Match List-I with List-II
| List-I (Function on \([0, \pi]\)) | List-II (Point of absolute maximum) | ||
| (A) | \(\sin 2x\) | (I) | \(\displaystyle\frac{\pi}{3}\) |
| (B) | \(\sin x - \cos x\) | (II) | \(\pi\) |
| (C) | \(\sqrt{3}\sin x + \cos x\) | (III) | \(\displaystyle\frac{\pi}{4}\) |
| (D) | \(x - \sin x\) | (IV) | \(\displaystyle\frac{3\pi}{4}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
On a closed interval the absolute maximum is at a critical point or an end point, so each function is checked at both.
(A) \(\sin 2x\)
This matches (III) \(\frac{\pi}{4}\).
\( f'(x) = 2\cos 2x = 0 \) at \( \frac{\pi}{4}, \frac{3\pi}{4} \)
\( f\left(\frac{\pi}{4}\right) = 1 \), \( f\left(\frac{3\pi}{4}\right) = -1 \), \( f(0) = f(\pi) = 0 \)
(B) \(\sin x - \cos x\)
This matches (IV) \(\frac{3\pi}{4}\).
\( f'(x) = \cos x + \sin x = 0 \) at \( \frac{3\pi}{4} \)
\( f\left(\frac{3\pi}{4}\right) = \sqrt{2} \), \( f(0) = -1 \), \( f(\pi) = 1 \)
(C) \(\sqrt{3}\sin x + \cos x\)
This matches (I) \(\frac{\pi}{3}\).
\( f'(x) = \sqrt{3}\cos x - \sin x = 0 \) when \( \tan x = \sqrt{3} \), at \( \frac{\pi}{3} \)
\( f\left(\frac{\pi}{3}\right) = \frac{3}{2} + \frac{1}{2} = 2 \), \( f(0) = 1 \), \( f(\pi) = -1 \)
(D) \(x - \sin x\)
This matches (II) π.
\( f'(x) = 1 - \cos x \gt 0 \) for \(0 \lt x \le \pi\), so f is increasing
The maximum is at the right end point: \( f(\pi) = \pi \)