Questions: What is the largest interval containing x=2π on which sin(x) is one-to-one? Select one: a. [π, 3π] b. [π/2, 5π/2] c. [3π/2, 7π/2] d. [3π/2, 5π/2]

Transcript text: What is the largest interval containing $x=2 \pi$ on which $\sin (x)$ is one-to-one? Select one: a. $[\pi, 3 \pi]$ b. $\left[\frac{\pi}{2}, \frac{5 \pi}{2}\right]$ c. $\left[\frac{3 \pi}{2}, \frac{7 \pi}{2}\right]$ d. $\left[\frac{3 \pi}{2}, \frac{5 \pi}{2}\right]$

Solution

Solution Steps

Step 1: Understand the problem

We need to find the largest interval containing $ x = 2\pi $ on which the function $ \sin(x) $ is one-to-one. A function is one-to-one on an interval if it is either strictly increasing or strictly decreasing on that interval.

Step 2: Analyze the sine function

The sine function, $ \sin(x) $, is periodic with a period of $ 2\pi $. It is one-to-one on intervals where it is either strictly increasing or strictly decreasing. The sine function is strictly increasing on intervals like $ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $ and strictly decreasing on intervals like $ \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] $.

Step 3: Identify the interval containing $ x = 2\pi $

We need to find the largest interval containing $ x = 2\pi $ where $ \sin(x) $ is one-to-one. The sine function is strictly decreasing on the interval $ \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] $ and strictly increasing on the interval $ \left[\frac{3\pi}{2}, \frac{5\pi}{2}\right] $. Since $ x = 2\pi $ lies in the interval $ \left[\frac{3\pi}{2}, \frac{5\pi}{2}\right] $, this is the largest interval containing $ x = 2\pi $ where $ \sin(x) $ is one-to-one.

Step 4: Match the interval with the given options

The interval $ \left[\frac{3\pi}{2}, \frac{5\pi}{2}\right] $ corresponds to option d.