ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.12

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.12

Verify Rolle’s theorem for the following (1 to 5) functions and find point (or points) in the interval where derivative is zero :

Question 1.
(i) f(x) = x² – 5x + 6 in [1, 4]
(ii) = x² + 2x – 8, x ∈ [- 4, 2]
(iii) f(x) = x³ – 3x in [- √3, 0] (NCERT)
Solution:
(i) Given f(x) = x² – 5x + 6 …………(1)
Since f(x) be a polynomial function and hence continuous everywhere and
∴ f(x) be continuous in [1, 4].
Also f(x) being a polynomial function so it is differentiable everywhere.
∴ f(x) is derivable in (1, 4).
Also, f(1) = 1² – 5 + 6 = 2 ;
f(4) = 16 – 20 + 6 = 2
f(1) = f(4)
Thus, all the three conditions of Rolle’s Theorem are satisfied.
So ∃ atleast one real number
c ∈ (1, 4) s.t f’(c) = 0
Diff. (1) w.r.t. x, we have
f’(x) = 2x – 5
Now f'(c) = 0
⇒ 2c – 5 = 0.
⇒ c = \(\frac{5}{2}\) ∈ (1, 4)
So there exists \(\frac{5}{2}\) ∈ (1, 4) s.t. f'(\(\frac{5}{2}\)) = 0
Hence Rolle’s theorem is verified and c = \(\frac{5}{2}\).

(ii) Given, f(x) = x² + 2x – 8.
Since f(x) is polynomial in x so it is continuous in [- 4, 2]
also f’(x) = 2x + 2 exists ∀ x ∈ (- 4, 2)
∴ f is derivable in (- 4, 2).
also, f(- 4) = 16 – 8 – 8 = 0
and f(2) = 4 + 4 – 8 = 0
∴ f (- 4) = f(2)
Thus all the three conditions of Roile’s theorem are satisfied.
∴ ∃ one real number c ∈ (- 4, 2) s.t.f’ (c) = 0
i.e. 2c + 2 = 0
⇒ c = – 1 ∈ (- 4, 2)
Hence Rolle’s theorem is verified.

(iii) Given f(x) = x³ – 3x ………( 1)
Clearly f(x) be a polynomial in x.
∴ f (x) be contriuous in [- √3, 0].
Also f(x) be derivable in (- √3, 0).
Since f’(x) = 3x² – 3 which exists in (- √3, 0).
f(- √3) = (- √3)³ – 3 (- √3)
= – 3√3 + 3√3 = 0
f(0) = 0
Thus, f(- √3) = f(0)
Thus, all the three conditions of Rolle’s theorem are satisfied.
So ∃ atleast one real number c ∈ (- √3, 0) s.t. f’(c) = 0
Diff. eqn. (1) both sides w.r.t. x, we have
f’(x) = 3x² – 3
Now f’(c) = 0
⇒ 3c²– 3 = 0
⇒ c = ± 1
Clearly c = 1 ∉ (- √3, 0)
So there exists c = – 1 ∈ (- √3, 0)
s.t. f’(c) = 0
Hence Roll&s theorem is verified and c = – 1.

Question 1 (old).
(ii) f(x) = x² + 2 on [- 2, 2] (NCERT)
Solution:
We have f(x) = x² + 2
Since f(x) is polynormal in x
∴ it is continuous in [- 2, 2]
also f'(x) = 2x exists ∀ x ∈ (2, 2)
∴ f is derivable in (- 2, 2).
also f(2) = 2² + 2 = 6
and f(- 2) = (- 2)² + 2 = 6
∴ f(2) = f(- 2)
Hence all the conditions of Rolle’s theorem are satisfied
∴ ∃ atleast one number c ∈ (- 2, 2) s.t. f'(c) = 0
i.e. 2c = 0
⇒ c = 0 ∈ (- 2, 2)
Hence Rolle’s theorem is verified.

