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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 - Mathematics

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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

  • \[\frac{3}{4}\]

MCQ
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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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Chapter 18: Maxima and Minima - Exercise 18.7 [Page 82]

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RD Sharma Mathematics [English] Class 12
Chapter 18 Maxima and Minima
Exercise 18.7 | Q 22 | Page 82

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