Question

# Let Q be the set of all rational numbers in [0, 1] and f: [0,1] $\to$ [0,1] be defined by f(x) = \left\{ \begin{align} & x\text{ for x}\in \text{Q} \\ & 1-x\text{ for x}\notin Q \\ \end{align} \right\}Then the set S= {x$\in$[0,1] :(fof) (x) = x} is equal to,(a) [0,1](b) Q(c) [0,1] – Q(d) (0,1)

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Hint: For solving this problem, we should be aware about the basic properties of functions and relations. In this case, we should be aware about how domains are associated as we carry out the function manipulations. Further, we should know that (fof)(x)=f(f(x)). Thus, this means that we can put the value of f(x) inside the input of f(x) to find the value.

To proceed further, we have,
f(x) = \left\{ \begin{align} & x\text{ for x}\in \text{Q} \\ & 1-x\text{ for x}\notin Q \\ \end{align} \right\}
Thus, to find (fof)(x)=f(f(x)), we have to solve this question in two cases.
Case 1: f(x) = x for x$\in$Q
fof(x) = f(f(x)) = f(x) = x
Since, f(x) = x in this case.
Thus, we get, fof(x) = x for x$\in$Q

Case 2: f(x) = 1-x for x$\notin$Q
fof(x) = f(f(x)) = f(1-x) = 1-(1-x) =x
Since, f(x) =1-x in this case.
Thus, we get, fof(x) = x for x$\notin$Q.
Thus, we have,
fof(x)= \left\{ \begin{align} & x\text{ for x}\in \text{Q} \\ & x\text{ for x}\notin Q \\ \end{align} \right\}
Thus, we have, fof(x) = x for all numbers in set S. Thus, fof(x) = x for the entire range of [0,1].
Hence, the correct answer is (a) [0,1].

Note: While solving the questions involving functions and relations, we should be careful to keep in mind the domain and range at every point of our calculation. For example, although not observed in this problem, if in one of the intermediate steps, we come across a place where we get $\dfrac{1}{1-x}$ term, then we have to remove x=1 from the solution set since, otherwise the denominator would be zero which is not possible.