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# If(X) = 1 4 X 2 + 2 X + 1 Then Its Maximum Value is (A) 4 3 (B) 2 3 (C) 1 (D) 3 4 - CBSE (Science) Class 12 - Mathematics

#### 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

$\frac{4}{3}$

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

$\text{ Now, }$

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

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

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

$f'\left( x \right) = 0$

$\Rightarrow 8x + 2 = 0$

$\Rightarrow 8x = - 2$

$\Rightarrow x = \frac{- 1}{4}$

$\text { Now, }$

$f''\left( x \right) = 8$

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

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

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

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

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

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Solution If(X) = 1 4 X 2 + 2 X + 1 Then Its Maximum Value is (A) 4 3 (B) 2 3 (C) 1 (D) 3 4 Concept: Graph of Maxima and Minima.
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