A) ? 2
B) 0
C) 3
D) None of these
Correct Answer: D
Solution :
\[f(x)=x+\frac{1}{x}\] Þ \[{f}'(x)=1-\frac{1}{{{x}^{2}}}\] Put \[{f}'(x)=0\], \ \[x=-1,\,1\] Since x > 0, so no maximum value can be found.
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