2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced AIEEE Solved Paper-2003

  • question_answer
    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     AIEEE  Solved  Paper-2003

    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.
You will be redirected in 3 sec spinner