Question

Let $S$ be set of all rational numbers. The functions $f:R\to R,g:R\to R$ are defined as
$f\left(x\right)=\{\begin{array}{cc}0,& x\in S\\ 1,& x\notin S\end{array}$
$g\left(x\right)=\{\begin{array}{cc}-1& x\in S\\ 0& x\notin S\end{array}$
then, $\left(fog\right)\left(\pi \right)+\left(gof\right)\left(e\right)=$

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

Answer: (A)

$-1$

$f\left(x\right)=\{\begin{array}{cc}0,& x\text{is rational}\\ 1,& x\text{is irrational}\end{array}$
$g\left(x\right)=\{\begin{array}{cc}-1& x\text{is rational}\\ 0& x\text{is irrational}\end{array}$
$fog\left(\pi \right)=f\left(g\right(\pi \left)\right)=f\left(0\right)=0$
so $gof\left(e\right)=g\left(f\right(e\left)\right)=g\left(1\right)=-1$
$\therefore fog\left(\pi \right)+gof\left(e\right)=-1$