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JEE Main & Advanced Mathematics Functions Question Bank Limits

  • question_answer
    Given that\[f'\](2)=6 and \[{f}'(1)=4)=\], then \[\underset{h\to 0}{\mathop{\lim }}\,\frac{f(2h+2+{{h}^{2}})-f(2)}{f(h-{{h}^{2}}+1)-f(1)}=\] [IIT Screening 2003]

    A)                 Does not exist

    B)                 Is equal to ? 3/2

    C)                 Is equal to 3/2

    D)                 Is equal to 3

    Correct Answer: D

    Solution :

                       \[\underset{h\to 0}{\mathop{\lim }}\,\frac{f(2h+2+{{h}^{2}})-f(2)}{f(h-{{h}^{2}}+1)-f(1)}=\underset{h\to 0}{\mathop{\lim }}\,\frac{{f}'(2h+2+{{h}^{2}})(2+2h)}{{f}'(h-{{h}^{2}}+1)(1-2h)}\]                                                                           \[=\frac{6\times 2}{4\times 1}=3\].


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