| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2005 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polynomial Division & Manipulation |
| Type | Stationary Points of Rational Functions |
| Difficulty | Standard +0.3 This is a structured multi-part question on rational functions requiring standard FP1 techniques: identifying asymptotes, analyzing discriminants for intersections/roots, and finding stationary points via the equal roots condition. While it involves several steps and the equal roots approach is slightly less routine than direct differentiation, each part follows predictable methods with clear guidance, making it slightly easier than average for an A-level question. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02v Inverse and composite functions: graphs and conditions for existence1.07n Stationary points: find maxima, minima using derivatives4.02i Quadratic equations: with complex roots |
| Answer | Marks |
|---|---|
| Asymptotes \(x = 0\), \(y = 1\) | B1, B1 (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Delta = 4 - 8 < 0\), so numerator never 0 | E2,1 (2 marks) | OE; E1 for incomplete explanation |
| Answer | Marks | Guidance |
|---|---|---|
| Method for solving quadratic | M1 | "i" must appear |
| Roots \(-1 \pm i\) | A2,1 (3 marks) | A1 if one error made |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x) = k \Rightarrow x^2 + 2x + 2 = kx^2\) | M1 | |
| \(\Rightarrow (1-k)x^2 + 2x + 2 = 0\) | m1 | |
| Equal roots \(\Rightarrow 4 - 8(1-k) = 0\) | A1 (3 marks) | Convincingly shown (AG) |
| Answer | Marks | Guidance |
|---|---|---|
| \(k = \frac{1}{2}\) | B1 | |
| \(y = \frac{1}{2}\) at SP | B1ft | ft wrong value for \(k\) |
| So \(\frac{1}{2}x^2 + 2x + 2 = 0\) | M1 | |
| and \(x = -2\) at SP | A1 (4 marks) |
## Question 9:
**Part (a)**
Asymptotes $x = 0$, $y = 1$ | B1, B1 (2 marks) |
**Part (b)(i)**
$\Delta = 4 - 8 < 0$, so numerator never 0 | E2,1 (2 marks) | OE; E1 for incomplete explanation
**Part (b)(ii)**
Method for solving quadratic | M1 | "i" must appear
Roots $-1 \pm i$ | A2,1 (3 marks) | A1 if one error made
**Part (c)(i)**
$f(x) = k \Rightarrow x^2 + 2x + 2 = kx^2$ | M1 |
$\Rightarrow (1-k)x^2 + 2x + 2 = 0$ | m1 |
Equal roots $\Rightarrow 4 - 8(1-k) = 0$ | A1 (3 marks) | Convincingly shown (AG)
**Part (c)(ii)**
$k = \frac{1}{2}$ | B1 |
$y = \frac{1}{2}$ at SP | B1ft | ft wrong value for $k$
So $\frac{1}{2}x^2 + 2x + 2 = 0$ | M1 |
and $x = -2$ at SP | A1 (4 marks) |9 The function f is defined by
$$f ( x ) = \frac { x ^ { 2 } + 2 x + 2 } { x ^ { 2 } }$$
\begin{enumerate}[label=(\alph*)]
\item Write down the equations of the two asymptotes to the curve $y = \mathrm { f } ( x )$.
\item By considering the expression $x ^ { 2 } + 2 x + 2$ :
\begin{enumerate}[label=(\roman*)]
\item show that the graph of $y = \mathrm { f } ( x )$ does not intersect the $x$-axis;
\item find the non-real roots of the equation $\mathrm { f } ( x ) = 0$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Show that, if the equation $\mathrm { f } ( x ) = k$ has two equal roots, then
$$4 - 8 ( 1 - k ) = 0$$
\item Deduce that the graph of $y = \mathrm { f } ( x )$ has exactly one stationary point and find its coordinates.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2005 Q9 [11]}}