Let $f'(\sin x)<0$ and $f''(\sin x)>0$ for all $x \in(0, \pi / 2)$ and $g(x)=f(\sin x)+f(\cos x)$, then $g(x)$ is decreasing in
Answer & explanation
Correct answer: option 2
We have, $g(x)=f(\sin x)+f(\cos x)$
$\Rightarrow g'(x)=f'(\sin x) \cos x-f'(\cos x) \sin x$
$\Rightarrow g''(x)=f''(\sin x) \cos ^2 x+f''(\cos x) \sin ^2 x -f'(\sin x) \sin x-f'(\cos x) \cos x $
$\Rightarrow g''(x)>0$ for all $x \in(0, \pi / 2) $
⇒ g'(x) is increasing on $(0, \pi / 2)$
Also, $g'\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} f'\left(\frac{1}{\sqrt{2}}\right)-\frac{1}{\sqrt{2}} f'\left(\frac{1}{\sqrt{2}}\right)=0$
∴ g'(x) < 0 for all $x \in\left(0, \frac{\pi}{4}\right)$ and g'(x) > 0 for all $x \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$
⇒ g(x) is decreasing on $\left(0, \frac{\pi}{4}\right)$ and increasing on $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$