Locate the points of discontinuity (if any) of the function: f x = x 3 - 3 , if x ≤ 2 x 2 + 1 , if x > 2

Given function is f x = x 3 - 3 , if x ≤ 2 x 2 + 1 , if x > 2 We note that the function f is defined at all the points of the real line. Let c be a point on the real line. Case I If c < 2 then f c = c 3 - 3 and lim x → c f x = lim x → c x 3 - 3 = c 3 - 3 ∴ lim x → c f x = f c Thus, f is continuous at all c < 2 . Case II If c = 2 then f c = c 3 - 3 LHL = lim x → 2 - f x = lim x → 2 x 3 - 3 = 2 3 - 3 = 5 RHL = lim x → 2 + f x = lim x → 2 x 2 + 1 = 2 2 + 1 = 5 Here, we get LHL = RHL ∴ Function f is continuous at x = 2 . Case III If c > 2 then f c = c 2 + 1 and lim x → c f x = lim x → c x 2 + 1 = c 2 + 1 ∴ lim x → c f x = f c Thus, f is continuous for all c > 2 . Hence, f is continuous everywhere on R . Hence, it has no point of discontinuity.

Locate the points of discontinuity (if any) of the function: f x = x 4 - 16 x - 2 , if x ≠ 2 16 , if x = 2

Given function is f x = x 4 - 16 x - 2 , if x ≠ 2 16 , if x = 2 We note that the function f is defined at all the points of the real line.We have to examine the continuity for all x ∈ R . Let c be a point on the real line. lim x → 2 x 4 - 16 x - 2 = lim x → 2 x 2 - 4 x 2 + 4 x - 2 ∵ a 2 - b 2 = a + b a - b = lim x → 2 x + 2 x - 2 x 2 + 4 x - 2 = lim x → 2 x + 2 x 2 + 4 = 2 + 2 2 2 + 4 = 4 × 8 = 32 Also given that f 2 = 16 ∴ lim x → 2 f x ≠ f 2 Thus, f x is discontinuous at x = 2 . At c ≠ 2 , then f c = c 4 - 16 c - 2 lim x → c f x = lim x → c x 4 - 16 x - 2 = c 4 - 16 c - 2 lim x → c f x = f c ∴ f is continuous at x = c ≠ 2 ∴ f x has only one point of discontinuity at x = 2 .

Locate the points of discontinuity (if any) of the function: f x = sin 3 x x + cos x , if x > 0 4 - 3 x , if x ≤ 0 .

The given function is f x = sin 3 x x + cos x , if x > 0 4 - 3 x , if x ≤ 0 . We note that the function f is defined at all the points of the real line. Let c be a point on the real line. Case I If c < 0 then f c = 4 - 3 c and lim x → c f x = lim x → c 4 - 3 x = 4 - 3 c ∴ lim x → c f x = f c Thus, f is continuous for all c < 0 . Case II If c = 0 then f c = f ( 0 ) = 4 - 3 × 0 = 4 LHL = lim x → 0 - f x = lim x → 0 4 - 3 x = 4 - 3 × 0 = 4 RHL = lim x → 0 + f x = lim x → 0 sin 3 x x + cos x = lim x → 0 3 . sin 3 x 3 x + lim x → 0 cos x [multiplying numerator and denominator by 3 in the first fraction] = 3 × lim x → 0 sin 3 x 3 x + 1 ∵ lim θ → 0 cos θ = 1 = 3 × 1 + 1 ∵ lim θ → 0 sin θ θ = 1 = 4 Here, we get LHL = RHL = f ( 0 ) ∴ Function f is continuous at x = 0 . Case III If c > 0 then f c = sin 3 c c + cos c and lim x → c f x = lim x → c sin 3 x x + cos x = lim x → c sin 3 x x + lim x → c cos x = sin 3 c c + cos c ∴ lim x → c f x = f c Thus, f is continuous ∀ x > 0 . Hence, f is a continuous function . Hence, f has no point of discontinuity.

Locate the points of discontinuity (if any) of the function: f x = 2 x ; if x < 0 0 ; if 0 ≤ x ≤ 1 4 x ; if x > 1

The given function is f x = 2 x ; if x < 0 0 ; if 0 ≤ x ≤ 1 4 x ; if x > 1 When x < 0 then f x = 2 x and when x > 1 then f x = 4 x which are polynomials in x and hence continuous everywhere. [Since a polynomial function is continuous everywhere.] ∴ f x is continuous for each x < 0 and x > 1 . When 0 < x < 1 ; f x = 0 , which is a constant function and hence continuous everywhere. ∴ f x is continuos for each x ∈ 0 , 1 . When x = 0 then f 0 = 0 LHL = lim x → 0 - f x = lim x → 0 2 x = 2 × 0 = 0 RHL = lim x → 0 + f x = 0 ∴ LHL = RHL = f 0 ∴ Function f is continuous at x = 0 . When x = 1 then LHL = lim x → 1 - f x = lim x → 1 0 = 0 RHL = lim x → 1 + f x = lim x → 1 4 x = 4 × 1 = 4 ∴ LHL ≠ RHL Thus, f has a point of discontinuity at x = 1 .

