A) \[\sqrt{ab}\]
B) \[2\sqrt{\frac{a}{b}}\]
C) \[2\sqrt{\frac{b}{a}}\]
D) \[2\sqrt{ab}\]
Correct Answer: D
Solution :
[d]: Given, \[f(x)=ax+\frac{b}{x}\Rightarrow f'(x)=a-\frac{b}{{{x}^{2}}}\] For minima, \[f'(x)=0\Rightarrow a=\frac{b}{{{x}^{2}}}\Rightarrow x=\sqrt{\frac{b}{a}}\] \[f''(x)=\frac{2b}{{{x}^{3}}}\Rightarrow f''\left( \sqrt{\frac{b}{a}} \right)=\frac{2b}{{{\left( \frac{b}{a} \right)}^{3/2}}}>0\] Hence f(x) is minimum at \[x=\sqrt{\frac{b}{a}}\]and its minimum value is \[a\times \sqrt{\frac{b}{a}}+\frac{b}{\sqrt{\frac{b}{a}}}=\sqrt{ab}+\sqrt{ab}=2\sqrt{ab}\]
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