Question 2.
(i) f(x) = x³ + 3x² – 24x – 80 in [- 4, 5]
(ii) f(x) = x (x – 1)² in O, 1 (NCERT Excmplar)
(iii) f(x) = (x – 1) (x – 2) (x – 3) in [1, 3]
(iv) f(x) = \(\sqrt{4-x^2}\) in [- 2, 2] (NCERT Exemplar)
Solution:
(i) Given f (x) = x³ + 3x² – 24x – 80 in [- 4, 5]
Since f(x) is polynomial in x
∴ Continuous everywhere and differentiable.
Hence f is continuous in [- 4, 5] and derivable in (- 4, 5).
Also, f(- 4) = – 64 – 48 + 96 – 80 = 0
and f(5) = 125 + 75 – 120 – 80 = 0
∴ All the three conditions of rolle’s theorem are satisfied.
∴ ∃ atleast one real no. c ∈ (- 4, 5) s.t. f’(c) = 0
i.e. 3c² + 6c – 24 = 0
⇒ c² + 2c – 8 = 0
i.e. c = \(\frac{-2 \pm 6}{2}\)
= 2, – 4
but c = – 4 ∉ (- 4, 5)
∴ c = 2 ∈ (- 4, 5)
Hence Rolle’s theorem is verified and e 2.

(ii) Given f(x) = x (x – 1)² on [0, 1]
and f(x) = x (x² – 2x + 1)
Since f(x) ¡s polynomial in x
∴ continuous in [0, 1] and derivable in (0, 1) as f’(x) = 3x² – 4x + 1 exists ∀ x ∈ (0, 1)
also, f(0) = 0 = f(1)
∴ all the three conditions of RoBe’s theorem are satisfied.
∴ ∃ atleast one real number c ∈ (0, 1) s.t. f’(c) = 0
i.e. 3c² – 4c + 1 = 0
⇒ c = 1, \(\frac{1}{3}\)
but c = 1 ∉ (0, 1)
∴ c = \(\frac{1}{3}\) ∈ (0, 1)
Hence Rolle’s theorem is verified and c = \(\frac{1}{3}\).

(iii) Given f(x) = (x- 1) (x – 2) (x -3)
= (x – 1) (x² – 5x + 6)
= (x³ – 6x² + 11x – 6)
f(x) = x³ – 6x² + 11x – 6
Since f(x) is polynomial in x
∴ it is continuous and derivable everywhere.
∴ f(x) is continuous in [1, 3] and derivable in (1, 3)
Also f(1) = 0 = f(3)
All the three conditions of Rolle’s theorem are satisfied.
∴∃ atleast one real number c ∈ (1, 3) s.t. f’(c) = 0
i.e. 3c² – 12c + 11 = 0
⇒ c = \(\frac{12 \pm \sqrt{12}}{6} \)
= \(\frac{6 \pm \sqrt{3}}{3}\)
= 2 + \(\frac{1}{\sqrt{3}}\), 2 – \(\frac{1}{\sqrt{3}}\)
∴ c = \(\frac{6 \pm \sqrt{3}}{3}\) ∈ (1, 3)
Hence there are more than one c ∈ (1, 3) s.t.f’ (c) = 0.

(iv) Given f(x) = \(\sqrt{4-x^2}\) …………..(1)
Clearly D_F = [- 2, 2].
So f(x) is clearly continuous in its domain i.e. [- 2, 2]
Also,f’ (x) = \(\frac{1}{2 \sqrt{4-x^2}}\) (- 2x)
= \(\frac{-x}{\sqrt{4-x^2}}\)
Clearly f’ (x) exists in (- 2, 2).
So f(x) is clearly derivable in (- 2, 2).
Also f(- 2) – 0 = F (2)
So all the three conditions of Rolle’s theorem are satisfied.
So 3 atleast one real number c ∈ (- 2, 2) s.t f’ (C) = 0
Now, f’ (c) = 0
\(\frac{-c}{\sqrt{4-c^2}}\) = 0
c = 0 ∈ (- 2, 2)
So there exists a real number c ∈ (- 2, 2) s.t f’(c) = 0
Hence Rolle’s Theorem verified and c = 0.

