2022 H2 Maths Paper 1, Question 10

The curve has equation

Differentiating,

At a stationary point, $\frac{dy}{dx} = 0$:

Since $x \neq 1$, $(x - 1)^{2}$ takes every value in $(0,\infty)$ as $x$ ranges over its domain, so this equation has a solution (a stationary point exists) exactly when $\frac{a + 2b}{a} > 0$, and has none when $\frac{a + 2b}{a} \leq 0$. For $C$ to have no stationary points we therefore need $\frac{a + 2b}{a} \leq 0$; since $a > 0$,

We are given $b \neq - \frac{1}{2}a$, i.e. $a + 2b \neq 0$, so equality is excluded:

With $b = - 2a$: $a + 2b = a - 4a = - 3a$, so

Since the curve is undefined at $x = 1$, $x = 1$ is a vertical asymptote. As $x \rightarrow \pm \infty$, $\frac{3a}{x - 1} \rightarrow 0$, so

is the oblique asymptote. At $x = 0$:

so $C$ crosses the $y$-axis at $(0,a)$. Setting $y = 0$ and dividing by $a$ (since $a > 0$):

Multiplying through by $(x - 1)$:

$\therefore$ $C$ has asymptotes $x = 1$ and $y = a(x - 2)$, crosses the $y$-axis at $(0,a)$, and crosses the $x$-axis at $x = \frac{3 \pm \sqrt{13}}{2}$.

The sketch below is drawn for $a = 1$ (it also shows the line from part (c), in gold).

The graph of $y = ax - a = a(x - 1)$ is a straight line through $(1,0)$ with gradient $a$, shown in gold on the sketch above.

Setting $a = 1$ in part (b), $C$ becomes $y = (x - 2) - \frac{3}{x - 1} = x - 2 - \frac{3}{x - 1}$, the left side of the inequality; setting $a = 1$ in part (c), the line becomes $y = x - 1$, the right side. So solving

is the same as finding where $C$ (with $a = 1$) lies on or below the line $y = x - 1$.

To find where they meet, set $a(x - 2) - \frac{3a}{x - 1} = a(x - 1)$ and divide by $a$:

This is the only intersection (the equation reduced to a linear one, not quadratic), matching $( - 2, - 3)$ on the sketch.

From the sketch: for $x > 1$, the curve descends from $- \infty$ near the asymptote and stays below the line for all $x > 1$. For $x < 1$, the curve descends from $\text{+}\infty$ near the asymptote, crosses the line once at $x = - 2$, and lies above the line for $- 2 < x < 1$ and below it for $x < - 2$.