Locate the points of discontinuity (if any) of the function: f x = x + 2 ; if x ≤ 1 x - 2 ; if 1 < x < 2 0 ; if x ≥ 2

Given function is f x = x + 2 ; if x ≤ 1 x - 2 ; if 1 < x < 2 0 ; if x ≥ 2 We note that the function f is defined at all the points of the real line. So we examine f for all value of x ∈ R . Let c be a point on the real line. Case I If c < 1 then f c = c + 2 and lim x → c f x = lim x → c x + 2 = c + 2 ∴ lim x → c f x = f c Thus, f is continuous ∀ c < 1 . Case II At c = 1 , then f c = f 1 = 1 + 2 = 3 LHL = lim x → 1 - f x = lim x → 1 - x + 2 = 1 + 2 = 3 RHL = lim x → 1 + f x = lim x → 1 + x - 2 = 1 - 2 = - 1 LHL ≠ RHL Thus, f is discontinuous at c = 1 . Case III If 1 < c < 2 then f c = c - 2 lim x → c f x = lim x → c x - 2 = c - 2 ∴ lim x → c f x = f c Thus, f is continuous ∀ c where 1 < c < 2 . Case IV If c > 2 then f c = 0 lim x → c f x = lim x → c 0 = 0 ∴ f is continuous ∀ c > 2 . Case V If c = 2 , then f c = 0 LHL = lim x → 2 - f x = lim x → 2 - x - 2 = 2 - 2 = 0 RHL = lim x → 2 + f x = lim x → 2 + 0 = 0 Here, we get LHL = RHL = f 0 Thus, f is continuous at c = 2 . On combining all five cases, f has a point of discontinuity at x = 1 .

Find the value of the constant k so that the function f defined by f x = k x 2 , x ≤ 2 x - 3 , x > 2 may be continuous.

Given function is f x = k x 2 , x ≤ 2 x - 3 , x > 2 Here f 2 = k x 2 = k × 2 2 = 4 k RHL = lim x → 2 + f x = lim x → 2 x - 3 = 2 - 3 = - 1 LHL = lim x → 2 - f x = lim x → 2 k x 2 = k × 2 2 = 4 k For the function f x to be continuos function at x = 2 we must have ∴ lim x → 2 + f x = lim x → 2 - f x = f 2 ⇒ - 1 = 4 k ⇒ k = - 1 4

For what choice of a and b is the function continuous: f x = - 5 ; if x ≤ - 1 a x - b ; if - 1 < x < 3 7 ; if x ≥ 3

Given function is f x = - 5 ; if x ≤ - 1 a x - b ; if - 1 < x < 3 7 ; if x ≥ 3 Since it is, given that f be a continuous function. Thus f must be continuous at x = - 1 and x = 3 . When x = - 1 , then f - 1 = - 5 LHL = lim x → - 1 - f x = lim x → - 1 - 5 = - 5 RHL = lim x → - 1 + f x = lim x → - 1 a x - b = - a - b Since f is continuous at x = - 1 . ∴ lim x → - 1 - f x = lim x → - 1 + f x = f - 1 ∴ - 5 = - a - b ⇒ a + b = 5 . . . . . . . . . . . . . 1 When x = 3 , then f 3 = 7 LHL = lim x → 3 - f x = lim x → 3 a x - b = 3 a - b RHL = lim x → 3 + f x = lim x → 3 7 = 7 Since f is continuous at x = 3 . ∴ 3 a - b = 7 . . . . . . . . . . . . . . . . . 2 On solving equations 1 and 2 simultaneously, we get a = 3 and b = 2 .

For what choice of a and b is the function continuous: f x = x 2 , x ≤ 0 a x + b , x > 0

Given function is f x = x 2 , x ≤ 0 a x + b , x > 0 Since it is given that f be a continuous function. Thus, f must be continuous at x = 0 . When x = 0 , then f 0 = 0 LHL = lim x → 0 - f x = lim x → 0 x 2 = 0 RHL = lim x → 0 + f x = lim x → 0 a x + b = a × 0 + b = b ∵ The function f is continuous everywhere. ∴ function is also continuous at x = 0 . ∴ lim x → 0 - f x = lim x → 0 + f x = f 0 ∴ b = 0 and a can have any value.

E M B I B E

Understanding ISC Mathematics Class 12 Volume 1>Chapter 5 - Continuity and Differentiability>EXERCISE 5.2

M L Aggarwal Solutions for Chapter: Continuity and Differentiability, Exercise 2: EXERCISE 5.2

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Continuity and Differentiability, Exercise 2: EXERCISE 5.2

Attempt the practice questions on Chapter 5: Continuity and Differentiability, Exercise 2: EXERCISE 5.2 with hints and solutions to strengthen your understanding. Understanding ISC Mathematics Class 12 Volume 1 solutions are prepared by Experienced Embibe Experts.