`f(x)={(x-1",",-1 le x le 0),(x^(2)",",0le x le 1):} and g(x)=sinx`
Consider the functions `h_(1)(x)=f(|g(x)|) and h_(2)(x)=|f(g(x))|.`
Which of the following is not true about `h_(2)(x)`?

Options

A. The domain is R
B. It is periodic with period `2pi`.
C. The range is [0, 1].
D. None of these

Explanation

To solve the problem, we need to analyze the functions \( h_1(x) \) and \( h_2(x) \) based on the given functions \( f(x) \) and \( g(x) \).

### Step 1: Define the functions

We have:
- \( f(x) = \begin{cases} x - 1 & \text{for } -1 \leq x \leq 0 \\ x^2 & \text{for } 0 < x \leq 1 \end{cases} \)
- \( g(x) = \sin x \)

### Step 2: Analyze \( h_1(x) = f(|g(x)|) \)

1. **Calculate \( |g(x)| \)**:
- Since \( g(x) = \sin x \), we have \( |g(x)| = |\sin x| \).
- The range of \( |\sin x| \) is from 0 to 1.

2. **Determine the piece of \( f(x) \) to use**:
- For \( 0 \leq |\sin x| \leq 1 \), we will use the second piece of \( f(x) \): \( f(x) = x^2 \).
- Thus, \( h_1(x) = f(|g(x)|) = (|\sin x|)^2 = \sin^2 x \).

### Step 3: Analyze \( h_2(x) = |f(g(x))| \)

1. **Calculate \( f(g(x)) \)**:
- For \( g(x) = \sin x \), we need to determine where \( \sin x \) lies in the domain of \( f(x) \).
- The range of \( \sin x \) is from -1 to 1.

2. **Evaluate \( f(g(x)) \)**:
- For \( -1 \leq \sin x \leq 0 \), we use the first piece of \( f(x) \): \( f(x) = x - 1 \).
- Thus, \( f(\sin x) = \sin x - 1 \).
- For \( 0 < \sin x \leq 1 \), we use the second piece of \( f(x) \): \( f(x) = x^2 \).
- Thus, \( f(\sin x) = (\sin x)^2 \).

3. **Combine the results**:
- Therefore, we can write:
\[ h_2(x) = \begin{cases} |\sin x - 1| & \text{for } -1 \leq \sin x \leq 0 \\ |(\sin x)^2| & \text{for } 0 < \sin x \leq 1 \end{cases} \]
- Since \( |(\sin x)^2| = (\sin x)^2 \), we can simplify:
\[ h_2(x) = \begin{cases} 1 - \sin x & \text{for } -1 \leq \sin x < 0 \\ (\sin x)^2 & \text{for } 0 < \sin x \leq 1 \end{cases} \]

### Step 4: Determine the properties of \( h_2(x) \)

1. **Periodicity**:
- Since \( \sin x \) is periodic with period \( 2\pi \), both pieces of \( h_2(x) \) will also be periodic with period \( 2\pi \).

2. **Range**:
- The range of \( h_2(x) \) is from 0 to 2, since \( |\sin x - 1| \) can reach a maximum of 2 and \( (\sin x)^2 \) can reach a maximum of 1.

### Conclusion: Identify which statement is not true about \( h_2(x) \)

- The statements regarding \( h_2(x) \) could include:
1. It is periodic.
2. Its range is [0, 2].
3. It is continuous.
4. It is not defined for some \( x \).

Given our analysis, we can conclude that all statements about \( h_2(x) \) are true except for any statement that suggests it is not defined for some \( x \), as \( h_2(x) \) is defined for all \( x \).