Find the values of $a$ and $b$ such that the function $f$ defined by $f(x) = \begin{cases} \frac{x - 4}{|x - 4|} + a, & \text{if } x < 4 \\ a + b, & \text{if } x = 4 \\ \frac{x - 4}{|x - 4|} + b, & \text{if } x > 4 \end{cases}$ is a continuous function at $x = 4$.

$a = 1, b = -1$

The correct answer is Option (3) → $a = 1, b = -1$ ##

We have, $f(x) = \begin{cases} \frac{x - 4}{|x - 4|} + a, & \text{if } x < 4 \\ a + b, & \text{if } x = 4 \\ \frac{x - 4}{|x - 4|} + b, & \text{if } x > 4 \end{cases}$

At $x = 4$:

$\text{LHL} = \lim\limits_{x \to 4^-} \left( \frac{x - 4}{|x - 4|} + a \right)$

Put $x = 4 - h$:

$= \lim\limits_{h \to 0} \left( \frac{-h}{|-h|} + a \right) = \lim\limits_{h \to 0} \left( \frac{-h}{h} + a \right) = -1 + a$

$\text{RHL} = \lim\limits_{x \to 4^+} \left( \frac{x - 4}{|x - 4|} + b \right)$

Put $x = 4 + h$:

$= \lim\limits_{h \to 0} \left( \frac{h}{|h|} + b \right) = \lim\limits_{h \to 0} (1 + b) = 1 + b$

For continuity, $\text{LHL} = \text{RHL} = f(4)$:

$⇒-1 + a = 1 + b = a + b$

Now, $-1 + a = a + b$ and $1+b=a+b$

Hence, $a = 1$ and $b = -1$.