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

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

Given:

$f(x)=sin\: x (1+cos \: x ),0

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

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}