Suppose f, g, and h be three real valued function defined on R.
Let `f(x) = 2x + |x|, g(x) = (1)/(3)(2x-|x|)` and `h(x) = f(g(x))`
The domain of definition of the function ` l (x) = sin^(-1) ( f(x) - g (x) )` is equal to

Options

A. `(3/8, infty]`
B. `( - infty, 1]`
C. ` [ -1, 1]`
D. `( - infty, 3/8]`

Answer

To find the domain of the function \( l(x) = \sin^{-1}(f(x) - g(x)) \), we first need to determine the expressions for \( f(x) \) and \( g(x) \) based on the definitions provided.

### Step 1: Define \( f(x) \) and \( g(x) \)

1. **For \( f(x) \)**:
\[ f(x) = 2x + |x| \]
- If \( x \geq 0 \), then \( |x| = x \) and:
\[ f(x) = 2x + x = 3x \]
- If \( x < 0 \), then \( |x| = -x \) and:
\[ f(x) = 2x - x = x \]

2. **For \( g(x) \)**:
\[ g(x) = \frac{1}{3}(2x - |x|) \]
- If \( x \geq 0 \), then \( |x| = x \) and:
\[ g(x) = \frac{1}{3}(2x - x) = \frac{1}{3}x \]
- If \( x < 0 \), then \( |x| = -x \) and:
\[ g(x) = \frac{1}{3}(2x + x) = x \]

### Step 2: Calculate \( f(x) - g(x) \)

Now we will find \( f(x) - g(x) \) for both cases.

- **Case 1: \( x \geq 0 \)**
\[ f(x) - g(x) = 3x - \frac{1}{3}x = \frac{9x}{3} - \frac{1}{3}x = \frac{8x}{3} \]

- **Case 2: \( x < 0 \)**
\[ f(x) - g(x) = x - x = 0 \]

### Step 3: Define \( l(x) \)

Now we can express \( l(x) \):
\[ l(x) = \sin^{-1}(f(x) - g(x)) \]
- For \( x \geq 0 \):
\[ l(x) = \sin^{-1}\left(\frac{8x}{3}\right) \]
- For \( x < 0 \):
\[ l(x) = \sin^{-1}(0) = 0 \]

### Step 4: Determine the domain of \( l(x) \)

The function \( \sin^{-1}(y) \) is defined for \( y \) in the range \([-1, 1]\). Therefore, we need to find when \( f(x) - g(x) \) lies within this interval.

1. **For \( x \geq 0 \)**:
\[ \frac{8x}{3} \leq 1 \implies 8x \leq 3 \implies x \leq \frac{3}{8} \]
Thus, for \( x \geq 0 \), the valid range is \( 0 \leq x \leq \frac{3}{8} \).

2. **For \( x < 0 \)**:
\( l(x) = 0 \) is always valid since \( \sin^{-1}(0) \) is defined.

### Final Domain

Combining both cases, the domain of \( l(x) \) is:
\[ (-\infty, 0) \cup \left[0, \frac{3}{8}\right] \]