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

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

Answers (1)

Answer:

 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}

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