A) 3
B) 1
C) 2
D) 1/2
Correct Answer: C
Solution :
Since, \[f(x)=2{{x}^{3}}-9a{{x}^{2}}+12{{a}^{2}}x+1\] \[f'(x)=6{{x}^{2}}-18ax+12{{a}^{2}}\] For maxima and minima put \[f'(x)=0\] \[\Rightarrow \] \[6\,({{x}^{2}}-3ax+2{{a}^{2}})=0\] \[\Rightarrow \] \[{{x}^{2}}-3ax+2{{a}^{2}}=0\] \[\Rightarrow \] \[{{x}^{2}}-2ax-ax+2{{a}^{2}}=0\] \[\Rightarrow \] \[x(x-2a)\,-a(x-2a)=0\] \[\Rightarrow \] \[(x-a)\,(x-2a)=0\] \[\Rightarrow \] \[x=a\], \[x=2a\] Now. \[f''(x)=12x-18a\] At \[x=a\], \[f''(x)=12a-18a=6a\] \[\therefore \,f(x)\] will be maximum at \[x=a\] i.e., p = a At \[x=2a\], \[f''(x)=24a-18a=6a\] \[\therefore \,\,f(x)\] will be minimum at \[x=2a\] i.e., \[q=2a\] Given, \[{{p}^{2}}=q\] \[\Rightarrow \] \[{{a}^{2}}=2a\] \[\Rightarrow \] \[a=2\]You need to login to perform this action.
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