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  • Explain solution RD Sharma class 12 chapter Maxima and Minima exercise Multiple choice question, question 23 maths.

Explain solution RD Sharma class 12 chapter Maxima and Minima exercise Multiple choice question, question 23 maths

Answers (1)

Answer: option (b) 2

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

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

Solution:

We have,

xy =1

\Rightarrow y=\frac{1}{x}

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=1

\Rightarrow \frac{1}{y}=1

\Rightarrow y=1

Now,

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

f''(1)=\frac{2}{1^3}=2>0

So, x = 1 is a local minima.

Minimum Value of f(1)=1+\frac{1}{1}=1+1=2

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