If `f(x)` is defined on `[0, 1]` by the rule `f(x)={x,` if `x` is rational ,`1-x,` if `x` is rational ' then for all `x in [0,1]` ,`f(f(x))` is

Options

A. constant
B. 1+x
C. x
D. None of these

Explanation

To solve the problem, we need to find \( f(f(x)) \) for the function defined as follows:

\[ f(x) = \begin{cases} x & \text{if } x \text{ is rational} \\ 1 - x & \text{if } x \text{ is irrational} \end{cases} \]

### Step 1: Determine \( f(f(x)) \) when \( x \) is rational.

If \( x \) is a rational number, then by the definition of the function:

\[ f(x) = x \]

Now we need to find \( f(f(x)) \):

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

### Step 2: Determine \( f(f(x)) \) when \( x \) is irrational.

If \( x \) is an irrational number, then:

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

Next, we need to find \( f(f(x)) \):

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

Now, we need to check if \( 1 - x \) is rational or irrational. Since \( x \) is irrational, \( 1 - x \) is also irrational. Therefore, we use the second case of the function:

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

### Conclusion

In both cases (whether \( x \) is rational or irrational), we find that:

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

Thus, for all \( x \in [0, 1] \):

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