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JCECE Engineering JCECE Engineering Solved Paper-2013

  • 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 to

    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,         \[{{p}^{2}}=q\]  \[\Rightarrow \]       \[{{a}^{2}}=2a\] \[\Rightarrow \]               \[a(a-2)=0\] \[\therefore \]  \[a=2\]


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