A) 1
B) 2
C) 3
D) 4
Correct Answer: B
Solution :
\[f(x)=x+\frac{1}{x}\Rightarrow f'(x)=1-\frac{1}{{{x}^{2}}}\] Þ \[f'(x)=0\Rightarrow {{x}^{2}}-1=0\Rightarrow x=1,-1\] But given that x is positive, hence at \[x=1,\,\,f(x)=1+\frac{1}{1}=2\].
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