Let the function \(\mathrm{g}: \mathrm{R} \rightarrow\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) be given by \(g(t)=\frac{\pi}{2}-2 \cot ^{-1}\left(3^{-t}\right) .\) Then \(g\) is - (1) even and is strictly increasing in \((-\infty, \infty)\) (2) odd and is strictly decreasing in \((-\infty, \infty)\) (3) even and is strictly decreasing in \((-\infty, \infty)\) (4) odd and is strictly increasing is \((-\infty, \infty)\) (5) \(g(0)=\frac{\pi}{2}\)

Inverse trigonometric functions are the inverses of the usual trigonometric functions. For example, \(\tan^{-1}(x)\) is the inverse of \(\tan(x)\), and it gives the angle whose tangent is \(x\).
In our exercise, we encountered the inverse cotangent function, \(\cot^{-1}(x)\). It's helpful to know the identity:
\[ \cot^{-1}(x) = \frac{\pi}{2} - \tan^{-1}(x) \]
This identity is key because it allows us to convert between different inverse trigonometric functions, making it easier to simplify and analyze expressions. In the given function, we used this identity to rewrite and simplify the original function \(g(t)=\frac{\pi}{2} - 2 \cot^{-1}(3^{-t})\). This made further manipulations possible.

Monotonicity refers to whether a function is strictly increasing or decreasing or neither over a certain interval. To determine this property, we can use the first derivative of the function.
The function is strictly increasing if its first derivative is always positive. Conversely, it is strictly decreasing if the first derivative is always negative.
In the exercise, we found the first derivative of the function \(g(t) = -\frac{\pi}{2} + 2 \tan^{-1}(3^{-t})\):
\[ g^{'}(t) = -\frac{2 \cdot 3^{-t} \ln(3)}{1 + 3^{-2t}} \]
Since \(3^{-t} \ln(3)\) is always positive and the denominator is also positive, \(g^{'}(t)\) is always negative, which means the function \(g(t)\) is strictly decreasing over the entire interval \((-\infty, \infty)\).

A function is considered even if \(f(t) = f(-t)\). It has symmetrical properties about the y-axis. Conversely, a function is odd if \(f(t) = -f(-t)\), and it has rotational symmetry about the origin.
In our problem, we needed to determine whether \(g(t)\) was even or odd.
We first calculated \(g(-t)\):
\[ g(-t) = -\frac{\pi}{2} + 2 \tan^{-1}(3^{t}) \]
Then, we compared it with \(-g(t)\):
\[ -g(t) = \frac{\pi}{2} - 2 \tan^{-1}(3^{-t}) \]
Since \(g(-t) = -g(t)\), we concluded that the function \(g(t)\) is odd.

The derivative of a function measures how the function's value changes as its input changes. It's a fundamental concept in calculus, and it helps us understand the behavior of functions.
For a function \(f(t)\), the first derivative \(f'(t)\) gives the rate of change of \(f(t)\) with respect to \(t\).
In the given exercise, we needed to find the derivative of \(g(t) = -\frac{\pi}{2} + 2 \tan^{-1}(3^{-t})\).
Using the chain rule, we derived:
\[ g^{'}(t) = 2 \cdot \frac{1}{1+(3^{-t})^2} \cdot (-3^{-t} \ln(3)) \]
Simplifying that, we found:
\[ g^{'}(t) = -\frac{2 \cdot 3^{-t} \ln(3)}{1 + 3^{-2t}} \]
This negative derivative indicated that \(g(t)\) is strictly decreasing. Understanding derivatives is crucial for analyzing the behavior of functions, such as identifying intervals of increase and decrease.