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JEE Main & Advanced Mathematics Functions Question Bank Limits

  • question_answer
    \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\log (a+x)-\log a}{x}+k\underset{x\to e}{\mathop{\lim }}\,\frac{\log x-1}{x-e}=1,\]then                              [IIT Screening]

    A)                 \[k=e\left( 1-\frac{1}{a} \right)\]

    B)                 \[k=e(1+a)\]

    C)                 \[k=e(2-a)\]      

    D)                 The equality is not possible

    Correct Answer: A

    Solution :

                       Let \[f(x)=\log x\,\,\Rightarrow \,\,{f}'\,(x)=\frac{1}{x}\]            Therefore, given function\[={f}'(a)+k{f}'(e)=1\]            \[\Rightarrow \,\,\frac{1}{a}+\frac{k}{e}=1\,\,\Rightarrow \,\,k=e\,\left( \frac{a-1}{a} \right)\]                 Aliter : Apply L-Hospital?s rule to find both the limits.


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