Question

Let $f : \mathbb{R} \to \mathbb{R}$ be a strictly decreasing function with $|f(t)| < \pi/2$ for all $t \in \mathbb{R}$. Let $g : [0, \pi] \to \mathbb{R}$ be a function defined by $g(t) = \sin(f(t))$. Which one of the following statements is Correct?

• A. $g$ is decreasing on $[0, \pi]$.
• B. $g$ is increasing on $[0, \pi]$.
• C. $g$ is increasing on $(0, \pi/2)$ and decreasing on $(\pi/2, \pi)$.
• D. $g$ is decreasing on $(0, \pi/2)$ and increasing on $(\pi/2, \pi)$.

Hint:

Remember this simple composition rule:
Increasing \(\circ\) Decreasing = Decreasing.
Since \(\sin(x)\) is increasing on \((-\pi/2, \pi/2)\). and \(f(t)\) is decreasing, the composition is decreasing on the entire domain.

Answer 1

The Correct Option is A

To solve the given problem, let's analyze the functions \(f(t)\) and \(g(t) = \sin(f(t))\).

1. Understanding the Function \(f(t)\):
   • The function \(f : \mathbb{R} \to \mathbb{R}\) is strictly decreasing. This means that for any \(t_1, t_2 \in \mathbb{R}\), if \(t_1 < t_2\), then \(f(t_1) > f(t_2)\).
   • We are given that \(|f(t)| < \pi/2\) for all \(t \in \mathbb{R}\). Hence, the values of \(f(t)\) lie within the interval \((-\pi/2, \pi/2)\).
2. Behavior of \(g(t) = \sin(f(t))\):
   • Since \(f(t)\) is strictly decreasing, as \(t\) increases, \(f(t)\) decreases.
   • The sine function, \(\sin(x)\), is an increasing function on the interval \((-\pi/2, \pi/2)\).
3. Analysis of \(\sin(f(t))\):
   • Since \(f(t)\) decreases as \(t\) increases and \(\sin(x)\) is increasing in \((-\pi/2, \pi/2)\), the composition \(g(t) = \sin(f(t))\) should decrease as \(t\) increases over \([0, \pi]\).
4. Conclusion:
   • The function \(g(t)\) is decreasing on the interval \([0, \pi]\).

Therefore, the correct answer is: \(g\) is decreasing on \([0, \pi]\).