Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(f(x) = |x| + 1\) | (I) | Not differentiable at \(x = 3\) only |
| (B) | \(f(x) = |x - 3|\) | (II) | Not differentiable at \(x = -3\) only |
| (C) | \(f(x) = |x + 3|\) | (III) | Not differentiable at \(x = 3, -3\) only |
| (D) | \(f(x) = |x^2 - 9|\) | (IV) | Not differentiable at \(x = 0\) only |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
The graph of \(|x - a|\) has a sharp corner at x = a: the left derivative there is −1 and the right derivative is 1. It is differentiable everywhere else.
(A) \(f(x) = |x| + 1\)
This matches (IV) not differentiable at x = 0 only: adding 1 only shifts the graph of \(|x|\) up, keeping its corner at x = 0.
(B) \(f(x) = |x - 3|\)
This matches (I) not differentiable at x = 3 only, the corner at x = 3.
(C) \(f(x) = |x + 3|\)
This matches (II) not differentiable at x = −3 only, the corner at x = −3.
(D) \(f(x) = |x^2 - 9|\)
This matches (III) not differentiable at x = 3, −3 only. Here \(x^2 - 9\) changes sign at x = 3 and x = −3. Near x = 3:
\(x \gt 3:\ f(x) = x^2 - 9,\quad f'(x) = 2x \to 6\)
\(x \lt 3:\ f(x) = 9 - x^2,\quad f'(x) = -2x \to -6\)
The left and right derivatives differ, so f is not differentiable at x = 3; in the same way (−6 and 6 swapped) it is not differentiable at x = −3. Elsewhere, including x = 0, \(f(x) = \pm(x^2 - 9)\) is a polynomial and is differentiable.
So the matching is (A)-(IV), (B)-(I), (C)-(II), (D)-(III).