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$(0,\,3\pi /4)$
$ \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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