Match List-I with List-II

Choose the correct answer from the options given below:

(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

The correct answer is Option (1) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

(A) $f(x)=\begin{cases}\frac{x}{|x|},&x\ne 0\\0,&x=0\end{cases}$

Left and right limits at $0$ are $-1$ and $1$, so $f$ is discontinuous at $0$.

So (A) → (III).

(B) $f(x)=|x|$ is continuous everywhere but not differentiable at $x=0$.

So (B) → (I).

(C) $f(x)=|x^{2}-1|$ is continuous everywhere but not differentiable where $x^{2}-1=0$, i.e., $x=1,-1$.

So (C) → (IV).

(D) $f(x)=|x-1|$ is continuous everywhere but not differentiable at $x=1$.

So (D) → (II).

Final answer: (A)–(III), (B)–(I), (C)–(IV), (D)–(II)