Question

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

• A. Even and is strictly increasing in $(0,\infty )$
• B. Odd and is strictly decreasing in $(–\infty ,\infty )$
• C. Odd and is strictly increasing in $(–\infty ,\infty )$
• D. Neither even nor odd, but is strictly increasing in $(–\infty ,\infty )$

Answer 1

The correct option is C

Odd and is strictly increasing in $(–\infty ,\infty )$

Explanation for the correct option:

Determine the relationship of $g$:

We have, $g\left(u\right)=2{\tan }^{-1}\left(e^u\right)–\left(\frac{\pi }{2}\right)$

Checking if the function is odd or even function.

Putting $u\mathrm{as}-\mathrm{u}$, in the equation.

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

So, $g\left(u\right)$ is an odd function.

Now,

$$\begin{gathered} g\text{'}\left(u\right)=\frac{2}{1+e^{2u}}·e^u\left[\mathrm{Differentiating}g\left(u\right)\right] \\ =\frac{2e^u}{1+e^{2u}} \end{gathered}$$

Therefore, $g\text{'}\left(u\right)>0$for all $u\in \mathrm{ℝ}$

Thus, $g\left(u\right)$ is an increasing function for all $u\in \mathrm{ℝ}$

Therefore, option (C) is the correct answer.