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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    If the function \[f(x)={{x}^{3}}-12a{{x}^{2}}+36{{a}^{2}}x\] \[-4(a>0)\]attains its maximum and minimum at\[x=p\]and\[x=g\]respectively and if\[3p={{q}^{2}},\]then a is equal to

    A)  \[\frac{1}{6}\]                                  

    B)  \[\frac{1}{36}\]

    C)  \[\frac{1}{3}\]                  

    D)         \[18\]

    E)  \[6\]

    Correct Answer: A

    Solution :

    Given,\[f(x)={{x}^{3}}-12a{{x}^{2}}+36{{a}^{2}}x-4\] On differentiating w.r.t.\[x,\]we get \[f(x)=3{{x}^{2}}-24ax+36{{a}^{2}}\] Put \[f(x)=0\Rightarrow 3(3{{x}^{2}}-8ax+12{{a}^{2}})=0\] \[\Rightarrow \]               \[(x-6a)(x-2a)=0\] \[\Rightarrow \]               \[x=2a,6a\] Now,     \[f(x)=6x-24a\] At \[x=2a,f(x)=12a-24a=-12a<0,\] maxima ie, \[p=2a\] At\[x=6a,f(x)=36a-24a=12a>0,\]minima ie, \[q=6a\] Also, given      \[3p={{q}^{2}}\] \[\Rightarrow \]                               \[3\times 2a={{(6a)}^{2}}\] \[\Rightarrow \]                               \[a=\frac{1}{6}\]


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