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

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

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)=[x^(2) if x is irrational,1 if x is rational then

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

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

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

f(x)={x if x is rational -x if x is irrational then Lt_(x rarr0)f(x)=

If the functions f(x) and g(x) are defined on R rarr 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

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