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  • Explain solution RD Sharma class 12 chapter Maxima and Minima exercise Multiple choice question, question 26 maths

Explain solution RD Sharma class 12 chapter Maxima and Minima exercise Multiple choice question, question 26 maths

Answers (1)

Answer: option(c)  \frac{1}{6}

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

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

Solution:

We have,

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

 f'(x)=\frac{4+x+x^2-x(1-2x)}{(4+x+x^2)^2}                                           

For maxima and minima f'(x)=0

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

\begin{aligned} &\Rightarrow 4+x+x^{2}-x-2 x^{2}=0 \\ &\Rightarrow 4-x^{2}=0 \\ &\Rightarrow x^{2}=4 \\ &\Rightarrow x=\pm 2 \notin[-1,1] \\ &f(1)=\frac{1}{4+1+1}=\frac{1}{6} \\ &f(-1)=\frac{-1}{4-1+1}=\frac{-1}{4} \end{aligned}

So, \frac{1}{6}  is the maximum value.

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