Get Answers to all your Questions

header-bg qa

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

Answers (1)


 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.


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


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


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}

Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support