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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2011

  • question_answer
    The area bounded by the x-axis, the curve\[y=f(x)\]and the lines\[x=1\]and\[x=b\]is equal to\[\sqrt{{{b}^{2}}+1}-\sqrt{2}\]for all\[b>1,\]then\[f(x)\]is

    A)  \[\sqrt{x-1}\]         

    B)  \[\sqrt{x+1}\]

    C)  \[\sqrt{{{x}^{2}}+1}\]       

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

    Correct Answer: D

    Solution :

     Area\[=\int_{x=1}^{b}{f(x)}dx=\sqrt{{{b}^{2}}+1}-\sqrt{2}\] \[=\sqrt{{{b}^{2}}+1}-\sqrt{1+1}\] \[=|\sqrt{{{x}^{2}}+1}|_{1}^{b}\] \[\Rightarrow \] \[f(x)=\frac{d}{dx}(\sqrt{{{x}^{2}}+1})\] \[=\frac{x}{\sqrt{{{x}^{2}}+1}}\]


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