Practicing Success
If $f'\left(x^2-4 x+3\right)>0$ for all $x \in(2,3)$, then f(sin x) is increasing on |
$\underset{n \in Z}{\cup}\left(2 n \pi,(4 n+1) \frac{\pi}{2}\right)$ $\underset{n \in Z}{\cup}\left((4 n-1) \frac{\pi}{2}, 2 n \pi\right)$ R none of these |
$\underset{n \in Z}{\cup}\left(2 n \pi,(4 n+1) \frac{\pi}{2}\right)$ |
Let $\alpha=x^2-4 x+3$. Clearly, $x^2-4 x+3$ is an increasing function on (2, 3) and $\alpha=x^2-4 x+3 \in(-1,0)$. Therefore, $f'\left(x^2-4 x+3\right)>0$ $\Rightarrow f'(\alpha)>0 $ for all $ \alpha \in(-1,0)$ Let $g(x)=f(\sin x)$. Then, $g'(x)=f'(\sin x) \cos x$ For g(x) to be increasing, we must have g'(x) > 0 $\Rightarrow f'(\sin x) \cos x>0$ $\Rightarrow \sin x \in(-1,0) $ and $ \cos x>0$ $\Rightarrow x \in((2 n-1) \pi, 2 n \pi)$ and, $x \in\left(2 n \pi, 2 n \pi+\frac{\pi}{2}\right) \cup\left(2 n \pi, (2 n+1) \frac{\pi}{2}\right),$ where $n \in Z$ $\Rightarrow x \in\left(2 n \pi,(2 n+1) \frac{\pi}{2}\right), n \in Z$ $\Rightarrow x \in \underset{n \in Z}{\cup}\left(2 n \pi,(2 n+1) \frac{\pi}{2}\right)$ |