Question

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) \(g(0)=\frac{\pi}{2}\)

Short answer

Option (2) is correct: g is odd and strictly decreasing on (-∞, ∞).

Step-by-step solution

Simplify the Function

Given function: g(t) = \frac{\pi}{2} - 2 \cot^{-1}(3^{-t}). We need to analyze the properties of this function. First, recall the identity for the inverse cotangent function: \[ \cot^{-1} x = \frac{\pi}{2} - \tan^{-1} x \] Using this identity, we can rewrite the function g(t).

Apply the Identity

Rewrite g(t) using the inverse cotangent identity: \[ g(t) = \frac{\pi}{2} - 2 \left(\frac{\pi}{2} - \tan^{-1}(3^{-t})\right) \] Simplify the expression: \[ g(t) = \frac{\pi}{2} - \pi + 2 \tan^{-1}(3^{-t}) \] \[ g(t) = -\frac{\pi}{2} + 2 \tan^{-1}(3^{-t}). \]

Analyze Even or Odd

To determine if the function is even or odd: - A function f(t) is even if f(t) = f(-t).- A function f(t) is odd if f(t) = -f(-t). Evaluate g(-t): \[ g(-t) = -\frac{\pi}{2} + 2 \tan^{-1}(3^{t}). \] Compare g(-t) to -g(t): \[ -g(t) = \frac{\pi}{2} - 2 \tan^{-1}(3^{-t}). \] Since g(-t) = -g(t), the function g(t) is odd.

Determine Monotonicity

To determine if the function is strictly increasing or decreasing, consider the derivative of g(t): \[ g'(t) = \frac{d}{dt} \left(-\frac{\pi}{2} + 2 \tan^{-1}(3^{-t})\right). \] Using the chain rule: \[ g'(t) = 2 \cdot \frac{1}{1+(3^{-t})^2} \cdot (-3^{-t} \ln(3)), \] \[ g'(t) = -\frac{2 \cdot 3^{-t} \ln(3)}{1 + 3^{-2t}}. \] Since 3^{-t} \ln(3) is always positive and the denominator is also positive: \[ g'(t) < 0, \] so g(t) is strictly decreasing over (-\infty, \infty).

Verify g(0)

Finally, calculate g(0): \[ g(0) = -\frac{\pi}{2} + 2 \tan^{-1}(3^{-0}) \] \[ g(0) = -\frac{\pi}{2} + 2 \tan^{-1}(1). \] Since \tan^{-1}(1) = \frac{\pi}{4}: \[ g(0) = -\frac{\pi}{2} + 2 \left(\frac{\pi}{4}\right) \] \[ g(0) = -\frac{\pi}{2} + \frac{\pi}{2} = 0. \]

Key concepts

Inverse Trigonometric Functions

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

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 \((-\infty, \infty)\).

Even and Odd Functions

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.

Derivatives

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.