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 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 .

Find the values of a and b so that the function f x = x 2 + 3 x + a , if x ≤ 1 b x + 2 , if x > 1 is differentiable at each x ∈ R .

Given: f x = x 2 + 3 x + a , if x ≤ 1 b x + 2 , if x > 1 It is given that the function is differentiable at each x ∈ R and every differentiable function is continuous. So f ( x ) is continuous at x = 1 Therefore, lim x → 1 - f x = lim x → 1 + f x = f 1 ⇒ lim x → 1 x 2 + 3 x + a = lim x → 1 b x + 2 = a + 4 ⇒ a + 4 = b + 2 = a + 4 → 1 Since, f ( x ) is differentiable at x = 1 . So, ( LHD at x = 1 ) = ( RHD at x = 1 ) Thus, lim x → 1 - f ( x ) - f ( 1 ) x - 1 = lim x → 1 + f ( x ) - f ( 1 ) x - 1 ⇒ lim x → 1 x 2 + 3 x + a - a - 4 x - 1 = lim x → 1 b x + 2 - 4 - a x - 1 ⇒ lim x → 1 x 2 + 3 x - 4 x - 1 = lim x → 1 b x + 2 - 4 - a x - 1 ⇒ lim x → 1 x 2 + 4 x - x - 4 x - 1 = lim x → 1 b x + 2 - 4 - a x - 1 ⇒ lim x → 1 ( x + 4 ) ( x - 1 ) x - 1 = lim x → 1 b x - 2 - a x - 1 ⇒ lim x → 1 ( x + 4 ) ( x - 1 ) x - 1 = lim x → 1 b x - b x - 1 [Using 1 ] ⇒ lim x → 1 x + 4 = lim x → 1 b ⇒ 5 = b From 1 , we have: a + 4 = b + 2 ⇒ a + 4 = 5 + 2 ⇒ a = 7 - 4 ⇒ a = 3 Hence, a = 3 , b = 5 .

Show that the function f x = | 2 x - 3 | [ x ] , x ≥ 1 sin π x 2 , x < 1 is continuous but not differentiable at x = 1 .

Given: f x = | 2 x - 3 | [ x ] , x ≥ 1 sin π x 2 , x < 1 Now, check continuity at x = 1 ( LHL at x = 1 ) = lim x → 1 - f x = lim h → 0 f 1 - h = lim h → 0 sin π ( 1 - h ) 2 = sin π 2 = 1 And ( RHL at x = 1 ) = lim x → 1 + f x = lim h → 0 f 1 + h = lim h → 0 | 2 ( 1 + h ) - 3 | [ 1 + h ] = lim h → 0 | 2 ( 1 + h ) - 3 | = 1 And also, f 1 = 2 1 - 3 1 = 1 Hence, L H L = R H L = f 1 Similarly, check differentiability at x = 1 Now, ( LHD at x = 1 ) = lim x → 1 - f ( x ) - f ( 1 ) x - 1 ( LHD at x = 1 ) = lim h → 0 f ( 1 - h ) - f ( 1 ) 1 - h - 1 ( LHD at x = 1 ) = lim h → 0 f ( 1 - h ) - f ( 1 ) - h ( LHD at x = 1 ) = lim h → 0 sin π ( 1 - h ) 2 - 1 - h ( LHD at x = 1 ) = lim h → 0 cos π h 2 - 1 - h ( LHD at x = 1 ) = - π 2 lim h → 0 cos π h 2 - 1 π h 2 = 0 Similarly, ( RHD at x = 1 ) = lim x → 1 + f ( x ) - f ( 1 ) x - 1 ( RHD at x = 1 ) = lim h → 0 f ( 1 + h ) - f ( 1 ) 1 + h - 1 ( RHD at x = 1 ) = lim h → 0 f ( 1 + h ) - f ( 1 ) h ( RHD at x = 1 ) = lim h → 0 - ( 2 ( 1 + h ) - 3 ) - 1 h ( RHD at x = 1 ) = lim h → 0 - 2 h h = - 2 Thus, LHD ≠ RHD Hence, the function is continuous but not differentiable at x = 1 .

If f x = a x 2 - b , if | x | < 1 1 | x | , if | x | ≥ 1 is differentiable at x = 1 , find a , b .

Given, f x = a x 2 - b , if | x | < 1 1 | x | , if | x | ≥ 1 ⇒ f x = - 1 x , if x < - 1 a x 2 + b , if - 1 < x < 1 1 x , if x ≥ 1 It is given that the given function is differentiable at x = 1 . We know every differentiable function is continuous. Therefore, it is continuous at x = 1 . Then, lim x → 1 - f x = lim x → 1 + f x ⇒ lim x → 1 a x 2 - b = lim x → 1 1 x ⇒ a - b = 1 … ( i ) It is also differentiable at x = 1 . Therefore, L H D = R H D ⇒ lim x → 1 - f ( x ) - f ( 1 ) x - 1 = lim x → 1 + f ( x ) - f ( 1 ) x - 1 ⇒ lim x → 1 a x 2 - b - 1 x - 1 = lim x → 1 1 x - 1 x - 1 ⇒ lim x → 1 a x 2 + 1 - a - 1 x - 1 = lim x → 1 - ( x - 1 ) x x - 1 [Using ( i ) ] ⇒ lim x → 1 a x + 1 = lim x → 1 - 1 ⇒ 2 a = - 1 ⇒ a = - 1 2 From ( i ) , we have: a - b = 1 ⇒ - 1 2 - b = 1 ⇒ b = - 3 2 Hence, when a = - 1 2 and b = - 3 2 , the function is differentiable at x = 1 .

E M B I B E

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

R D Sharma Solutions for Chapter: Differentiability, Exercise 1: EXERCISE

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Differentiability, Exercise 1: EXERCISE

Attempt the free practice questions on Chapter 10: Differentiability, Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. MATHEMATICS CLASS XII VOLUME-1 solutions are prepared by Experienced Embibe Experts.

Questions from R D Sharma Solutions for Chapter: Differentiability, Exercise 1: EXERCISE with Hints & Solutions

HARD

12th CBSE

IMPORTANT

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

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

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 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 $m \leq 0$ .

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 10 - Differentiability>EXERCISE>Q 8

Find the values of $a$ and $b$ so that the function $f (x) = \{\begin{cases} x^{2} + 3 x + a, & if x \leq 1 \\ b x + 2, & if x > 1 \end{cases}$ is differentiable at each $x \in R$ .

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 10 - Differentiability>EXERCISE>Q 9

Show that the function $f (x) = \{\begin{cases} | 2 x - 3 | [x], & x \geq 1 \\ \sin (\frac{π x}{2}), & x < 1 \end{cases}$ is continuous but not differentiable at $x = 1$ .