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11th Class Mathematics Sample Paper Mathematics Olympiad - Sample Paper-8

  • question_answer
    If \[\mathbf{f}\left( \mathbf{2} \right)=\mathbf{2},\mathbf{f}'\left( \mathbf{2} \right)=1\]then \[\underset{x\to 2}{\mathop{Lt}}\,\frac{2{{x}^{2}}-4f(x)}{x-2}\]is equal to

    A)  4                                

    B)  \[-2\]             

    C) \[-4\]                            

    D)  2

    Correct Answer: A

    Solution :

    [a] \[\because f\left( 2 \right)=2\] \[f'\left( 2 \right)=1\] \[\because \underset{x\to 2}{\mathop{lim}}\,\frac{2{{x}^{2}}-4.f(x)}{x-2}\]Putting\[x=2\], then which is\[\frac{0}{0}\]form. \[=\underset{x\to 2}{\mathop{lim}}\,\frac{4x-4.f(x)}{1}\] [By L' Hospital Rule] \[=\frac{4\times 2-4\times 1}{1}=4\]. Hence, option [a] is correct.


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