Let $ f:(0,\frac{\pi}{2})\rightarrow\mathbb{R}$ be given by $f(x)=(\sin x)^\pi-\pi \sin x+\pi$. Then which of the following are true?
a. f is an increasing function.
b. f is a decreasing function.
c. $f(x)>0 \ \ \forall x\in (0,\frac{\pi}{2})$
d. $f(x)<0$ for some $x\in (0,\frac{\pi}{2})$
My attempt: I tried doing $f'(x)=0$.
I got $\pi (\sin x)^{\pi-1} \cos x -\pi \cos x=0$. Did not know how to proceed after this. Kindly help.
$$f'(x) = \pi(\sin x)^{\pi - 1}\cos x - \pi \cos x$$$$ = \pi\cos x\left((\sin x)^{\pi - 1} - 1\right)$$ Now note that since in the domain $\displaystyle \bigg(0, \dfrac{\pi}{2}\bigg)$, $\sin x$ and $\cos x$ are both always positive and less than $1$, hence $(\sin x)^{\pi - 1} < 1$ (because $\pi - 1 >1$) which implies that $f'(x) < 0$, or that $f(x)$ is decreasing.