1.
(d) f : R+ → R+
2.
(a) For x, y ∈ N,
f(x) = f(y) ⇒ x3 = y3 ⇒ x = y
Therefore, f is injective
For 2 ∈ N, there does not exist x in the domain N such that f(x) = x3 = 2.
∴ f is not surjective.
(b) f : R → R given by f(x) = [x]
It seen that f(1.1) = 1 and f(1.8) = 1;
But 1.1 ≠ 1.8;
∴ f is not injective
There does not exist any element x ∈ R
such that f(x) = 0.7
∴ f is not surjective.