Let f(x) = |sin x| Then

The correct option is: "f is everywhere continuous but not differentiable at x = nπ, n ∈ Z"

Let us analyze the function $f\left(x\right)=|\sin x|$ step-by-step to understand its continuity and differentiability.

Step 1: Analyzing Continuity
The sine function, $g\left(x\right)=\sin x$, is continuous everywhere on the real number line.
The absolute value function, $h\left(u\right)=\left|u\right|$, is also a continuous function everywhere.
Since the composition of two continuous functions is always continuous, the function $f\left(x\right)=(h\circ g)\left(x\right)=|\sin x|$ is continuous everywhere on the set of real numbers $\mathbb{R}$.

Step 2: Identifying Potential Points of Non-differentiability
The absolute value function $\left|u\right|$ is differentiable everywhere except where its argument $u=0$.
Therefore, the function $f\left(x\right)=|\sin x|$ is differentiable everywhere except possibly at points where:
$\sin x=0$
We know that $\sin x=0$ when:
$x=n\pi ,n\in \mathbb{Z}$

Step 3: Checking Differentiability at $x=n\pi$
Let us evaluate the left-hand derivative (LHD) and the right-hand derivative (RHD) at these points.
For any integer $n$, let $c=n\pi$. Note that $f\left(c\right)=|\sin (n\pi \left)\right|=0$.

The right-hand derivative at $x=c$ is given by:
$$f^{\prime }\left(c^{+}\right)=\underset{h\to 0^{+}}{\lim }\frac{f(c+h)-f\left(c\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{|\sin (n\pi +h\left)\right|-0}{h}$$
Using the trigonometric identity $\sin (n\pi +h)=(-1{)}^n\sin h$:
$$f^{\prime }\left(c^{+}\right)=\underset{h\to 0^{+}}{\lim }\frac{\left|\right(-1{)}^n\sin h|}{h}=\underset{h\to 0^{+}}{\lim }\frac{|\sin h|}{h}$$
Since $h>0$ and close to 0, $\sin h>0$, so $|\sin h|=\sin h$:
$$f^{\prime }\left(c^{+}\right)=\underset{h\to 0^{+}}{\lim }\frac{\sin h}{h}=1$$

The left-hand derivative at $x=c$ is given by:
$$f^{\prime }\left(c^{-}\right)=\underset{h\to 0^{-}}{\lim }\frac{f(c+h)-f\left(c\right)}{h}=\underset{h\to 0^{-}}{\lim }\frac{|\sin h|}{h}$$
Since $h<0$ and close to 0, $\sin h<0$, so $|\sin h|=-\sin h$:
$$f^{\prime }\left(c^{-}\right)=\underset{h\to 0^{-}}{\lim }\frac{-\sin h}{h}=-1$$

Since the left-hand derivative ($-1$) does not equal the right-hand derivative ($1$) at $x=n\pi$, the function is not differentiable at these points. Geometrically, the graph of $|\sin x|$ has sharp corners or cusps at every integer multiple of $\pi$.

Thus, $f\left(x\right)=|\sin x|$ is everywhere continuous but not differentiable at $x=n\pi ,n\in \mathbb{Z}$.