Let f: ℝ arrow ℝ be defined as f(x) = (2x^2 - 3x + 2)/(3x^2 + x +...

Given $f(x) = \dfrac{2x^2 - 3x + 2}{3x^2 + x + 3}$

To check for one-one, let $f(x) = f(y)$

$\dfrac{2x^2 - 3x + 2}{3x^2 + x + 3} = \dfrac{2y^2 - 3y + 2}{3y^2 + y + 3}$

Cross-multiplying and simplifying, we get:

$11x^2y - 11xy^2 - 11x + 11y = 0$

$11xy(x - y) - 11(x - y) = 0$

$11(x - y)(xy - 1) = 0$

This implies $x = y$ or $xy = 1$.

Since $f(x) = f(1/x)$ for $x \neq 0$ (for example, $f(2) = f(1/2) = \dfrac{4}{17}$), the function is many-one (not one-one).

To check for onto, let $y = \dfrac{2x^2 - 3x + 2}{3x^2 + x + 3}$

$(3y - 2)x^2 + (y + 3)x + (3y - 2) = 0$

For $x \in \mathbb{R}$, the discriminant $D \ge 0$:

$(y + 3)^2 - 4(3y - 2)^2 \ge 0$

$(y + 3 - 6y + 4)(y + 3 + 6y - 4) \ge 0$

$(7 - 5y)(7y - 1) \ge 0$

$(5y - 7)(7y - 1) \le 0$

$y \in \left[\dfrac{1}{7}, \dfrac{7}{5}\right]$

Since the range is a proper subset of the codomain $\mathbb{R}$, the function is not onto.

Thus, $f$ is neither one-one nor onto.

Answer: neither one-one nor onto