Problem Statement
Let $$f : \mathbb{R} \to \mathbb{R}$$ be defined by $$f(x) = x^2 + 1$$. Is $$f$$ one-one? Is it onto? Justify your answers with proper mathematical arguments.
Verified Solution & Marking Scheme
Test for Injectivity (One-One)
Consider $$x_1 = 1$$ and $$x_2 = -1$$. Both are real numbers in the domain. We have $$f(1) = 1^2 + 1 = 2$$ and $$f(-1) = (-1)^2 + 1 = 2$$. Since $$f(1) = f(-1)$$ but $$1 \neq -1$$, $$f$$ is NOT one-one.
Test for Surjectivity (Onto)
For any $$x \in \mathbb{R}$$, $$x^2 \geq 0 \implies x^2 + 1 \geq 1$$. Thus the range of $$f$$ is $$[1, \infty)$$. However, the co-domain is $$\mathbb{R}$$. Since the range is a proper subset of the co-domain (e.g., negative reals like $$-2$$ have no pre-image), $$f$$ is NOT onto.