Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Function | Points of discontinuity | ||
| (A) | \(f(x) =\displaystyle\frac{x^2+1}{x}\) | (I) | \(x=4\) |
| (B) | \(f(x)=\displaystyle\frac{|x-1|}{x-1}\) | (II) | \(x = 2\) |
| (C) | \(f(x)=\begin{cases}x-1, & x \lt 2\\x+1, & x \ge 2\end{cases}\) | (III) | \(x = 0\) |
| (D) | \(f(x)=\displaystyle\frac{1-x}{(x-4)}\) | (IV) | \(x = 1\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
As is usual in such lists, a point where the graph breaks is taken as a point of discontinuity: either the function is not defined there (a zero denominator), or its value jumps.
(A) \(f(x) =\displaystyle\frac{x^2+1}{x}\)
This matches (III) x = 0: the denominator x is zero there, and \(|f(x)| \to \infty\) as \(x \to 0\).
(B) \(f(x)=\displaystyle\frac{|x-1|}{x-1}\)
This matches (IV) x = 1: f is not defined at 1, and it jumps there:
\(f(x) = -1\) for \(x \lt 1\)
\(f(x) = 1\) for \(x \gt 1\)
(C) \(f(x)=\begin{cases}x-1, & x \lt 2\\x+1, & x \ge 2\end{cases}\)
This matches (II) x = 2, where the left and right limits differ:
\(\displaystyle \lim_{x \to 2^-} f(x) = 2 - 1 = 1\)
\(\displaystyle \lim_{x \to 2^+} f(x) = 2 + 1 = 3\)
(D) \(f(x)=\displaystyle\frac{1-x}{(x-4)}\)
This matches (I) x = 4: the denominator \(x - 4\) is zero there, while the numerator is \(-3 \ne 0\).
So the matching is (A)-(III), (B)-(IV), (C)-(II), (D)-(I).