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Question
If(x) = \[\frac{1}{4x^2 + 2x + 1}\] then its maximum value is _________________ .
Options
\[\frac{4}{3}\]
\[\frac{2}{3}\]
1
\[\frac{3}{4}\]
Solution
\[\frac{4}{3}\]
\[\text { Maximum value of }\frac{1}{4 x^2 + 2x + 1}= \text { Minimum value of }4 x^2 + 2x + 1 \]
\[\text{ Now, }\]
\[f\left( x \right) = 4 x^2 + 2x + 1\]
\[ \Rightarrow f'\left( x \right) = 8x + 2\]
\[\text { For a local maxima or a local minima, we must have } \]
\[f'\left( x \right) = 0\]
\[ \Rightarrow 8x + 2 = 0\]
\[ \Rightarrow 8x = - 2\]
\[ \Rightarrow x = \frac{- 1}{4}\]
\[\text { Now, } \]
\[f''\left( x \right) = 8\]
\[ \Rightarrow f''\left( 1 \right) = 8 > 0\]
\[\text { So, x } = \frac{- 1}{4} \text { is a local minima } . \]
\[\text { Thus },\frac{1}{4 x^2 + 2x + 1}\text { is maximum at x } = \frac{- 1}{4} . \]
\[ \Rightarrow \text { Maximum value of } \frac{1}{4 x^2 + 2x + 1} = \frac{1}{4 \left( \frac{- 1}{4} \right)^2 + 2\left( \frac{- 1}{4} \right) + 1}\]
\[ = \frac{1}{\frac{4}{16} - \frac{1}{2} + 1} = \frac{16}{12} = \frac{4}{3}\]
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