If f(x)= x, g(x)= 2 x, h(x)= 3 x and I(x)= x , then which of the...

(a) Let $f(x)=\cos x$, then $f^{\prime}(x)=-\sin x$.
In interval \left(0, \frac{\pi}{2}\right), f^{\prime}(x) < 0$<br />Therefore,$f(x)$is strictly decreasing on \left(0, \frac{\pi}{2}\right)$
(b) Let $f(x)=\cos 2 x \Rightarrow f^{\prime}(x)=-2 \sin x 2 x$
In interval \left(0, \frac{\pi}{2}\right), f^{\prime}(x) < 0$<br />Because$\sin 2 x$will either lie in the first or second quadrant which will give a positive value.<br />Therefore,$f(x)$is strictly decreasing on \left(0, \frac{\pi}{2}\right)$
(c) Let $f^{\prime}(x)=\cos 3 x$
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=-3 \sin 3 \mathrm{x}$. In Interval \left(0, \frac{\pi}{3}\right), \mathrm{f}^{\prime}(\mathrm{x}) < 0$<br />Because$\sin 3 \mathrm{x}$will either lie in the first or second quadrant which will give a positive value.<br />Therefore,$\mathrm{f}(\mathrm{x})$irstrictly decreasing on \left(0, \frac{\pi}{3}\right)$.
When $\mathrm{x} \in\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$, then $\mathrm{f}^{\prime}(\mathrm{x})>0$
Because $\sin 3 x$ will lie in the third quadrant. Therefore, $f(x)$ is not strictly decreasing on
\left(0, \frac{\pi}{2}\right)$<br />(d) Let$f(x)=\tan x \Rightarrow f^{\prime}(x)=\sec ^2 x$.<br />In Interval$\mathrm{x} \in\left(0, \frac{\pi}{2}\right), \mathrm{f}^{\prime}(\mathrm{x})>0$<br />Therefore,$f(x)$is not strictly decreasing on<br />\left(0, \frac{\pi}{2}\right)$