Let `f : [0, 1] rarr [0, 1] `be a continuous function such that `f (f (x))=1 for all x in[0,1]`then:

Options

A. `f(x)=x` for at least one `x in (0,1)`
B. f(x) will be differential in [0,1]
C. f(x)+x=0 for at least one x such that `0 le xle 1`
D. none of these

Explanation

To solve the problem, we need to analyze the given function \( f: [0, 1] \to [0, 1] \) which satisfies the condition \( f(f(x)) = 1 \) for all \( x \in [0, 1] \).

### Step-by-Step Solution:

1. **Understanding the Condition**:
The condition \( f(f(x)) = 1 \) implies that applying the function \( f \) twice to any \( x \) in the interval [0, 1] results in the constant value 1. This means that \( f(x) \) must be such that when we apply \( f \) again, we reach 1.

2. **Finding Possible Values for \( f(x) \)**:
Since \( f(f(x)) = 1 \), we can deduce that \( f(x) \) must be a value in the interval [0, 1] such that when \( f \) is applied to it, the result is 1. The only possible value for \( f(x) \) that satisfies \( f(f(x)) = 1 \) is \( f(x) = 1 \).

3. **Checking the Continuity**:
Since \( f \) is continuous on the closed interval [0, 1], and we have established that \( f(x) = 1 \) for all \( x \), we can conclude that the function is constant.

4. **Conclusion**:
Therefore, the only continuous function \( f: [0, 1] \to [0, 1] \) that satisfies the condition \( f(f(x)) = 1 \) for all \( x \in [0, 1] \) is:
\[ f(x) = 1 \quad \text{for all } x \in [0, 1]. \]

### Final Answer:
The function \( f(x) \) is \( f(x) = 1 \) for all \( x \in [0, 1] \).