Question:

Step 1: Define the function and its domain/codomain. Since the specific function was not provided in the question, we consider a common example of a function from real numbers to real numbers that is neither one-one nor onto. Let the function $f$ be defined as $f: \mathbb{R} \to \mathbb{R}$ by $f(x) = x^2$. Step 2: Check for injectivity (one-one property). A function $f: A \to B$ is one-one if for any $x_1, x_2 \in A$, $f(x_1) = f(x_2)$ implies $x_1 = x_2$. To show that a function is *not* one-one, we need to find at least two distinct elements in the domain that map to the same element in the codomain. Consider two distinct values from the domain $\mathbb{R}$, for example, $x_1 = 2$ and $x_2 = -2$. We evaluate the function at these points: $$f(2) = 2^2 = 4$$ $$f(-2) = (-2)^2 = 4$$ Since $f(2) = f(-2)$ but $2 \neq -2$, the function $f(x) = x^2$ is not one-one. Step 3: Check for surjectivity (onto property). A function $f: A \to B$ is onto 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. For $f(x) = x^2$, the codomain is $\mathbb{R}$. The range of $f(x) = x^2$ is the set of all non-negative real numbers, which can be written as $[0, \infty)$. Consider an element from the codomain that is not in the range, for example, $y = -1 \in \mathbb{R}$. We try to find an $x \in \mathbb{R}$ such that $f(x) = -1$: $$x^2 = -1$$ There is no real number $x$ whose square is $-1$. Therefore, $-1$ (and any other negative real number) in the codomain $\mathbb{R}$ does not have a pre-image in the domain $\mathbb{R}$. Since the range of $f$ ($[0, \infty)$) is not equal to the codomain $\mathbb{R}$, the function $f(x) = x^2$ is not onto. Step 4: Conclude the mapping nature. Based on the analysis in Step 2 and Step 3, the function $f(x) = x^2$ (from $\mathbb{R} \to \mathbb{R}$) is neither one-one nor onto. The final answer is $\boxed{\text{Neither one-one nor onto}}$.