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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 34

  • question_answer
    \[\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{h}\int\limits_{x}^{x+h}{\frac{dz}{z+\sqrt{{{z}^{2}}+1}}}\] is equal to

    A) 0                     

    B)        \[\frac{1}{x+\sqrt{{{x}^{2}}+1}}\]

    C) \[\frac{1}{\sqrt{{{x}^{2}}+1}}\]             

    D)        None of these

    Correct Answer: B

    Solution :

    [b] \[\underset{h\to 0}{\mathop{\lim }}\,\frac{\int\limits_{x}^{x+h}{\frac{dx}{z+\sqrt{{{z}^{2}}+1}}}}{h},\frac{0}{0}\] Use L' Hospital Rule \[\underset{h\to 0}{\mathop{\lim }}\,\frac{\frac{1}{x+h\sqrt{{{(x+h)}^{2}}+1}}.(0+1)-0}{1}\] \[=\frac{1}{x+\sqrt{{{x}^{2}}+1}}\]


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