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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2009

  • question_answer
    If\[f(1)=2\]and\[f'(1)=1,\]then the value of\[\underset{x\to 1}{\mathop{\lim }}\,\frac{2x-f(x)}{x-1}\]is

    A)  \[-1\]            

    B)  0

    C)  \[1\]              

    D)  2

    Correct Answer: C

    Solution :

     \[\underset{x\to 1}{\mathop{\lim }}\,\frac{2x-f(x)}{x-1}\] \[=\underset{x\to 1}{\mathop{\lim }}\,\frac{2-f'(x)}{1}\]             [by L' Hospital rule] \[=2-f'(1)=2-(1)=1\]


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