A) 1
B) \[\frac{2}{3}\]
C) \[\frac{4}{3}\]
D) \[\frac{8}{3}\]
Correct Answer: B
Solution :
We have the function \[f(x)={{\cot }^{-1}}x+x\] On differentiating both sides w.r.t.\[x,\]we get \[f(x)=-\frac{1}{1+{{x}^{2}}}+1\] \[\Rightarrow \] \[f(x)=\frac{-1+1+{{x}^{2}}}{1+{{x}^{2}}}\] \[\Rightarrow \] \[f(x)=\frac{{{x}^{2}}}{1+{{x}^{2}}}\] Clearly,\[f(x)>0\]for all\[x\in R\]. Hence,\[f(x)\]increases in the interval\[(-\infty ,\infty )\].
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