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 f(x)={x, when x is rational,1-x, when x is irrational,then

The functions RR,g:R R are defined as f(x)={0, whenxis rational and 1 when x is irrational } and g(x)={-1 when x is rational and 0 when x~ is~ irrational}~ find~ fog (pi) and gof(e)

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

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

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

Two mappings f:R to R and g:R to R are defined as follows: f(x)={(0,"when x is rational"),(1,"when x is irrational"):} and g(x)={(-1,"wnen x is rational"),(0,"when x is irrational"):} then the value of [(gof)(e)+(fog)(pi)] is -

Two mappings f:R to R and g:R to R are defined as follows: f(x)={(0,"when x is rational"),(1,"when x is irrational"):} and g(x)={(-1,"wnen x is rational"),(0,"when x is irrational"):} then the value of [(gof)(e)+(fog)(pi)] 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

If f(x)={x,x is rational 1-x,x is irrational , then f(f(x)) is