A) \[\frac{1}{6}\]
B) \[\frac{1}{36}\]
C) \[\frac{1}{3}\]
D) \[18\]
E) \[6\]
Correct Answer: A
Solution :
Given,\[f(x)={{x}^{3}}-12a{{x}^{2}}+36{{a}^{2}}x-4\] On differentiating w.r.t.\[x,\]we get \[f(x)=3{{x}^{2}}-24ax+36{{a}^{2}}\] Put \[f(x)=0\Rightarrow 3(3{{x}^{2}}-8ax+12{{a}^{2}})=0\] \[\Rightarrow \] \[(x-6a)(x-2a)=0\] \[\Rightarrow \] \[x=2a,6a\] Now, \[f(x)=6x-24a\] At \[x=2a,f(x)=12a-24a=-12a<0,\] maxima ie, \[p=2a\] At\[x=6a,f(x)=36a-24a=12a>0,\]minima ie, \[q=6a\] Also, given \[3p={{q}^{2}}\] \[\Rightarrow \] \[3\times 2a={{(6a)}^{2}}\] \[\Rightarrow \] \[a=\frac{1}{6}\]You need to login to perform this action.
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