A) 2
B) \[-2\]
C) 0
D) None of these
Correct Answer: A
Solution :
Let \[y=x+\frac{1}{x}\] \[\frac{dy}{dx}=1-\frac{1}{{{x}^{2}}}\] For maxima and minima \[\frac{dy}{dx}=0\] \[\Rightarrow \] \[1-\frac{1}{{{x}^{2}}}=0\] \[\Rightarrow \] \[{{x}^{2}}=1\] \[\Rightarrow \] \[x=\pm 1\] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{2}{{{x}^{3}}}\] At \[x=1,\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{2}{1}>0\] At \[x=-1,\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{2}{1}<0\] \[\therefore \]y is minimum at\[x=1\] and minimum value \[=1+\frac{1}{1}\] \[=2\] Alternative Method \[\frac{x+\frac{1}{x}}{2}\ge \sqrt{x\frac{1}{x}}\] \[\Rightarrow \] \[x+\frac{1}{x}\ge 2\]
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