Consider the function `f(x)` satisfying the identity `f(x) +f((x-1)/(x))=1+x AA x in R -{0,1}, and g(x)=2f(x)-x+1.`
The domain of `y=sqrt(g(x))` is

Options

A. `(-oo,(1-sqrt(5))/(2)] cup [1,(1+sqrt(5))/(2)]`
B. `(-oo,(1-sqrt(5))/(2)] cup (0,1)cup [(1+sqrt(5))/(2),oo)`
C. `[(-1-sqrt(5))/(2),0] cup [(-1+sqrt(5))/(2),1)`
D. None of these

Explanation

To solve the problem, we need to find the domain of the function \( y = \sqrt{g(x)} \), where \( g(x) = 2f(x) - x + 1 \) and \( f(x) \) satisfies the identity \( f(x) + f\left(\frac{x-1}{x}\right) = 1 + x \).

### Step 1: Understand the identity involving \( f(x) \)

We start with the identity:
\[ f(x) + f\left(\frac{x-1}{x}\right) = 1 + x \]
This identity will help us find a general form for \( f(x) \).

### Step 2: Replace \( x \) in the identity

We can replace \( x \) with \( \frac{x-1}{x} \) in the identity:
\[ f\left(\frac{x-1}{x}\right) + f\left(\frac{\frac{x-1}{x}-1}{\frac{x-1}{x}}\right) = 1 + \frac{x-1}{x} \]
Calculating \( \frac{\frac{x-1}{x}-1}{\frac{x-1}{x}} \):
\[ \frac{x-1-x}{x-1} = \frac{-1}{x-1} \]
Thus, the identity becomes:
\[ f\left(\frac{x-1}{x}\right) + f\left(-\frac{1}{x-1}\right) = 1 + \frac{x-1}{x} \]

### Step 3: Subtract the two identities

Now, we can subtract the two identities:
\[ f(x) - f\left(-\frac{1}{x-1}\right) = x - \left(1 + \frac{x-1}{x}\right) \]
This simplifies to:
\[ f(x) - f\left(-\frac{1}{x-1}\right) = x - 1 - \frac{x-1}{x} \]

### Step 4: Find \( f(x) \)

Continuing from the previous steps, we can derive \( f(x) \) in terms of \( x \). After some algebra, we find:
\[ f(x) = \frac{x^3 - x^2 - 1}{2x(x-1)} \]

### Step 5: Substitute \( f(x) \) into \( g(x) \)

Now, substitute \( f(x) \) into the expression for \( g(x) \):
\[ g(x) = 2f(x) - x + 1 = 2\left(\frac{x^3 - x^2 - 1}{2x(x-1)}\right) - x + 1 \]
This simplifies to:
\[ g(x) = \frac{x^3 - x^2 - 1}{x(x-1)} - x + 1 \]

### Step 6: Simplify \( g(x) \)

After simplifying \( g(x) \), we find:
\[ g(x) = \frac{x^2 - x - 1}{x(x-1)} \]

### Step 7: Find the domain of \( y = \sqrt{g(x)} \)

For \( y = \sqrt{g(x)} \) to be defined, \( g(x) \) must be greater than or equal to zero:
\[ \frac{x^2 - x - 1}{x(x-1)} \geq 0 \]

### Step 8: Determine the critical points

The critical points are found by solving \( x^2 - x - 1 = 0 \) and \( x(x-1) = 0 \). The roots of \( x^2 - x - 1 = 0 \) are:
\[ x = \frac{1 \pm \sqrt{5}}{2} \]
The roots of \( x(x-1) = 0 \) are \( x = 0 \) and \( x = 1 \).

### Step 9: Test intervals

We need to test the intervals determined by the critical points:
1. \( (-\infty, 0) \)
2. \( (0, 1 - \frac{\sqrt{5}}{2}) \)
3. \( (1 - \frac{\sqrt{5}}{2}, 1) \)
4. \( (1, 1 + \frac{\sqrt{5}}{2}) \)
5. \( (1 + \frac{\sqrt{5}}{2}, \infty) \)

### Step 10: Identify the valid intervals

After testing these intervals, we find that \( g(x) \geq 0 \) in the intervals:
\[ (-\infty, 1 - \frac{\sqrt{5}}{2}) \cup (1 + \frac{\sqrt{5}}{2}, \infty) \]

### Conclusion

Thus, the domain of \( y = \sqrt{g(x)} \) is:
\[ (-\infty, 1 - \frac{\sqrt{5}}{2}) \cup (1 + \frac{\sqrt{5}}{2}, \infty) \]