Question

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

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

Answer 1

Answer: (D)

increases in $(-\infty ,\infty )$

We have $f\left(x\right)=2x+{\cot }^{-1}x+\log (\sqrt{1+x^2}-x)$

$\therefore f^{\prime }\left(x\right)=2-\frac{1}{1+x^2}+\frac{1}{\sqrt{1+x^2}-x}(\frac{x}{\sqrt{1+x^2}}-1)$

$=\frac{2+2x^2-1}{1+x^2}+\frac{1}{\sqrt{1+x^2}-x}\left(\frac{x-\sqrt{1+x^2}}{\sqrt{1+x^2}}\right)$

$=\frac{1+2x^2}{1+x^2}-\frac{1}{\sqrt{1+x^2}}=\frac{1+2x^2}{1+x^2}-\frac{\sqrt{(1+x^2)}}{1+x^2}$

$=\frac{x^2+\sqrt{1+x^2}(\sqrt{1+x^2}-1)}{1+x^2}>0$ for all $x$

Hence, $f\left(x\right)$ is an increasing function on $(-\infty ,\infty )$.