Match List-I with List-II

List-I	List-II
Function	Property
(A) $f(x) = \begin{cases} \frac{x}{\vert x \vert} & : x \neq 0 \\ 0 & : x = 0 \end{cases}$	(I) continuous but not differentiable at $x= 0$
(B) $f(x) = \vert x \vert$	(II) continuous but not differentiable at $x=1$
(C) $f(x) = \vert x^2 - 1 \vert$	(III) discontinuous at $x = 0$
(D) $f(x) = \vert x - 1 \vert$	(IV) continuous but not differentiable at $x = 1, -1$

Choose the correct answer from the options given below:

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$f(x) = \begin{cases} \frac{x}{|x|} & : x \neq 0 \\ 0 & : x = 0 \end{cases}$

When $x > 0$, $|x| = x$, so $\dfrac{x}{|x|} = 1$

When $x < 0$, $|x| = -x$, so $\dfrac{x}{|x|} = -1$

So this simplifies to:

$f(x) = \begin{cases} +1 & : x > 0 \\ 0 & : x = 0 \\ -1 & : x < 0 \end{cases}$

Left-hand limit $= \lim_{x \to 0^-} f(x) = -1$

Right-hand limit $= \lim_{x \to 0^+} f(x) = +1$

Since $-1 \neq +1$, the function is discontinuous at $x = 0$.

$(A) \to (III)$

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$f(x) = |x| = \begin{cases} x & : x \geq 0 \\ -x & : x < 0 \end{cases}$

Both sides approach $0$ as $x \to 0$, and $f(0) = 0$, so the function is continuous at $x = 0$.

Left-hand derivative $= \lim_{h \to 0^-} \dfrac{|h| - 0}{h} = \dfrac{-h}{h} = -1$

Right-hand derivative $= \lim_{h \to 0^+} \dfrac{|h| - 0}{h} = \dfrac{h}{h} = 1$

Since $-1 \neq 1$, the function is not differentiable at $x = 0$.

Continuous but not differentiable at $x = 0$.

$(B) \to (I)$

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$f(x) = |x^2 - 1|$

The expression inside the modulus becomes zero at $x = 1$ and $x = -1$.

$f(x) = \begin{cases} x^2 - 1 & : x \leq -1 \text{ or } x \geq 1 \\ 1 - x^2 & : -1 < x < 1 \end{cases}$

Since $|x^2 - 1|$ is the absolute value of a continuous function, it is continuous everywhere.

At $x = 1$:

Left-hand derivative $= \dfrac{d}{dx}(1 - x^2)\big|_{x=1} = -2(1) = -2$

Right-hand derivative $= \dfrac{d}{dx}(x^2 - 1)\big|_{x=1} = 2(1) = 2$

Since $-2 \neq 2$, not differentiable at $x = 1$.

At $x = -1$:

Left-hand derivative $= \dfrac{d}{dx}(x^2 - 1)\big|_{x=-1} = 2(-1) = -2$

Right-hand derivative $= \dfrac{d}{dx}(1 - x^2)\big|_{x=-1} = -2(-1) = 2$

Since $-2 \neq 2$, not differentiable at $x = -1$.

Continuous but not differentiable at $x = 1, -1$.

$(C) \to (IV)$

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$f(x) = |x - 1| = \begin{cases} x - 1 & : x \geq 1 \\ 1 - x & : x < 1 \end{cases}$

Both sides approach $0$ as $x \to 1$, and $f(1) = 0$, so the function is continuous at $x = 1$.

Left-hand derivative $= \lim_{h \to 0^-} \dfrac{|1 + h - 1| - 0}{h} = \dfrac{-h}{h} = -1$

Right-hand derivative $= \lim_{h \to 0^+} \dfrac{|1 + h - 1| - 0}{h} = \dfrac{h}{h} = 1$

Since $-1 \neq 1$, the function is not differentiable at $x = 1$.

Continuous but not differentiable at $x = 1$.

$(D) \to (II)$

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$(A) \to (III),\; (B) \to (I),\; (C) \to (IV),\; (D) \to (II)$