Show that f x = x 1 3 is differentiable at x = 0 .

Given: f x = x 1 3 We have, L H D = lim x → 0 - f x - f 0 x - 0 = lim h → 0 f 0 - h - f 0 0 - h - 0 = lim h → 0 0 - h 1 3 - 0 1 3 - h = lim h → 0 - h 1 3 - h = lim h → 0 - h - 2 3 = 0 Similarly, R H D = lim x → 0 + f x - f 0 x - 0 = lim h → 0 f 0 + h - f 0 0 + h - 0 = lim h → 0 0 + h 1 3 - 0 1 3 h = lim h → 0 h 1 3 h = lim h → 0 h - 2 3 = 0 Thus, L H D = R H D = 0 Hence, f x = x 1 3 is differentiable at x = 0 .

Show that f x = 12 x - 3 , if x ≤ 3 2 x 2 + 5 if x > 3 is differentiable at x = 3 . Also, find f ' ( 3 ) .

Given that, the function is f x = 12 x - 3 , if x ≤ 3 2 x 2 + 5 if x > 3 Now, L H D = lim x → 3 - f x - f 3 x - 3 = lim x → 3 12 x - 13 - 23 x - 3 = lim x → 3 12 x - 36 x - 3 = lim x → 3 12 ( x - 3 ) x - 3 = lim x → 3 12 = 12 Similarly, R H D = lim x → 3 f x - f 3 x - 3 = lim x → 3 2 x 2 + 5 - 23 x - 3 = lim x → 3 2 ( x 2 - 9 ) x - 3 = lim x → 3 2 x - 3 x + 3 x - 3 [Since 0 0 form, so apply factorisation method] = lim x → 3 2 ( x + 3 ) = 2 × 6 = 12 Thus, L H D = R H D = 12 . So, f ( x ) is differentiable at x = 3 and f ' ( 3 ) = 12 .

Show that the function f x = x m sin 1 x , x ≠ 0 0 , x = 0 is differentiable at x = 0 , if m > 1 .

Given: f x = x m sin 1 x , x ≠ 0 0 , x = 0 Let m = 2 , then the function becomes f x = x 2 sin 1 x , x ≠ 0 0 , x = 0 Checking differentiability at x = 0 Now, L H D = lim x → 0 - f ( x ) - f ( 0 ) x - 0 and R H D = lim x → 0 + f ( x ) - f ( 0 ) x - 0 ⇒ L H D = lim h → 0 f ( 0 - h ) - f ( 0 ) 0 - h - 0 and R H D = lim h → 0 f ( 0 + h ) - f ( 0 ) 0 + h - 0 ⇒ L H D = lim h → 0 f ( - h ) - f ( 0 ) - h and R H D = lim h → 0 f ( h ) - f ( 0 ) h ⇒ L H D = lim h → 0 - h 2 sin 1 - h - 0 - h and R H D = lim h → 0 h 2 sin 1 h - 0 h ⇒ L H D = lim h → 0 h sin 1 h and R H D = lim h → 0 h sin 1 h And we know that, lim h → 0 h sin 1 h = 0 × (an oscillating function between - 1 and 1 ) = 0 Thus, L H D = R H D = 0 Hence, the given function is differentiable at x = 0 .

Show that the function f x = x m sin 1 x , x ≠ 0 0 , x = 0 is continuous but not differentiable at x = 0 , if 0 < m < 1 .

