Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Function on R | Number of points where it is not differentiable | ||
| (A) | \(|x| + |x - 1| + |x - 2|\) | (I) | 0 |
| (B) | \(|x^{2} - 1|\) | (II) | 1 |
| (C) | \(|x|^{3}\) | (III) | 2 |
| (D) | \(x|x - 2|\) | (IV) | 3 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
A modulus can create a corner only where the expression inside it is zero; there we compare the left and right derivatives.
(A) \(|x| + |x - 1| + |x - 2|\)
This matches (IV) 3. At each of x = 0, 1, 2 exactly one term has a corner (its slope jumps from −1 to 1) while the other two terms are smooth there, so each of the three points is a corner.
(B) \(|x^{2} - 1|\)
This matches (III) 2. The inside is zero at x = ±1, and its derivative 2x is ±2 ≠ 0 there:
\(Lf'(1) = -2\), \(Rf'(1) = 2\)
\(Lf'(-1) = -2\), \(Rf'(-1) = 2\)
(C) \(|x|^{3}\)
This matches (I) 0. At x = 0:
\( \displaystyle \frac{f(h) - f(0)}{h} = \frac{|h|^{3}}{h} \to 0 \) from both sides
(D) \(x|x - 2|\)
This matches (II) 1. Removing the modulus:
\(f(x) = 2x - x^{2}\) for \(x \lt 2\), slope at 2: \(2 - 4 = -2\)
\(f(x) = x^{2} - 2x\) for \(x \ge 2\), slope at 2: \(4 - 2 = 2\)
The slopes differ only at x = 2.