Question

Let the function \( g: (-\infty, 0) \rightarrow \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \) be given by \( g(u) = 2 \tan^{-1}(e^u) - \frac{\pi}{2} \). Determine the properties of \( g \).

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

Hint:

Remember, for a function to be odd, \( f(-x) = -f(x) \) must hold true, and the function's derivative should be positive for increasing nature.

Answer 1

The Correct Option is C

Step 1: {Analyze evenness or oddness}
\[ g(-u) = 2 \tan^{-1}(e^{-u}) - \frac{\pi}{2} \] \[ = 2 \left( \frac{\pi}{2} - \tan^{-1}(e^u) \right) - \frac{\pi}{2} \] \[ = \pi - 2 \tan^{-1}(e^u) - \frac{\pi}{2} = -2 \tan^{-1}(e^u) + \frac{\pi}{2} \] \[ = -(2 \tan^{-1}(e^u) - \frac{\pi}{2}) = -g(u) \] Thus, \( g \) is an odd function. 
Step 2: {Verify increasing nature}
The derivative \( g'(u) = 2\frac{1}{1+e^{2u}}e^u>0 \) for all \( u \), indicating \( g \) is strictly increasing over \( (-\infty, \infty) \).

Answer 2

Step 1: Analyze the function \( g(u) \)
The given function is:
\[ g(u) = 2 \tan^{-1}(e^u) - \frac{\pi}{2}. \] We are asked to determine the properties of \( g \), specifically whether it is odd, and whether it is increasing or decreasing.

Step 2: Check if the function is odd
A function \( g(u) \) is odd if \( g(-u) = -g(u) \) for all \( u \). Let's check if this holds for the given function. First, calculate \( g(-u) \):
\[ g(-u) = 2 \tan^{-1}(e^{-u}) - \frac{\pi}{2}. \] Now, recall the identity for the arctangent function: \[ \tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2} \text{ for } x > 0. \] For \( e^u > 0 \), we have: \[ \tan^{-1}(e^{-u}) = \frac{\pi}{2} - \tan^{-1}(e^u). \] Thus, \[ g(-u) = 2 \left(\frac{\pi}{2} - \tan^{-1}(e^u)\right) - \frac{\pi}{2} = \pi - 2 \tan^{-1}(e^u) - \frac{\pi}{2} = -2 \tan^{-1}(e^u) + \frac{\pi}{2}. \] This simplifies to: \[ g(-u) = -\left(2 \tan^{-1}(e^u) - \frac{\pi}{2}\right) = -g(u). \] Therefore, \( g(u) \) is an odd function.

Step 3: Check if the function is strictly increasing
To determine whether \( g(u) \) is strictly increasing, we need to compute its derivative and check if it is positive for all \( u \). The derivative of \( g(u) \) is:
\[ g'(u) = 2 \cdot \frac{d}{du} \left( \tan^{-1}(e^u) \right). \] Using the chain rule, we have: \[ \frac{d}{du} \left( \tan^{-1}(e^u) \right) = \frac{1}{1 + (e^u)^2} \cdot \frac{d}{du}(e^u) = \frac{e^u}{1 + e^{2u}}. \] Thus, \[ g'(u) = 2 \cdot \frac{e^u}{1 + e^{2u}}. \] Since \( e^u > 0 \) for all \( u \), we have \( g'(u) > 0 \) for all \( u \). Therefore, \( g(u) \) is strictly increasing.

Step 4: Conclusion
Since \( g(u) \) is both odd and strictly increasing for all \( u \in (-\infty, \infty) \), the correct answer is:

Odd and is strictly increasing in \( (-\infty, \infty) \)