Define f(x) = frac{1}{2}[|sin x| + sin x],0

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(C) Given,$f(x) = \frac{1}{2}[|\sin x| + \sin x]$ for $0 < x \leq 2\pi$. Case $I$: When $0 < x \leq \pi$,$\sin x \geq 0$,so $|\sin x| = \sin x$. Thus,

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increasing in $\left(0, \frac{\pi}{2}\right)$ and decreasing in $\left(\frac{\pi}{2}, \pi\right)$

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(C) Given,$f(x) = \frac{1}{2}[|\sin x| + \sin x]$ for $0 Case $I$: When $0 Thus,$f(x) = \frac{1}{2}[\sin x + \sin x] = \sin x$. The derivative is $f'(x) = \cos x$. For $0  0$,so $f(x)$ is increasing. For $\frac{\pi}{2} Case $II$: When $\pi Thus,$f(x) = \frac{1}{2}[-\sin x + \sin x] = 0$. Therefore,$f(x)$ is increasing in $\left(0, \frac{\pi}{2}\right)$ and decreasing in $\left(\frac{\pi}{2}, \pi\right)$.

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

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