Question
Find the intervals in which f(x) is increasing or decreasing:
$\text{f}(\text{x})=\sin\text{x}+|\sin\text{x}|,0<\text{x}\leq2\pi$
$\text{f}(\text{x})=\sin\text{x}+|\sin\text{x}|,0<\text{x}\leq2\pi$
Case I:
When $\text{x}\in(0,\pi)$ $\text{f}(\text{x})=\sin\text{x}+\sin\text{x}=2\sin\text{x}$ $\Rightarrow\text{f}'(\text{x})=2\cos\text{x}$ As, $\cos\text{x}>0$ for $\text{x}\in\Big(0,\frac{\pi}{2}\Big)$ and $\cos\text{x}<0$ for $\text{x}\in\Big(\frac{\pi}{2},\pi\Big)$ So, $\text{f}'(\text{x})>0$ for $\text{x}\in\Big(0,\frac{\pi}{2}\Big)$ and $\text{f}'(\text{x})<0$ for $\text{x}\in\Big(\frac{\pi}{2},\pi\Big)$ $\therefore$ f(x) is increasing on $\Big(0,\frac{\pi}{2}\Big)$ and f(x) is decreasing on $\Big(\frac{\pi}{2},\pi\Big).$Case II:
When $\text{x}\in(\pi,2\pi)$ $\text{f}(\text{x})=\sin\text{x}-\sin\text{x}=0$ $\Rightarrow\text{f}'(\text{x})=0$ $\therefore$ f(x) is neither increasing nor decreasing on $(\pi,2\pi).$Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.