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

Answers (1)

Answer:

\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}

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