The domain of definition of `f(x)=cos^(- 1)(x+[x])` is

Options

A. `[0, 1) `
B. `R-Z`
C. `(0, oo)`
D. none of these

Explanation

To find the domain of the function \( f(x) = \cos^{-1}(x + [x]) \), where \([x]\) denotes the greatest integer function, we need to determine the values of \(x\) for which the expression \(x + [x]\) lies within the range of the inverse cosine function, which is \([-1, 1]\).

### Step 1: Understand the range of \( \cos^{-1} \)
The function \( \cos^{-1}(y) \) is defined for \( y \) in the interval \([-1, 1]\). Therefore, we need to ensure that:
\[ -1 \leq x + [x] \leq 1 \]

### Step 2: Analyze the expression \( x + [x] \)
The greatest integer function \([x]\) gives the largest integer less than or equal to \(x\). We can express \(x\) as:
\[ x = n + f \]
where \(n = [x]\) (an integer) and \(f\) is the fractional part of \(x\) such that \(0 \leq f < 1\). Thus, we have:
\[ x + [x] = (n + f) + n = 2n + f \]

### Step 3: Set up inequalities
We need to solve the inequalities:
1. \( -1 \leq 2n + f \)
2. \( 2n + f \leq 1 \)

### Step 4: Analyze the first inequality
From \( -1 \leq 2n + f \):
- Since \(f \geq 0\), we have:
\[ -1 \leq 2n \implies n \geq -\frac{1}{2} \]
Thus, \(n\) must be at least \(0\) (since \(n\) is an integer).

### Step 5: Analyze the second inequality
From \(2n + f \leq 1\):
- Since \(0 \leq f < 1\), we can deduce:
\[ 2n \leq 1 \implies n \leq \frac{1}{2} \]
Thus, \(n\) can only be \(0\).

### Step 6: Substitute \(n = 0\)
If \(n = 0\), then:
\[ x + [x] = 2(0) + f = f \]
So we need:
\[ -1 \leq f \leq 1 \]
Since \(0 \leq f < 1\), this condition is satisfied.

### Step 7: Determine the range of \(x\)
Since \(n = 0\), we have:
\[ 0 \leq x < 1 \]
Thus, the domain of \(f(x)\) is:
\[ [0, 1) \]

### Conclusion
The domain of definition of \(f(x) = \cos^{-1}(x + [x])\) is:
\[ \boxed{[0, 1)} \]