If the function f defined by f x = 1 - tan x 4 x - π , 0 < x < π 2 , x ≠ π 4 k , x = π 4 is continuous at x = π 4 , find the value of k .

Given function is f x = 1 - tan x 4 x - π , 0 < x < π 2 , x ≠ π 4 k , x = π 4 is continuous at x = π 4 . We have to find the value of k . Now, lim x → π 4 f x = lim x → π 4 1 - tan x 4 x - π put x = π 4 + h so that as x → π 4 ⇒ h → 0 ∴ lim x → π 4 f x = lim h → 0 1 - tan π 4 + h 4 π 4 + h - π = lim h → 0 1 - tan π 4 + h π + 4 h - π = lim h → 0 1 - tan π 4 + tan h 1 - tan π 4 · tan h 4 h = lim h → 0 1 - 1 + tan h 1 - tan h 4 h = [ ∵ tan π 4 = 1 ] = lim h → 0 1 - tan h - 1 + tan h 1 - tan h 4 h = lim h → 0 1 - tan h - 1 + tan h 4 h 1 - tan h = lim h → 0 1 - tan h - 1 - tan h 4 h 1 - tan h = lim h → 0 - 2 tan h 4 h 1 - tan h = - 2 4 lim h → 0 tan h h · lim h → 0 1 1 - tan h = - 1 2 × 1 × 1 1 - tan 0 ∵ lim θ → 0 tan θ θ = 1 = - 1 2 Also, f π 4 = k ∵ f x is continuous at x = π 4 ∴ lim x → π 4 f x = f π 4 ⇒ - 1 2 = k Hence, k = - 1 2

Discuss the continuity of the function f defined by f x = 3 4 5 , i f 0 ≤ x ≤ 1 , i f 1 < x < 3 , i f 3 ≤ x ≤ 10 .

Given function is f x = 3 4 5 , i f 0 ≤ x ≤ 1 , i f 1 < x < 3 , i f 3 ≤ x ≤ 10 . We have to discuss the continuity of the function. Since the constant function is continuous everywhere. ∴ When 0 ≤ x < 1 ; f x = 3 , which is a constant function and hence f x is continuous in 0 ≤ x < 1 . When 1 < x < 3 ; f x = 4 , which is a constant function and hence f x is continuous in 1 < x < 3 . When 3 < x ≤ 10 ; f x = 5 , which is a constant function and hence f x is continuous in 3 < x ≤ 10 . Continuity at x = 1 : LHL = lim x → 1 - f x = lim x → 1 - 3 = 3 RHL = lim x → 1 + f x = lim x → 1 + 4 = 4 ∴ LHL ≠ RHL Thus, f is discontinuous at x = 1 . Continuity at x = 3 : LHL = lim x → 3 - f x = lim x → 3 - 4 = 4 RHL = lim x → 3 + f x = lim x → 3 + 5 = 5 ∴ LHL ≠ RHL Thus, f is discontinuous at x = 3 . Hence, f is continuous in 0 , 10 except at x = 1 and x = 3 .

Find the derivative of the function w. r. t. x : a + x a - x .

Let y = a + x a - x Differentiating both sides with respect to x , we have d y d x = a - x d d x a + x - a + x d d x a - x a - x 2 [Using quotient rule] = a - x d d x a + d d x x - a + x d d x a - d d x x a - x 2 = a - x 0 + d d x x - a + x 0 - d d x x a - x 2 = a - x 1 2 x - a + x - 1 2 x a - x 2 = a - x + a + x 2 x a - x 2 ∵ d d x x = 1 2 x ; d d x x n = n x n - 1 = a - x + a + x 2 x a - x 2 = a - x + a + x 2 x a - x 2 = 2 a 2 x a - x 2 = a x a - x 2

E M B I B E

Understanding ISC Mathematics Class 12 Volume 1>Chapter 5 - Continuity and Differentiability>CHAPTER TEST

EASY

12th ICSE

IMPORTANT

Earn 100

Is the function defined by $f (x) = x^{2} - \sin x + 5$ continuous at $x = π ?$

Important Questions on Continuity and Differentiability

EASY

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 5 - Continuity and Differentiability>CHAPTER TEST

Find the value of

k

so that

f (x) = \{\begin{cases} \frac{\sin k x}{x}, if x < 0 \\ 8 - 3 x, if x \geq 0 \end{cases}

may be continuous at

x = 0 .

EASY

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 5 - Continuity and Differentiability>CHAPTER TEST

If the function

f

defined by

f (x) = \{\begin{cases} \frac{1 - \cos 2 x}{2 x^{2}}, if x \neq 0 \\ k, if x = 0 \end{cases}

is continuous at

x = 0

, find the value of

k .

EASY

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 5 - Continuity and Differentiability>CHAPTER TEST

If the function

f

defined by

f (x) = \{\begin{cases} \frac{1 - \tan x}{4 x - π}, 0 < x < \frac{π}{2}, x \neq \frac{π}{4} \\ k, x = \frac{π}{4} \end{cases}

is continuous at

x = \frac{π}{4}

, find the value of

k .

EASY

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 1>Chapter 5 - Continuity and Differentiability>CHAPTER TEST

Discuss the continuity of the function $f$ defined by $f (x) = \{\begin{cases} \begin{matrix} 3 \\ 4 \\ 5 \end{matrix} & \begin{matrix} , i f 0 \leq x \leq 1 \\ , i f 1 < x < 3 \\ , i f 3 \leq x \leq 10 . \end{matrix} \end{cases}$