Question 3.
(i) f(x) = cos 2x in [- \(\frac{\pi}{4}\), \(\frac{\pi}{4}\)]
(ii) f(x) = sin x – 1 in [\(\frac{\pi}{2}\), \(\frac{5 \pi}{2}\)]
Solution:
(i) Given f(x) = cos 2x in [- \(\frac{\pi}{4}\), \(\frac{\pi}{4}\)]
Since cosine function is continuous and derivable everywhere in its domain.
∴ f(x) is cintinuous in [- \(\frac{\pi}{4}\), \(\frac{\pi}{4}\)] and derivable in (- \(\frac{\pi}{4}\), \(\frac{\pi}{4}\))
also f(- \(\frac{\pi}{4}\)) = 0 = f(\(\frac{\pi}{4}\))
∴ all the three conditions of Roll’s theorem satisfied.
∃ atleast one real number c ∈ (- \(\frac{\pi}{4}\), \(\frac{\pi}{4}\))
s.t. f'(c) = 0 i.e. – 2 sin 2c = 0
⇒ sin 2c = 0
⇒ c = 0 ∈ (- \(\frac{\pi}{4}\), \(\frac{\pi}{4}\))
Hence Roll’s theorem is verified and c = 0.

(ii) Since sine function and constant functions are continuous and differentiable everywhere.
∴ f(x) is continuous in [\(\frac{\pi}{2}\), \(\frac{5 \pi}{2}\)]
and derivable in (\(\frac{\pi}{2}\), \(\frac{5 \pi}{2}\)).
Also f(\(\frac{\pi}{2}\)) = 1 – 1 = o
and f(\(\frac{5 \pi}{2}\)) = sin \(\frac{5 \pi}{2}\) – 1
= 1 – 1 = 0
∴ All the three conditions of Rolle’s theorem are satisfied.
∴ ∃ atleast one real no c ∈ (\(\frac{\pi}{2}\), \(\frac{5 \pi}{2}\))
s.t. f'(c) = 0
i.e. cos c = 0
⇒ c = nπ + \(\frac{\pi}{2}\) ∀ n ∈ I
Thus, c = \(\frac{\pi}{2}\), \(\frac{3 \pi}{2}\), \(\frac{5 \pi}{2}\)
But c = – \(\frac{\pi}{2}\), \(\frac{\pi}{2}\) ∉ (\(\frac{\pi}{2}\), \(\frac{5 \pi}{2}\))
∴ c = \(\frac{3 \pi}{2}\) [\(\frac{\pi}{2}\), \(\frac{5 \pi}{2}\)]
Hence Rolls theorem is verified and
c = \(\frac{3 \pi}{2}\).

Question 4.
(i) f(x) = sin 2x in [0, \(\frac{\pi}{2}\)] (NCERT Exampler)
(ii) f(x) = sin 3x in [0, π].
Solution:
(i) Given f(x) = sin 2x in [0, \(\frac{\pi}{2}\)]
Since sin function is continuous and derivable everywhere
∴ f(x) is continuous in [0, \(\frac{\pi}{2}\)] and derivable in (0, \(\frac{\pi}{2}\))
Also, f(0) = 0 = f(\(\frac{\pi}{2}\))
∴ All the three conditions of Roll’s theorem are sarisfied.
∴ ∃ atleast one real no. c ∈ (0, \(\frac{\pi}{2}\)) s.t. f'(c) = 0
i.e. 2 cos 2c = 0
⇒ cos 2c = 0
⇒ 2c = \(\frac{\pi}{2}\)
⇒ c = \(\frac{\pi}{4}\) ∈ (0, \(\frac{\pi}{2}\))
Hence Roll’s theorem is verified and c = \(\frac{\pi}{4}\).

(ii) Given f(x) = sin 3x
since sin 3x trigonometric sine function, is continuous and differentiable everywhere.
∴ f(x) is continuous on [0, π] and f (x) is differentiable on (0, π).
f(0) = 0 ; f(π) = sin 3π = 0
∴ f(0) = f(π)
Thus, all the three conditions of Rolle’s theorem arc satisfied.
Now we want to show that ∃ atleast one real number c ∈ (0, π) s.t. f’ (c) = 0
we have, f(x) = sin 3x
∴ f’ (x) = 3 cos 3x
Now f’(c) = 0
⇒ 3 cos 3c = 0
⇒ 3c = \(\frac{\pi}{2}\)
⇒ c = \(\frac{\pi}{6}\)
i.e. c = π/6 ∈ (0, π) s.t.f’ (c) = 0
Hence Rolle’s theorem verified.

