| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2008 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polynomial Division & Manipulation |
| Type | Stationary Points of Rational Functions |
| Difficulty | Challenging +1.2 This FP1 question requires finding stationary points without calculus by using the discriminant condition (b²-4ac=0), which is a non-standard technique requiring insight beyond routine differentiation. Part (a) is trivial, part (c) is standard curve sketching, but part (b) elevates this above average difficulty as students must recognize that at a stationary point, a rearranged quadratic in x must have exactly one solution. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02n Sketch curves: simple equations including polynomials1.02o Sketch reciprocal curves: y=a/x and y=a/x^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Asymptotes \(x = 0,\; x = 4,\; y = 0\) | \(B1 \times 3\) | |
| Subtotal | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(y = k \Rightarrow 2 = kx(x-4)\) | M1 | |
| \(\Rightarrow 0 = kx^2 - 4kx - 2\) | A1 | |
| Discriminant \(= (4k)^2 + 8k\) | m1 | |
| At SP \(y = -\frac{1}{2}\) | A1 | not just \(k = -\frac{1}{2}\) |
| \(\Rightarrow 0 = -\frac{1}{2}x^2 + 2x - 2\) | m1 | |
| So \(x = 2\) | A1 | |
| Subtotal | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Curve with three branches approaching vertical asymptotes correctly | B1 | |
| Outer branches correct | B1 | |
| Middle branch correct | B1 | |
| Subtotal | 3 | |
| Total | 12 | |
| TOTAL | 75 |
## Question 9(a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Asymptotes $x = 0,\; x = 4,\; y = 0$ | $B1 \times 3$ | |
| **Subtotal** | **3** | |
## Question 9(b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $y = k \Rightarrow 2 = kx(x-4)$ | M1 | |
| $\Rightarrow 0 = kx^2 - 4kx - 2$ | A1 | |
| Discriminant $= (4k)^2 + 8k$ | m1 | |
| At SP $y = -\frac{1}{2}$ | A1 | not just $k = -\frac{1}{2}$ |
| $\Rightarrow 0 = -\frac{1}{2}x^2 + 2x - 2$ | m1 | |
| So $x = 2$ | A1 | |
| **Subtotal** | **6** | |
## Question 9(c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Curve with three branches approaching vertical asymptotes correctly | B1 | |
| Outer branches correct | B1 | |
| Middle branch correct | B1 | |
| **Subtotal** | **3** | |
| **Total** | **12** | |
| **TOTAL** | **75** | |9 A curve $C$ has equation
$$y = \frac { 2 } { x ( x - 4 ) }$$
\begin{enumerate}[label=(\alph*)]
\item Write down the equations of the three asymptotes of $C$.
\item The curve $C$ has one stationary point. By considering an appropriate quadratic equation, find the coordinates of this stationary point.\\
(No credit will be given for solutions based on differentiation.)
\item Sketch the curve $C$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2008 Q9 [12]}}