| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | January |
| 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 understanding of asymptotes, algebraic manipulation, discriminant analysis for real roots, and connecting the range of k-values to stationary points without calculus. The multi-step reasoning linking parts (c) to (d) and the constraint against differentiation elevates this above standard A-level questions, though each individual step uses accessible techniques. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02n Sketch curves: simple equations including polynomials1.02q Use intersection points: of graphs to solve equations |
| Answer | Marks | Guidance |
|---|---|---|
| 7(a) Denom never zero, so no vert asymp | E1 | |
| Horizontal asymptote is \(y = 0\) | B1 | 2 marks |
| 7(b) \(x - 4 = k(x^2 + 9)\) | M1 | 2 marks |
| Hence result clearly shown | A1 | |
| 7(c) Real roots if \(b^2 - 4ac \geq 0\) | E1 | |
| Discriminant \(= 1 - 4k(9k + 4)\) | M1 | 5 marks |
| \(\ldots = -(36k^2 + 16k - 1)\) | m1 | |
| \(\ldots = -(18k - 1)(2k + 1)\) | m1 | |
| Result (AG) clearly justified | A1 | |
| 7(d) \(k = -\frac{1}{2} \Rightarrow -\frac{1}{2}x^2 - x - \frac{1}{2} = 0\) | M1A1 | |
| \(\ldots \Rightarrow (x + 1)^2 = 0 \Rightarrow x = -1\) | A1 | 6 marks |
| \(k = \frac{1}{18} \Rightarrow \frac{1}{18}x^2 - x + \frac{2}{9} = 0\) | A1 | |
| \(\ldots \Rightarrow (x - 9)^2 = 0 \Rightarrow x = 9\) | A1 | |
| SPs are \(\left(-1, -\frac{1}{2}\right), \left(9, \frac{1}{18}\right)\) | A1 |
**7(a)** Denom never zero, so no vert asymp | E1 | |
Horizontal asymptote is $y = 0$ | B1 | 2 marks
**7(b)** $x - 4 = k(x^2 + 9)$ | M1 | 2 marks | AG
Hence result clearly shown | A1 |
**7(c)** Real roots if $b^2 - 4ac \geq 0$ | E1 | | PI (at any stage)
Discriminant $= 1 - 4k(9k + 4)$ | M1 | 5 marks | m1 for expansion
$\ldots = -(36k^2 + 16k - 1)$ | m1 | | m1 for correct factorisation
$\ldots = -(18k - 1)(2k + 1)$ | m1 | | eg by sketch or sign diagram
Result (AG) clearly justified | A1 |
**7(d)** $k = -\frac{1}{2} \Rightarrow -\frac{1}{2}x^2 - x - \frac{1}{2} = 0$ | M1A1 | | or equivalent using $k = \frac{1}{18}$
$\ldots \Rightarrow (x + 1)^2 = 0 \Rightarrow x = -1$ | A1 | 6 marks
$k = \frac{1}{18} \Rightarrow \frac{1}{18}x^2 - x + \frac{2}{9} = 0$ | A1 |
$\ldots \Rightarrow (x - 9)^2 = 0 \Rightarrow x = 9$ | A1 |
SPs are $\left(-1, -\frac{1}{2}\right), \left(9, \frac{1}{18}\right)$ | A1 | | correctly paired
**Total: 15 marks**
---7 A graph has equation
$$y = \frac { x - 4 } { x ^ { 2 } + 9 }$$
\begin{enumerate}[label=(\alph*)]
\item Explain why the graph has no vertical asymptote and give the equation of the horizontal asymptote.
\item Show that, if the line $y = k$ intersects the graph, the $x$-coordinates of the points of intersection of the line with the graph must satisfy the equation
$$k x ^ { 2 } - x + ( 9 k + 4 ) = 0$$
\item Show that this equation has real roots if $- \frac { 1 } { 2 } \leqslant k \leqslant \frac { 1 } { 18 }$.
\item Hence find the coordinates of the two stationary points on the curve.\\
(No credit will be given for methods involving differentiation.)
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2011 Q7 [15]}}