To determine whether a function is injective or not, we need to check whether different values in the domain map to the same value in the codomain. If this does not happen, then the function is injective.
A) f
= n^2 + 1
Let's consider two different integers, say a and b, such that f(a) = f(b).
This implies that a^2 + 1 = b^2 + 1.
Simplifying the above equation, we get (a^2 - b^2) = 0.
Factoring, we get (a-b)(a+b) = 0.
This implies that either a = b or a = -b.
Therefore, the function is not injective.
B) f
= n^3
Let's consider two different integers, say a and b, such that f(a) = f(b).
This implies that a^3 = b^3.
Taking the cube root of both sides, we get a = b.
Therefore, the function is injective.
C) f
= n^3 - 4
Let's consider two different integers, say a and b, such that f(a) = f(b).
This implies that a^3 - 4 = b^3 - 4.
Simplifying the above equation, we get (a^3 - b^3) = 0.
Factoring, we get (a-b)(a^2 + ab + b^2) = 0.
Since the discriminant of a^2 + ab + b^2 is negative, it cannot be zero.
Therefore, the only possibility is a = b.
Therefore, the function is injective.
D) f
= n/2
Let's consider two different integers, say a and b, such that f(a) = f(b).
This implies that a/2 = b/2.
Simplifying the above equation, we get a = b.
Therefore, the function is injective.
In conclusion, the injective functions are B) f
= n^3 and D) f
= n/2.