Question

If the function $g:\left(-\infty ,\infty \right)\to \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$ be given by $g\left(u\right)=2{\tan }^{-1}\left(e^u\right)-\frac{\mathrm{\pi }}{2}$. Then $g$ is

• A. even and strictly increasing in $\left(0,\infty \right)$.
• B. odd and strictly increasing in $\left(-\infty ,\infty \right)$.
• C. odd and strictly decreasing in $\left(-\infty ,\infty \right)$.
• D. neither even nor odd but is strictly increasing in $\left(-\infty ,\infty \right)$.

Answer 1

The correct option is B

odd and strictly increasing in $\left(-\infty ,\infty \right)$.

Explanation for the correct option

Step 1: Solve to identify whether the function is odd or even

The given function $g\left(u\right)=2{\tan }^{-1}\left(e^u\right)-\frac{\mathrm{\pi }}{2}$.

Put $u=-u$.

$$\begin{gathered} \Rightarrow g\left(-u\right)=2{\tan }^{-1}\left(e^{-u}\right)-\frac{\mathrm{\pi }}{2} \\ \Rightarrow g\left(-u\right)=2{\tan }^{-1}\left(\frac{1}{e^u}\right)-\frac{\mathrm{\pi }}{2} \\ \Rightarrow g\left(-u\right)=2{\cot }^{-1}\left(e^u\right)-\frac{\mathrm{\pi }}{2}\left[\because {\tan }^{-1}\left(\frac{1}{\theta }\right)={\cot }^{-1}\left(\theta \right)\right] \\ \Rightarrow g\left(-u\right)=2\left[\frac{\mathrm{\pi }}{2}-{\tan }^{-1}\left(e^u\right)\right]-\frac{\mathrm{\pi }}{2}\left[\because {\tan }^{-1}\left(\theta \right)+{\cot }^{-1}\left(\theta \right)=\frac{\pi }{2}\right] \\ \Rightarrow g\left(-u\right)=\pi -2{\tan }^{-1}\left(e^u\right)-\frac{\mathrm{\pi }}{2} \\ \Rightarrow g\left(-u\right)=-2{\tan }^{-1}\left(e^u\right)+\frac{\mathrm{\pi }}{2} \\ \Rightarrow g\left(-u\right)=-\left[2{\tan }^{-1}\left(e^u\right)-\frac{\mathrm{\pi }}{2}\right] \\ \Rightarrow g\left(-u\right)=-g\left(u\right) \end{gathered}$$

Thus, the given function is an odd function.

Step 2: Solve to identify whether the function is increasing or decreasing

Differentiate the given function with respect to $u$.

$$\begin{gathered} \frac{d}{du}\left[g\left(u\right)\right]=\frac{d}{du}\left[2{\tan }^{-1}\left(e^u\right)-\frac{\mathrm{\pi }}{2}\right] \\ \Rightarrow g\text{'}\left(u\right)=2\times \frac{1}{1+{\left(e^u\right)}^2}\times \frac{d}{du}\left(e^u\right)\left[\because \frac{d}{dx}{\tan }^{-1}\left(x\right)=\frac{1}{1+x^2}\right] \\ \Rightarrow g\text{'}\left(u\right)=\frac{2e^u}{1+e^{2u}} \end{gathered}$$

As, $e^u>0$, thus, $\frac{2e^u}{1+e^{2u}}>0$.

Therefore, $g\text{'}\left(u\right)>0$ in the region $\left(-\infty ,\infty \right)$.

Thus, the given function is increasing in $\left(-\infty ,\infty \right)$.

Therefore, the function $g\left(u\right)=2{\tan }^{-1}\left(e^u\right)-\frac{\mathrm{\pi }}{2}$ is odd and strictly increasing in $\left(-\infty ,\infty \right)$.

Hence, option(B) is the correct answer.