A) 2
B) 3
C) 4
D) 0
Correct Answer: A
Solution :
\[xy=1\Rightarrow y=\frac{1}{x}\]and let\[z=x+y\] \[z=x+\frac{1}{x}\Rightarrow \frac{dz}{dx}=1-\frac{1}{{{x}^{2}}}\] Now \[\frac{dz}{dx}=0\Rightarrow 1-\frac{1}{{{x}^{2}}}=0\] \[x=-1,\,+1\]and \[\frac{{{d}^{2}}z}{d{{x}^{2}}}=\frac{2}{{{x}^{3}}}\] \[{{\left( \frac{{{d}^{2}}z}{d{{x}^{2}}} \right)}_{x=1}}=\frac{2}{1}=2=+ive\] \[\therefore \]Hence \[x=1\]is point of minima and \[x=1\]and \[y=1\] \[\therefore \]Minimum value \[=x+y=2\].
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