Show that the function f x = sin x x , when x ≠ 0 2 , when x = 0 is continuous at each point except 0 .

Given, f x = sin x x , when x ≠ 0 2 , when x = 0 We know that a function f x is continuous at a point x = a of its domain if lim x → a f x = f a Now, lim x → 0 f x = lim x → 0 sin x x = 1 And, f ( 0 ) = 2 ∴ lim x → 0 f x ≠ f 0 So, f ( x ) is discontinuous at x = 0 . Now, for x < 0 or x > 0 f x = sin x x We know that trigonometric functions are continuous at all points in its domain. So, sin x is a continuous function and polynomial functions are also continuous functions. So, x is also continuous function. ∴ sin x x is continuous function at each point except x = 0 Hence, the given function is continuous at each point except 0 .

Show that f x = ( 2 x - 1 ) , if x < 2 3 x 2 , if x ≥ 2 is continuous.

Given function f x = ( 2 x - 1 ) , if x < 2 3 x 2 , if x ≥ 2 We know that a function f x is said to be continuous at a point x = a of its domain if lim x → a f x = f a Here, we need to check the continuity of the function at x = 2 Left hand limit at x = 2 lim x → 2 - f x = lim x → 2 - ( 2 x - 1 ) = 2 2 – 1 = 3 Right hand limit at x = 2 lim x → 2 + f x = lim x → 2 + 3 x 2 = 3 ( 2 ) 2 = 3 Also, f ( 2 ) = 3 ( 2 ) 2 = 3 As lim x → 2 - f x = lim x → 2 + f x = f 2 = 3 Therefore, The function f ( x ) is continuous at x = 2 . Hence, the given function is continuous function.

Show that f x = x , if x ≠ 0 1 , if x = 0 is continuous at each point except 0 .

Given function is f x = x , if x ≠ 0 1 , if x = 0 Left hand limit at x = 0 lim x → 0 - f x = lim h → 0 f 0 - h = lim h → 0 f - h = l i m h → 0 - h = 0 Right hand limit at x = 0 lim x → 0 + f x = lim h → 0 f 0 + h = lim h → 0 f h = lim h → 0 h = 0 Also, f ( 0 ) = 1 As, lim x → 0 - f x = lim x → 0 + f x ≠ f 0 And, we know that polynomial functions are continuous functions. So, f ( x ) = x is continuous for other values of x except 0 Hence, f ( x ) is continuous everywhere expect at x = 0 .

Locate the point of discontinuity of the function f x = ( x 3 - x 2 + 2 x - 2 ) , if x ≠ 1 4 , if x = 1

Given function f x = ( x 3 - x 2 + 2 x - 2 ) , if x ≠ 1 4 , if x = 1 We know that a function f x is continuous at a point x = a of its domain if lim x → a f x = f a Now, consider lim x → 1 f x = lim x → 1 x 3 - x 2 + 2 x - 2 = 1 - 1 + 2 - 2 = 0 And, f ( 1 ) = 4 Since lim x → 1 f ( x ) ≠ f ( 1 ) Therefore, f ( x ) is not continuous at x = 1 .

E M B I B E

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Differential Calculus>Chapter 9 - Continuity and Differentiability>EXERCISE

R S Aggarwal Solutions for Exercise 2: EXERCISE

Author:R S Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Exercise 2: EXERCISE

Attempt the practice questions from Exercise 2: EXERCISE with hints and solutions to strengthen your understanding. SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12 solutions are prepared by Experienced Embibe Experts.

Questions from R S Aggarwal Solutions for Exercise 2: EXERCISE with Hints & Solutions

MEDIUM

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 9 - Continuity and Differentiability>EXERCISE>Q 4

Discuss the continuity of the function $f (x) = \sec x$ .

MEDIUM

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 9 - Continuity and Differentiability>EXERCISE>Q 5

Show that $\cos | x |$ is a continuous function.

MEDIUM

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 9 - Continuity and Differentiability>EXERCISE>Q 6

Show that the function $f (x) = \{\begin{matrix} \frac{\sin x}{x}, when x \neq 0 \\ 2, when x = 0 \end{matrix}$ is continuous at each point except $0$ .