A) ? 1
B) 1
C) 0
D) a
Correct Answer: C
Solution :
\[f(x)=\sqrt{ax}+\frac{{{a}^{2}}}{\sqrt{ax}},\] then Þ \[{f}'(x)=\frac{\sqrt{a}}{2\sqrt{x}}+\frac{{{a}^{2}}}{\sqrt{a}}\left( \frac{-1}{2}{{x}^{-3/2}} \right)\] Þ \[{f}'(x)=\frac{\sqrt{a}}{2\sqrt{x}}-\frac{{{a}^{2}}}{2\sqrt{a}}{{x}^{-3/2}}\] Þ \[{f}'(a)=\frac{\sqrt{a}}{2\sqrt{a}}-\frac{{{a}^{2}}}{2\sqrt{a}\,.\,{{a}^{3/2}}}\]Þ\[{f}'(a)=\frac{1}{2}-\frac{{{a}^{2}}}{2{{a}^{2}}}=0\].
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