If `f(x)={x,` when `x` is rational and `0,` when `x` is irrational `g(x)={0,` when `x` is rational and `x,` when `x` is irrational then `(f-g)` is

A. one-one and into
B. neither one-one nor onto
C. many one and onto
D. one-one and onto

Options

A. one-one and into
B. neither one-one nor onto
C. many-one and onto
D. one-one and onto

Answer

D

Explanation

We have
`(f-g)(x)=f(x)-g(x)`
`Rightarrow (f-g)(x)={{:(,x-0=x,"if x is rational"),(,0-x=-x,"if x is irratioanl"):}`
Let h=f-g
Let x,y be any two distinct real numbers. Then
`x ne y Rightarrow -x ne -y`
`therefore x ne yRightarrow h(x) ne h(y) Rightarrow (f-g)(x) ne (f-g)(y)`
`Rightarrow (f-g)(-y)=y`
Thus, every `y in R ("Co-domain")` has its pre-image in R (domain). So `f-g:R to R` is onto.
Hence, `f-g:R to R` is both one-one and onto