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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2009

  • question_answer
        If\[f(x)=a\log |x|+b{{x}^{2}}+x\]has its extremum value at\[x=-1\]and\[x=2,\]then

    A)  \[a=2,b=-1\]                     

    B)  \[a=2,b=\frac{-1}{2}\]

    C)  \[a=-2,b=\frac{1}{2}\]  

    D)  None of these

    Correct Answer: B

    Solution :

                    We have, \[f(x)=a\log |x|+b{{x}^{2}}+x\] \[\therefore \]  \[f(x)=\frac{a}{x}+2bx+1\] Since,\[f(x)\]attains its extremum value at \[x=-1,2\]. \[\therefore \]  \[f(-1)\frac{-a}{1}-2b+1=0\] \[\Rightarrow \]               \[f(-1)=-a-2b+1=0\] and        \[f(2)=\frac{a}{2}+4b+1=0\] \[\Rightarrow \]               \[a=2,b=-\frac{1}{2}\]


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