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प्रश्न
The sum of two non-zero numbers is 8, the minimum value of the sum of the reciprocals is ______________ .
पर्याय
\[\frac{1}{4}\]
\[\frac{1}{2}\]
\[\frac{1}{8}\]
none of these
उत्तर
\[\frac{1}{2}\]
\[\text { Let the two non - zero numbers be x and y . Then,} \]
\[x + y = 8\]
\[ \Rightarrow y = 8 - x ............\left( 1 \right)\]
\[\text { Now,} \]
\[f\left( x \right) = \frac{1}{x} + \frac{1}{y}\]
\[ \Rightarrow f\left( x \right) = \frac{1}{x} + \frac{1}{8 - x} ..................\left[ \text { From eq. } \left( 1 \right) \right]\]
\[ \Rightarrow f'\left( x \right) = \frac{- 1}{x^2} + \frac{1}{\left( 8 - x \right)^2}\]
\[\text { For a local minima or a local maxima, we must have } \]
\[f'\left( x \right) = 0\]
\[ \Rightarrow \frac{- 1}{x^2} + \frac{1}{\left( 8 - x \right)^2} = 0\]
\[ \Rightarrow \frac{- \left( 8 - x \right)^2 + x^2}{\left( x \right)^2 \left( 8 - x \right)^2} = 0\]
\[ \Rightarrow - 64 - x^2 + 16x + x^2 = 0\]
\[ \Rightarrow 16x - 64 = 0\]
\[ \Rightarrow x = 4\]
\[f''\left( x \right) = \frac{2}{x^3} - \frac{2}{\left( 8 - x \right)^3}\]
\[ \Rightarrow f''\left( 4 \right) = \frac{2}{4^3} - \frac{2}{\left( 8 - 4 \right)^3}\]
\[ \Rightarrow f''\left( 4 \right) = \frac{2}{64} - \frac{2}{64} = 0\]
\[ \therefore \text { Minimum value }= \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\]
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