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

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 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).


 f(x)=sin x+ |sin x |,0<x\leq 2\pi

To prove:

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


  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


Here we have

f(x)=sin x+ |sin x |,0<x\leq 2\pi

We know that

f(x)=\{2 \sin x, \quad \text { if } 0<x \leq \pi \& 0, \text { if } \pi<x>2 \pi

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

f^{\prime}(x)=\{2 \cos x, \quad \text { if } 0<x \leq \pi \& 0, \quad \text { if } \pi<x<2 \pi

\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)

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Gurleen Kaur

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