A) 17
B) 177
C) 77
D) None of these
Correct Answer: B
Solution :
Let \[f(x)={{x}^{3}}-12{{x}^{2}}+36x+17\] \[\therefore f'(x)=3{{x}^{2}}-24x+36=0\] at \[x=2,\,6\] Again \[{f}''(x)=6x-24\] is \[-ve\]at \[x=2\] So that \[f(6)=17,\ \ f(2)=49\] At the end points \[=f(1)=42,\,\,f(10)=177\] So that \[f(x)\] has its maximum value as 177.You need to login to perform this action.
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