Question

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

• A. [0, 1]
• B. Q
• C. [0, 1] - Q
• D. (0, 1)

Answer 1

Answer: (A)

[0, 1]

Let $x\in Q$, then,

$f\left(x\right)=x$ where $x\in Q$

So, $fof\left(x\right)=f\left(f\right(x\left)\right)=f\left(x\right)=x$ as $x\in Q$

$\therefore fof\left(x\right)=x$ when $x\in Q$

Now,

Let $x\notin Q$ then

$f\left(x\right)=1-x$

$\therefore fof\left(x\right)=1-(1-x)=x$

as $1-x\notin Q$ as $x\notin Q$

where $x\notin Q$

$fof\left(x\right)=\{\begin{array}{c}xwherex\in Q\&x\in [0,1]\\ xwherex\notin Q\&x\in [0,1]\end{array}$

$\therefore$ the set $S=[0,1]$