Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Function | Property | ||
| (A) | \(f(x) =\begin{cases}\displaystyle\frac{x}{|x|} & : x \ne 0\\ 0 & : x = 0\end{cases}\) | (I) | continuous but not differentiable at \(x= 0\) |
| (B) | \(f(x) = |x|\) | (II) | continuous but not differentiable at \(x=1\) |
| (C) | \(f(x) = |x^2 - 1|\) | (III) | discontinuous at \(x = 0\) |
| (D) | \(f(x) = |x - 1|\) | (IV) | continuous but not differentiable at \(x =1, -1\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
A function \(|g(x)|\) with g a polynomial is continuous everywhere. It has a corner (left and right derivatives differ) at a point where g(x) = 0 and \(g'(x) \ne 0\), so it is not differentiable there.
(A) \(f(x) =\begin{cases}\displaystyle\frac{x}{|x|} & : x \ne 0\\ 0 & : x = 0\end{cases}\)
This matches (III) discontinuous at x = 0. For x > 0, f(x) = 1 and for x < 0, f(x) = −1:
\(\displaystyle \lim_{x \to 0^-} f(x) = -1,\quad \lim_{x \to 0^+} f(x) = 1\)
The left and right limits differ, so the limit at 0 does not exist.
(B) \(f(x) = |x|\)
This matches (I) continuous but not differentiable at x = 0. At 0 the left derivative is −1 and the right derivative is 1.
(C) \(f(x) = |x^2 - 1|\)
This matches (IV) continuous but not differentiable at x = 1, −1. Here \(x^2 - 1 = 0\) at \(x = \pm 1\), and its derivative 2x is not 0 at either point. At x = 1, for example:
\(\text{left derivative} = -2x = -2\)
\(\text{right derivative} = 2x = 2\)
(D) \(f(x) = |x - 1|\)
This matches (II) continuous but not differentiable at x = 1. At 1 the left derivative is −1 and the right derivative is 1.
So the matching is (A)-(III), (B)-(I), (C)-(IV), (D)-(II).