1.Let Q\mathbb{Q}Q be the set of all rational numbers in [0,1][0, 1][0,1] and f:[0,1]→[0,1]f : [0, 1] \to [0, 1]f:[0,1]→[0,1] be defined by f(x)={xfor x∈Q1−xfor x∉Qf(x) = \begin{cases} x & \text{for } x \in \mathbb{Q} \\ 1-x & \text{for } x \notin \mathbb{Q} \end{cases}f(x)={x1−xfor x∈Qfor x∈/Q. Then, the set S={x∈[0,1]:(f∘f)(x)=x}S = \{x \in [0, 1] : (f \circ f)(x) = x\}S={x∈[0,1]:(f∘f)(x)=x} is equal toa.[0,1][0, 1][0,1]b.Q\mathbb{Q}Qc.[0,1]−Q[0, 1] - \mathbb{Q}[0,1]−Qd.(0,1)(0, 1)(0,1)Login to continueOnly logged in users canattempt or see the solution.
