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  • Explain solution RD Sharma class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 39 subquestion (iii) maths

Explain solution RD Sharma class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 39 subquestion (iii) maths

Answers (1)

Answer:

 f(x) \text{ is increasing in }(0,\frac{\pi }{3}) \\ \text {and} f(x) \text { is decreasing in } (\frac{\pi }{3},\frac{\pi }{2})

Given:

 f(x)=sin\: x (1+cos \: x ),0<x<\frac{\pi }{2}

To find:

 We have to find the intervals in which f(x) is 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:

We have,

f(x)=sin\: x (1+cos\, x ) 

Differentiating w.r.t. x we get,

\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}[\sin x(1+\cos x)] \\ &{\left[\therefore \frac{d}{d x} u v=v \frac{d u}{d x} \cdot u \frac{d v}{d x}\right]} \\ &\Rightarrow f^{\prime}(x)=\cos x-\sin x \cdot \sin x+\cos x \cdot \cos x \end{aligned}

\begin{aligned} &\Rightarrow f^{\prime}(x)=\cos x+\cos ^{2} x-\sin ^{2} x,\left[\therefore \sin ^{2} x=1-\cos ^{2} x\right] \\ &\Rightarrow f^{\prime}(x)=\cos x+\cos ^{2} x-1+\cos ^{2} x \\ &\Rightarrow f^{\prime}(x)=2 \cos ^{2} x+\cos x-1 \end{aligned}

\begin{aligned} &\Rightarrow f^{\prime}(x)=2 \cos ^{2} x+2 \cos x-\cos x-1 \\ &\Rightarrow f^{\prime}(x)=2 \cos x(\cos x+1)-1(\cos x+1) \\ &\Rightarrow f^{\prime}(x)=(2 \cos x-1)(\cos x+1) \end{aligned}

For f(x) to be increasing, we must have,

\begin{aligned} &f^{\prime}(x)>0 \\ &\Rightarrow f^{\prime}(x)=(2 \cos x-1)(\cos x+1)>0 \end{aligned}

This can only be possible when,

(2 \cos x-1)>0 \text { and }(\cos x+1)>0

\begin{aligned} &\Rightarrow 0<x<\frac{\pi}{3}\\ &\Rightarrow x \in\left(0, \frac{\pi}{3}\right)\\ &\text { So, } f(x) \text { is increasing in }\left(0, \frac{\pi}{3}\right) \end{aligned}

For f(x) to be decreasing we must have,\begin{aligned} &f^{\prime}(x)<0\\ &\Rightarrow f_{f}^{\prime}(x)=(2 \cos x-1)(\cos x+1)<0\\ &\text { This can only be possible when, }(2 \cos x-1)<0 \text { and }(\cos x+1)<0 \end{aligned}

\begin{aligned} &\Rightarrow \frac{\pi}{3}<x<\frac{\pi}{2}\\ &\Rightarrow x \in\left(\frac{\pi}{3}, \frac{\pi}{2}\right)\\ &\text { Hence } f(x) \text { is decreasing in }\left(\frac{\pi}{3}, \frac{\pi}{2}\right) \end{aligned}

Posted by

Gurleen Kaur

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