MCQ
The function $f(x) = {\tan ^{ - 1}}(\sin x + \cos x)$, $x > 0$ is always an increasing function on the interval
  • A
    $(0,\,\pi )$
  • B
    $(0,\,\pi /2)$
  • $(0,\pi /4)$
  • D
    $(0,\,3\pi /4)$

Answer

Correct option: C.
$(0,\pi /4)$
c
(c) $f(x) = y = {\tan ^{ - 1}}\left( {\sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right)} \right)$

$ \Rightarrow \,\,\tan y = \sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) \Rightarrow {\sec ^2}y\frac{{dy}}{{dx}} = \sqrt 2 \cos \left( {x + \frac{\pi }{4}} \right)$

$\frac{{dy}}{{dx}} > 0 \Rightarrow \cos \left( {x + \frac{\pi }{4}} \right) > 0$.

$\therefore \,\,\,x \in \left( {0,\,\,\frac{\pi }{4}} \right)$.

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