Match List-I with List-II. Here [x] denotes the greatest integer function

$\begin{array}{|l|l|} \hline \rule{0pt}{2.8ex}\text{List-I} & \text{List-II} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(A) } f(x) = [x] & \text{(I) is continuous everywhere but not differentiable at } x=-1 \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(B) } f(x) = |x-1| & \text{(II) is continuous everywhere except at all integral values} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(C) } f(x) = e^{|x|} & \text{(III) is continuous everywhere but not differentiable at } x=1 \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(D) } f(x) = |x+1| & \text{(IV) is continuous everywhere but not differentiable at } x=0 \\[1.2ex] \hline \end{array}$

Choose the correct answer from the options given below:

For $f(x) = [x]$ (Greatest Integer Function):

$[x]$ gives the largest integer $≤ x$

Example: $[2.7] = 2$, $[3] = 3$, $[-1.2] = -2$

The function has jumps at every integer value.

At $x = 1$: from the left $[0.9] = 0$, but $[1] = 1$ (discontinuous)

At $x = 2$: from the left $[1.9] = 1$, but $[2] = 2$ (discontinuous)

The function is discontinuous at all integer values.

(A) → (II)

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For $f(x) = |x - 1|$:

The function is V-shaped with a corner at $x = 1$.

When $x < 1$: $f(x) = -(x-1) = 1-x$

When $x ≥ 1$: $f(x) = x-1$

Absolute value functions are continuous everywhere.

At $x = 1$:

Left derivative: $f(x) = 1-x$, so $f'(x) = -1$

Right derivative: $f(x) = x-1$, so $f'(x) = +1$

Since $-1 ≠ +1$, the function is not differentiable at $x = 1$.

(B) → (III)

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For $f(x) = e^{|x|}$:

When $x < 0$: $f(x) = e^{-x}$

When $x ≥ 0$: $f(x) = e^x$

Both parts meet at $x = 0$: $e^0 = 1$ (continuous)

At $x = 0$:

Left derivative: $f(x) = e^{-x}$, so $f'(x) = -e^{-x}$, giving $f'(0^-) = -1$

Right derivative: $f(x) = e^x$, so $f'(x) = e^x$, giving $f'(0^+) = +1$

Since $-1 ≠ +1$, the function is not differentiable at $x = 0$.

(C) → (IV)

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For $f(x) = |x + 1|$:

The function is V-shaped with a corner at $x = -1$.

When $x < -1$: $f(x) = -(x+1) = -x-1$

When $x ≥ -1$: $f(x) = x+1$

Absolute value functions are continuous everywhere.

At $x = -1$:

Left derivative: $f(x) = -x-1$, so $f'(x) = -1$

Right derivative: $f(x) = x+1$, so $f'(x) = +1$

Since $-1 ≠ +1$, the function is not differentiable at $x = -1$.

(D) → (I)

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Final Matching:

(A) $f(x) = [x]$ → (II) continuous everywhere except at all integral values

(B) $f(x) = |x - 1|$ → (III) continuous everywhere but not differentiable at $x = 1$

(C) $f(x) = e^{|x|}$ → (IV) continuous everywhere but not differentiable at $x = 0$

(D) $f(x) = |x + 1|$ → (I) continuous everywhere but not differentiable at $x = -1$

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