A) 7
B) \[-7\]
C) 5
D) 6
Correct Answer: A
Solution :
Here \[f(x)={{x}^{3}}-3(a-2){{x}^{2}}+3ax+7\] \[\therefore \]\[f'(x)=3{{x}^{2}}-6(a-2)x+3a\] Since\[f'(x)\ge 0\forall x\in (0,1]\] and\[f'(x)\le 0\forall x\in [1,5)\] \[\therefore \]\[f'(x)=0\]at \[x=1\] \[\Rightarrow \]\[3-6(a-2)+3a=0\Rightarrow -3a+15=0\] \[\Rightarrow \]\[a=5\] So,\[f(x)={{x}^{3}}-9{{x}^{2}}+15x+7\] Now,\[\frac{f(x)-14}{{{(x-1)}^{2}}}=0\Rightarrow \frac{{{x}^{3}}-9{{x}^{2}}+15x-7}{{{(x-1)}^{2}}}=0\] \[\Rightarrow \]\[\frac{{{(x-1)}^{2}}(x-7)}{{{(x-1)}^{2}}}=0\]\[\Rightarrow \]\[x=7\]You need to login to perform this action.
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