is it's ans be a?

11 votes

Which one of the following functions is continuous at $x = 3?$

- $f(x) = \begin{cases} 2,&\text{if $x = 3$ } \\ x-1& \text{if $x > 3$}\\ \frac{x+3}{3}&\text{if $x < 3$ } \end{cases}$
- $f(x) = \begin{cases} 4,&\text{if $x = 3$ } \\ 8-x& \text{if $x \neq 3$} \end{cases}$
- $f(x) = \begin{cases} x+3,&\text{if $x \leq 3$ } \\ x-4& \text{if $x > 3$} \end{cases}$
- $f(x) = \begin{cases} \frac{1}{x^3-27}&\text{if $x \neq 3$ } \end{cases}$

3

$Properties\ of\ Discontinuity\\ A\ function\ f(x)\ will\ be\ discontinuous\ at\ x=a\ here\ x=3, if\ any\ one\ below\ hold\\ \\ 1. \lim_{x->a^{-}}f(x) \neq \lim_{x->a^{+}}f(x)\ (LHL\neq RHL)\\ 2. \lim_{x->a^{-}}f(x)\doteq \lim_{x->a^{+}}f(x)\neq f(a)\\ 3. f(a)\ is\ not\ defined\\ 4. At\ least\ one\ of\ the\ limit\ not\ exists$

36 votes

Best answer

For continuity, Left hand limit must be equal to right hand limit. For continuity at $x = 3$,

the value of $f(x)$ just above and just below $3$ must be the same.

A. $f(3) = 2. f(3+) = x - 1 = 2. f(3-) =\frac{(x+3)}{3} =\frac{6}{3} = 2.\text{ Hence continuous.}$

B. $f(3) = 4. f(3+) = f(3-) = 8 - 3 = 5. \text{ So, not continuous.}$

C. $f(3) = f(3-) = x + 3 = 6. f(3+) = x - 4 = -1.\text{ So, not continuous.}$

D. $f(3)\text{ is not existing. So, not continuous.}$

Correct Answer: $A$

the value of $f(x)$ just above and just below $3$ must be the same.

A. $f(3) = 2. f(3+) = x - 1 = 2. f(3-) =\frac{(x+3)}{3} =\frac{6}{3} = 2.\text{ Hence continuous.}$

B. $f(3) = 4. f(3+) = f(3-) = 8 - 3 = 5. \text{ So, not continuous.}$

C. $f(3) = f(3-) = x + 3 = 6. f(3+) = x - 4 = -1.\text{ So, not continuous.}$

D. $f(3)\text{ is not existing. So, not continuous.}$

Correct Answer: $A$