Question 5.
(i) f(x) = e^x sin x on [0, π] (ISC 2005)
(ii) f(x) = e^x cos x on [- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)]
(iii) f(x) = e^-x sin x in [0, π]
(iv) f(x) = e^2x (sin 2x – cos 2x) in [\(\frac{\pi}{8}\), \(\frac{5 \pi}{8}\)]
Solution:
(i) Given f(x) = e^x sinx
since, e^x and sin x are dif1erentiable and continuous everywhere.
Therefore, product of two functions
i.e. f(x) = e^x sin x is continuous on [0, π] and differentiable on (0, π).
Now f(0) = 0;
f(π) = e^π × 0 = 0
∴ f(0) = f(c)
So, all the three conditions of Rolle’s theorem are satisfied.
Now we want to show that, ∃ atleast one real number
c ∈ (0, π) such that f’ (c) = 0.
We have f (x) = e^x sin x
∴ f'(x) = e^x cos x + sin x e^x
= e^x (cos x + sin x)
Now f’ (c) = 0
⇒ e^c (cos c + sin c) = 0
⇒ cos c + sin c = 0 [∵ e^c > 0]
⇒ tan c = – 1
= tan \(\frac{3 \pi}{4}\)
⇒ c = \(\frac{3 \pi}{4}\) ∈ (0, π) such that f’ (c) = 0
Hence Rolle’s theorem verified.

(ii) Given f(x) = e^x cos x in [- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)]
Since cosine and exponential function are continuous everywhere.
Also product of two continuous functions is continuous.
∴ f (x) is continuous in [- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)] and derivable in (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)).
Also f’(x) = e^x cos x – e^x sin x
Also, f(- \(\frac{\pi}{2}\)) = 0
= f(\(\frac{\pi}{2}\))
Therefore all the three conditions of Rolle’s theorem are satisfied.
∴∃ atleast are real numbers c ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
s.t. f’(c) = 0, i.e. e^c (cos c – sin c) = 0
⇒ cos c – sin c = 0
[∵ e^c ≠ 0]
⇒ tan c = 1
⇒ c = \(\frac{\pi}{4}\) ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
∴ Rolle’s theorem is verified.

(iii) Given f(x) = e^-x sin x
Clearly e^-x is continuous and derivable everywhere and sin x is also continuous and differentiable in its domain.
Also product of two continuous functions is continuous.
Thus f(x) be continuous in [0, π]
and f(x) be derivable in (0, π).
Also f(0) = e^-0 sin 0 = 0;
f(π) = e^-π sin π
= e^-π × 0 = 0
∴ f(0) = f(π)
Thus, all the three conditions of Rolle’s theorem are satisfied.
So ∃ atleast one real number c ∈ (0, π) such that f’ (c) = 0
Diff. eqn. (1) both sides w.r.t. x, we have
⇒ f'(x) = e^-x cos x + sin x e^{– x} (- 1)
⇒ f’ (x) = e^{– x} (cos x – sin x)
Now f’ (c) = 0
⇒ e^{– c} (cos c – sin c) = 0
⇒ cos c – sin c = 0 (∵ e^{– c} > 0)
⇒ tan c = 1
= tan \(\frac{\pi}{4}\)
⇒ c = nπ + \(\frac{\pi}{4}\) ∀ n ∈ I
but c ∈ (0, π)
∴ c = \(\frac{\pi}{4}\)
Thus there exists \(\frac{\pi}{4}\) ∈ (0, π) s.t. f'(c) = 0
i.e. f'(\(\frac{\pi}{4}\)) = 0
Hence Rolle’s theorem is verified and c = \(\frac{\pi}{4}\).

