Target Exam

CUET

Subject

Section B1

Chapter

Continuity and Differentiability

Question:

Examine the continuity of the function $f(x) = \begin{cases} \frac{|x - 4|}{2(x - 4)}, & \text{if } x \neq 4 \\ 0, & \text{if } x = 4 \end{cases}$ at $x = 4$.

Options:

Continuous at $x = 4$ because $f(4) = 0$.

Discontinuous at $x = 4$ because the limit exists but is not equal to $0$.

Discontinuous at $x = 4$ because $\text{LHL} = -0.5$ and $\text{RHL} = 0.5$.

Continuous at $x = 4$ because the function is defined for all real numbers.

Correct Answer:

Discontinuous at $x = 4$ because $\text{LHL} = -0.5$ and $\text{RHL} = 0.5$.

Explanation:

The correct answer is Option (3) → Discontinuous at $x = 4$ because $\text{LHL} = -0.5$ and $\text{RHL} = 0.5$. ##

We have,

$f(x) = \begin{cases} \frac{|x - 4|}{2(x - 4)}, & \text{if } x \neq 4 \\ 0, & \text{if } x = 4 \end{cases} \text{ at } x = 4$

At $x = 4$,

$\text{LHL} = \lim\limits_{x \to 4^-} \frac{|x - 4|}{2(x - 4)}$

Put $x = 4 - h$,

$= \lim\limits_{h \to 0} \frac{|4 - h - 4|}{2[(4 - h) - 4]} = \lim\limits_{h \to 0} \frac{|-h|}{(8 - 2h - 8)}$

$= \lim\limits_{h \to 0} \frac{h}{-2h} = -\frac{1}{2} \quad \text{and } f(4) = 0 \neq \text{LHL [given]}$

So, $f(x)$ is discontinuous at $x = 4$.