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  • Need Solution for RD Sharma Maths Class 12 Chapter Maxima and Minima Exercise 17.2 Question 7

Need Solution for RD Sharma Maths Class 12 Chapter Maxima and Minima Exercise 17.2 Question 7

Answers (1)

Answer:

x=\frac{\pi}{4} and x=\frac{3\pi}{4} is the point of  local maxima and  local minima respectively. The value of local maxima and  local minima is 1 and -1 respectively.

Hint:

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

Given:

f\left ( x \right )=\sin2x, 0< x< \pi

Solution:

f\left ( x \right )=\sin2x

Differentiating f\left ( x \right ) with respect to ‘x’ then,

\begin{aligned} \frac{d}{d x}\{f(x)\} &=\frac{d}{d x}(\sin 2 x) \\ &=\cos 2 x \frac{d}{d x}(2 x) \quad\left[\because \frac{d}{d x}(\sin a x)=\cos a x \frac{d}{d x}(a x)\right] \\ &=\cos 2 x .2 \frac{d}{d x}(x)\left[\because \frac{d}{d x}(a x)=a \frac{d}{d x}(x)\right] \\ &=\cos 2 x .2 .1 \quad\left[\because \frac{d}{d x}(x)=1\right] \\ f^{\prime}(x) &=2 \cos 2 x \end{aligned}

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

\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 2 \cos 2 x=0 \\ &\Rightarrow \cos 2 x=0 \quad[\because 2 \neq 0] \\ &\Rightarrow 2 x=(2 n+1) \frac{\pi}{2} \quad ;\left[[\cos \theta=0] \Rightarrow \theta=(2 n+1) \frac{\pi}{2}\right] \\ &\Rightarrow x=(2 n+1) \frac{\pi}{4} ; n \in N \end{aligned}

\Rightarrow x=\frac{\pi}{4}, \frac{3 \pi}{4} \quad[\text { neglecting other value of } x \text { since } 0<x<\pi]

                                                           +                -                       + 

                                   -∞                            \frac{\pi}{4}                     \frac{3\pi}{4}                      ∞

 

 since {f}'\left ( x \right ) changes from +ve to -ve when x increases through \frac{\pi}{4}                    .

So, x=\frac{\pi}{4}    is the point of local maxima

The value of local maxima of f\left ( x \right ) at x=\frac{\pi}{4}   is

f\left(\frac{\pi}{4}\right)=\sin \left(2 \frac{\pi}{4}\right)=\sin \frac{\pi}{2}=1\left[\because \sin \frac{\pi}{2}=1\right]

Again since {f}'\left ( x \right )  changes from -ve to +ve when x increases through \frac{3\pi}{4}.

So,x=\frac{3 \pi}{4}   is the point of local minima

The value of local minima off\left ( x \right )  at x=\frac{3\pi}{4}is

\begin{aligned} f\left(\frac{3 \pi}{4}\right) &=\sin \left(2 \frac{3 \pi}{4}\right)=\sin \frac{3 \pi}{2} \\ &=\sin \left(\pi+\frac{\pi}{2}\right)=-\sin \frac{\pi}{2} \quad[\because \sin (\pi+\theta)=-\sin \theta] \\ &=-1 \quad\left[\because \sin \frac{\pi}{2}=1\right] \end{aligned}

 

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