(iv) Given f(x) = e^2x (sin 2x – cos 2x)
Clearly exponential function is differentiable everywhere and sine, cosine functions are differentiable everywhere in its domain,
f’(x) = e^2x (2 cos 2x + 2 sin 2x) + (sin 2x – cos 2x) e^2x . 2
= e^2x (4 sin 2x) which exists ∀ x ∈ R
Thus f(x) is continuous in [\(\frac{\pi}{8}\), \(\frac{5 \pi}{8}\)] and derivable in (\(\frac{\pi}{8}\), \(\frac{5 \pi}{8}\))
f(\(\frac{\pi}{8}\)) = \(e^{\frac{\pi}{4}\left(\sin \frac{\pi}{4}-\cos \frac{\pi}{4}\right)}\)
= 0

f(\(\frac{5 \pi}{8}\)) = \(e^{\frac{5 \pi}{4}}\left(\sin \frac{5 \pi}{4}-\cos \frac{5 \pi}{4}\right)\)
= \(e^{\frac{5 \pi}{4}}\left[\sin \left(\pi+\frac{\pi}{4}\right)-\cos \left(\pi+\frac{\pi}{4}\right)\right]\)
= \(e^{\frac{5 \pi}{4}}\left[-\sin \frac{\pi}{4}+\cos \frac{\pi}{4}\right]\)
= 0

Thus all the three conditions of Rolle’s Theorem are satisfied so atleast one real number C ∈ s.t.f'(c) = 0
⇒ 4e^2c sin 2c = 0
⇒ sin 2c = 0 [∵ e^2c > 0]
⇒ 2c = 0 π, 2π, 3π, ………….
⇒ c = 0, \(\frac{\pi}{2}\), π, \(\frac{3 \pi}{2}\), ………….
Clearly c = \(\frac{\pi}{2}\) ∈ (\(\frac{\pi}{8}\), \(\frac{5 \pi}{8}\)).
∴ Rolle’s theorem verified.

Question 5 (old).
(iii) f(x) = e^2x (sin 2x – cos 2x) in [\(\frac{\pi}{8}\), \(\frac{5 \pi}{8}\)]. (ISC 2008)
Solution:
Given f(x) = e^2x sin x
Clearly e^x is continuous and derivable everywhere and sin x is also continuous and differentiable in its domain.
Also product of two continuous functions is Continuous.
Thus f(x) be continuous in [0, π]
and f(x) be derivable in (0, π).
Also f(0) = e^-0 sin 0 = 0;
f(π) = e^{– π} sin π
= e^{– π} × 0 = 0
∴ f(0) = f( π)
Thus, all the three conditions of Rolle’s theorem are satisfied.
So ∃ atleast are real number c ∈ (0, π) such that f’ (c) = 0
Diff. eqn. (1) both sides w.r.t. x, we have
f’(x) = e^-x cos x + sin x e^-x (- 1)
⇒ f'(x) = e^-x (cos x – sin x)
Now f'(c) = 0
⇒ e^{– c} (cos c – sin c) = 0
⇒ cos c – sin c = 0 (∵ e^{– c} > 0)
⇒ tan c = 1
= tan \(\frac{\pi}{4}\)
⇒ c = nπ + \(\frac{\pi}{4}\) ∀ n ∈ I
but c ∈ (0, π)
∴ c = \(\frac{\pi}{4}\)
Thus there exists \(\frac{\pi}{4}\) ∈ (0, π) s.t. f'(c) = 0
i.e. f'(\(\frac{\pi}{4}\)) = 0
Hence Rolle’s theorem is verified and c = \(\frac{\pi}{4}\).

Question 6.
Apply Rolle’s theorem to find point (or points) on the following curves where the tangent is parallel to x-axis :
(i) y = x² in [- 2, 2]
(ii) y = – 1 + cos x on (0, 2π) (NCERT Exemplar)
Solution:
(i) Given y = x² in [- 2, 2]
Since f (x) is polynomial in x
∴ it is continuous and derivable everywhere
∴ f(x) is continuous in [- 2, 2] and derivable in (- 2, 2).
Also f(- 2) = 4 = f(2)
Hence all the three conditions of Rolle’s theorem are satisfied.
∴ ∃ atleast one real no. c ∈ (- 2, 2) s.t.f’(c) = 0
i.e. c = 0 ∈ (- 2, 2).
When x = 0,
y = 0²= 0
∴ (0, 0) be the required point at which tangent is || to x-axis.
[By Geometrical interpretation of Rolle’s Theorem]

