# Which of the following functions are strictly decreasing on

Question:

Which of the following functions are strictly decreasing on $\left(0, \frac{\pi}{2}\right) ?$

(A) cos (B) cos 2(C) cos 3(D) tan x

Solution:

(A) Let $f_{1}(x)=\cos x$.

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

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

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

(B) Let $f_{2}(x)=\cos 2 x$.

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

Now, $00 \Rightarrow-2 \sin 2 x<0$

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

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

(C) Let $f_{3}(x)=\cos 3 x$.

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

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

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

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

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

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

Now, in interval $\left(0, \frac{\pi}{3}\right), f_{3}(x)=-3 \sin 3 x<0\left[\right.$ as $\left.0$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 3 x>0\left[\right.$as$\left.\frac{\pi}{3}

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

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

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

Hence, the correct answers are A and B.