`f(x)= 0,` if `x` is irrational and `f(x)=x,` if `x` is rational. `g(x)=0,` if `x` is rational and `g(x)=x,` if `x` is irrational. Then `f-g` is

Let f(x)=[0, if x is rational and x if x is irrational and g(x)=[0, if x is irrational and x if x is rational then the function (f-g)x is

If f(x)={x^(2) if x is rational and 0, if x is irrational }, then

Let f(x)=0 for x is rational and x if x is irrational and g(x)=0 if x is irrational and x if x is rational then the function (f-g)x is

f(x)={(x", if x is rational"),(0", if x is irrational"):}, g(x)={(0", if x is rational"),(x", if x is irrational"):} Then, f-g is

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

f(x)={(x", if x is rational"),(0", if x is irrational"):}, g(x)={(0", if x is rational"),(x", if x is irrational"):} Then, f-g is (A) one-one and onto (B) one-one but nor onto (C) onto but not one-one (D) neither one-one nor onto

f(x)={(x", if x is rational"),(0", if x is irrational"):}, g(x)={(0", if x is rational"),(x", if x is irrational"):} Then, (f-g) is (A) one-one and onto (B) one-one but nor onto (C) onto but not one-one (D) neither one-one nor onto

If f(x) = {(x, if \x \ is \rational),(0,if \ x \ is \ irrational):} g(x) = {(0, if \x \ is \rational),(x,if \ x \ is \irrational):} then f-g is

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