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Question
If(x) = x+\[\frac{1}{x}\],x > 0, then its greatest value is _______________ .
Options
-2
0
3
none of these
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Solution
none of these
\[\text { Given }: f\left( x \right) = x + \frac{1}{x}\]
\[ \Rightarrow f'\left( x \right) = 1 - \frac{1}{x^2}\]
\[\text { For a local maxima or a local minima, we must have } \]
\[f'\left( x \right) = 0\]
\[ \Rightarrow 1 - \frac{1}{x^2} = 0\]
\[ \Rightarrow x^2 - 1 = 0\]
\[ \Rightarrow x^2 = 1\]
\[ \Rightarrow x = \pm 1\]
\[ \Rightarrow x = 1 ................\left( \text { Given }: x>0 \right)\]
\[\text { Now,} \]
\[f''\left( x \right) = \frac{2}{x^3}\]
\[ \Rightarrow f''\left( 1 \right) = 2 > 0\]
\[\text { So, x = 1 is a local minima } .\]
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