(ii) Let f(x) = y = – 1 + cos x
Clearly f(x) is continuous in [0, 2π] and
derivable in (0, 2π).
Since curve and constant functions are continuous and everywhere differentiable
Now f(0) = – 1 + 1 = 0;
f(2π) = – 1 + cos 2π
= – 1 + 1 = 0
∴ f(0) = f (2π)
Thus, all the three conditions of Rolle’s theorem are satisfied.
So ∃ atleast one real number c ∈ (0, 2π) s.t.f’ (c) = 0
Diff. eqn. (1) w.r.t. x, we get
f’ (x) = – sin x
Now f’(c) = 0
⇒ – sin c = 0
But π ∈ (0, 2π)
∴ c = π ∈ (0, 2π)
So there exists a real number c = π ∈ (0, 2π)
s.t f’ (π) = 0 i.e. at which the tangent is || to x-axis
When x = π
∴ from (1);
y = – 1 + cos π
= – 1 – 1
= – 2
Hence, the required point on given curve at which the tangent is parallel to x-axis be (π, – 2).

Question 7.
Verify the conditions of Rolle’s Theorem for the following function :
f(x) = log (x² + 2) – log 3 on [- 1, 1].
Find a point in the given interval where the tangent to the curve is parallel to x axis. (ISC 2016)
Solution:
Let f(x) = log (x² + 2) – log 3
Clearly f(x) is continuous in [- 1, 1].
Now f'(x) = \(\frac{2 x}{x^2+2}\) exists ∀ x ∈ (- 1, 1)
∴ f is derivable in (- 1, 1).
further f(- 1) = log 3 – log 3 = 0
and f(1) = log 3 – log 3 = 0
∴ f(- 1) = f (1) = 0
Thus all the three conditions of Rolle’s theorem are satisfied.
∴∃ atleast one real number c ∈ (- 1, 1) s.t.f’(c) = 0
i.e. \(\frac{2 c}{c^2+2}\) = 0
⇒ c = 0 ∈ (- 1, 1)
Hence Rolle’s theorem is verified.

Question 8.
If Rolle’s theorem holds for the function f(x) = x³ + ax² + bx in [1, 2] at the point x = \(\frac{4}{3}\), then find the values of a and b. (ISC 2009)
Solution:
Given f(x) = x³ + ax² + bx …………..(1)
Since it is given that, Rolle’s theorem holds for the function f(x) in [1, 2].
∴ f(1) = f(2)
⇒ 1 + a + b = 8 + 4a + 2b
⇒ 3a + b = – 7 ………….(2)
Also ∃ atleast one rai number x = c ∈ (1, 2) s.t f’(c) = 0
Also it is given that, Rolle’s theorem holds for f(x) is [1, 2] at the point x = \(\frac{4}{3}\).
∴ f'(\(\frac{4}{3}\)) = 0
Diff. (1) w.r.t. x; we have
f’(x) = 3x² + 2ax + b
Now f'(\(\frac{4}{3}\) = 0
0 = 3 × \(\frac{16}{9}\) + 2a × \(\frac{4}{3}\) + b
8a + 3b = – 16 …………(3)
On solving eqn. (2) and eqn. (3); we have a = – 5 ; b = 8.

Question 9.
Examine if Rolle’s theorcm is applicable to the function f(x) = |x| for x ∈ [- 2, 2] What can you say about the converse of Rolle’s theorem ? (NCERT)
Solution:
at x = 2
\(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) [x] = 2 and
\(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) [x]
= (2 – 1) = 1
∴ f(x) is not continuous at x = 2 ∈ [- 2, 2]
Hence Rolle’s theorem is not applicable.

Question 10.
What can you say about the applicability of Rolle’s theorem for the following functions in the indicated intervals ?
(i) f(x) = x^1/3 in [- 1, 1]
(ii) f(x) = x^2/3 in [- 2, 2]
(iii) f(x) = 2 + (x – 1)^2/3 in [0, 2]
(iv) f(x) = 1 + |x – 2| in [0, 4]
(v) f(x) = tan x in [0, π]
(vi) f(x) = sec x in [0, 2π]
(vii) f(x) = \(\frac{x(x-2)}{x-1}\) in [0, 2]
(viii) f(x) = x² + 1 in [- 1, 2]
Solution:
(i) Given f(x) = x^1/3
∴ f'(x) = \(\frac{1}{3} \frac{1}{x^{2 / 3}}\) does not exists at x = 0 ∈ (- 1, 1)
∴ function f(x) is not derivable in (- 1, 1)
Hence Rolle’s theorem is not applicable.

