(b) or not.

So f (0) = 2sin(0) + sin2(0)

⇒ f (0) = 2(0) + 0

⇒ f (0) = 0

Now, 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.

f'(x) = 2cos x + cos 2x

Now, f'(c) = 0

⇒ 2cos x + cos 2x = 0

⇒ (2cos c – 1) (cos c + 1) = 0

⇒ tan c = 1

⇒ c = π/3 in (0, π)

 ​Hence, Rolle’s theorem is verified for the given function f(x).

(xiii) f(x) = \frac{x}{2} - \sin[\frac{\pi x}{6}]       on [-1, 0]

Solution:

Since sin function is continuous and differentiable over R, f(x) is continuous on [-1, 0] and differentiable on (-1, 0).

f(-1) = \frac{-1}{2} - \sin[\frac{\pi (-1)}{6}]      = 0

f(0) = 0 – sin 0 = 0

Also f(-1) = f(0) = 0

Hence Rolle’s theorem is applicable so there exist a c belongs to (0, π) such that f’(c) = 0.

f'(x) = \frac{1}{2} - \frac{\pi}{6}\cos[\frac{\pi x}{6}]

Now, f'(c) = 0

⇒ \frac{1}{2} - \frac{\pi}{6}\cos[\frac{\pi c}{6}] = 0

⇒  \cos\frac{\pi c}{6} = \frac{3}{\pi}

⇒ c = (\frac{- 6}{\pi}) \cos^{- 1} [\frac{3}{\pi}]

Thus, c ϵ (-1, 0).

​Hence, Rolle’s theorem is verified for the given function f(x).

(xiv) f(x) = \frac{6x}{\pi} - 4 \sin^2 x       on [0, π/6]

Solution:

Since sin function is continuous and differentiable everywhere, f(x) is continuous on [0, π/6] and differentiable on (0, π/6).

f(0) = 0 – 0 = 0 

f(π/6) = 1 – 1 = 0 

Also, f(0) = f(π/6) = 0

Hence Rolle’s theorem is applicable so there exist a c belongs to (0, π/6) such that f’(c) = 0.

We have, f'(x) = 6/π – 8sinxcosx

 f'(x) = 6/π – 4sin2x

f'(c) = 0

6/π – 4sin2c = 0

4sin2c = 6/π

sin2c = 3/2π

2c = sin-1(21/44)

c = 1/2 sin-1(21/44)

c ∈ (-1/2, 1/2)

c ∈ (0, 11/21)

c ∈ (0, π/6)

​Hence, Rolle’s theorem is verified for the given function f(x).

(xv) f(x) = 4sin x on [0, π]

Solution:

Since sin and exponential functions are continuous and differentiable everywhere, f(x) is continuous on [0, π] and differentiable on (0, π).

f(0) = 4sin 0 = 1 

f(π) = 4sin π = 1 

Also, f(0) = f(π) = 1.

Hence Rolle’s theorem is applicable so there exist a c belongs to (0, π) such that f’(c) = 0.

We have, f'(x) = 4sin x cos x log 4

So, f’(c) = 0

 4sin c cos c log 4 = 0

 4sin c cos c = 0

 cos c = 0

 c = π/2 ∈ (0, π)

Hence, Rolle’s theorem is verified for the given function f(x).

(xvi) f (x) = x2 – 5x + 4 on [1, 4]

Solution:

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

We need to verify whether f(a) = f(b) or not.

So f (1) = 12 – 5(1) + 4

⇒ f (1) = 1 – 5 + 4

⇒ f (1) = 0

Now, 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 ∈  (1, 4) such that f’(c) = 0.

f'(x) = 2x – 5

Now, f'(c) = 2c – 5 = 0

Thus, c = 5 / 2 ∈ (1, 4).

Hence, Rolle’s theorem is verified for the given function f(x).

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

Solution:

Since sin and cos functions are continuous and differentiable everywhere, f(x) is continuous on [0, π/2] and differentiable on (0, π/2).

So, 

f(0) = sin4(0) + cos4(0) = 1

f(π/2) = sin4(π/2) + cos4(π/2) = 1

f(0) = f(π/2)

Hence Rolle’s theorem is applicable so there exist a c belongs to (0, π/2) such that f’(c) = 0.

⇒ f'(x) = 4sin3 x cos x – 4cos3 x sin x

so, f'(c) = 4sin3 x cos x – 4cos3 x sin x  = 0

4sin3 c cos c – 4cos3 c sin c  = 0

-sin4c = 0

4x = 0 or 4x = π

x = 0, or π/4 ∈ (0, π/2)

Hence, Rolle’s theorem is verified.

