A) \[x=2\]
B) \[x=4\]
C) \[x=0\]
D) \[x=3\]
Correct Answer: A
Solution :
\[f(x)=2{{x}^{3}}-15{{x}^{2}}+36x+4\]Þ \[{f}'(x)=6{{x}^{2}}-30x+36\] ?..(i) We know that for its maximum value \[{f}'(x)=0.\] \[6{{x}^{2}}-30x+36=0\] Þ \[(x-2)(x-3)=0\] Þ \[x=2,\,3.\] Again differentiating equation (i), we get \[{f}''(x)=12x-30\] Þ \[{f}''(2)=24-30=-6<0\]. Therefore \[f(x)\] is maximum at \[x=2.\]You need to login to perform this action.
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