| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2005 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Simple rational function analysis |
| Difficulty | Standard +0.3 This is a structured FP1 question on rational functions covering standard techniques: finding horizontal asymptotes by considering limits, explaining absence of vertical asymptotes via denominator analysis, and using the discriminant condition to find stationary points. While it requires multiple steps and connects the discriminant to stationary points (a moderately sophisticated insight for AS-level), each individual part follows well-established procedures taught in FP1, making it slightly easier than average overall. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02n Sketch curves: simple equations including polynomials1.07n Stationary points: find maxima, minima using derivatives |
| Surname | Other Names | ||||||||
| Centre Number | Candidate Number | ||||||||
| Candidate Signature | |||||||||
| Answer | Marks | Guidance |
|---|---|---|
| Asymptote is \(y = 1\) | B1 | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| Denominator never zero | E2,1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| — | — | — |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x) = k \Rightarrow (1-k)x^2 + 4x - 9k = 0\); Equal roots if \(16 + 36k(1-k) = 0\); ie if \(9k^2 - 9k - 4 = 0\) | M1A1, m1, A1 | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Solving quadratic for \(k\); \(k = -\frac{1}{3}\) or \(k = \frac{4}{3}\); \(4x^2 + 12x + 9 = 0\) or \(x^2 - 12x + 36 = 0\); SPs when \(x = -\frac{3}{2}, \ldots\) ;and when \(x = 6, \ldots\); SPs are \(\left(-\frac{3}{2}, -\frac{1}{3}\right)\) and \(\left(6, \frac{4}{3}\right)\) | M1, A1, m1, A1, A1, A1 | 6 marks |
**Part (a)(i)**
| Asymptote is $y = 1$ | B1 | 1 mark | — |
**Part (a)(ii)**
| Denominator never zero | E2,1 | 2 marks | E1 if incomplete |
**Part (a)(iii)**
| — | — | — | — |
**Part (b)**
| $f(x) = k \Rightarrow (1-k)x^2 + 4x - 9k = 0$; Equal roots if $16 + 36k(1-k) = 0$; ie if $9k^2 - 9k - 4 = 0$ | M1A1, m1, A1 | 4 marks | convincingly shown (AG) |
**Part (c)**
| Solving quadratic for $k$; $k = -\frac{1}{3}$ or $k = \frac{4}{3}$; $4x^2 + 12x + 9 = 0$ or $x^2 - 12x + 36 = 0$; SPs when $x = -\frac{3}{2}, \ldots$ ;and when $x = 6, \ldots$; SPs are $\left(-\frac{3}{2}, -\frac{1}{3}\right)$ and $\left(6, \frac{4}{3}\right)$ | M1, A1, m1, A1, A1, A1 | 6 marks | NMS 2/2 |
**Total: 13 marks**
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**GRAND TOTAL: 75 marks**9 The function f is defined by
$$f ( x ) = \frac { x ^ { 2 } + 4 x } { x ^ { 2 } + 9 }$$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item The graph of $y = \mathrm { f } ( x )$ has an asymptote which is parallel to the $x$-axis. Find the equation of this asymptote.
\item Explain why the graph of $y = \mathrm { f } ( x )$ has no asymptotes parallel to the $y$-axis.
\end{enumerate}\item Show that the equation $\mathrm { f } ( x ) = k$ has two equal roots if $9 k ^ { 2 } - 9 k - 4 = 0$.
\item Hence find the coordinates of the two stationary points on the graph of $y = \mathrm { f } ( x )$.
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Surname & \multicolumn{9}{|c|}{Other Names} \\
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\multicolumn{2}{|c|}{Centre Number} & & & \multicolumn{2}{|l|}{Candidate Number} & & & & \\
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\multicolumn{3}{|c|}{Candidate Signature} & & & & & & & \\
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General Certificate of Education\\
June 2005\\
Advanced Subsidiary Examination
MATHEMATICS\\
MFP1\\
Unit Further Pure 1
ASSESSMENT and\\
QUALIFICATIONS\\
ALLIANCE
Wednesday 22 June 2005 Afternoon Session
Insert for use in Question 7.\\
Fill in the boxes at the top of this page.\\
Fasten this insert securely to your answer book.
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2005 Q9 [13]}}