Question

f(x) = 2x − tan⁻¹ x − log $\left\{x+\sqrt{x^2+1}\right\}$ is monotonically increasing when
(a) x > 0
(b) x < 0
(c) x ∈ R
(d) x ∈ R − {0}

Answer 1

(c) x ∈ R

$$\begin{gathered} \mathrm{Given}:f\left(x\right)=2x-{\tan }^{-1}x-\log \left(x+\sqrt{x^2+1}\right) \\ f\text{'}\left(x\right)=2-\frac{1}{1+x^2}-\frac{1}{x+\sqrt{x^2+1}}\left(1+\frac{1}{2\sqrt{x^2+1}}.2x\right) \\ =2-\frac{1}{1+x^2}-\frac{1}{x+\sqrt{x^2+1}}\left(1+\frac{x}{\sqrt{x^2+1}}\right) \\ =2-\frac{1}{1+x^2}-\frac{1}{x+\sqrt{x^2+1}}\left(\frac{x+\sqrt{x^2+1}}{\sqrt{x^2+1}}\right) \\ =2-\frac{1}{1+x^2}-\frac{1}{\sqrt{x^2+1}} \\ =\frac{2+2x^2-1-\sqrt{x^2+1}}{1+x^2} \\ =\frac{1+2x^2-\sqrt{x^2+1}}{1+x^2} \\ \mathrm{For}f\left(x\right)\mathrm{to}\mathrm{be}\mathrm{monotonically}\mathrm{increasing},f\text{'}\left(x\right)>0. \\ \Rightarrow \frac{1+2x^2-\sqrt{x^2+1}}{1+x^2}>0 \\ \Rightarrow 1+2x^2-\sqrt{x^2+1}>0\left[\because \left(1+x^2\right)>0\right] \\ \Rightarrow 1+2x^2>\sqrt{x^2+1} \\ \Rightarrow {\left(1+2x^2\right)}^2>x^2+1 \\ \Rightarrow 1+4x^4+4x^2>x^2+1 \\ \Rightarrow 4x^4+3x^2>0 \\ \mathrm{Thus},f\left(x\right)\mathrm{is}\mathrm{monotonically}\mathrm{increasing}\mathrm{for}x\in R. \end{gathered}$$