If f is defined by f ( x ) = x 2 - 4 x + 7 , show that f ' 5 = 2 f ' 7 2 .

Given: f ( x ) = x 2 - 4 x + 7 Clearly, f ( x ) being a polynomial function, is everywhere differentiable. The derivative of f at x is given by f ' x = lim h → 0 f ( x + h ) - f ( x ) h ⇒ f ' x = lim h → 0 ( x + h ) 2 - 4 ( x + h ) + 7 - x 2 - 4 x + 7 h ⇒ f ' x = lim h → 0 x 2 + h 2 + 2 x h - 4 x - 4 h + 7 - x 2 + 4 x - 7 h ⇒ f ' x = lim h → 0 h 2 + 2 x h - 4 h h ⇒ f ' x = lim h → 0 h ( h + 2 x - 4 ) h ⇒ f ' x = 2 x - 4 Now, f ' 5 = 2 × 5 - 4 = 6 Similarly, f ' 7 2 = 2 × 7 2 - 4 = 3 Therefore, f ' 5 = 2 × 3 = 2 f ' 7 2 Hence, proved.

Examine the differentiability of the function f defined by f x = 2 x + 3 , if - 3 ≤ x ≤ - 2 x + 1 , if - 2 ≤ x < 0 x + 2 , if 0 ≤ x ≤ 1 .

Given, f x = 2 x + 3 , if - 3 ≤ x ≤ - 2 x + 1 , if - 2 ≤ x < 0 x + 2 , if 0 ≤ x ≤ 1 Since, polynomial functions are continuous and differentiable in their domain. We need to check the differentiability at x = - 2 & x = 0 Now, check the differentiability at x = - 2 Now, L H D = lim h → 0 f ( - 2 - h ) - f ( - 2 ) - h = lim h → 0 2 ( - 2 - h ) + 3 - ( - 2 + 1 ) - h = lim h → 0 2 h h = lim h → 0 2 = 2 Similarly, R H D = lim h → 0 f ( - 2 + h ) - f ( - 2 ) h = lim h → 0 ( - 2 + h ) + 1 - ( - 2 + 1 ) - h = lim h → 0 h - 1 - ( - 1 ) h = lim h → 0 h h = 1 Thus, L H D ≠ R H D ∴ f is not differentiable at x = - 2 Now, check differentiability at x = 0 So, L H D = lim h → 0 f ( 0 - h ) - f ( 0 ) - h = lim h → 0 ( 0 - h ) + 1 - ( 0 + 2 ) - h = lim h → 0 - h - 1 - h = lim h → 0 1 - 1 h , which does not exist. ∴ Hence f is not differentiable at x = 0 .

Discuss the continuity and differentiability of f x = x - c cos 1 x - c , x ≠ c 0 , x = c

Given that, we have: f x = x - c cos 1 x - c , x ≠ c 0 , x = c Now check continuity at x = c So, we have: L H L = lim x → c - f x = lim h → 0 f c - h = lim h → 0 c - h - c cos 1 c - h - c = lim h → 0 - h cos 1 h = 0 × [An oscillating number between - 1 and 1 ] = 0 Similarly, R H L = lim x → c + f x = lim h → 0 f c + h = lim h → 0 c + h - c cos 1 c + h - c = lim h → 0 h cos 1 h = 0 × [An oscillating number between - 1 and 1 ] = 0 Also, f 0 = c Thus, clearly, L H L = R H L = f 0 Hence, the function f x is continuous at x = c . Now, check differentiability at x = c So, we have: L H D = lim x → c - f ( x ) - f ( c ) x - c = lim h → 0 f ( c - h ) - f ( c ) c - h - c = lim h → 0 - h cos 1 - h - 0 - h = lim h → 0 cos 1 h Since, cos 1 h is an oscillating function which oscillates between - 1 and 1 ∴ L H D does not exist. Hence, f x is not differentiable at x = c .

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

If $f$ is defined by $f (x) = x^{2}$ , find $f^{'} (2)$ .

HARD

12th CBSE

IMPORTANT

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

If $f$ is defined by $f (x) = x^{2} - 4 x + 7$ , show that $f^{'} (5) = 2 f^{'} (\frac{7}{2})$ .

HARD

12th CBSE

IMPORTANT

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

If for the function $ϕ (x) = λ x^{2} + 7 x - 4, ϕ^{'} (5) = 97$ , find $λ$ .

HARD

12th CBSE

IMPORTANT

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

If $f (x) = x^{3} + 7 x^{2} + 8 x - 9$ , find $f^{'} (4)$ .

HARD

12th CBSE

IMPORTANT

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

Examine the differentiability of the function $f$ defined by $f (x) = \{\begin{matrix} 2 x + 3, if - 3 \leq x \leq - 2 \\ x + 1, if - 2 \leq x < 0 \\ x + 2, if 0 \leq x \leq 1 \end{matrix}$ .