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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2008

  • question_answer
        If the curve\[y=2{{x}^{3}}+a{{x}^{2}}+bx+c\]passes through the origin and the tangents drawn to it at\[x=-1\]and\[x=2\]are parallel to the\[x-\]axis, then the values of a, b and c are respectively

    A) \[12,-3\]and 0                   

    B) \[-3,-12\]and 0

    C)  \[-3,12\]and 0                  

    D) \[3,-12\]and 0

    Correct Answer: B

    Solution :

                    Given equation of curve is \[y=2{{x}^{3}}+a{{x}^{2}}+bx+c\]                    ...(i) Since, it is passes through (0, 0) \[\Rightarrow \]               \[0=2(0)+a(0)+b(0)+c\] \[\Rightarrow \]              \[c=0\]                                   ...(ii) On differentiating Eq. (i), we get \[\frac{dy}{dx}=6{{x}^{2}}+2ax+b\] Since, the tangents at\[x=-1\]and\[x=2\]are parallel to\[x-\]axis. \[\therefore \]  \[\frac{dy}{dx}=0\] At          \[x=-1\]                 \[6{{(-1)}^{2}}+2a(-1)+b=0\] \[\Rightarrow \]               \[6-2a+b=0\]                                      ?.(iii) At           \[x=2\]                 \[6{{(2)}^{2}}+2a(2)+b=0\] On solving Eqs. (iii) and (iv), we get \[a=-3,\text{ }b=-12\]and\[c=0\]


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