Practicing Success
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 |
$\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$ $\left(0, \frac{\pi}{4}\right)$ $\left(0, \frac{\pi}{2}\right)$ $\left(\frac{\pi}{6}, \frac{\pi}{2}\right)$ |
$\left(0, \frac{\pi}{4}\right)$ |
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 $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$ |