A) b
B) \[\sqrt{a}\]
C) \[\sqrt{b}\]
D) \[\sqrt{b/a}\]
Correct Answer: D
Solution :
\[f(x)=ax+\frac{b}{x}\] Þ \[{f}'(x)=a-\frac{b}{{{x}^{2}}}\] Þ \[{f}'(x)=0\Rightarrow x=\sqrt{\frac{b}{a}}\] Now \[{f}''(x)=\frac{2b}{{{x}^{3}}}\] Þ At \[x=\sqrt{\frac{b}{a}},\] \[{f}''(x)=+ve\] \ \[f(x)\] has the least value at \[x=\sqrt{\frac{b}{a}}\].
You need to login to perform this action.
You will be redirected in
3 sec
