`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))|.`
then

Options

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

Explanation

To solve the problem, we will analyze the functions \( f(x) \) and \( g(x) \) and then derive the functions \( h_1(x) \) and \( h_2(x) \).

### Step 1: Define the functions
The function \( f(x) \) is defined piecewise as follows:
\[ f(x) = \begin{cases} x - 1 & \text{for } -1 \leq x \leq 0 \\ x^2 & \text{for } 0 < x \leq 1 \end{cases} \]
The function \( g(x) \) is defined as:
\[ g(x) = \sin x \]

### Step 2: Analyze \( h_1(x) = f(|g(x)|) \)
First, we will compute \( |g(x)| \):
\[ |g(x)| = |\sin x| \]
Now, we need to find \( f(|g(x)|) \). The range of \( |\sin x| \) is from 0 to 1, so we will consider two cases based on the definition of \( f(x) \).

1. **When \( 0 \leq |\sin x| \leq 1 \)**:
- Since \( |\sin x| \) is always non-negative and can be at most 1, we will use the second part of the piecewise function:
\[ f(|g(x)|) = f(|\sin x|) = (|\sin x|)^2 \quad \text{for } 0 \leq |\sin x| \leq 1 \]
Thus,
\[ h_1(x) = |\sin x|^2 = \sin^2 x \]

### Step 3: Analyze \( h_2(x) = |f(g(x))| \)
Next, we compute \( f(g(x)) \):
\[ f(g(x)) = f(\sin x) \]
We need to consider the range of \( \sin x \):
- \( \sin x \) ranges from -1 to 1.

1. **When \( -1 \leq \sin x \leq 0 \)**:
- We use the first part of the piecewise function:
\[ f(\sin x) = \sin x - 1 \]

2. **When \( 0 < \sin x \leq 1 \)**:
- We use the second part of the piecewise function:
\[ f(\sin x) = (\sin x)^2 \]

Now, we need to find \( |f(g(x))| \):
- For \( -1 \leq \sin x \leq 0 \):
\[ |f(\sin x)| = | \sin x - 1 | = 1 - \sin x \quad (\text{since } \sin x \leq 0) \]
- For \( 0 < \sin x \leq 1 \):
\[ |f(\sin x)| = |(\sin x)^2| = (\sin x)^2 \]

Thus, we can summarize \( h_2(x) \):
\[ h_2(x) = \begin{cases} 1 - \sin x & \text{for } -1 \leq \sin x \leq 0 \\ (\sin x)^2 & \text{for } 0 < \sin x \leq 1 \end{cases} \]

### Summary of Results
- \( h_1(x) = \sin^2 x \)
- \( h_2(x) = \begin{cases} 1 - \sin x & \text{for } -1 \leq \sin x \leq 0 \\ (\sin x)^2 & \text{for } 0 < \sin x \leq 1 \end{cases} \)