Question

If $f\left(x\right)=2x+co{t}^{-1}\left(x\right)+\log [\sqrt{1+x^2}-x]$, 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 $[-\infty ,\infty ]$
• E. A and D are correct.

Answer 1

The correct option is E

A and D are correct.

Explanation for the correct option.

$f\left(x\right)=2x+co{t}^{-1}\left(x\right)+\log [\sqrt{1+x^2}-x]$ so,

$\begin{array}{rcl}f\text{'}\left(x\right)& =& 2-\frac{1}{1+x^2}+\left[\frac{1}{\sqrt{1+x^2}-x}\times \left(\frac{2x}{2\sqrt{1+x^2}}-1\right)\right]\\ & =& 2-\frac{1}{1+x^2}+\left[\frac{1}{\sqrt{1+x^2}-x}\times \left(\frac{x-\sqrt{1+x^2}}{\sqrt{1+x^2}}\right)\right]\\ & =& 2-\frac{1}{1+x^2}-\frac{1}{\sqrt{1+x^2}}\\ & =& \left(1-\frac{1}{1+x^2}\right)+\left(1-\frac{1}{\sqrt{1+x^2}}\right)\\ & & \end{array}$

Now, for all real numbers $x^2>0$, $1+x^2>1$ and $\sqrt{1+x^2}>1$

$\Rightarrow \begin{array}{rcl}& & \frac{1}{1+x^2}\end{array}\begin{array}{rcl}& \in & \end{array}\begin{array}{rcl}(0,1]\mathrm{and}1-\frac{1}{1+x^2}& \in & [0,1)\end{array}$ and $\begin{array}{l}\frac{1}{\sqrt{1+x^2}}\end{array}\begin{array}{rcl}& \in & \end{array}\begin{array}{rcl}(0,1]\Rightarrow 1-\frac{1}{\sqrt{1+x^2}}& \in & [0,1)\end{array}$

Therefore, $f\text{'}\left(x\right)>0\forall x\in \mathrm{ℝ}$, so it is increasing for all real numbers.

Hence option E is correct