Match List-I with List-II

List-I	List-II
Function	Points of discontinuity
(A) $f(x) = \frac{x^2 + 1}{x}$	(I) $x = 4$
(B) $f(x) = \frac{\vert x - 1 \vert}{x - 1}$	(II) $x = 2$
(C) $f(x) = \begin{cases} x - 1, & x < 2 \\ x + 1, & x \ge 2 \end{cases}$	(III) $x = 0$
(D) $f(x) = \frac{1 - x}{(x - 4)}$	(IV) $x = 1$

Choose the correct answer from the options given below:

A function is discontinuous at a point where it is not defined (like when a denominator becomes zero) or where there is a sudden jump in its value.

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(A) $f(x) = \dfrac{x^2 + 1}{x}$

The function is undefined when the denominator $= 0$, i.e., when $x = 0$

So, discontinuous at $x = 0$ → (III)

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(B) $f(x) = \dfrac{|x - 1|}{x - 1}$

The function is undefined when $x - 1 = 0$, i.e., when $x = 1$

Also, checking both sides of $x = 1$:

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

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

The function jumps from $-1$ to $+1$ and is undefined at $x = 1$

So, discontinuous at $x = 1$ → (IV)

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(C) $f(x) = \begin{cases} x - 1, & x < 2 \\ x + 1, & x \ge 2 \end{cases}$

We check continuity at the boundary point $x = 2$:

Left-hand limit $= \lim_{x \to 2^-}(x - 1) = 1$

Right-hand limit $= \lim_{x \to 2^+}(x + 1) = 3$

Since $1 \neq 3$, there is a jump at $x = 2$

So, discontinuous at $x = 2$ → (II)

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(D) $f(x) = \dfrac{1 - x}{x - 4}$

The function is undefined when $x - 4 = 0$, i.e., when $x = 4$

So, discontinuous at $x = 4$ → (I)

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