A) \[x=-2\]
B) \[x=0\]
C) \[x=1\]
D) \[x=2\]
Correct Answer: D
Solution :
Since, \[f(x)=\frac{x}{2}+\frac{2}{x}\] \[\therefore \] \[f'(x)=\frac{1}{2}-\frac{2}{{{x}^{2}}}\] For maxima or minima, put\[f'(x)=0\]. \[\frac{1}{2}-\frac{2}{{{x}^{2}}}=0\] \[\Rightarrow \]\[{{x}^{2}}=4\]\[\Rightarrow \]\[x=\pm 2\] Now, \[f''(x)=\frac{4}{{{x}^{3}}}\] \[\Rightarrow \]\[f''(2)=\frac{4}{8}=\frac{1}{2}>0\] and \[f''(-2)=-\frac{4}{8}=-\frac{1}{2}<0\] \[\therefore \]\[f(x)\]is minimum at\[x=2\]
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