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.You need to login to perform this action.
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