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

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

Answers (1)

Answer:

 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<x\leq 2\pi

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<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)

Posted by

Gurleen Kaur

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