#### Need solution for RD Sharma maths class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 7

$f(x) \text { is increasing on } (0, \frac{\pi }{2}) \text { and decreasing in } (\frac{\pi }{2},\pi ) \\ \text { Hence, } f(x) \text { is neither increasing nor decreasing in } ( 0,\pi )$

Given:

$f(x)=sin\: x$

To show:

$\text { We have to show that } f(x) \text { is increasing on } (0,\frac{\pi }{2}) \text { and decreasing on } (\frac{\pi }{2},\pi ) \text { and neither increasing nor decreasing on } (0,\pi ).$

Hint:

$1) \text { We know that if } f^{\prime}(x)>0 \text { for all } x \in(a, b) \text { then } f(x) \text { is increasing. } \\ 2) \text { We know that if } f^{\prime(x)}<0 \text { for all } x \in(a, b) \text { then } f (x) \text { is decreasing on } (a, b).$

Solution:

Given

$f(x)=sin\: x$

On differentiating both sides w.r.t x we get

\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}(\sin x) \\ &\Rightarrow f^{\prime}(x)=\cos x \end{aligned}

Taking different region from 0 to 2π.

Let

\begin{aligned} x\in (0,\frac{\pi }{2}) \end{aligned}

\begin{aligned} &\Longrightarrow \cos x<0\\ &\Longrightarrow f^{\prime}(x)<0\\ &\text { Thus } \mathrm{f}(\mathrm{x}) \text { is decreasing }\left(\frac{\pi}{2}, \pi\right) \text { . } \end{aligned}

Therefor the above condition we find that

$\Rightarrow f(x) \text { is increasing on }\left(0, \frac{\pi}{2}\right) \text { and decreasing in }\left(\frac{\pi}{2}, \pi\right)$

Hence,f(x) is neither increasing nor decreasing in  (0, π)