Answer: First, injective means that for every y, there is only one x such f(x) = y, and surjective means that if f is a function that goes from the set {x} to the set {y}, for every element y in the codomain there is at least one element x in the domain such f(x) = y
(a) f:R----> + R given by f(x) = x2 (i guess you written )
The function is not injective, because f(-2) = f(2), and is surjective, because the codomain is +R, and is always a positive real number.
(b) f:N----> + N given by f(n) = n2 (i guess you written )
As the naturals have no negative numbers, in this case the function is injective, but isn't surjective, because there is no number that when squared is equal to 5, for example.
(c) f: Zx Z → Z given by f(n, k) = n +k:
here the domain is of the form (n,k) and the function is n +k, so the numbers (z,w) and (w,z) return the same value when evaluated in f, then f is not injective. And is easy to see that f is surjective, because in the sum you can reach al the integers on the codomain.
Since and
, we can rewrite the right side of the equation as
Using the identity , we can subtract
from either side to obtain the identity
substituting that into our previous expression, the right side of our equation simply becomes
We can now write our whole equation as
Adding 2 to both sides:
dividing both sides by 3:
When 0 ≤ x ≤ π, tan x can only be equal to 1 when sin x = cos x, which happens at x = π/4, and it can only be equal to -1 when -sin x = cos x, which happens at x = 3π/4