If f is a continuous function and `x^3-(sqrt5+1)x^2+(sqrt5-2 + f(x))x+2 sqrt5-sqrt5 f(x)=0` satisfies for `x in R` then `f(sqrt5)` is equal to

Options

A. `2-sqrt5`
B. `5+sqrt5`
C. `3-sqrt5`
D. can not be determined

Answer

To solve the problem, we need to analyze the given equation and find the value of \( f(\sqrt{5}) \).

### Step 1: Substitute \( x = \sqrt{5} \) into the equation

The equation given is:
\[ x^3 - (\sqrt{5} + 1)x^2 + (\sqrt{5} - 2 + f(x))x + (2\sqrt{5} - \sqrt{5}f(x)) = 0 \]

Substituting \( x = \sqrt{5} \):
\[ (\sqrt{5})^3 - (\sqrt{5} + 1)(\sqrt{5})^2 + (\sqrt{5} - 2 + f(\sqrt{5}))(\sqrt{5}) + (2\sqrt{5} - \sqrt{5}f(\sqrt{5})) = 0 \]

### Step 2: Calculate each term

1. **Calculate \( (\sqrt{5})^3 \)**:
\[ (\sqrt{5})^3 = 5\sqrt{5} \]

2. **Calculate \( (\sqrt{5} + 1)(\sqrt{5})^2 \)**:
\[ (\sqrt{5} + 1)(\sqrt{5})^2 = (\sqrt{5} + 1)(5) = 5\sqrt{5} + 5 \]

3. **Calculate \( (\sqrt{5} - 2 + f(\sqrt{5}))(\sqrt{5}) \)**:
\[ (\sqrt{5} - 2 + f(\sqrt{5}))(\sqrt{5}) = 5 - 2\sqrt{5} + f(\sqrt{5})\sqrt{5} \]

4. **Calculate \( (2\sqrt{5} - \sqrt{5}f(\sqrt{5})) \)**:
\[ 2\sqrt{5} - \sqrt{5}f(\sqrt{5}) \]

### Step 3: Substitute back into the equation

Now substituting these calculated values back into the equation:
\[ 5\sqrt{5} - (5\sqrt{5} + 5) + (5 - 2\sqrt{5} + f(\sqrt{5})\sqrt{5}) + (2\sqrt{5} - \sqrt{5}f(\sqrt{5})) = 0 \]

### Step 4: Simplify the equation

Combining like terms:
\[ 5\sqrt{5} - 5\sqrt{5} - 5 + 5 - 2\sqrt{5} + f(\sqrt{5})\sqrt{5} + 2\sqrt{5} - \sqrt{5}f(\sqrt{5}) = 0 \]

This simplifies to:
\[ 0 + f(\sqrt{5})\sqrt{5} - \sqrt{5}f(\sqrt{5}) = 0 \]

### Step 5: Factor out \( f(\sqrt{5}) \)

This gives us:
\[ 0 = 0 \]

This means that the terms involving \( f(\sqrt{5}) \) cancel out, leading us to conclude that \( f(\sqrt{5}) \) can take any value without affecting the validity of the equation.

### Conclusion

Since \( f(\sqrt{5}) \) cannot be determined from the given equation, we conclude:
\[ f(\sqrt{5}) \text{ is indeterminate.} \]