If $f\left(x\right)=$$\{\begin{array}{cc}x,& \text{when x is rational}\\ 0,& \text{when x is irrational}\end{array}$; $g\left(x\right)=$$\{\begin{array}{cc}0,& \text{when x is rational}\\ x,& \text{when x is irrational}\end{array}$then $(f-g)$ is

The correct option is A (One-one onto )

$(f-g)\left(x\right)=$$\{\begin{array}{cc}x,& x\in Q\\ -x,& x\notin Q\end{array}$
As distinct elements of domain set are mapping to distinct elements of co-domain set. So f is one -one.
And every element of co-domain set has a pre-image element in its domain set.
Thus, $(f-g)\left(x\right)=$ is onto function as well.