#### Please solve RD Sharma class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 13 maths textbook solution

f(x) is decreasing on (0,$\pi$)  and increasing on (-$\pi$,0) and neither increasing nor decreasing on (-$\pi$,$\pi$).

Given:

$f(x)=cos\: x$

To show:

We have to show that f(x) is decreasing on (0,$\pi$)  and increasing on (-$\pi$,0) and neither increasing nor decreasing on (-$\pi$,$\pi$).

Hint:

1. For increasing function f'(x)>0 then f(x) is increasing.
2. For f(x) to be decreasing f'(x)<0

Solution:

Given

$f(x)=cos\: x$

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

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

Taking different region from -$\pi$ to $\pi$.

\begin{aligned} &\text { Let } x \in(0, \pi) \\ &\Rightarrow \sin x>0 \\ &\Rightarrow-\sin x<0 \end{aligned}

By applying negative sign change comparison sign.

\begin{aligned} &\Rightarrow f^{\prime}(x)<0\\ \end{aligned}

Thus f(x) is decreasing (0,$\pi$)

\begin{aligned} &\text { Let } x \in(-\pi, 0)\\ &\Longrightarrow \sin x<0\\ &\Rightarrow-\sin x>0 \end{aligned}

By applying negative sign change comparison sign.

$\Rightarrow f'(x)> 0$

Thus f(x) is increasing (-$\pi$,0).

Therefore from above condition we find that

?f(x) is decreasing in (0, $\pi$) and increasing in (-$\pi$,0)

Hence, f(x) is neither increasing nor decreasing in  (-$\pi$, $\pi$)