The function \(f(x) = \tan^{-1}(\sin x + \cos x)\), \(0 \lt x \lt \displaystyle\frac{\pi}{2}\), is strictly increasing in
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\left(0, \displaystyle\frac{\pi}{4}\right)\)
Using \(\frac{d}{du}\tan^{-1}u = \frac{1}{1 + u^{2}}\) and the chain rule:
\( \displaystyle f'(x) = \frac{\cos x - \sin x}{1 + (\sin x + \cos x)^{2}} \)
The denominator is always positive, so f′(x) has the sign of cos x − sin x. In \(\left(0, \frac{\pi}{2}\right)\):
\(\cos x \gt \sin x\) for \( \displaystyle 0 \lt x \lt \frac{\pi}{4} \)
\(\cos x \lt \sin x\) for \( \displaystyle \frac{\pi}{4} \lt x \lt \frac{\pi}{2} \)
So f is strictly increasing in \(\left(0, \frac{\pi}{4}\right)\) and strictly decreasing in \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\).
The intervals \(\left(0, \frac{\pi}{2}\right)\) and \(\left(\frac{\pi}{6}, \frac{\pi}{3}\right)\) both contain points to the right of \(\frac{\pi}{4}\), where f is decreasing.