#### Provide solution RD Sharma maths class 12 chapter Maxima and Minima exercise 17.2 question 10 maths textbook solution

$x=\frac{\pi}{6}$ and $x=-\frac{\pi}{6}$is the point of  local maxima and  local minima respectively. The value of

local maxima and  local minima is  $\frac{\sqrt{3}}{2}-\frac{\pi}{6} \text { and } \frac{-\sqrt{3}}{2}+\frac{\pi}{6}$ respectively.

Hint:

Use first derivative test to find the point and value of local maxima or local minima.

Given:

$f(x)=\sin 2 x-x, \frac{-\pi}{2} \leq x \leq \frac{\pi}{2}$

Solution:

$f(x)=\sin 2 x-x$

Differentiating $f(x)$ with respect to ‘x’ then,

\begin{aligned} \frac{d}{d x}\left\{f^{\prime}(x)\right\} &=\frac{d}{d x}(\operatorname{Sin} 2 x-x)=\frac{d}{d x} \operatorname{Sin} 2 x-\frac{d}{d x}(x)\left[\because \frac{d}{d x}(h(x) \pm g(x))=\frac{d}{d x} h(x) \pm \frac{d}{d x} g(x)\right] \\ &=\operatorname{Cos} 2 x .2-1 \quad\left[\because \frac{d}{d x}(\operatorname{Sin} a x)=\operatorname{Cos} a x \cdot a\right] \\ f^{\prime}(x) &=2 \operatorname{Cos} 2 x-1 \quad\left[\because \frac{d}{d x}(x)=1\right] \\ \Rightarrow f^{\prime}(x) &=-(1-2 \operatorname{Cos} 2 x) \end{aligned}

By first derivative test, for local maxima and local minima ,we have

\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow-(1-2 \operatorname{Cos} 2 x)=0 \Rightarrow 1-2 \operatorname{Cos} 2 x=0 \\ &\Rightarrow 2 \operatorname{Cos} 2 x-1=0 \Rightarrow 2 \operatorname{Cos} 2 x=1 \\ &\Rightarrow \operatorname{Cos} 2 x=\frac{1}{2} \Rightarrow \operatorname{Cos} 2 x=\operatorname{Cos} \frac{\pi}{3} \\ &\Rightarrow 2 x=2 n \pi \pm \frac{\pi}{3} ; \mathrm{n} \in \mathbb{Z}[\cos \theta=\cos \alpha \Rightarrow \theta=2 n \pi \pm \alpha, n \in \mathbb{Z}] \\ \end{aligned}

$\Rightarrow \mathrm{x}=\pm \frac{\pi}{6} ; \mathrm{n} \in \mathbb{Z} \\$

$\Rightarrow x=\frac{\pi}{6}, \pi \pm \frac{\pi}{6} \cdots \\$

$\Rightarrow \mathrm{x}=\frac{\pi}{6},-\frac{\pi}{6}\left[\text { neglecting other values of } \mathrm{x} \text { since }-\frac{\pi}{2} \leq \mathrm{x} \leq \frac{\pi}{2}\right]$

-               +                      -

-∞                           $\frac{-\pi}{6}$                     $\frac{\pi}{6}$

since ${f}'\left ( x \right )$ changes from +ve to –ve  when $x$ increases through $\frac{\pi}{6}$                    .

So, $x=\frac{\pi}{6}$    is the point of local maxima

The value of local maxima of $f\left ( x \right )$ at $x=\frac{\pi}{6}$   is

\begin{aligned} &f\left(\frac{\pi}{6}\right)=\sin 2 \frac{\pi}{6}-\frac{\pi}{6} \\ &=\sin \frac{\pi}{3}-\frac{\pi}{6} \\ &=\frac{\sqrt{3}}{2}-\frac{\pi}{6} \end{aligned}

Again since${f}'\left ( x \right )$ changes from –ve  to +ve when $x$ increases through $\frac{-\pi}{6}$ .

So, $x=\frac{-\pi}{6}$  is the point of local minima

The value of local minima of $f\left ( x \right )$ at $x=\frac{-\pi}{6}$is

\begin{aligned} &f\left(-\frac{\pi}{6}\right)=\sin 2 \times\left(-\frac{\pi}{6}\right)-\left(-\frac{\pi}{6}\right) \\ &=-\sin \frac{\pi}{3}+\frac{\pi}{6} \quad[\because \sin (-\theta)=-\sin \theta] \\ &=-\frac{\sqrt{3}}{2}+\frac{\pi}{6} \quad\left[\because \sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\right] \end{aligned}