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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2007

  • question_answer
    If\[\underset{x\to 1}{\mathop{\lim }}\,\frac{a{{x}^{2}}+bx+c}{{{(x-1)}^{2}}}=2,\]then (a, b, c) is

    A)  \[(2,-4,\text{ }2)\]                         

    B)  (2, 4, 2)

    C)  \[(2,4,-2)\]        

    D)         \[(2,-4,-2)\]

    E)  \[(-2,4,2)\]

    Correct Answer: A

    Solution :

    Given that,\[\underset{x\to 1}{\mathop{\lim }}\,\,\frac{ax\text{ }+bx+c}{{{(x-1)}^{2}}}=2\] This limit will exist, if \[a{{x}^{2}}+bx+c=2{{(x-1)}^{2}}\] \[\Rightarrow \]               \[a{{x}^{2}}+bx+c=2{{x}^{2}}-4x+2\] \[\Rightarrow \]               \[a=2,b=-4,c=2\]


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