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  • Please solve RD Sharma class 12 Chapter Maxima and Minima exercise Multiple choice question, question 4 maths textbook solution.

Please solve RD Sharma class 12 Chapter Maxima and Minima exercise Multiple choice question, question 4 maths textbook solution.

Answers (1)

Answer: (d)Maximum Value < Minimum Value      

Hint: For local maxima or minima, we must havef'(x) =0.

Given: f(x)=x+\frac{1}{x}

Solution:

We have,

f(x)=x+\frac{1}{x}

f'(x)=1-\frac{1}{x^2}

For maxima and minima f'(x)=0

\Rightarrow 1-\frac{1}{x^2}=0

\Rightarrow x^2-1=0

\Rightarrow x^2=1

\Rightarrow x=\pm 1

Now,

f''(x)=\frac{2}{x^3}

f''(1)=2>0

So, x =1 is a point of local minima.

Also, f''(-1)=-2<0

So, x = -1 is a point of local maxima.

\therefore Maximum Value < Minimum Value      

Posted by

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