(ii) Given f(x) = x^2/3
∴ f'(x) = \(\frac{2}{3}\) x^-1/3
= \(\frac{2}{3 x^{1 / 3}}\)
which does not exists at x = 0 ∈ (- 1, 1)
∴ function f(x) is not derivable in (- 1, 1)
Hence Rolle’s theorem is not verified.

(iii) Given f(x) = 2 + (x -1)^2/3 in [0, 2]
∴ f'(x) = \(\frac{2}{3}\) (x – 1)^-1/3
= \(\frac{2}{3(x-1)^{1 / 3}}\)
which does not exists at x = 1.
∴ function f(x) is not derivable in x = 1 ∈ (0, 2).
Hence Rolle’s theorem is not applicable.

(iv) Given f(x) = 1 + |x – 2|
Clearly f(x) is continuous in [0, 4]

Differentiability at x = 2
Lf'(2) = \(\ {Lt}_{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2}\)

[∵ as x → 2⁺
⇒ x > 2
⇒ x – 2 > 0
∴ |x – 2| = x – 2]
∴ Lf'(2) ≠ Rf'(2)
Thus f is not derivable or differentiable at x = 2.
Also, f'(x) = \(\frac{x-2}{|x-2|}\) ; x ≠ 2
Thus derivable of f(x) does not exists at x = 2.
∴ f(x) is not derivable at x = 2 ∈ (0, 4)
Hence f(x) is not derivable in (0, 4).
So condition (ii) of Rolle’s theorem is not satisfied.
Thus Rolle’s theorem is not applicable for f(x) = 1 + |x – 2| in [0, 4].

(v) Given f(x) = tan x in [0, 4]
Since f'(x) = sec² x does not exists at x = \(\frac{\pi}{2}\)
Thus f(x) is not differentiable at x = \(\frac{\pi}{2}\) ∈ [0, π]
∴ f is not derivable in (0, π)
Hence Rolle’s theorem is not applicable.

(vi) Given, f(x) = sec x
Clearly at x = \(\frac{\pi}{2}\) ∈ [0, 2π]
f(\(\frac{\pi}{2}\)) = sec \(\frac{\pi}{2}\) which does not exists
as \(\underset{x \rightarrow \frac{\pi^{-}}{2}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow \frac{\pi^{-}}{2}}{\mathrm{Lt}}\) sec x → + ∞
and \(\underset{x \rightarrow \frac{\pi^{+}}{2}}{\mathrm{Lt}}\) f(x) = latex]\underset{x \rightarrow \frac{\pi^{+}}{2}}{\mathrm{Lt}}[/latex] sec x → – ∞
Hence f(x) is continuous at x = \(\frac{\pi}{2}\) ∈ [0, 2π]
∴ f(x) is discontinuous in [0, 2π]
Hence the condition (i) of Roll’s theorem is not satisfied.
Thus, Rolle’s theorem is not appicable to functionf(x) = sec x in [0, 2π].

(vii) We have ;
f(x) = \(\frac{x(x-2)}{x-1}\)
Now f'(x) = \(\frac{(x-1)(2 x-2)-\left(x^2-2 x\right)}{(x-1)^2}\)
= \(\frac{2\left(x^2-2 x+1\right)-\left(x^2-2 x\right)}{(x-1)^2}\)
= \(\frac{x^2-2 x+2}{(x-1)^2}\)
Clearly f’( x) does not exists at x = 1 ∈ (0, 2)
Hence Rolle’s theorem is not applicable.

(viii) Given f(x) = x² + 1
Clearly f(x) be a polynomial function
∴ f(x) be continuous in [- 1, 2] and derivable in (- 1, 2).
But f(- 1) = 1 + 1 = 2;
f(2) = 2²+ 1 = 5
∴ f(- 1) ≠ f(2)
So condition (iii) of Rolle’s theorem is not satisfied.
Hence Rolle’s theorem is not applicable to function f(x) = 1 + x² in [- 1, 2].