Question

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

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

Answer 1

Answer: (C)

Given,
$f(x)=\frac{1}{2}[|\sin x|+\sin x], 0< x \leq 2 \pi$
Case I, when, $0 < x \leq \pi$
$f(x)=\frac{1}{2}[\sin x+\sin x]=\sin x $
$f^{'}(x)=\cos x $
$\cos x>0, $ for $ 0< x< \frac{\pi}{2} $ (increasing)
$\cos x<0, $ for $ \frac{\pi}{2}< x< \pi $ (decreasing)
Case II When $\pi< x \leq 2 \pi$
$f(x)=\frac{1}{2}[-\sin x+\sin x]=0$