Question

A function $f:N \rightarrow N$ is defined as $f(x)=x^2+17$. What is the type of function?

• A. Injective
• B. Surjective
• C. Bijective
• D. neither surjective nor injective

Answer 1

Answer: (A)

Injective

Answer 2

To determine the type of function $f: N \rightarrow N$ defined as $f(x) = x^2 + 17$, we need to check for injectivity (one-to-one) and surjectivity (onto). The domain $N$ and codomain $N$ represent the set of natural numbers, typically $N = \{1, 2, 3, \ldots\}$.

1. Check for Injectivity (One-to-one)

A function $f: A \rightarrow B$ is injective if for any two distinct elements $x_1, x_2$ in the domain $A$, their images under $f$ are also distinct, i.e., $x_1 \neq x_2 \implies f(x_1) \neq f(x_2)$. Equivalently, if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

Let's assume $f(x_1) = f(x_2)$ for $x_1, x_2 \in N$. $x_1^2 + 17 = x_2^2 + 17$ Subtracting 17 from both sides: $x_1^2 = x_2^2$ Taking the square root of both sides: $\sqrt{x_1^2} = \sqrt{x_2^2}$ $|x_1| = |x_2|$ Since $x_1, x_2 \in N$, they are positive integers. Therefore, $|x_1| = x_1$ and $|x_2| = x_2$. So, $x_1 = x_2$.

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$, the function $f(x)$ is injective.

2. Check for Surjectivity (Onto)

A function $f: A \rightarrow B$ is surjective if for every element $y$ in the codomain $B$, there exists at least one element $x$ in the domain $A$ such that $f(x) = y$. In other words, the range of the function must be equal to its codomain.

The codomain is $N = \{1, 2, 3, \ldots\}$. Let's find the range of $f(x)$. Since $x \in N$, the smallest value $x$ can take is 1. If $x=1$, $f(1) = 1^2 + 17 = 1 + 17 = 18$. If $x=2$, $f(2) = 2^2 + 17 = 4 + 17 = 21$. If $x=3$, $f(3) = 3^2 + 17 = 9 + 17 = 26$.

For any $x \in N$, $x \ge 1$. Therefore, $x^2 \ge 1^2 = 1$. And $x^2 + 17 \ge 1 + 17 = 18$.

This means that the minimum value in the range of $f(x)$ is 18. The range of $f$ is $\{18, 21, 26, \ldots\}$. The codomain is $N = \{1, 2, 3, \ldots\}$.

Since the range $\{18, 21, 26, \ldots\}$ is a proper subset of the codomain $\{1, 2, 3, \ldots\}$, there are elements in the codomain (e.g., $1, 2, \ldots, 17$) that are not images of any element from the domain. Therefore, the function $f(x)$ is not surjective.

Conclusion

The function $f(x) = x^2 + 17$ is injective but not surjective. A function is bijective if it is both injective and surjective. Since $f(x)$ is not surjective, it is not bijective. Given the options:

• Injective: This is true.
• Surjective: This is false.
• Bijective: This is false (since not surjective).
• neither surjective nor injective: This is false (since it is injective).

The most accurate description among the given options is "Injective".