#### Please solve RD Sharma class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 25 maths textbook solution

$f(x) \text { is increasing on } (-\frac{\pi }{3},\frac{\pi }{3})$

Given:

$f(x)=-\frac{x}{2}+sin\: x$

To prove:

$\text { We have to determine whether } f(x) \text { is increasing or decreasing on } (-\frac{\pi }{3},\frac{\pi }{3})$

Hint:

For f(x) to be increasing we must have f ’(x) > 0 and f ‘(x) < 0 for decreasing.

Solution:

We have

$f(x)=-\frac{x}{2}+sin\: x$

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

\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(-\frac{x}{2}+\sin x\right) \\ &\Rightarrow f^{\prime}(x)=-\frac{1}{2}+\cos x \\ &\text { Now, } x \in\left(-\frac{\pi}{3}, \frac{\pi}{3}\right) \\ &\Rightarrow-\frac{\pi}{3}

\begin{aligned} &\Rightarrow \cos \left(-\frac{\pi}{3}\right)<\cos x<\cos \frac{\pi}{3} \\ &\Rightarrow \frac{1}{2}<\cos x,\left[\therefore \cos \left(-\frac{\pi}{3}\right)=\cos \frac{\pi}{3}=\frac{1}{2}\right] \\ &\Rightarrow-\frac{1}{2}+\cos x>0 \\ &\Rightarrow f^{\prime}(x)>0 \end{aligned}

$\text { Hence } f(x) \text { is increasing function for x } \in \: (-\frac{\pi }{3},\frac{\pi }{3})$