Mathematics - Application of Derivative Question with Solution | TestHub

(a) Let $f(x)=\cos x$, then $f^{\prime }(x)=-\sin x$.
In interval $\left(0,\frac{\pi }{2}\right),f^{\prime }(x)<0$
Therefore, $f(x)$ is strictly decreasing on $\left(0,\frac{\pi }{2}\right)$
(b) Let $f(x)=\cos 2x\Rightarrow f^{\prime }(x)=-2\sin x2x$
In interval $\left(0,\frac{\pi }{2}\right),f^{\prime }(x)<0$
Because $\sin 2x$ will either lie in the first or second quadrant which will give a positive value.
Therefore, $f(x)$ is strictly decreasing on $\left(0,\frac{\pi }{2}\right)$
(c) Let $f^{\prime }(x)=\cos 3x$
$\Rightarrow {\mathrm{f}}^{\prime }(\mathrm{x})=-3\sin 3\mathrm{x}$. In Interval $\left(0,\frac{\pi }{3}\right),{\mathrm{f}}^{\prime }(\mathrm{x})<0$
Because $\sin 3\mathrm{x}$ will either lie in the first or second quadrant which will give a positive value.
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 $\sin 3x$ will lie in the third quadrant. Therefore, $f(x)$ is not strictly decreasing on
$\left(0,\frac{\pi }{2}\right)$
(d) Let $f(x)=\tan x\Rightarrow f^{\prime }(x)={\sec }^{2}x$.
In Interval $\mathrm{x}\in \left(0,\frac{\pi }{2}\right),{\mathrm{f}}^{\prime }(\mathrm{x})>0$
Therefore, $f(x)$ is not strictly decreasing on
$\left(0,\frac{\pi }{2}\right)$