If `f:[0,1]to[0,1]` is defined by `f(x)={{:(x," if x is rational"),(1-x," if x is irrational"):}` , then (fof)(x) is

Options

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

Answer

To solve the problem, we need to find \( (f \circ f)(x) \), which means we need to apply the function \( f \) to itself. The function \( f \) is 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 \( f(f(x)) \)

We will consider two cases based on whether \( x \) is rational or irrational.

#### Case 1: \( x \) is rational

If \( x \) is rational, then:
\[ f(x) = x \]
Now we need to find \( f(f(x)) \):
\[ f(f(x)) = f(x) = f(x) = x \]
So, if \( x \) is rational, then \( (f \circ f)(x) = x \).

#### Case 2: \( x \) is irrational

If \( x \) is irrational, then:
\[ f(x) = 1 - x \]
Now we need to find \( f(f(x)) \):
\[ f(f(x)) = f(1 - x) \]
Now we need to determine whether \( 1 - x \) is rational or irrational. Since \( x \) is irrational, \( 1 - x \) is also irrational. Therefore:
\[ f(1 - x) = 1 - (1 - x) = x \]
So, if \( x \) is irrational, then \( (f \circ f)(x) = x \).

### Conclusion

In both cases, whether \( x \) is rational or irrational, we find that:
\[ (f \circ f)(x) = x \]

Thus, the final result is:
\[ (f \circ f)(x) = x \]

### Final Answer
The answer is \( x \).