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  • Need solution for RD Sharma maths class 12 chapter 17 Maxima and Minima exercise 17.3 question 6

 Need solution for RD Sharma maths class 12 chapter  17 Maxima and Minima  exercise 17.3 question 6

Answers (1)

Answer:

x=\frac{\pi}{4}  is a point of local minima and its local minimum value will be f\left(\frac{\pi}{4}\right)=1-\frac{\pi}{2}

x=\frac{3\pi}{4} is a point local maxima and its local maximum value will be f\left(\frac{3\pi}{4}\right)=-1-\frac{3\pi}{2}

Hint:

\text { Put } f^{\prime} x=0 

Given:

f(x)=\tan x-2 x

Explanation:

Differentiating f(x) with respect to x

\begin{aligned} &f(x)=\sec ^{2} x-2\\ &\text { Put } \mathrm{f}^{\prime}(\mathrm{x})=0\\ &\sec ^{2} x-2=0\\ &\Rightarrow \sec ^{2} \mathrm{x}=2\\ &\Rightarrow \sec \mathrm{x}=\pm \sqrt{2}\\ &=\sec x=\sqrt{2}\\ &\Rightarrow \mathrm{x}=\frac{\pi}{4}\\ &\text { or, } \sec x=-\sqrt{2}\\ &\text { or, } \sec \mathrm{x}=\pi-\frac{\pi}{4}\\ &=\frac{3 \pi}{4} \end{aligned}

Thus x=\frac{\pi}{4}  or x=\frac{3\pi}{4} is possible points of local maxima and minima

Differentiating f’(x)  with respect to x

\begin{aligned} &\mathrm{f}^{\mathrm{u}}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left[\sec ^{2} \mathrm{x}-2\right] \\ &=2 \sec \mathrm{x}(\sec \mathrm{xtan} \mathrm{x}) \\ &=2 \sec ^{2} \mathrm{x} \tan \mathrm{x} \\ &\text { Put } \mathrm{x}=\frac{\pi}{4} \text { in } \mathrm{f}^{\prime \prime}(\mathrm{x}) \\ &\begin{aligned} \mathrm{f}=\left(\frac{\pi}{4}\right) &=2 \sec ^{2}\left(\frac{\pi}{4}\right) \tan \left(\frac{\pi}{4}\right) \\ &=2(2)(1) \\ &=4>0 \end{aligned} \end{aligned}

So ,

\mathrm{x}=\frac{\pi}{4} is a point of Local Minima and its local maximum value will be \mathrm{f}\left(\frac{\pi}{4}\right)=1-\frac{\pi}{2}

\begin{aligned} &\text { And Put, }\\ &\mathrm{x}=\frac{3 \pi}{4} \text { in } \mathrm{f}^{\prime \prime}(\mathrm{x})\\ &\mathrm{f}^{\prime \prime}\left(\frac{3 \pi}{4}\right)=2 \sec ^{2}\left(\frac{3 \pi}{4}\right) \tan \left(\frac{3 \pi}{4}\right)\\ &=2(-\sqrt{2})^{2} \quad(-1)\\ &=-4<0 \end{aligned}

So x=\frac{3 \pi}{4}  is a point local maxima and its local maximum value will be f\left(\frac{3 \pi}{4}\right)=-1-\frac{3 \pi}{2}

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