#### Question

If(x) = \[\frac{1}{4x^2 + 2x + 1}\] then its maximum value is _________________ .

\[\frac{4}{3}\]

\[\frac{2}{3}\]

1

\[\frac{3}{4}\]

#### Solution

\[\frac{4}{3}\]

\[\text { Maximum value of }\frac{1}{4 x^2 + 2x + 1}= \text { Minimum value of }4 x^2 + 2x + 1 \]

\[\text{ Now, }\]

\[f\left( x \right) = 4 x^2 + 2x + 1\]

\[ \Rightarrow f'\left( x \right) = 8x + 2\]

\[\text { For a local maxima or a local minima, we must have } \]

\[f'\left( x \right) = 0\]

\[ \Rightarrow 8x + 2 = 0\]

\[ \Rightarrow 8x = - 2\]

\[ \Rightarrow x = \frac{- 1}{4}\]

\[\text { Now, } \]

\[f''\left( x \right) = 8\]

\[ \Rightarrow f''\left( 1 \right) = 8 > 0\]

\[\text { So, x } = \frac{- 1}{4} \text { is a local minima } . \]

\[\text { Thus },\frac{1}{4 x^2 + 2x + 1}\text { is maximum at x } = \frac{- 1}{4} . \]

\[ \Rightarrow \text { Maximum value of } \frac{1}{4 x^2 + 2x + 1} = \frac{1}{4 \left( \frac{- 1}{4} \right)^2 + 2\left( \frac{- 1}{4} \right) + 1}\]

\[ = \frac{1}{\frac{4}{16} - \frac{1}{2} + 1} = \frac{16}{12} = \frac{4}{3}\]