Match List-I with List-II

List-I	List-II
(A) $f(x) = x \sin x$	(I) is not continuous at $x = -3$
(B) $f(x) = \frac{\vert x\vert }{x}, x \neq 0$ and $f(x) = 1 \text{ at } x = 0$	(II) is continuous everywhere
(C) $f(x) = x - [x]$, $[x]$ denotes greatest integer function	(III) is not differentiable at $x = 1$
(D) $f(x) = e^{\vert x - 1\vert }$	(IV) is not continuous at $x = 0$

Choose the correct answer from the options given below:

$f(x) = x \sin x$

Both $x$ and $\sin x$ are continuous and differentiable everywhere. The product of two continuous functions is always continuous.

So $x \sin x$ is continuous everywhere.

(A) → (II)

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$f(x) = \dfrac{|x|}{x},\ x \neq 0$ and $f(0) = 1$

When $x > 0$: $f(x) = \dfrac{x}{x} = 1$

When $x < 0$: $f(x) = \dfrac{-x}{x} = -1$

At $x = 0$: $f(0) = 1$

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

$\lim_{x \to 0^+} f(x) = +1$

Since LHL $\neq$ RHL, the limit does not exist, so $f$ is not continuous at $x = 0$.

(B) → (IV)

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$f(x) = x - [x]$ is the fractional part function $\{x\}$, which is discontinuous at every integer.

At $x = -3$:

$\lim_{x \to -3^-} f(x)$: here $[x] = -4$, so $f(x) = x + 4 \to -3 + 4 = 1$

$\lim_{x \to -3^+} f(x)$: here $[x] = -3$, so $f(x) = x + 3 \to -3 + 3 = 0$

$f(-3) = -3 - [-3] = -3 + 3 = 0$

Since LHL $(= 1) \neq$ RHL $(= 0)$, $f$ is not continuous at $x = -3$.

(C) → (I)

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The function is given by:

$f(x) = e^{|x-1|}$

Since $|x-1|$ is continuous everywhere, $e^{|x-1|}$ is also continuous everywhere.

However, $|x-1|$ has a sharp corner point at $x = 1$, so differentiability must be checked specifically at that point using first principles.

Left-Hand Derivative (LHD):

$\lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^-} \frac{e^{|h|} - e^0}{h}$

Since $h \to 0^-$, $h$ is negative, which means $|h| = -h$:

$\lim_{h \to 0^-} \frac{e^{-h} - 1}{h} = -1$

Right-Hand Derivative (RHD):

$\lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^+} \frac{e^{|h|} - e^0}{h}$

Since $h \to 0^+$, $h$ is positive, which means $|h| = h$:

$\lim_{h \to 0^+} \frac{e^{h} - 1}{h} = +1$

Since the Left-Hand Derivative ($-1$) $\neq$ Right-Hand Derivative ($+1$), the function $f(x)$ is not differentiable at $x = 1$.

Therefore, (D) $\rightarrow$ (III)

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(A) → (II)

(B) → (IV)

(C) → (I)

(D) → (III)