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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}


We have,



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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