Let `y = f (x)` such that `xy = x+y +1, x in R-{1} and g (x) =x f (x)`
There exist two values of `x, x_(1) and x _(2)` where `g '(x) =1/2,` then `|x _(1)| + |x_(2)|=`

Options

A. 1
B. 2
C. 4
D. 5

Answer

To solve the problem step by step, we will follow the given information and derive the necessary equations.

### Step 1: Express \( y \) in terms of \( f(x) \)
We start with the equation given in the problem:

\[ xy = x + y + 1 \]

Substituting \( y = f(x) \):

\[ x f(x) = x + f(x) + 1 \]

### Step 2: Rearranging the equation
Now, we can rearrange the equation to isolate \( f(x) \):

\[ x f(x) - f(x) = x + 1 \]

Factoring out \( f(x) \):

\[ f(x)(x - 1) = x + 1 \]

### Step 3: Solve for \( f(x) \)
Now, we can solve for \( f(x) \):

\[ f(x) = \frac{x + 1}{x - 1} \]

### Step 4: Define \( g(x) \)
Next, we define \( g(x) \):

\[ g(x) = x f(x) = x \cdot \frac{x + 1}{x - 1} \]

This simplifies to:

\[ g(x) = \frac{x(x + 1)}{x - 1} = \frac{x^2 + x}{x - 1} \]

### Step 5: Differentiate \( g(x) \)
To find \( g'(x) \), we will use the quotient rule:

\[ g'(x) = \frac{(x - 1)(2x + 1) - (x^2 + x)(1)}{(x - 1)^2} \]

### Step 6: Simplify \( g'(x) \)
Expanding the numerator:

\[ g'(x) = \frac{(x - 1)(2x + 1) - (x^2 + x)}{(x - 1)^2} \]

Calculating the first part:

\[ = \frac{(2x^2 + x - 2x - 1) - (x^2 + x)}{(x - 1)^2} \]

This simplifies to:

\[ = \frac{2x^2 - x - 1 - x^2 - x}{(x - 1)^2} = \frac{x^2 - 2x - 1}{(x - 1)^2} \]

### Step 7: Set \( g'(x) = \frac{1}{2} \)
Now we set the derivative equal to \( \frac{1}{2} \):

\[ \frac{x^2 - 2x - 1}{(x - 1)^2} = \frac{1}{2} \]

Cross-multiplying gives:

\[ 2(x^2 - 2x - 1) = (x - 1)^2 \]

### Step 8: Expand and rearrange
Expanding both sides:

\[ 2x^2 - 4x - 2 = x^2 - 2x + 1 \]

Rearranging gives:

\[ 2x^2 - 4x - 2 - x^2 + 2x - 1 = 0 \]

This simplifies to:

\[ x^2 - 2x - 3 = 0 \]

### Step 9: Factor the quadratic equation
Factoring the quadratic:

\[ (x - 3)(x + 1) = 0 \]

Thus, we have:

\[ x = 3 \quad \text{or} \quad x = -1 \]

### Step 10: Find \( |x_1| + |x_2| \)
Let \( x_1 = 3 \) and \( x_2 = -1 \):

\[ |x_1| + |x_2| = |3| + |-1| = 3 + 1 = 4 \]

### Final Answer
Thus, the final answer is:

\[ \boxed{4} \]