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A function $f: A \to B$ is one-one if $f(x_1) = f(x_2)$ implies $x_1 = x_2$. Let us check for $g(x) = x^2 - 4x - 5$. Consider $g(x_1) = g(x_2)$. $$x_1^2 - 4x_1 - 5 = x_2^2 - 4x_2 - 5$$ $$x_1^2 - x_2^2 - 4(x_1 - x_2) = 0$$ $$(x_1 - x_2)(x_1 + x_2) - 4(x_1 - x_2) = 0$$ $$(x_1 - x_2)(x_1 + x_2 - 4) = 0$$ This implies $x_1 = x_2$ or $x_1 + x_2 = 4$. Since there exist distinct values (e.g., $x=1$ and $x=3$ both give $g(1) = -8$ and $g(3) = -8$), the function is not one-one.
A function is onto if its range equals its codomain. Assuming the codomain is $\mathbb{R}$, we find the range of $g(x) = x^2 - 4x - 5$. Completing the square: $$g(x) = (x^2 - 4x + 4) - 4 - 5 = (x - 2)^2 - 9$$ Since $(x - 2)^2 \ge 0$ for all real $x$, the minimum value is $-9$. Thus, the range is $[-9, \infty)$. Since the range is not equal to the codomain $\mathbb{R}$, the function is not onto.
Final Answer: The function is neither one-one nor onto.
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