Need solution for RD Sharma maths class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 39 subquestion (ii)

$f(x)=\sin x+|\sin x| \text { is } \\ \text { an increasing in interval } \left(0, \frac{\pi}{2}\right), f(x) \text { is decreasing in } \left(\frac{\pi}{2}, \pi\right) \\ \text { and neither increasing nor decreasing in }(\pi, 2 \pi).$

Given:

$f(x)=sin x+ |sin x |,0

To prove:

We have to find the interval in which f(x) is an increasing or decreasing.

Hint:

1. for f(x) to be increasing we must have f'(x)>0
2. for f(x) to be decreasing we must have f'(x)<0

Solution:

Here we have

$f(x)=sin x+ |sin x |,0

We know that

$f(x)=\{2 \sin x, \quad \text { if } 02 \pi$

On differentiating f(x) w.r.t x we get

$f^{\prime}(x)=\{2 \cos x, \quad \text { if } 0

$\text { The function } 2cos\: x \text { will be positive in }(0,\frac{\pi }{2})$

$\text { Hence the function is increasing in the interval }(0,\frac{\pi }{2})$

$\text { The function 2cos x will be negative between }(\frac{\pi }{2},\pi )$

$\text { Hence the function } f(x) \text { is decreasing in the interval }(\frac{\pi }{2},\pi ). \text { the value of } f'(x)=0 \text { when } \pi \: \leq x\:< 2\pi .$

Therefore the function fx is neither increasing nor decreasing in the interval ($\pi$,2$\pi$)