A) \[a=-11\]
B) \[a=-6\]
C) \[a=6\]
D) \[a=11\]
Correct Answer: D
Solution :
\[f(x)={{x}^{3}}-6{{x}^{2}}+ax+b\] Þ \[{f}'(x)=3{{x}^{2}}-12x+a\] Þ \[{f}'(c)=0\] Þ \[{f}'\left( 2+\frac{1}{\sqrt{3}} \right)=0\] Þ \[3{{\left( 2+\frac{1}{\sqrt{3}} \right)}^{2}}-12\left( 2+\frac{1}{\sqrt{3}} \right)+a=0\] Þ \[3\left( 4+\frac{1}{3}+\frac{4}{\sqrt{3}} \right)-12\left( 2+\frac{1}{\sqrt{3}} \right)+a=0\] \[12+1+4\sqrt{3}-24-4\sqrt{3}+a=0\] Þ \[a=11\].You need to login to perform this action.
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