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  • Provide solution RD Sharma maths class 12 chapter Maxima and Minima exercise 17.2 question 11 maths textbook solution

Provide solution RD Sharma maths class 12 chapter Maxima and Minima exercise 17.2 question 11 maths textbook solution

Answers (1)

Answer

x= \frac{\pi}{3} and x= -\frac{\pi}{3}is the point of local maxima an local minima respectively. The value of the local

maxima and local minima is \sqrt{3} -\frac{\pi}{3} and  --\sqrt{3} +\frac{\pi}{3} respectively

Hint

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

Given

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

Solution:

f(x)=2 \sin x-x

Differentially f(x) w \cdot r, t^{\prime} x^{\prime} \text { then }

\begin{aligned} &\begin{aligned} \frac{d}{d x}\{f(x)\} &=\frac{d}{d x}(2 \operatorname{Sin} x-x)=\frac{d}{d x}(2 \operatorname{Sin} x)-\frac{d}{d x}(x) \\ \end{aligned} \end{aligned}          \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] \\

                      \begin{aligned} &\begin{aligned} \\ &=2 \frac{d}{d x} \sin x-\frac{d}{d x}(x) & \end{aligned} \end{aligned}                                                      \left[\because \frac{d}{d x} a f(x)=a \frac{d}{d x}\{f(x)\}\right]                              

     \therefore f^{\prime}(x)=-(1-2 \operatorname{Cos} x)

    \Rightarrow f^{\prime}(x)=-(1-2 \operatorname{Cos} x)                                                         \left[\because \frac{d}{d x} \operatorname{Sin} x=\operatorname{Cos} x, \frac{d}{d x}(x)=1\right]   

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

\begin{aligned} &f^{\prime}(x)=0 \quad \Rightarrow-(1-2 \operatorname{Cos} x)=0 \Rightarrow 1-2 \operatorname{Cos} x=0 \\ &\Rightarrow 2 \cos x-1=0 \Rightarrow 2 \cos x=1 \\ &\Rightarrow \cos x=\frac{1}{2} \quad \Rightarrow \cos x=\cos \frac{\pi}{3} \quad\left[\because \frac{1}{2}=\cos \frac{\pi}{3}\right] \\ &\Rightarrow x=2 n \pi \pm \frac{\pi}{3} \quad, n \in \mathbb{Z}[\because \operatorname{Cos} \theta=\operatorname{Cos} \alpha \Rightarrow \theta=2 n \pi \pm \propto, n \in \mathbb{Z}] \\ &x=\pm \frac{\pi}{3}, 2 \pi \pm \frac{\pi}{3},-2 \pi \pm \frac{\pi}{3} \\ &x=\frac{\pi}{3},-\frac{\pi}{3}\left[\text { Neglecting other values of } x \text { since }-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\right] \end{aligned}

                                                         -               +                      - 

                                   -∞                       -\frac{\pi}{3}                  \frac{\pi}{3}                           ∞

 

since {f}'\left ( x \right ) changes from +ve to -ve when x increases through \frac{\pi}{3}                    .

So, x=\frac{\pi}{3}    is the point of local maxima

The value of local maxima of f\left ( x \right ) at x=\frac{\pi}{3}  is

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

 

Again since {f}'\left ( x \right ) changes from -ve to +ve    when x increases through \frac{-\pi}{3} .

So, x=\frac{-\pi}{3}  is the point of local minima.

The value of local minima of f\left ( x \right ) at x=\frac{-\pi}{3}is

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

\begin{aligned} &=-2 \cdot \frac{\sqrt{3}}{2}+\frac{\pi}{3}\left[\because \sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\right] \\ &=-\sqrt{3}+\frac{\pi}{3} \end{aligned}

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