If the function `g:(-oo,oo)->(-pi/2,pi/2)` is given by `g(u)=2tan^-1(e^u)-pi/2.` Then, `g` is

Options

A. even and is strictly increasing in `(0,oo)`
B. odd and is strictly decreasing in `(-oo,oo)`
C. odd is strictly increasing in `(-oo,oo)`
D. neither even nor odd but is strictly increasing in `(-oo,oo)`

Answer

To solve the problem, we will analyze the function \( g(u) = 2 \tan^{-1}(e^u) - \frac{\pi}{2} \) and determine its properties, including whether it is even or odd, and its monotonicity (increasing or decreasing).

### Step 1: Differentiate the function

We start by differentiating \( g(u) \) with respect to \( u \):

\[ g'(u) = \frac{d}{du} \left( 2 \tan^{-1}(e^u) - \frac{\pi}{2} \right) \]

Using the chain rule, we differentiate \( \tan^{-1}(e^u) \):

\[ g'(u) = 2 \cdot \frac{1}{1 + (e^u)^2} \cdot \frac{d}{du}(e^u) \]

Since the derivative of \( e^u \) is \( e^u \), we have:

\[ g'(u) = 2 \cdot \frac{1}{1 + e^{2u}} \cdot e^u \]

Thus,

\[ g'(u) = \frac{2 e^u}{1 + e^{2u}} \]

### Step 2: Analyze the sign of the derivative

Next, we analyze the expression \( g'(u) = \frac{2 e^u}{1 + e^{2u}} \).

- The term \( e^u \) is always positive for all \( u \in \mathbb{R} \).
- The denominator \( 1 + e^{2u} \) is also always positive.

Since both the numerator and denominator are positive, we conclude that:

\[ g'(u) > 0 \quad \text{for all } u \in \mathbb{R} \]

This indicates that the function \( g(u) \) is strictly increasing on the entire interval \( (-\infty, \infty) \).

### Step 3: Check if the function is even or odd

To check if \( g(u) \) is even or odd, we compute \( g(-u) \):

\[ g(-u) = 2 \tan^{-1}(e^{-u}) - \frac{\pi}{2} \]

Using the identity \( \tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2} \) for \( x > 0 \):

\[ \tan^{-1}(e^{-u}) + \tan^{-1}(e^u) = \frac{\pi}{2} \]

Thus,

\[ g(-u) = 2 \left( \frac{\pi}{2} - \tan^{-1}(e^u) \right) - \frac{\pi}{2} \]
\[ = \pi - 2 \tan^{-1}(e^u) - \frac{\pi}{2} \]
\[ = \frac{\pi}{2} - 2 \tan^{-1}(e^u) \]

Now, we can express \( g(u) + g(-u) \):

\[ g(u) + g(-u) = \left( 2 \tan^{-1}(e^u) - \frac{\pi}{2} \right) + \left( 2 \tan^{-1}(e^{-u}) - \frac{\pi}{2} \right) \]
\[ = 2 \tan^{-1}(e^u) + 2 \tan^{-1}(e^{-u}) - \pi \]
\[ = 0 \]

Since \( g(u) + g(-u) = 0 \), we conclude that \( g(u) \) is an odd function.

### Conclusion

From our analysis, we have determined that:

1. The function \( g(u) \) is strictly increasing on the interval \( (-\infty, \infty) \).
2. The function \( g(u) \) is an odd function.

Thus, the final answer is that \( g \) is an odd function and is strictly increasing in \( (-\infty, \infty) \).