If `f(x)=|sin x|` then domain of f for the existence of inverse of

Options

A. `[0,pi]`
B. `[0,pi//2]`
C. `[-pi//4, pi//4]`
D. none of these

Answer

To find the domain of the function \( f(x) = |\sin x| \) for the existence of its inverse, we need to ensure that the function is one-to-one (injective) on that domain. Let's go through the steps to determine the appropriate domain.

### Step 1: Understand the function
The function \( f(x) = |\sin x| \) takes the absolute value of the sine function. The sine function oscillates between -1 and 1, so the absolute value will oscillate between 0 and 1.

### Step 2: Identify the intervals
To find a domain where \( f(x) \) is one-to-one, we need to restrict \( x \) to intervals where \( \sin x \) does not repeat values. The sine function is one-to-one in the intervals:
- \( \left[-\frac{\pi}{2}, 0\right] \)
- \( \left[0, \frac{\pi}{2}\right] \)

### Step 3: Analyze the intervals
1. **Interval \( [0, \frac{\pi}{2}] \)**:
- In this interval, \( \sin x \) is non-negative and increasing. Therefore, \( |\sin x| = \sin x \) and the function is one-to-one.

2. **Interval \( [-\frac{\pi}{2}, 0] \)**:
- In this interval, \( \sin x \) is non-positive and decreasing. Therefore, \( |\sin x| = -\sin x \) and the function is also one-to-one.

### Step 4: Conclusion
Since \( f(x) = |\sin x| \) is one-to-one in both intervals \( [0, \frac{\pi}{2}] \) and \( [-\frac{\pi}{2}, 0] \), we can choose either of these intervals as the domain for which the inverse exists.

The most common choice is \( [0, \frac{\pi}{2}] \) since it covers the positive values of \( |\sin x| \).

Thus, the domain of \( f(x) \) for the existence of its inverse is:
\[ \text{Domain: } [0, \frac{\pi}{2}] \]