Question

Define $f\left(x\right)=\frac{1}{2}[\left|\sin x\right|+\sin x],0<x<2\pi$
Then, $f$ is

• A. increasing in $(\frac{\pi }{2},\frac{3\pi }{2})$
• B. decreasing in $(0,\frac{\pi }{2})$ and increasing in $(\frac{\pi }{2},\pi )$
• C. increasing in $(0,\frac{\pi }{2})$ and decreasing in $(\frac{\pi }{2},\pi )$
• D. increasing in $(0,\frac{\pi }{4})$ and decreasing in $(\frac{\pi }{4},\pi )$

Answer 1

The correct option is C increasing in $(0,\frac{\pi }{2})$ and decreasing in $(\frac{\pi }{2},\pi )$

$\left|sinx\right|=sinx$ , $xϵ[0,\pi ]$

$=-sinx$ , $xϵ[\pi ,2\pi ]$

So, $f\left(x\right)=\frac{1}{2}[\left|sinx\right|+sinx]$

$=\frac{1}{2}[sinx+sinx]=sinx$ $xϵ[0,\pi ]$

$=\frac{1}{2}[sinx+sinx]=0$ $xϵ[\pi ,2\pi ]$

$sinx$ is increasing in $(0,\frac{\pi }{2})$ and is decreasing in $(\frac{\pi }{2},\pi )$

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

And $f\left(x\right)$is constant in $(\pi ,2\pi )$.

Hence, answer is option (C).