if `f(x)=sqrt(x+sqrt(x+)sqrt(x+)sqrt(…oo)),`

Options

A. `1/(1-2f(x)`
B. `1/(2f(x)-1)`
C. `1/(1+2f(x))`
D. `1/(2+f(x))`

Explanation

To solve the equation \( f(x) = \sqrt{x + \sqrt{x + \sqrt{x + \ldots}}} \), we can follow these steps:

### Step 1: Set up the equation
Since the expression inside the square root is repeating itself, we can set:
\[ f(x) = \sqrt{x + f(x)} \]

### Step 2: Square both sides
To eliminate the square root, we square both sides of the equation:
\[ f(x)^2 = x + f(x) \]

### Step 3: Rearrange the equation
Rearranging the equation gives us:
\[ f(x)^2 - f(x) - x = 0 \]

### Step 4: Solve the quadratic equation
This is a quadratic equation in the form \( ax^2 + bx + c = 0 \), where \( a = 1, b = -1, c = -x \). We can use the quadratic formula:
\[ f(x) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values of \( a, b, \) and \( c \):
\[ f(x) = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-x)}}{2 \cdot 1} \]
\[ f(x) = \frac{1 \pm \sqrt{1 + 4x}}{2} \]

### Step 5: Determine the correct solution
Since \( f(x) \) represents a square root function, we take the positive root:
\[ f(x) = \frac{1 + \sqrt{1 + 4x}}{2} \]

### Conclusion
The function \( f(x) \) is given by:
\[ f(x) = \frac{1 + \sqrt{1 + 4x}}{2} \]