Let `f:[-2, 2]rarrR` be a continuous function such that f(x) assumes only irrationlal values. If `f(sqrt2)=sqrt2`. Then

Options

A. `f(0)=0`
B. `f(sqrt2-1)=sqrt2-1`
C. `f(sqrt2-1)=sqrt2+1`
D. `f(sqrt2-1)=sqrt2`

Answer

To solve the problem, we need to analyze the properties of the function \( f \) given the constraints.

### Step-by-step Solution:

1. **Understanding the Function**:
We are given a continuous function \( f: [-2, 2] \to \mathbb{R} \) that assumes only irrational values. This means for every \( x \in [-2, 2] \), \( f(x) \) must be an irrational number.

**Hint**: Recall that a function is continuous if small changes in \( x \) result in small changes in \( f(x) \).

2. **Given Condition**:
We know that \( f(\sqrt{2}) = \sqrt{2} \). Since \( \sqrt{2} \) is an irrational number, this condition is consistent with the property of the function.

**Hint**: Identify that \( \sqrt{2} \) is within the interval \([-2, 2]\).

3. **Continuity and the Intermediate Value Theorem**:
By the Intermediate Value Theorem, since \( f \) is continuous on the interval \([-2, 2]\) and takes the value \( \sqrt{2} \) at \( x = \sqrt{2} \), it must take on all values between \( f(-2) \) and \( f(2) \).

**Hint**: Consider what values \( f(-2) \) and \( f(2) \) can take, given that \( f(x) \) can only be irrational.

4. **Bounding the Function**:
If \( f \) takes only irrational values, it cannot jump to any rational number. Therefore, if \( f(\sqrt{2}) = \sqrt{2} \), then \( f(x) \) must be constrained around this value.

**Hint**: Think about the implications of continuity on the values that \( f(x) \) can take near \( \sqrt{2} \).

5. **Conclusion of Constancy**:
Since \( f \) cannot take any rational values and must remain continuous, the only way for \( f \) to satisfy both the condition of being continuous and only taking irrational values is to be constant. Therefore, \( f(x) \) must be equal to \( \sqrt{2} \) for all \( x \) in the interval \([-2, 2]\).

**Hint**: Reflect on the definition of a constant function and how it relates to continuity.

6. **Final Result**:
Thus, we conclude that:
\[ f(x) = \sqrt{2} \quad \text{for all } x \in [-2, 2]. \]

### Summary:
The function \( f \) must be constant and equal to \( \sqrt{2} \) across the entire interval \([-2, 2]\) due to its continuity and the requirement that it only takes irrational values.