The user wants me to determine the properties (symmetry and monotonicity) of the function $g(u) = 2{\tan ^{ - 1}}\left( {{e^u}} \right) - {\pi \over 2}$.

My plan is to:

1. Check if the function is even, odd, or neither by evaluating $g(-u)$.
2. Check if the function is strictly increasing or decreasing by finding the derivative $g'(u)$ and analyzing its sign.
3. Combine these findings to select the correct option.

Step 1: Check for Symmetry (Even/Odd)

The given function is: $g(u) = 2{\tan ^{ - 1}}(e^u) - {\pi \over 2}$

To check for symmetry, we evaluate $g(-u)$: $g( - u) = 2{\tan ^{ - 1}}(e^{-u}) - {\pi \over 2}$ Since $e^{-u} = {1 \over e^u}$, we have: $g( - u) = 2{\tan ^{ - 1}}\left( {{1 \over e^u}} \right) - {\pi \over 2}$

We use the identity for inverse trigonometric functions: ${\tan ^{ - 1}}(x) + {\tan ^{ - 1}}\left( {{1 \over x}} \right) = {\pi \over 2}$ for $x > 0$. Since $e^u > 0$ for all real $u$, this identity is applicable. Therefore, ${\tan ^{ - 1}}\left( {{1 \over e^u}} \right) = {\pi \over 2} - {\tan ^{ - 1}}(e^u)$.

Substituting this back into the expression for $g(-u)$: $g( - u) = 2\left( {{\pi \over 2} - {\tan ^{ - 1}}(e^u)} \right) - {\pi \over 2}$ $g( - u) = \pi - 2{\tan ^{ - 1}}(e^u) - {\pi \over 2}$ $g( - u) = {\pi \over 2} - 2{\tan ^{ - 1}}(e^u)$

Now, let's compare this with $-g(u)$. $- g(u) = - \left( {2{\tan ^{ - 1}}(e^u) - {\pi \over 2}} \right) = - 2{\tan ^{ - 1}}(e^u) + {\pi \over 2}$

We can see that $g(-u) = -g(u)$. This means the function $g(u)$ is an odd function. This eliminates options A and D.

Step 2: Check for Monotonicity (Increasing/Decreasing)

To determine if the function is increasing or decreasing, we find its derivative, $g'(u)$, with respect to $u$. $g(u) = 2{\tan ^{ - 1}}(e^u) - {\pi \over 2}$

Using the chain rule, d\over{dx}}(\tan^{-1}(f(x))) = {1 \over {1 + (f(x))^2}} \cdot f'(x): $g'(u) = {d \over {du}}\left( {2{\tan ^{ - 1}}(e^u) - {\pi \over 2}} \right)$ $g'(u) = 2 \cdot {d \over {du}}({\tan ^{ - 1}}(e^u)) - 0$ $g'(u) = 2 \cdot \left( {{1 \over {1 + {{(e^u)}^2}}}} \right) \cdot {d \over {du}}(e^u)$ $g'(u) = 2 \cdot \left( {{1 \over {1 + e^{2u}}}} \right) \cdot e^u$ $g'(u) = {{2e^u} \over {1 + e^{2u}}}$

Now, we analyze the sign of $g'(u)$ for all $u \in ( - \infty ,\infty )$.

• The numerator, $2e^u$, is always positive since $e^u > 0$ for all real $u$.
• The denominator, $1 + e^{2u}$, is also always positive since $e^{2u} = (e^u)^2 > 0$, so $1 + e^{2u} > 1$.

Since both the numerator and the denominator are positive, $g'(u) > 0$ for all $u \in ( - \infty ,\infty )$.

A function whose derivative is positive over its entire domain is strictly increasing over that domain.

Step 3: Conclusion

From Step 1, we found that the function $g(u)$ is odd. From Step 2, we found that the function $g(u)$ is strictly increasing in $( - \infty ,\infty )$.

Combining these two results, we look for the option that states the function is odd and strictly increasing in $( - \infty ,\infty )$.

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

Thus, the correct option is C.