TS EAMCET · Maths · Application of Derivatives
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 & Solution
Correct Answer
(C) increasing in \(\left(0, \frac{\pi}{2}\right)\) and decreasing in \(\left(\frac{\pi}{2}, \pi\right)\)
Step-by-step Solution
Detailed explanation
Given, \[ f(x)=\frac{1}{2}[|\sin x|+\sin x], 0 0, \text { for } 0 < x < \frac{\pi}{2} \text { (increasing) } \\ \cos x & < 0, \text { for } \frac{\pi}{2} < x < \pi \text { (decreasing) } \end{aligned} \] Case II When \(\pi < x \leq 2 \pi\) \[ f(x)=\frac{1}{2}[-\sin x+\sin x]=0 \]
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