Question

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

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

Answer 1

Answer: (C), (D)

$\cos x$

Checking the options :

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

Differentiating w.r.t. $x,$

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

As, $\sin x>0$ for $xϵ(0,\frac{\pi }{2})$

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

So, $\cos x$ is decreasing in $(0,\frac{\pi }{2})$

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

Differentiating w.r.t.$x,$

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

For $2xϵ(0,\pi )i.e.xϵ(0,\frac{\pi }{2});\sin 2x>0$

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

So, $\cos 2x$ is decreasing in $(0,\frac{\pi }{2})$

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

Differentiating w.r.t.$x,$

$\Rightarrow f^{\prime }\left(x\right)=-3\sin 3x$
For $3xϵ(0,\pi )\Rightarrow xϵ(0,\frac{\pi }{3})$
$\Rightarrow \sin 3x>0$
so, $\sin 3x>0$ for $xϵ(0,\frac{\pi }{3})$

$\Rightarrow f^{\prime }\left(x\right)<0,i.e.$ decreasing in $xϵ(0,\frac{\pi }{3})$
For,
$f\left(x\right)=\tan x$

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

As ${\sec }^{2}x>0$ for $xϵ(0,\frac{\pi }{2})$

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

So, $\tan x$ is increasing in $(0,\frac{\pi }{2})$

Hence, option $\left(A\right)$ and $\left(B\right)$ are correct.