The function $f(x)=\tan ^{-1}(\sin x+\cos x)$ is an increasing function in
The function $f(x)=\tan ^{-1}(\sin x+\cos x)$ is an increasing function in
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$\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$
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$\left(-\frac{\pi}{2}, \frac{\pi}{4}\right)$
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$\left(0, \frac{\pi}{2}\right)$
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$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
Solution
$f^{\prime}(x)=\frac{1}{1+(\sin x+\cos x)^2}(\cos x-\sin x)$
$=\frac{\sqrt{2} \cos \left(x+\frac{\pi}{4}\right)}{1+(\sin x+\cos x)^2}$
$f(x)$ is increasing if $-\frac{\pi}{2} < x+\frac{\pi}{4} < \frac{\pi}{2}$
$-\frac{3 \pi}{4} < x < \frac{\pi}{4}$
hence $f(x)$ is increasing when $x \in\left(-\frac{\pi}{2}, \frac{\pi}{4}\right)$.
Asked in: JEE Main 2007
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