Given: f x = x m sin 1 x , x ≠ 0 0 , x = 0 Let m = 1 2 , 0 < m < 1 . Then the function becomes f x = x 1 2 sin 1 x , x ≠ 0 0 , x = 0 Checking Continuity at x = 0 , Now, L H L = lim f x x → 0 - = lim f 0 - h h → 0 = lim h → 0 ( 0 - h ) 1 2 sin 1 0 - h = lim h → 0 h 1 2 sin 1 h = 0 × [An oscillating number between - 1 and 1 ] = 0 Similarly, R H L = lim f x x → 0 + = lim f 0 + h h → 0 = lim h → 0 ( 0 + h ) 1 2 sin 1 0 + h = lim h → 0 h 1 2 sin 1 h = 0 × [An oscillating number between - 1 and 1 ] = 0 And f ( 0 ) = 0 Thus, L H L = R H L = f 0 Hence, the given function is continuous at x = 0 , if 0 < m < 1 . Checking differentiability at x = 0 , Now, L H D = lim x → 0 - f ( x ) - f ( 0 ) x - 0 and R H D = lim x → 0 + f ( x ) - f ( 0 ) x - 0 ⇒ L H D = lim h → 0 f ( 0 - h ) - f ( 0 ) 0 - h - 0 and R H D = lim h → 0 f ( 0 + h ) - f ( 0 ) 0 + h - 0 ⇒ L H D = lim h → 0 f ( - h ) - f ( 0 ) - h and R H D = lim h → 0 f ( h ) - f ( 0 ) h ⇒ L H D = lim h → 0 - h 1 2 sin 1 - h - 0 - h and R H D = lim h → 0 h 1 2 sin 1 h - 0 h ⇒ L H D = lim h → 0 h - 1 2 sin 1 h and R H D = lim h → 0 h - 1 2 sin 1 h And we know that, lim h → 0 h - 1 2 sin 1 h = 0 × (an oscillating number between - 1 and 1 ) = 0 Thus, L H D = R H D = 0 Hence, the given function is differentiable at x = 0 .

Show that the function f x = x m sin 1 x , x ≠ 0 0 , x = 0 neither continuous nor differentiable, if m ≤ 0 .

Given: f x = x m sin 1 x , x ≠ 0 0 , x = 0 Let m = 0 , m ≤ 0 . Then the function becomes f x = x 0 sin 1 x , x ≠ 0 0 , x = 0 ⇒ f x = sin 1 x , x ≠ 0 0 , x = 0 Since we know during finding continuity and differentiability we will come across the term sin 1 0 and it is not defined. So given function is neither continuous nor differentiable at x = 0 if m ≤ 0 .

E M B I B E

MATHEMATICS CLASS XII VOLUME-1>Chapter 10 - Differentiability>EXERCISE>Q 1

HARD

12th CBSE

IMPORTANT

Earn 100

Show that $f (x) = | x - 3 |$ is continuous but not differentiable at $x = 3$ .

Important Questions on Differentiability

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 10 - Differentiability>EXERCISE>Q 2

Show that

f (x) = x^{\frac{1}{3}}

is differentiable at

x = 0

.

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 10 - Differentiability>EXERCISE>Q 3

Show that

f (x) = \{\begin{cases} 12 x - 3, & if x \leq 3 \\ 2 x^{2} + 5 & if x > 3 \end{cases}

is differentiable at

x = 3

. Also, find

f^{'} (3)

.

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 10 - Differentiability>EXERCISE>Q 4

Show that the function

f

defined as follows, is continuous at

x = 2

, but not differentiable at

x = 2

, where

f (x) = \{\begin{cases} 3 x - 2, & 0 < x \leq 1 \\ 2 x^{2} - x, & 1 < x \leq 2 \\ 5 x - 4, & x > 2 \end{cases}

.

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 10 - Differentiability>EXERCISE>Q 5

Discuss the continuity and differentiability of the

f (x) = |x| + |x - 1|

in the interval

(- 1, 2)

.

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 10 - Differentiability>EXERCISE>Q 6

Find whether the function is differentiable at

x = 1

and

x = 2

, where

f (x) = \{\begin{matrix} x, & x \leq 1 \\ 2 - x, & 1 \leq x \leq 2 \\ - 2 + 3 x - x^{2}, & x > 2 \end{matrix}

.

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 10 - Differentiability>EXERCISE>Q 7

Show that the function $f (x) = \{\begin{cases} x^{m} \sin (\frac{1}{x}) & , x \neq 0 \\ 0 & , x = 0 \end{cases}$ is differentiable at $x = 0$ , if $m > 1$ .

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 10 - Differentiability>EXERCISE>Q 7

Show that the function $f (x) = \{\begin{cases} x^{m} \sin (\frac{1}{x}), & x \neq 0 \\ 0, & x = 0 \end{cases}$ is continuous but not differentiable at $x = 0$ , if $0 < m < 1$ .

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 10 - Differentiability>EXERCISE>Q 7

Show that the function $f (x) = \{\begin{cases} x^{m} \sin (\frac{1}{x}) & , x \neq 0 \\ 0 & , x = 0 \end{cases}$ neither continuous nor differentiable, if $m \leq 0$ .