A) \[x=2\]
B) \[x=4\]
C) \[~x=0\]
D) \[x=3\]
Correct Answer: A
Solution :
\[f(x)=2{{x}^{3}}-15{{x}^{2}}+36x+4\]put \[f'(x)=0\] \[f'(x)=6{{x}^{2}}-30x+36\]\[6{{x}^{2}}-30x+36=0\] \[f''=12x-30\] \[{{x}^{2}}-5x+6=0\] put the value of \[x\]in \[f''(x)\] \[{{x}^{2}}-3x-2x+6=0\] \[=12\times 2-30\] \[=-6\] \[x(x-3)-2(x-3)=0\] \[(x-2)(x-3)=0\] \[=12\times 03-30=+\,6\]is not maximumYou need to login to perform this action.
You will be redirected in
3 sec