A) \[1\]
B) \[2\]
C) \[\frac{1}{2}\]
D) \[3\]
Correct Answer: B
Solution :
\[f(x)=2{{x}^{3}}-9a{{x}^{2}}+12{{a}^{2}}x+1\] \[\therefore \] \[f'(x)=6{{x}^{2}}-18ax+12{{a}^{2}}\] and \[f''(x)=12x-18a\] For \[\max /\min ,\,\,6{{x}^{2}}-18ax+12{{a}^{2}}=0\] \[\Rightarrow \] \[{{x}^{2}}-3ax+2{{a}^{2}}=0\] \[\Rightarrow \] \[(x-a)(x-2a)=0\] \[\Rightarrow \] \[x=a\]or\[x=2a\] Now,\[f''(a)=12a-18a=-6a<0\] and \[f''(2a)=24a-18a=6a>0\] \[\therefore \]\[f(x)\]maximum at x = a and minimum at \[x=2a\]. \[\Rightarrow \] \[p=a\] and \[q=2a\] Given that, \[{{p}^{2}}=q\] \[\Rightarrow \] \[{{a}^{2}}=2a\] \[\Rightarrow \] \[a(a-2)=0\] \[\therefore \] \[a=2\]You need to login to perform this action.
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