Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Function and point | Derivative at the point | ||
| (A) | \(|x + 1| - |x - 1|\) at \(x = 3\) | (I) | 4 |
| (B) | \(|x^{2} - 2x - 3|\) at \(x = 2\) | (II) | \(\displaystyle\frac{1}{2}\) |
| (C) | \(x|x - 4|\) at \(x = 0\) | (III) | 0 |
| (D) | \(|\sin x|\) at \(x = \displaystyle\frac{4\pi}{3}\) | (IV) | −2 |
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)-(I), (D)-(II)
None of the points is a corner, so near each point the modulus is removed by the sign of the expression inside, and the result is differentiated as usual.
(A) \(|x + 1| - |x - 1|\) at \(x = 3\)
This matches (III) 0. For x > 1 both insides are positive:
\(f(x) = (x + 1) - (x - 1) = 2\), so \(f'(3) = 0\)
(B) \(|x^{2} - 2x - 3|\) at \(x = 2\)
This matches (IV) −2. Since \(x^{2} - 2x - 3 = (x - 3)(x + 1) \lt 0\) for −1 < x < 3:
\(f(x) = 3 + 2x - x^{2}\), \(f'(x) = 2 - 2x\), \(f'(2) = -2\)
(C) \(x|x - 4|\) at \(x = 0\)
This matches (I) 4. For x < 4, \(|x - 4| = 4 - x\):
\(f(x) = x(4 - x) = 4x - x^{2}\), \(f'(x) = 4 - 2x\), \(f'(0) = 4\)
(D) \(|\sin x|\) at \(x = \frac{4\pi}{3}\)
This matches (II) \(\frac{1}{2}\). For \(\pi \lt x \lt 2\pi\), \(\sin x \lt 0\):
\(f(x) = -\sin x\), \(f'(x) = -\cos x\)
\( \displaystyle f'\left(\frac{4\pi}{3}\right) = -\left(-\frac{1}{2}\right) = \frac{1}{2} \)