Question

If the function $ f(x)=2{{x}^{3}}-9a{{x}^{2}}+12{{a}^{2}}x+1 $ , where $ a>0 $ attains its maximum and minimum at $ p $ and $ q $ respectively, such that $ {{p}^{2}}=q $ , then a equals to

• A. $ 1 $
• B. $ 2 $
• C. $ \frac{1}{2} $
• D. $ 3 $

Answer 1

Answer: (B)

$ 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 $