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

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Step 1 : Understanding the Question:
We are given a strictly decreasing function \(f(t)\). on \(\mathbb{R}\). such that its values always lie within the open interval \((-\pi/2, \pi/2)\).
We are asked to determine the monotonicity of the function \(g(t) = \sin(f(t))\) on the interval \([0, \pi]\).

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Step 2 : Key Formula or Approach:
The composition of two functions \(h(f(t))\) depends on the monotonicity of both functions.
If \(f\) is strictly decreasing, and \(h\) is strictly increasing on the range of \(f\)., then their composition \(h \circ f\) is strictly decreasing.
Alternatively, we can use the chain rule for differentiation if \(f\) is differentiable: \(g'(t) = \cos(f(t)) \cdot f'(t)\).

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Step 3 : Detailed Explanation:
Let us use the definition of monotonicity:
Let \(t_1, t_2 \in [0, \pi]\) such that \(t_1 < t_2\).
Since the function \(f\) is strictly decreasing on \(\mathbb{R}\)., we have:
\[ f(t_1) > f(t_2) \] We are given that \(|f(t)| < \pi/2\) for all \(t \in \mathbb{R}\)., which means:
\[ f(t_1), f(t_2) \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \] Let us consider the sine function, \(h(x) = \sin(x)\).
The derivative of \(h(x)\) is \(h'(x) = \cos(x)\).
Since \(\cos(x) > 0\) for all \(x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)., the sine function is strictly increasing on this interval.
Since \(h(x) = \sin(x)\) is strictly increasing, it preserves the order of its inputs:
\[ \sin(f(t_1)) > \sin(f(t_2)) \] This implies:
\[ g(t_1) > g(t_2) \] Since \(t_1 g(t_2)\)., the function \(g(t)\) is strictly decreasing on the entire interval \([0, \pi]\).

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Step 4 : Final Answer:
The function \(g\) is decreasing on \([0, \pi]\).
This corresponds to option (A).