STD 12 - 6. Application of derivatives — Mathematics STD 12 Science — Question

$\therefore f_{1}^{\prime}(x)=-\sin x$

In interval $\left(0, \frac{\pi}{2}\right), f_{1}(x)=-\sin x<0$

$\therefore f_{1}(x)=\cos x$ is strictly decreasing in interval $\left(0, \frac{\pi}{2}\right)$

let $f_{2}(x)=\cos 2 x$

$\therefore f_{2}^{\prime}(x)=-2 \sin 2 x$

Now, $0 < x < \frac{\pi }{2} \Rightarrow 0 < 2x < \pi  \Rightarrow \sin \,2x > 0 \Rightarrow  - 2\,\sin \,2x < 0$

$\therefore f_{2}^{\prime}(x)=-2 \sin 2 x<0$ on $\left(0, \frac{\pi}{2}\right)$

$\therefore f_{2}(x)=\cos 2 x$ is strictly decreasing in interval $\left(0, \frac{\pi}{2}\right).$

let $f_{3}(x)=\cos 3 x$

$\therefore f_{3}^{\prime}(x)=-3 \sin 3 x$

Now, $f^{\prime} 3(x)=0$

$\Rightarrow \sin 3 x=0 \Rightarrow 3 x=\pi, a s x \in\left(0, \frac{\pi}{2}\right)$

$\Rightarrow x=\frac{\pi}{3}$

The point $x=\frac{\pi}{3}$ divides the interval $\left(0, \frac{\pi}{2}\right)$ into two disjoint intervals

i.e., $\left(0, \frac{\pi}{3}\right)$ and $\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$

Now, in interval $\left( {0,\frac{\pi }{3}} \right),{f_3}(x) =  - 3\sin 3x < 0[{\text{as }}0 < x < \frac{\pi }{3} \Rightarrow 0 < 3x < \pi ]$

$\therefore f_{3}$ is strictly decreasing in interval $\left(0, \frac{\pi}{3}\right)$

However, in interval $\left( {\frac{\pi }{3},\frac{\pi }{2}} \right),{f_3}(x) =  - 3\sin 3x > 0\,[{\text{ as }}\frac{\pi }{3} < x < \frac{\pi }{2} \Rightarrow \pi  < 3x < \frac{{3\pi }}{2}]$

$\therefore f_{3}$ is strictly increasing in interval $\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$

Hence, $f_{3}$ is neither increasing nor decreasing in interval $\left(0, \frac{\pi}{2}\right)$

let $f_{4}(x)=\tan x$

$\therefore f_{4}^{\prime}(x)=\sec ^{2} x$

In interval $\left(0, \frac{\pi}{2}\right), f_{4}^{\prime}(x)=\sec ^{2} x>0$

$\therefore f_{4}$ is strictly increasing in interval $\left(0, \frac{\pi}{2}\right)$

Therefore, function $\cos x$ and $\cos 2 x$ are strictly decreasing in $\left(0, \frac{\pi}{2}\right)$