Examine the continuity of the function $f(x) = |x| + |x - 1|$ at $x = 1$.

Continuous at $x = 1$ because $\text{LHL} = \text{RHL} = f(1) = 1$.

The correct answer is Option (2) → Continuous at $x = 1$ because $\text{LHL} = \text{RHL} = f(1) = 1$. ##

We have, $f(x) = |x| + |x - 1|$ at $x = 1$

At $x = 1$,

$\text{LHL} = \lim\limits_{x \to 1^-} [|x| + |x - 1|]$

Put $x = 1 - h$,

$= \lim\limits_{h \to 0} [|1 - h| + |1 - h - 1|] = 1 + 0 = 1$

and $\text{RHL} = \lim\limits_{x \to 1^+} [|x| + |x - 1|]$

Put $x = 1 + h$,

$= \lim\limits_{h \to 0} [|1 + h| + |1 + h - 1|] = 1 + 0 = 1 \text{ and } f(1) = |1| + |0| = 1$

$∴\text{LHL} = \text{RHL} = f(1)$

Hence, $f(x)$ is continuous at $x = 1$.