If $f(x) = |\sin x|$, then

$f$ is everywhere continuous but not differentiable at $x = n\pi, n \in Z$

The correct answer is Option (2) → $f$ is everywhere continuous but not differentiable at $x = n\pi, n \in Z$ ##

We have, $f(x) = |\sin x|$

Let $f(x) = v \circ u(x) = v[u(x)] \quad [\text{where, } u(x) = \sin x \text{ and } v(x) = |x|]$

$= v(\sin x) = |\sin x|$

where, $u(x)$ and $v(x)$ are both continuous.

Hence, $f(x) = v \circ u$, $u(x)$ is also a continuous function but $v(x)$ is not differentiable at $x = 0$.

So, $f(x)$ is not differentiable where $\sin x = 0 ⇒x = n\pi, n \in \mathbb{Z}$

Hence, $f(x)$ is continuous everywhere but not differentiable at $x = n\pi, n \in \mathbb{Z}$.