Question

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

• A. $\cos x$
• B. $\cos 2x$
• C. $\cos 3x$
• D. $\tan x$

Answer 1

Answer: (A), (B)

The correct options are
A $\cos x$
B $\cos 2x$

A function $f\left(x\right)$ is said to be strictly decreasing on $(a,b)$ if

$f^{\prime }\left(x\right)<0$ for all $x\in (a,b)$

Option A

$f\left(x\right)=\cos x$

$f^{\prime }\left(x\right)=-\sin x$

For $x\in (0,\frac{\pi }{2})$, $\sin x$ is positive

i.e. $\sin x>0$ for $x\in (0,\frac{\pi }{2})$

$-\sin x<0$ for $x\in (0,\frac{\pi }{2})$

$\Rightarrow f^{\prime }\left(x\right)<0$ for $x\in (0,\frac{\pi }{2})$

Hence, $f\left(x\right)=\cos x$ is strictly decreasing on $(0,\frac{\pi }{2})$

Option B,

$f\left(x\right)=\cos 2x$

$f^{\prime }\left(x\right)=-2\sin 2x$

Since, $0<x<\frac{\pi }{2}$

$\Rightarrow 0<2x<\pi$

We know that $\sin t>0$ for $t\in (0,\pi )$

So, $\sin 2x>0$ for $2x\in (0,\pi )$

So, $\sin 2x>0$ for $x\in (0,\frac{\pi }{2})$

$\Rightarrow -2\sin 2x<0$ for $x\in (0,\frac{\pi }{2})$

$\Rightarrow f^{\prime }\left(x\right)<0$ for $x\in (0,\frac{\pi }{2})$

Hence, $f\left(x\right)=\cos 2x$ is strictly decreasing on $(0,\frac{\pi }{2})$

Option C,

$f\left(x\right)=\cos 3x$

$f^{\prime }\left(x\right)=-3\sin 3x$

Since, $0<x<\frac{\pi }{2}$

$\Rightarrow 0<3x<\frac{3\pi }{2}$

$\Rightarrow 3x\in (0,\pi )\cup \pi \cup (\pi ,\frac{3\pi }{2})$

We know that $\sin t>0$ for $t\in (0,\pi )$ and $\sin t<0$ for $t\in (\pi ,\frac{3\pi }{2})$

So, $\sin 3x>0$ for $3x\in (0,\pi )$ and $\sin 3x<0$ for $3x\in (\pi ,\frac{3\pi }{2})$

$-3\sin 3x<0$ for $x\in (0,\frac{\pi }{3})$ and $-3\sin 3x>0$ for $3x\in (\pi ,\frac{3\pi }{2})$

Hence, $f\left(x\right)$ is strictly decreasing in $x\in (0,\frac{\pi }{3})$ and increasing in $(\frac{\pi }{3},\frac{\pi }{2})$

Hence, $f\left(x\right)=\cos 3x$ is not strictly decreasing on $(0,\frac{\pi }{2})$

Option D,

$f\left(x\right)=\tan x$

$f^{\prime }\left(x\right)={\sec }^{2}x$

For $0<x<\frac{\pi }{2}$, $se{c}^{2}x>0$

$\Rightarrow f^{\prime }\left(x\right)>0$

Hence, $f\left(x\right)$ is strictly increasing on $(0,\frac{\pi }{2})$