Q61 / 120 · JEE Main 2007 · Mathematics

The function $f(x)=\tan ^{-1}(\sin x+\cos x)$ is an increasing function in

  • A
    $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$
  • Correct answer
    B
    $\left(-\frac{\pi}{2}, \frac{\pi}{4}\right)$
  • C
    $\left(0, \frac{\pi}{2}\right)$
  • D
    $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

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 $x \in\left(-\frac{\pi}{2}, \frac{\pi}{4}\right)$.

Hence, the correct answer is option: 'B'.

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