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

Options

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

Answer

To solve the problem, we need to find the composition of the function \( f \) with itself, denoted as \( (f \circ f)(x) \). 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 analyze two cases based on whether \( x \) is rational or irrational.

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

If \( x \) is rational, then by the definition of \( f \):

\[ f(x) = x \]

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

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

So, if \( x \) is rational, \( f(f(x)) = x \).

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

If \( x \) is irrational, then by the definition of \( f \):

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

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

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

Next, we need to determine whether \( 1 - x \) is rational or irrational. Since \( x \) is irrational, \( 1 - x \) will also be irrational (the sum of a rational number and an irrational number is irrational). Therefore:

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

So, if \( x \) is irrational, \( f(f(x)) = x \).

### Conclusion

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

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

Thus, we conclude that:

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

### Final Answer

The final answer is:

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

This corresponds to option 3.

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