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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 9

  • question_answer
    If \[f(x)={{\log }_{e}}(1+x)-\frac{{{\tan }^{-1}}x}{1+x},\]\[x>0,\] then the number of points where sgn \[\left( f(x) \right)\] is discontinuous is

    A) \[0\]

    B) \[1\]                     

    C) \[2\]                     

    D) None of these

    Correct Answer: A

    Solution :

    [a] \[f(x)={{\log }_{e}}(1+x)-\frac{{{\tan }^{-1}}x}{1+x}\] \[\Rightarrow \,\,f'(x)=\frac{1}{1+x}-\frac{\frac{1}{1+{{x}^{2}}}(1+x)-{{\tan }^{-1}}x}{{{(1+x)}^{2}}}\] \[=\frac{1+x-\frac{1+x}{1+{{x}^{2}}}+{{\tan }^{-1}}x}{{{(1+x)}^{2}}}\] \[=\frac{\frac{{{x}^{2}}(1+x)}{1+{{x}^{2}}}+{{\tan }^{-1}}x}{{{(1+x)}^{2}}}>0\,\,\forall \,\,x>0\] i.e., \[f(x)\] is an increasing function. \[f(x)>f(0)\Rightarrow f(x)>0\forall x>0\] \[\therefore sgn \,(f(x))=1\,\forall \,x\,\,>\,0\] Hence, there is no point of discontinuity.  


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