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JCECE Engineering JCECE Engineering Solved Paper-2012

  • question_answer
    The function \[y={{x}^{2}}+ax+b\] has a minimum at \[x=3\] and minimum value is\[5\]. Find\[a+b\].

    A) \[-6\]                                    

    B) \[14\]

    C) \[20\]                                   

    D) \[8\]

    Correct Answer: D

    Solution :

    We have,\[y={{x}^{2}}+ax+b\] \[\Rightarrow \]               \[\frac{dy}{dx}=2x+a\] Given,   \[\frac{dy}{dx}=0\]at\[x=3\] \[\therefore \]  \[2x+a=0\] \[\Rightarrow \]               \[a=-2x=-6\] Also, at\[x=3\]and\[y=5\] \[\therefore \]  \[9-6(3)+b=5\]                 \[b=5-9+18=14\] Hence,\[a+b=-6+14=-8\]


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