Function `f : [(pi)/(2), (3pi)/(2)] rarr [-1, 1], f(x) = sin x` is

Options

A. Many -one onto
B. Onto
C. One-one onto
D. Many-one into

Answer

To analyze the function \( f : \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \to [-1, 1] \) defined by \( f(x) = \sin x \), we need to determine its properties: whether it is one-to-one (1-1) and onto.

### Step 1: Determine the range of the function
The first step is to find the range of \( f(x) = \sin x \) over the interval \( \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \).

- At \( x = \frac{\pi}{2} \), \( f\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1 \).
- At \( x = \frac{3\pi}{2} \), \( f\left(\frac{3\pi}{2}\right) = \sin\left(\frac{3\pi}{2}\right) = -1 \).

Since \( \sin x \) is a continuous function and it decreases from \( 1 \) to \( -1 \) in the interval \( \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \), the range of \( f \) is indeed \( [-1, 1] \).

### Step 2: Check if the function is one-to-one (1-1)
A function is one-to-one if it never takes the same value twice for different inputs.

- In the interval \( \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \), \( \sin x \) is strictly decreasing. This means for any two values \( x_1 \) and \( x_2 \) in this interval, if \( x_1 < x_2 \), then \( f(x_1) > f(x_2) \).

Thus, \( f \) is one-to-one.

### Step 3: Check if the function is onto
A function is onto if every element in the codomain has a pre-image in the domain.

- The codomain of \( f \) is \( [-1, 1] \). Since \( f(x) = \sin x \) achieves every value from \( 1 \) to \( -1 \) as \( x \) varies from \( \frac{\pi}{2} \) to \( \frac{3\pi}{2} \), every \( y \) in \( [-1, 1] \) has a corresponding \( x \) in the domain that maps to it.

Thus, \( f \) is onto.

### Conclusion
Since \( f \) is both one-to-one and onto, we conclude that:

- The function \( f(x) = \sin x \) defined from \( \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \) to \( [-1, 1] \) is a bijection (1-1 and onto).

### Final Answer
The correct option is: **1-1 and onto**.

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