Find all points of discontinuity of the function f ( t ) = 1 t 2 + t - 2 , where t = 1 x - 1 .

Given, f ( t ) = 1 t 2 + t - 2 ⇒ f ( t ) = 1 1 ( x - 1 ) 2 + 1 ( x - 1 ) - 2 [Putting t = 1 x - 1 ] = 1 1 + x - 1 - 2 ( x - 1 ) 2 ( x - 1 ) 2 = ( x - 1 ) 2 x - 2 x 2 - 2 + 4 x = ( x - 1 ) 2 - 2 x 2 - 5 x + 2 = ( x - 1 ) 2 - 2 x 2 - 4 x - x + 2 = ( x - 1 ) 2 - ( x - 2 ) ( 2 x - 1 ) = ( x - 1 ) 2 ( 2 - x ) ( 2 x - 1 ) Now, the function will not be defined and discontinuous when the denominator tends to zero. ⇒ 2 - x = 0 ⇒ x = 2 and 2 x - 1 = 0 ⇒ x = 1 2 Therefore, the required points of discontinuity for the given function are 1 , 2 and 1 2 .

Find all points of discontinuity of the function f ( t ) = 1 t 2 + t - 2 , where t = 1 x - 1 .

Given, f ( t ) = 1 t 2 + t - 2 ⇒ f ( t ) = 1 1 ( x - 1 ) 2 + 1 ( x - 1 ) - 2 [Putting t = 1 x - 1 ] = 1 1 + x - 1 - 2 ( x - 1 ) 2 ( x - 1 ) 2 = ( x - 1 ) 2 x - 2 x 2 - 2 + 4 x = ( x - 1 ) 2 - 2 x 2 - 5 x + 2 = ( x - 1 ) 2 - 2 x 2 - 4 x - x + 2 = ( x - 1 ) 2 - ( x - 2 ) ( 2 x - 1 ) = ( x - 1 ) 2 ( 2 - x ) ( 2 x - 1 ) Now, the function will not be defined and discontinuous when the denominator tends to zero. ⇒ 2 - x = 0 ⇒ x = 2 and 2 x - 1 = 0 ⇒ x = 1 2 Therefore, the required points of discontinuity for the given function are 1 , 2 and 1 2 .

Examine the differentiability of f , where f is defined by f x = x [ x ] , , if 0 ≤ x < 2 ( x - 1 ) x , if 2 ≤ x < 3 at x = 2

Given that, f x = x [ x ] , , if 0 ≤ x < 2 ( x - 1 ) x , if 2 ≤ x < 3 We know that, a function f is differentiable at a point ' a ' in its domain if Lf ' ( a ) = Rf ' ( a ) Where L f ' ( a ) = lim h → 0 f ( a - h ) - f ( a ) - h and R f ' ( a ) = lim h → 0 f ( a + h ) - f ( a ) h Left-hand differentiability: L f ' ( 2 ) = lim h → 0 f ( 2 - h ) - f ( 2 ) - h = lim h → 0 ( 2 - h ) [ 2 - h ] - ( 2 - 1 ) 2 - h = lim h → 0 ( 2 - h ) · 1 - 2 - h [ ∵ [ 2 - h ] = 1 ] = lim h → 0 2 - h - 2 - h = 1 Thus, Lf ' ( 2 ) = 1 . Right-hand differentiability: R f ' ( 2 ) = lim h → 0 f ( 2 + h ) - f ( 2 ) h = lim h → 0 ( 2 + h - 1 ) ( 2 + h ) - ( 2 - 1 ) · 2 h = lim h → 0 ( 1 + h ) ( 2 + h ) - 2 h = lim h → 0 2 + h + 2 h + h 2 - 2 h = lim h → 0 3 h + h 2 h = lim h → 0 h ( 3 + h ) h = 3 Thus, Rf ' ( 2 ) = 3 . Hence, L f ' ( 2 ) ≠ R f ' ( 2 ) Therefore, f ( x ) is not differentiable at x = 2

Examine the differentiability of f , where f is defined by f x = x 2 sin 1 x , if x ≠ 0 0 , if x = 0 at x = 0

Given, f x = x 2 sin 1 x , if x ≠ 0 0 , if x = 0 at x = 0 We know that, a function f is differentiable at a point x = a if, L f ' ( a ) = R f ' ( a ) where L f ' ( a ) = lim h → 0 f ( a - h ) - f ( a ) - h and R f ' ( a ) = lim h → 0 f ( a + h ) - f ( a ) h Left hand differentiability: ∴ L f ' ( 0 ) = lim h → 0 f ( 0 - h ) - f ( 0 ) - h = lim h → 0 ( 0 - h ) 2 sin 1 ( 0 - h ) - 0 - h = h 2 · sin - 1 h - h = lim h → 0 h · sin 1 h = 0 - 1 ≤ sin 1 h ≤ 1 Thus, L f ' ( 0 ) = 0 . Right hand differentiability: R f ' ( 0 ) = lim h → 0 f ( 0 + h ) - f ( 0 ) h = lim h → 0 ( 0 + h ) 2 sin 1 0 + h - 0 h = lim h → 0 h 2 sin 1 h h = lim h → 0 h · sin 1 h = 0 - 1 ≤ sin 1 h ≤ 1 Thus, R f ' ( 0 ) = 0 Hence, L f ' ( 0 ) = R f ' ( 0 ) = 0 Therefore, f ( x ) is differentiable at x = 0

Examine the differentiability of f , where f is defined by f x = 1 + x , if x ≤ 2 5 - x , if x > 2 at x = 2 .

