A) one-one and onto
B) neither one-one nor onto
C) one-one but not onto
D) onto but not one-one
Correct Answer: A
Solution :
We are given that \[f:R\to R\,\,such\,\,that\,\,f(x)=\left\{ \begin{matrix} 0,\,\,\,\,x\in rational \\ x,\,\,\,\,\,\,x\,\,\,\in \,\,irrational \\ \end{matrix} \right.\]i \[g:R\to R\,\,such\,\,that\,\,g(x)=\left\{ \begin{matrix} 0,\,\,\,\,x\in irrational \\ x,\,\,\,\,\,\,x\,\,\,\in \,\,rational \\ \end{matrix} \right.\] \[\therefore \,\,\,\,(f\,-g):R\to R\,\,such\,\,that\] \[(f\,-g)\,(x)\,\,=\,\,\left\{ \begin{matrix} -x,\,\,if\,\,x\in rational \\ x,\,\,if\,\,x\in irrational \\ \end{matrix} \right.\] Since \[f\,-g:R\to R\] for any x there is only one value of \[\left( f(x)-g(x) \right)\] whether x is rational or irrational. Moreover as \[x\in R,\,\,f(x)-g\,(x)\] also belongs to R. Therefore, \[(f-g)\] is one-one onto.You need to login to perform this action.
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