`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 ontoA. one-one and intoB. one-one but nor ontoC. onto but not one-oneD. one-one nor onto

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

1.	`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 ontoA. one-one and intoB. one-one but nor ontoC. onto but not one-oneD. one-one nor onto
Answer» Correct Answer - D Let `phi (x)=f(x)-g(x)={(x", "x in Q),(-x", "x notin Q):}` Now, to check one-one. Take any straight line parallel to X-axis which will intersect `phi(x)` only at one point. `rArr phi(x)` is one-one. To check onto. As `f(x)={(x", "x in Q),(-x", "x notin Q):}`, which shows y = x and y = -x for rational and irrational values `rarr y in ` real numbers. `therefore="Codomain"=rArr "onto" ` Thus, `f-g` is one-one and onto.