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  • Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise Multiple choice question, question 14.

Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise Multiple choice question, question 14.

Answers (1)

Answer: \frac{1}{4}

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

Given: f(x)=\frac{x}{4-x+x^2}

Solution:

We have,

\begin{aligned} &f(x)=\frac{x}{4-x+x^{2}} \\ &f^{\prime}(x)=\frac{4-x+x^{2}-x(-1+2 x)}{\left(4-x+x^{2}\right)^{2}} \end{aligned}                             

For maxima and minima f'(x)=0

\Rightarrow \frac{4-x+x^{2}-x(-1+2 x)}{\left(4-x+x^{2}\right)^{2}}

\Rightarrow 4-x+x^2-x(-1+2x)=0

\Rightarrow 4-x^2=0

\Rightarrow x^2=4

\Rightarrow x=\pm 2\euro [-1,1]

f(-1)=\frac{-1}{4+1+1}=\frac{-1}{6}

f(1)=\frac{1}{4-1+1}=\frac{1}{4}

Hence, the maximum value is  \frac{1}{4}.

Note: option is not matching with the answer given in the book.

Posted by

infoexpert24

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