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Let $Q$ be the set of all rational numbers in $[0,1]$ and $f:[0,1] \rightarrow[0,1]$ be defined by
$$
f(x)=\left\{\begin{array}{r}
x \text { for } x \in Q \\
1-x \text { for } x \notin Q
\end{array}\right.
$$
Then, the set $S=\{x \in[0,21]:(f o f)(x)\}$ is equal to
Options:
$$
f(x)=\left\{\begin{array}{r}
x \text { for } x \in Q \\
1-x \text { for } x \notin Q
\end{array}\right.
$$
Then, the set $S=\{x \in[0,21]:(f o f)(x)\}$ is equal to
Solution:
2781 Upvotes
Verified Answer
The correct answer is:
$[0,1]$
Given, $f(x)=\left\{\begin{array}{cc}x & \text { for } x \in Q \\ 1-x & \text { for } x \notin Q\end{array}\right.$ is defined for
$$
f:[0,1] \rightarrow[0,1]
$$
If $x$ is rational, then
$$
\begin{array}{rlrl}
f(x) & =x \\
\therefore \quad & f(f(x)) & =f(x)=x
\end{array}
$$
If $x$ is irrational, then
$$
\begin{aligned}
f(x) & =1-x \\
\therefore \quad f \circ f(x) & =f(1-x)=1-(1-x) \\
& =x
\end{aligned}
$$
$\therefore f \circ f(x)=x$ is possible for all values of domain $[0,1]$.
$$
f:[0,1] \rightarrow[0,1]
$$
If $x$ is rational, then
$$
\begin{array}{rlrl}
f(x) & =x \\
\therefore \quad & f(f(x)) & =f(x)=x
\end{array}
$$
If $x$ is irrational, then
$$
\begin{aligned}
f(x) & =1-x \\
\therefore \quad f \circ f(x) & =f(1-x)=1-(1-x) \\
& =x
\end{aligned}
$$
$\therefore f \circ f(x)=x$ is possible for all values of domain $[0,1]$.
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