Question

If $f\left(x\right)=2x+{\cot }^{-1}x+\log (\sqrt{1+x^2}-x)$, then $f\left(x\right)$

• A. Increases in $[0,\infty )$
• B. Decreases in $[0,\infty )$
• C. Neither increases nor decreases in $[0,\infty )$
• D. Increases in $(-\infty ,\infty )$

Answer 1

Answer: (A), (D)

Increases in $[0,\infty )$
Increases in $(-\infty ,\infty )$

$f^{\prime }\left(x\right)=2+\left(\frac{-1}{1+x^2}\right)+\frac{\frac{2x}{2\sqrt{1+x^2}}-1}{\sqrt{1+x^2}-x}=2-\frac{1}{1+x^2}+\frac{x-\sqrt{1+x^2}}{\sqrt{1+x^2}(\sqrt{1+x^2}-x)}=2-\frac{1}{1+x^2}-\frac{1}{\sqrt{1+x^2}}{x}^2\in [0,\infty );1+x^2\in [1,\infty )\Rightarrow \frac{1}{1+x^2}\in (0,1]\Rightarrow 1-\frac{1}{1+x^2}\in [0,1)\Rightarrow f^{\prime }\left(x\right)=(1-\frac{1}{1+x^2})+(1-\frac{1}{\sqrt{1+x^2}})\Rightarrow f^{\prime }\left(x\right)>0,\forall x\in R$

Hence $f\left(x\right)$ increases $\forall x\in R$