Given that, f x = 1 + x , if x ≤ 2 5 - x , if x > 2 We know that, a function f is differentiable at a point x = a if Lf ' ( a ) = Rf ' ( a ) where, L f ' ( a ) = lim h → 0 f ( a - h ) - f ( a ) - h and R f ' ( a ) = lim h → 0 f ( a + h ) - f ( a ) h Now, Left hand differentiability: L f ' ( 2 ) = lim h → 0 f ( 2 - h ) - f ( 2 ) - h = lim h → 0 ( 1 + 2 - h ) - ( 1 + 2 ) - h = lim h → 0 3 - h - 3 - h = lim h → 0 - h - h = lim h → 0 1 = 1 Thus, L f ' ( 2 ) = 1 Now, right hand differentiability: R f ' ( 2 ) = lim h → 0 f ( 2 + h ) - f ( 2 ) h = lim h → 0 [ 5 - ( 2 + h ) ] - ( 1 + 2 ) h = lim h → 0 3 - h - 3 h = lim h → 0 - h h = lim h → 0 - 1 = - 1 Thus, R f ' ( 2 ) = - 1 So, L f ' ( 2 ) ≠ R f ' ( 2 ) Thus, f ( x ) is not differentiable at x = 2 .

A function f : R → R satisfies the equation f ( x + y ) = f ( x ) f ( y ) for all x , y ∈ R , f ( x ) ≠ 0 . Suppose that the function is differentiable at x = 0 and f ' ( 0 ) = 2 Prove that f ' ( x ) = 2 f ( x ) .

Given, f : R → R satisfies the equation f ( x + y ) = f ( x ) f ( y ) for all x , y ∈ R , f ( x ) ≠ 0 Now, f ( 0 + 0 ) = f ( 0 ) × f ( 0 ) ⇒ f ( 0 ) f ( 0 ) - 1 = 0 ⇒ f ( 0 ) = 1 Let us take any point x = 0 at which the function f ( x ) is differentiable. So, f ' ( 0 ) = lim h → 0 f ( 0 + h ) - f ( 0 ) h ⇒ 2 = lim h → 0 f ( 0 ) · f ( h ) - f ( 0 ) h ⇒ 2 = lim h → 0 f ( 0 ) [ f ( h ) - 1 ] h ⇒ 2 = lim h → 0 [ f ( h ) - 1 ] h ........... i Now, f ' ( x ) = lim h → 0 f ( x + h ) - f ( x ) h = lim h → 0 f ( x ) · f ( h ) - f ( x ) h [ ∵ f ( x + y ) = f ( x ) · f ( y ) ] = lim h → 0 f ( x ) [ f ( h ) - 1 ] h = 2 f ( x ) from equation ( i ) Therefore, f ' ( x ) = 2 f ( x ) .

Differentiate the following function w.r.t. x 8 x x 8

Let y = 8 x x 8 Taking log on both sides, we get, log y = log 8 x x 8 ⇒ log y = log 8 x - log x 8 ⇒ log y = x log 8 - 8 log x log a b = b log a Differentiating both sides w.r.t. x , 1 y · d y d x = log 8 · 1 - 8 x ⇒ d y d x = y log 8 - 8 x Thus, d y d x = 8 x x 8 log 8 - 8 x .

E M B I B E

NCERT Exemplar Mathematics - Class 12>Chapter 5 - Continuity and Differentiability>EXERCISE>Q 18

MEDIUM

12th CBSE

IMPORTANT

Earn 100

Find all points of discontinuity of the function $f (t) = \frac{1}{t^{2} + t - 2}$ , where $t = \frac{1}{x - 1}$ .

Important Questions on Continuity and Differentiability

EASY

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 5 - Continuity and Differentiability>EXERCISE>Q 19

Show that the function

f (x) = | \sin x + \cos x |

is continuous at

x = π

.

HARD

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 5 - Continuity and Differentiability>EXERCISE>Q 20

Examine the differentiability of $f$ , where $2 - x = 0 \Rightarrow x = 2$ is defined by

$f (x) = \{\begin{cases} x [x], & , if 0 \leq x < 2 \\ (x - 1) x, & if 2 \leq x < 3 \end{cases}$ at $x = 2$

HARD

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 5 - Continuity and Differentiability>EXERCISE>Q 21

Examine the differentiability of $f$ , where $f$ is defined by

$f (x) = \{\begin{cases} x^{2} \sin \frac{1}{x} & , if x \neq 0 \\ 0 & , if x = 0 \end{cases}$ at $x = 0$

HARD

12th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 12>Chapter 5 - Continuity and Differentiability>EXERCISE>Q 22

Examine the differentiability of $f$ , where $f$ is defined by $f (x) = \{\begin{cases} 1 + x & , if x \leq 2 \\ 5 - x & , if x > 2 \end{cases}$ at $x = 2$ .