(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 ∈ (0, π) such that f’(c) = 0.

f'(x) = cos x – 2cos 2x 

so, f'(c) = 0

f'(c) = cos c – 2cos 2c = 0

cos c – (2cos2c + 1) = 0

2cos2 c – cosc – 1 = 0

(cos c – 1)(2cos c + 1) = 0

cos c = 1, -1/2

c = cos(π/2), cos(2π/3)

So, c = π/2, 2π/3 ∈ (0, π)

Hence, Rolle’s theorem is verified.

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

Solution:

Given: y = 16 – x2, x ϵ [– 1, 1]

⇒ y (– 1) = 16 – (– 1)2

⇒ y (– 1) = 16 – 1

⇒ y (– 1) = 15

Now, 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).

⇒ 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

Hence, 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).

Question 5. At what points on the following curves is the tangent parallel to the x-axis?

(i) y = x2 on [-2, 2]

Solution:

Given: f(x) = x2

 f'(x) = 2x

Also, f(-2) = f(2) = 22 = (-2)2 = 4

So, there exists a c ϵ (-2, 2) such that f’(c) = 0.

⇒ 2c = 0

⇒ c = 0

Therefore the required point is (0, 0).

(ii) y = e^{1 - x^2}       on [−1, 1]

Solution:

Since exponential function is continuous and differentiable everywhere.

Here f(1) = f(-1) = 1

Thus, all the conditions of Rolle’s theorem are satisfied.

Consequently, there exists at least one point c in (-1, 1) for which f'(c) = 0.

⇒ f'(c) = 0

⇒ -2c e^{1 – c^2} = 0

⇒ c = 0

Hence, (0, e) is the required point.

(iii) y = 12 (x + 1) (x − 2) on [−1, 2]

Solution:

Since polynomial function is continuous and differentiable everywhere.

Here f(2) = f(-1) = 0.

Thus, all the conditions of Rolle’s theorem are satisfied.

Consequently, there exists at least one point c in (-1, 2) for which f'(c) = 0.

But f'(c) = 0

⇒ 24c – 12 = 0

⇒ c = 1/2

Hence, (1/2, -27) is the required point.

Question 6. If f: [5, 5] to R is a differentiable function and if f'(x) does not vanish anywhere, prove that f(-5) ≠ f(5).

Solution:

Here f(x) is differentiable on (5,5).

By Mean Value Theorem,

f'(c) = \frac{f(5)-f(-5)}{5-(-5)}

⇒ 10 f'(c) = f(5) – f(-5)

It is also given that f'(x) does not vanish anywhere.

⇒ f'(c) ≠ 0

⇒ 10 f'(c) ≠ 0

⇒  f(5) – f(-5) ≠ 0

⇒  f(5) ≠ f(-5)

Hence Proved

Question 7. Examine if Rolle’s theorem is applicable to any one of the following functions.

(i) f (x) = [x] for x ∈ [5, 9]

(ii) f (x) = [x] for x ∈ [−2, 2]

Can you say something about the converse of Rolle’s Theorem from these functions?

Solution:

Rolle’s theorem states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) such that f(a) = f(b), then f′(x) = 0 for some x with a ≤ x ≤ b.

(i) f(x) = [x] for x ∈ [ 5 , 9 ]

 f(x) is not continuous at x = 5 and x = 9. Thus, f (x) is not continuous on [5, 9].

Also , f (5) = [5] = 5 and f (9) = [9] = 9

∴ f (5) ≠ f (9)

The left hand limit of  f at x = n is ∞ and the right hand limit of f at x = n is 0.

Since the left and the right hand limits of f at x = n are not equal, f is not differentiable on (5, 9).

Hence, Rolle’s theorem is not applicable on function f(x) for x ∈ [5 , 9].  

(ii)  f (x) = [x] for x ∈  [-2 , 2]

f(x) is not continuous at x = −2 and x = 2. Thus, f (x) is not continuous on [−2, 2].

Also, f (-2) = [-2] = -2 and f (2) = [2] = 2

∴ f (-2) ≠ f (2)

The left hand limit of f at x = n is ∞ and the right hand limit of f at x = n is ∞.

Since the left and the right hand limits of f at x = n are not equal, thus, f is not differentiable on (−2, 2).

Hence, Rolle’s theorem is not applicable on function f(x) = [x] for x ∈ [ -2, 2].

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.

Leave a Comment

Your email address will not be published.