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#### Please Solve R.D.Sharma class 12 Chapter 17 Maxima and Minima  Exercise 17.3 Question 6 Maths Textbook Solution.

$\mathrm{x}=\frac{\pi}{4}$ 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:

Put,$\mathrm{f}^{\prime} \mathrm{x}=0$

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 } f^{\prime}(x)=0 \\ &\sec ^{2} x-2=0 \\ &\Rightarrow \sec ^{2} x=2 \\ &\Rightarrow \sec x=\pm \sqrt{2} \\ &=\sec x=\sqrt{2} \\ &\Rightarrow x=\frac{\pi}{4} \end{aligned}

\begin{aligned} &\text { or, } \sec x=-\sqrt{2} \\ &\text { or, } \sec x=\pi-\frac{\pi}{4} \end{aligned}

$=\frac{3 \pi}{4}$

Thus $\mathrm{x}=\frac{\pi}{4}$ 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 \times \operatorname{san} \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}) \end{aligned}

\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}

So ,

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

And Put,

\begin{aligned} &x=\frac{3 \pi}{4} \text { in } f^{\prime \prime}(x) \\ &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}(-1) \\ &=-4<0 \end{aligned}

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