Problem Statement
Consider the rational function $f(x) = \frac{2x^2 - 3x + 5}{x - 2}$ for $x \in \mathbb{R}, x \neq 2$.
**(a)** [3 marks] Show by algebraic long division that $f(x) = 2x + 1 + \frac{7}{x - 2}$, and hence write down the equations of both asymptotes.
**(b)** [3 marks] Find the exact coordinates of the two stationary points on the graph of $y = f(x)$.
**(c)** [2 marks] State the range of values of $k \in \mathbb{R}$ for which the equation $f(x) = k$ has no real solutions.
Verified Solution & Marking Scheme
Part (a) [M1]
Dividing: $(2x^2 - 3x + 5) \div (x - 2) = 2x + 1$ remainder $7$
Part (a) [A1]
$f(x) = 2x + 1 + \frac{7}{x - 2}$ (AG)
Part (a) [A1]
Vertical asymptote: $x = 2$; Oblique asymptote: $y = 2x + 1$
Part (b) [M1]
$f'(x) = 2 - \frac{7}{(x - 2)^2} = 0 \implies (x - 2)^2 = \frac{7}{2} \implies x = 2 \pm \sqrt{3.5}$
Part (b) [A1]
$x_1 = 2 + \sqrt{3.5} \approx 3.87, \quad x_2 = 2 - \sqrt{3.5} \approx 0.129$
Part (b) [A1]
$y_1 = 2(2 + \sqrt{3.5}) + 1 + \frac{7}{\sqrt{3.5}} = 5 + 4\sqrt{3.5} \approx 12.48$; $y_2 = 5 - 4\sqrt{3.5} \approx -2.48$
Part (c) [R1]
The graph has local minimum at $y_1 \approx 12.48$ and local maximum at $y_2 \approx -2.48$
Part (c) [A1]
No solutions for $5 - 4\sqrt{3.5} < k < 5 + 4\sqrt{3.5}$ (or $-2.48 < k < 12.5$)