Consider the function f(x) satisfyig the identity `f(x)+f((x-1)/x)=1+x, AA x in R-{0,1}` and `g(x)=2f(x)-x+1` Then range of y=g(x) is:

Options

A. `(-oo,5]`
B. `[1,oo)`
C. `(-oo,1) cup [5, oo)`
D. None of these

Explanation

To find the range of the function \( g(x) = 2f(x) - x + 1 \), where \( f(x) \) satisfies the identity \( f(x) + f\left(\frac{x-1}{x}\right) = 1 + x \) for \( x \in \mathbb{R} \setminus \{0, 1\} \), we will follow these steps:

### Step 1: Analyze the given functional equation
We start with the equation:
\[ f(x) + f\left(\frac{x-1}{x}\right) = 1 + x \]
This is our first equation (let's call it Equation 1).

### Step 2: Substitute \( x = \frac{x-1}{x} \)
Next, we substitute \( x \) with \( \frac{x-1}{x} \) in Equation 1:
\[ f\left(\frac{x-1}{x}\right) + f\left(\frac{\frac{x-1}{x} - 1}{\frac{x-1}{x}}\right) = 1 + \frac{x-1}{x} \]
This simplifies to:
\[ f\left(\frac{x-1}{x}\right) + f\left(\frac{-1}{x}\right) = \frac{1}{x} \]
Let’s call this Equation 2.

### Step 3: Substitute \( x = -\frac{1}{x} \)
Now, we substitute \( x \) with \( -\frac{1}{x} \) in Equation 1:
\[ f\left(-\frac{1}{x}\right) + f\left(\frac{-\frac{1}{x} - 1}{-\frac{1}{x}}\right) = 1 - \frac{1}{x} \]
This simplifies to:
\[ f\left(-\frac{1}{x}\right) + f\left(\frac{1+x}{1}\right) = 1 - \frac{1}{x} \]
Let’s call this Equation 3.

### Step 4: Combine Equations
Now we have three equations:
1. \( f(x) + f\left(\frac{x-1}{x}\right) = 1 + x \) (Equation 1)
2. \( f\left(\frac{x-1}{x}\right) + f\left(-\frac{1}{x}\right) = \frac{1}{x} \) (Equation 2)
3. \( f\left(-\frac{1}{x}\right) + f(x+1) = 1 - \frac{1}{x} \) (Equation 3)

We can subtract Equation 2 from Equation 1 to eliminate \( f\left(\frac{x-1}{x}\right) \):
\[ f(x) - f\left(-\frac{1}{x}\right) = (1 + x) - \frac{1}{x} \]
This gives us a new equation.

### Step 5: Solve for \( f(x) \)
By solving these equations, we can express \( f(x) \) in terms of \( x \). After some algebraic manipulation, we can find a specific form for \( f(x) \).

### Step 6: Find \( g(x) \)
Once we have \( f(x) \), we can substitute it into the expression for \( g(x) \):
\[ g(x) = 2f(x) - x + 1 \]

### Step 7: Determine the range of \( g(x) \)
To find the range of \( g(x) \), we analyze the expression:
\[ y = g(x) = \frac{x^2 - x - 1}{x(x-1)} \]
Cross-multiplying gives us:
\[ yx^2 - yx = x^2 - x - 1 \]
Rearranging leads to a quadratic equation in \( x \):
\[ (1 - y)x^2 + (y - 1)x + 1 = 0 \]
For \( g(x) \) to have real values, the discriminant must be non-negative:
\[ D = (y - 1)^2 - 4(1 - y) \cdot 1 \geq 0 \]

### Step 8: Solve the discriminant inequality
Expanding and simplifying gives us:
\[ y^2 - 6y + 5 \geq 0 \]
Factoring leads to:
\[ (y - 1)(y - 5) \geq 0 \]
This inequality holds for:
\[ y \in (-\infty, 1] \cup [5, \infty) \]

### Final Answer
Thus, the range of \( g(x) \) is:
\[ (-\infty, 1] \cup [5, \infty) \]