If `f(x)=2x+abs(x),g(x)=1/3(2x-abs(x))` and h(x)=f(g(x)), domain of `underbrace(sin^(-1)(h(h(h(h...h(x)...)))))_("n times")` is

Options

A. [-1,1]
B. `[-1,-1/2] cup [1/2,1]`
C. `[-1,-1/2]`
D. `[1/2,1]`

Answer

To solve the problem step by step, we will analyze the functions \( f(x) \), \( g(x) \), and \( h(x) \) and then determine the domain of \( \sin^{-1}(h(h(h(...h(x)...)))) \) for \( n \) times.

### Step 1: Define the functions
Given:
- \( f(x) = 2x + |x| \)
- \( g(x) = \frac{1}{3}(2x - |x|) \)

### Step 2: Analyze \( f(x) \)
The absolute value function \( |x| \) behaves differently based on the sign of \( x \):
- If \( x \geq 0 \): \( |x| = x \)
\[ f(x) = 2x + x = 3x \]
- If \( x < 0 \): \( |x| = -x \)
\[ f(x) = 2x - x = x \]

### Step 3: Analyze \( g(x) \)
Similarly, we analyze \( g(x) \):
- If \( x \geq 0 \): \( |x| = x \)
\[ g(x) = \frac{1}{3}(2x - x) = \frac{1}{3}x \]
- If \( x < 0 \): \( |x| = -x \)
\[ g(x) = \frac{1}{3}(2x + x) = x \]

### Step 4: Define \( h(x) \)
Now we define \( h(x) = f(g(x)) \):
- For \( x \geq 0 \):
\[ g(x) = \frac{1}{3}x \implies h(x) = f\left(\frac{1}{3}x\right) = 3\left(\frac{1}{3}x\right) = x \]
- For \( x < 0 \):
\[ g(x) = x \implies h(x) = f(x) = x \]

Thus, we conclude:
\[ h(x) = x \quad \text{for all } x \]

### Step 5: Analyze \( \sin^{-1}(h(h(h(...h(x)...)))) \)
Since \( h(x) = x \), applying \( h \) any number of times will still yield \( x \):
\[ h(h(h(...h(x)...))) = x \]

### Step 6: Determine the domain of \( \sin^{-1}(x) \)
The function \( \sin^{-1}(x) \) is defined for:
\[ -1 \leq x \leq 1 \]

### Conclusion
Thus, the domain of \( \sin^{-1}(h(h(h(...h(x)...)))) \) is:
\[ \text{Domain: } [-1, 1] \]