| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2009 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Rational curve analysis with turning points and range restrictions |
| Difficulty | Standard +0.8 This FP1 question requires multiple techniques: identifying asymptotes from rational functions, algebraic manipulation to show no intersection, deriving conditions for intersection with horizontal lines, using discriminant for equal roots, and connecting this to stationary points. While each individual step is accessible, the multi-part structure requiring sustained reasoning across parts (c) and (d), plus the non-obvious connection between equal roots and stationary points, elevates this above standard A-level questions. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02d Quadratic functions: graphs and discriminant conditions1.02f Solve quadratic equations: including in a function of unknown1.02k 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 |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = 1\) | B1 | |
| \(x = 5\) | B1 | |
| \(y = 0\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Set \(\frac{x^2}{(x-1)(x-5)} = -1\) | M1 | |
| \(x^2 = -(x-1)(x-5) = -(x^2 - 6x + 5)\) | M1 | |
| \(2x^2 - 6x + 5 = 0\), discriminant \(= 36 - 40 = -4 < 0\) | A1 | Hence no intersection |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{x^2}{(x-1)(x-5)} = k \Rightarrow x^2 = k(x-1)(x-5)\) | M1 | |
| \(x^2 = k(x^2 - 6x + 5) \Rightarrow (k-1)x^2 - 6kx + 5k = 0\) | A1 | Shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Equal roots: discriminant \(= 0\): \((6k)^2 - 4(k-1)(5k) = 0\) | M1 | |
| \(36k^2 - 20k(k-1) = 0 \Rightarrow 36k^2 - 20k^2 + 20k = 0\) | ||
| \(16k^2 + 20k = 0 \Rightarrow 4k(4k+5) = 0\) | A1 | Hence \(k(4k+5) = 0\) shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(k = 0\) or \(k = -\frac{5}{4}\) | B1 | |
| \(k=0\): \(x = 0\), point \((0, 0)\) | B1 | |
| \(k = -\frac{5}{4}\): substitute into \((k-1)x^2 - 6kx + 5k = 0\) | M1 | |
| \(-\frac{9}{4}x^2 + \frac{30}{4}x - \frac{25}{4} = 0 \Rightarrow 9x^2 - 30x + 25 = 0 \Rightarrow (3x-5)^2 = 0\) | A1 | |
| \(x = \frac{5}{3}\), point \(\left(\frac{5}{3}, -\frac{5}{4}\right)\) | A1 | Both coordinates correct |
# Question 8:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = 1$ | B1 | |
| $x = 5$ | B1 | |
| $y = 0$ | B1 | |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Set $\frac{x^2}{(x-1)(x-5)} = -1$ | M1 | |
| $x^2 = -(x-1)(x-5) = -(x^2 - 6x + 5)$ | M1 | |
| $2x^2 - 6x + 5 = 0$, discriminant $= 36 - 40 = -4 < 0$ | A1 | Hence no intersection |
## Part (c)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{x^2}{(x-1)(x-5)} = k \Rightarrow x^2 = k(x-1)(x-5)$ | M1 | |
| $x^2 = k(x^2 - 6x + 5) \Rightarrow (k-1)x^2 - 6kx + 5k = 0$ | A1 | Shown |
## Part (c)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Equal roots: discriminant $= 0$: $(6k)^2 - 4(k-1)(5k) = 0$ | M1 | |
| $36k^2 - 20k(k-1) = 0 \Rightarrow 36k^2 - 20k^2 + 20k = 0$ | | |
| $16k^2 + 20k = 0 \Rightarrow 4k(4k+5) = 0$ | A1 | Hence $k(4k+5) = 0$ shown |
## Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $k = 0$ or $k = -\frac{5}{4}$ | B1 | |
| $k=0$: $x = 0$, point $(0, 0)$ | B1 | |
| $k = -\frac{5}{4}$: substitute into $(k-1)x^2 - 6kx + 5k = 0$ | M1 | |
| $-\frac{9}{4}x^2 + \frac{30}{4}x - \frac{25}{4} = 0 \Rightarrow 9x^2 - 30x + 25 = 0 \Rightarrow (3x-5)^2 = 0$ | A1 | |
| $x = \frac{5}{3}$, point $\left(\frac{5}{3}, -\frac{5}{4}\right)$ | A1 | Both coordinates correct |8 A curve has equation
$$y = \frac { x ^ { 2 } } { ( x - 1 ) ( x - 5 ) }$$
\begin{enumerate}[label=(\alph*)]
\item Write down the equations of the three asymptotes to the curve.
\item Show that the curve has no point of intersection with the line $y = - 1$.
\item \begin{enumerate}[label=(\roman*)]
\item Show that, if the curve intersects the line $y = k$, then the $x$-coordinates of the points of intersection must satisfy the equation
$$( k - 1 ) x ^ { 2 } - 6 k x + 5 k = 0$$
\item Show that, if this equation has equal roots, then
$$k ( 4 k + 5 ) = 0$$
\end{enumerate}\item Hence find the coordinates of the two stationary points on the curve.
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2009 Q8 [15]}}