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JEE Main & Advanced AIEEE Solved Paper-2010

  • question_answer
    Let\[f:R\to R\]be a positive increasing function with \[\underset{x\to \infty }{\mathop{\lim }}\,\frac{f(3x)}{f(x)}=1\].Then \[\underset{x\to \infty }{\mathop{\lim }}\,\frac{f(2x)}{f(x)}=\]       AIEEE  Solved  Paper-2010

    A) 1             

    B)                        \[\frac{2}{3}\]                   

    C)        \[\frac{3}{2}\]                   

    D)        3

    Correct Answer: A

    Solution :

    Function\[(\uparrow )\] \[f(x)\le f(2x)\le f(3x)\] \[1\le \frac{f(2x)}{f(x)}\le \frac{f(3x)}{f(x)}\] given that \[\frac{f(3x)}{f(x)}=1\] hence \[1\le \frac{f(2x)}{f(x)}\le 1\] hence \[\underset{x\to \infty }{\mathop{\lim }}\,\frac{f(2x)}{f(x)}=1\](by sandwich theorem)


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