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#### Please Solve R.D. Sharma class 12 Chapter 17 Maxima and Minima  Exercise 17.3 Question 1 Sub Question 10 Maths Textbook Solution.

Point of local minima value is a and it’s local minimum value is 2a. Also point of local maxima is -a and it’s local maximum value is  -2a

Hint:

First find critical values of f(x) by solving f'(x) =0 then find f''(x).

If f''(c) >0 then c is point of local minima.

If f’’(c)<0 then c is point of local maxima .

where c &c are critical points.

Put c and c in f(x) to get minimum value & maximum value.

Given:

$f(x)=x+\frac{a^{2}}{x}, a>0$

Explanation:

We have,

\begin{aligned} &f(x)=x+\frac{a^{2}}{x} \\ &f^{\prime}(x)=1-\frac{a^{2}}{x^{2}} \\ &f^{\prime \prime}(x)=\frac{2 a^{2}}{x^{3}} \end{aligned}

For maxima and minima ,f’(x)=0

\begin{aligned} 1-\frac{a^{2}}{x^{2}} &=0 \\ x &=\pm a \end{aligned}

At x=a,

$f^{\prime \prime}(a)=\frac{2}{a}>0, \text { as } a>0$

So, x=a is point of local minima

at x=-a,

$f^{\prime \prime}(-a)=-\frac{2}{a}<0$

∴x =-a is point of local maxima.

$\mathrm{f}(\mathrm{a})=x+\frac{a^{2}}{x}=a+\frac{a^{2}}{a}=2 a \text { is the minimum value. }$

$\mathrm{f}(-\mathrm{a})=x+\frac{a^{2}}{x}=-a+\frac{a^{2}}{-a}=-2 a \text { is the maximum value }$

Hence, local max. value =f(-a)=-2a & local min. value =f(a)=2a.