A) \[-122~\]
B) \[-222\]
C) 222
D) 122
Correct Answer: D
Solution :
Here, \[{{x}^{2}}-11x+30\le 0\] \[\Rightarrow \]\[{{x}^{2}}-5x-6x+30\le 0\] \[\Rightarrow \]\[(x-5)(x-6)\le 0\] \[\Rightarrow \]\[5\le x\le 6\] \[\therefore \]\[S=\{x\in R,5\le x\le 6\}\] Now,\[f(x)=3{{x}^{3}}-18{{x}^{2}}+27x-40\] \[\therefore \]\[f'(x)=9{{x}^{2}}-36x+27=9(x-1)(x-3),\]which is positive in [5, 6]. So, f(x) is increasing in [5,6] Hence, maximum value of f(x) = f(6) =122You need to login to perform this action.
You will be redirected in
3 sec