If the functions `f(x)` and `g(x)` are defined on `R -> R` such that `f(x) = (x+3, x in` ratioal; `4x, x in ` irrational `` , and `g(x) = x + sqrt5, 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

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)= {(0,"x" in " rational"),(x, x in " irrational"):} and g(x)= {(0,"x" in " irrational"),(x, x in " rational"):} then (f - g) (x) is a)one - one and onto b)neither one - one nor onto c)one - one but not onto d)onto but not one - one

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

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

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