Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Function | Derivative with respect to x | ||
| (A) | \(\tan^{-1}\left(\displaystyle\frac{\sin x}{1 + \cos x}\right),\ -\pi \lt x \lt \pi\) | (I) | 1 |
| (B) | \(\cos^{-1}(\sin x),\ -\displaystyle\frac{\pi}{2} \lt x \lt \frac{\pi}{2}\) | (II) | \(\displaystyle\frac{2}{1 + x^{2}}\) |
| (C) | \(\tan^{-1}\left(\displaystyle\frac{1 + \tan x}{1 - \tan x}\right),\ -\displaystyle\frac{\pi}{4} \lt x \lt \frac{\pi}{4}\) | (III) | −1 |
| (D) | \(\cos^{-1}\left(\displaystyle\frac{1 - x^{2}}{1 + x^{2}}\right),\ 0 \lt x \lt 1\) | (IV) | \(\displaystyle\frac{1}{2}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
Each function first simplifies to a simple expression; the given interval keeps the angle in the principal range.
(A) \(\tan^{-1}\left(\frac{\sin x}{1 + \cos x}\right),\ -\pi \lt x \lt \pi\)
This matches (IV) \(\frac{1}{2}\). Using the half-angle formulas:
\( \displaystyle \frac{\sin x}{1 + \cos x} = \frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^{2}\frac{x}{2}} = \tan\frac{x}{2} \)
Since \(\frac{x}{2}\) lies in \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), the function is \(\frac{x}{2}\), with derivative \(\frac{1}{2}\).
(B) \(\cos^{-1}(\sin x),\ -\frac{\pi}{2} \lt x \lt \frac{\pi}{2}\)
This matches (III) −1. Since \(\sin x = \cos\left(\frac{\pi}{2} - x\right)\) and \(\frac{\pi}{2} - x\) lies in \((0, \pi)\):
\( \displaystyle \cos^{-1}(\sin x) = \frac{\pi}{2} - x \)
Its derivative is −1.
(C) \(\tan^{-1}\left(\frac{1 + \tan x}{1 - \tan x}\right),\ -\frac{\pi}{4} \lt x \lt \frac{\pi}{4}\)
This matches (I) 1. Since \(\tan\frac{\pi}{4} = 1\):
\( \displaystyle \frac{1 + \tan x}{1 - \tan x} = \tan\left(\frac{\pi}{4} + x\right) \)
Since \(\frac{\pi}{4} + x\) lies in \(\left(0, \frac{\pi}{2}\right)\), the function is \(\frac{\pi}{4} + x\), with derivative 1.
(D) \(\cos^{-1}\left(\frac{1 - x^{2}}{1 + x^{2}}\right),\ 0 \lt x \lt 1\)
This matches (II) \(\frac{2}{1 + x^{2}}\). Let \(x = \tan\theta\) with \(0 \lt \theta \lt \frac{\pi}{4}\):
\( \displaystyle \frac{1 - \tan^{2}\theta}{1 + \tan^{2}\theta} = \cos 2\theta \)
Since \(2\theta\) lies in \(\left(0, \frac{\pi}{2}\right)\), the function is \(2\theta = 2\tan^{-1}x\), with derivative \(\frac{2}{1 + x^{2}}\).