If the functions `f(x) and g(x)` are defined on `R -> R` such that `f(x)={0, x in` retional and `x, x in` irrational ; `g(x)={0, x in` irratinal and `x,x in` rational then `(f-g)(x)` is

If the functions f(x) and g(x) are defined on R rarr R such that f(x)=(x+3,x in ratioal 4x,x in irrational,and g(x)=x+sqrt(5),x in irrational;-x,x in rational Then (f-g)(x) is

If the functions f(x) and g(x) are defined on R rarr R such that f(x) ={:{(0, x in " rational "),( x, x in " irrational ") :} g(x) ={:{(0, x in " irrational ") ,(x, x in " rational "):} then (f-g) (x) 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

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

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)=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

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, 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