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JEE Main & Advanced Mathematics Functions Question Bank Limits

  • question_answer
    The value of the constant \[\alpha \] and \[\beta \] such that \[\underset{x\to \infty }{\mathop{\lim }}\,\left( \frac{{{x}^{2}}+1}{x+1}-\alpha x-\beta  \right)=0\] are respectively [Orissa JEE 2005]

    A)                 (1, 1)

    B)                 (?1, 1)

    C)                 (1, ?1)

    D)                 (0, 1)

    Correct Answer: C

    Solution :

                       \[\underset{x\to \infty }{\mathop{\lim }}\,\left( \frac{{{x}^{2}}+1}{x+1}-2x-\beta  \right)=0\]                    Þ \[\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{x}^{2}}(1-\alpha )-x(\alpha +\beta )+1-b}{x+1}=0\]                    Since the limit of the given expression is zero, therefore degree of the polynomial in numerator must be less than denominator. \ \[1-\alpha =0\] and \[\alpha +\beta =0\] Þ \[\alpha =1\] and \[\beta =-1\].


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