A) 2
B) \[\sqrt{2}\]
C) \[{{2}^{1/3}}\]
D) 1
Correct Answer: D
Solution :
[d] \[f(a)=\int\limits_{a}^{2a}{\left( \frac{x}{6}+\frac{1}{{{x}^{2}}} \right)dx=\left( \frac{{{x}^{2}}}{12}-\frac{1}{x} \right)_{a}^{2a}}\] \[=\left( \frac{4{{a}^{2}}}{12}-\frac{1}{2a}-\frac{{{a}^{2}}}{12}+\frac{1}{a} \right)=\frac{{{a}^{2}}}{4}+\frac{1}{2a}\] Let \[f'(a)=\frac{2a}{4}-\frac{1}{2{{a}^{2}}}=0\] \[\Rightarrow a=1\] which is a point of minima.
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