Examine the function f x = 1 + x , i f x ≤ 2 5 - x , i f x > 2 for differentiability at x = 2 .

Given function is f x = 1 + x , i f x ≤ 2 5 - x , i f x > 2 We have to examine the function for differentiability at x = 2 . Here the domain of the function is R . Now, L f ' 2 = lim h → 0 f 2 - h - f 2 - h = lim h → 0 1 + 2 - h - 3 - h = lim h → 0 3 - h - 3 - h = lim h → 0 - h - h = lim h → 0 1 = 1 R f ' 2 = lim h → 0 f 2 + h - f 2 h = lim h → 0 5 - 2 + h - 3 h = lim h → 0 5 - 2 - h - 3 h = lim h → 0 3 - h - 3 h = lim h → 0 - h h = lim h → 0 - 1 = - 1 ∴ L f ' 2 ≠ R f ' 2 Hence, f is not derivable at x = 2 .

Show that the function f is continuous at x = 1 for all values of a where f x = a x 2 + 1 , x ≥ 1 x + a , x < 1 . Find its right and left-hand derivatives at x = 1 . Hence, find the condition for the existence of the derivative at x = 1 .

Given function is f x = a x 2 + 1 , x ≥ 1 x + a , x < 1 . Firstly, we have to show that the function f is continuous at x = 1 for all values of a . Continuity at x = 1 : LHL = lim x → 1 - f x = lim x → 1 - x + a = 1 + a RHL = lim x → 1 + f x = lim x → 1 + a x 2 + 1 = a 1 2 + 1 = a + 1 Also, f 1 = a x 2 + 1 = a 1 2 + 1 = a + 1 LHL = RHL = f ( 1 ) Thus, f is continuous at x = 1 , for all values of a . Differentiability at x = 1 : L f ' 1 = lim h → 0 f 1 - h - f 1 - h = lim h → 0 1 - h + a - a 1 2 + 1 - h = lim h → 0 1 - h + a - a - 1 - h = lim h → 0 - h - h = lim h → 0 1 = 1 R f ' 1 = lim h → 0 f 1 + h - f 1 h = lim h → 0 a 1 + h 2 + 1 - a 1 2 + 1 h = lim h → 0 a 1 + 2 h + h 2 + 1 - a + 1 h = lim h → 0 a + 2 a h + a h 2 + 1 - a - 1 h = lim h → 0 a + 2 a h + a h 2 + 1 - a - 1 h = lim h → 0 + 2 a h + a h 2 h = lim h → 0 + h 2 a + a h h = lim h → 0 2 a + a h = lim h → 0 2 a + a · lim h → 0 h ∵ lim x → a f x + g x = lim x → a f x + lim x → a g x = 2 a + 0 = 2 a Now, since given that the function f x is differentiable at x = 1 . ∴ L f ' 1 = R f ' 1 ⇒ 1 = 2 a ⇒ a = 1 2

Prove that the function for f x = x x , if x ≠ 0 0 , if x = 0 . is not differentiable at x = 0 .

The given function is f x = x x , if x ≠ 0 0 , if x = 0 . which can be written as f x = x x = 1 ; if x > 0 x - x = - 1 ; if x < 0 0 ; x = 0 [The absolute value or modulus of a real number x ,denoted by x ,is the non-negative value of x without regard to its sign. Namely x = x if x is positive, x = - x if x is negative and 0 = 0 ] Continuity at x = 0 ; LHL = lim x → 0 - f x = lim x → 0 - 1 = - 1 RHL = lim x → 0 + f x = lim x → 0 1 = 1 Here, we get LHL ≠ RHL Thus, function f is discontinuous at x = 0 . ∴ f is not differentiable at x = 0 .[ ∵ Every differentiable function is continuous. So if a function is discontinuous at a point then it can not be differentiable at that point.]

E M B I B E

Understanding ISC Mathematics Class 12 Volume 1>Chapter 5 - Continuity and Differentiability>EXERCISE 5.4

M L Aggarwal Solutions for Chapter: Continuity and Differentiability, Exercise 4: EXERCISE 5.4

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Continuity and Differentiability, Exercise 4: EXERCISE 5.4

Attempt the practice questions on Chapter 5: Continuity and Differentiability, Exercise 4: EXERCISE 5.4 with hints and solutions to strengthen your understanding. Understanding ISC Mathematics Class 12 Volume 1 solutions are prepared by Experienced Embibe Experts.

Questions from M L Aggarwal Solutions for Chapter: Continuity and Differentiability, Exercise 4: EXERCISE 5.4 with Hints & Solutions

EASY

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 5 - Continuity and Differentiability>EXERCISE 5.4

Examine the function $f (x) = \{\begin{cases} 1 + x, i f & x \leq 2 \\ 5 - x, i f & x > 2 \end{cases}$ for differentiability at $x = 2$ .

EASY

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 5 - Continuity and Differentiability>EXERCISE 5.4

If $f (2) = 4$ and $f' (2) = 4$ , then evaluate $\lim_{x \to 2} \frac{x f (2) - 2 f (x)}{x - 2}$ .

EASY

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 5 - Continuity and Differentiability>EXERCISE 5.4

Examine the function for continuity at $x = 1$ and differentiability at $x = 2$ .

$f (x) = \{\begin{matrix} 5 x - 4, \\ 4 x^{2} - 3 x, \\ 3 x + 4, \end{matrix} \begin{matrix} 0 < x < 1 \\ 1 < x < 2 \\ x \geq 2 . \end{matrix}$

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 5 - Continuity and Differentiability>EXERCISE 5.4

Show that the function $f (x) = 2 x - |x|$ is continuous at $x = 0$ but not differentiable at $x = 0$ .

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 5 - Continuity and Differentiability>EXERCISE 5.4

Show that the function $f (x) = |x - 5|$ is continuous but not differentiable at $x = 5$ .

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 5 - Continuity and Differentiability>EXERCISE 5.4

If the function $f (x) = |x - 3| + |x - 4|$ , then show that $f$ is not differentiable at $x = 3$ and $x = 4$ .

MEDIUM

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 5 - Continuity and Differentiability>EXERCISE 5.4

Show that the function $f$ is continuous at $x = 1$ for all values of $a$ where $f (x) = \{\begin{cases} a x^{2} + 1 & , x \geq 1 \\ x + a & , x < 1 \end{cases}$ .