| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2006 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Simple rational function analysis |
| Difficulty | Standard +0.8 This FP1 question requires multiple techniques: finding asymptotes, algebraic manipulation to derive an inequality condition, interpreting that condition geometrically to locate a stationary point without calculus, and synthesizing all information into a sketch. The non-differentiation constraint in (b)(ii) requires insight that the stationary point occurs at the boundary of the range, making this more challenging than routine curve sketching. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.02t Solve modulus equations: graphically with modulus function |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Intersections at \((-1, 0)\), \((3, 0)\) | B1B1 (2) | Allow \(x=-1,\ x=3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Asymptotes \(x=0,\ x=2,\ y=1\) | \(\text{B1}\times 3\) (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(y=k \Rightarrow ky^2 - 2kx = x^2 - 2x - 3\) | M1A1 | M1 for clearing denominator |
| \(\Rightarrow (k-1)x^2 + (-2k+2)x + 3 = 0\) | A1\(\checkmark\) | ft numerical error |
| \(\Delta = 4(k-1)(k-4)\), hence result | m1A1 (5) | Convincingly shown AG |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(y=4\) at SP | B1 | |
| \(3x^2 - 6x + 3 = 0\), so \(x=1\) | M1A1 (3) | A0 if other point(s) given |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Curve with three branches | B1 | Approaching vertical asymptotes |
| Middle branch correct | B1 | Coordinates of SP not needed |
| Other two branches correct | B1 (3) | 3 asymptotes shown |
## Question 9:
### Part (a)(i)
| Working | Marks | Guidance |
|---------|-------|----------|
| Intersections at $(-1, 0)$, $(3, 0)$ | B1B1 (2) | Allow $x=-1,\ x=3$ |
### Part (a)(ii)
| Working | Marks | Guidance |
|---------|-------|----------|
| Asymptotes $x=0,\ x=2,\ y=1$ | $\text{B1}\times 3$ (3) | |
### Part (b)(i)
| Working | Marks | Guidance |
|---------|-------|----------|
| $y=k \Rightarrow ky^2 - 2kx = x^2 - 2x - 3$ | M1A1 | M1 for clearing denominator |
| $\Rightarrow (k-1)x^2 + (-2k+2)x + 3 = 0$ | A1$\checkmark$ | ft numerical error |
| $\Delta = 4(k-1)(k-4)$, hence result | m1A1 (5) | Convincingly shown AG |
### Part (b)(ii)
| Working | Marks | Guidance |
|---------|-------|----------|
| $y=4$ at SP | B1 | |
| $3x^2 - 6x + 3 = 0$, so $x=1$ | M1A1 (3) | A0 if other point(s) given |
### Part (c)
| Working | Marks | Guidance |
|---------|-------|----------|
| Curve with three branches | B1 | Approaching vertical asymptotes |
| Middle branch correct | B1 | Coordinates of SP not needed |
| Other two branches correct | B1 (3) | 3 asymptotes shown |9 A curve $C$ has equation
$$y = \frac { ( x + 1 ) ( x - 3 ) } { x ( x - 2 ) }$$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Write down the coordinates of the points where $C$ intersects the $x$-axis. (2 marks)
\item Write down the equations of all the asymptotes of $C$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Show that, if the line $y = k$ intersects $C$, then
$$( k - 1 ) ( k - 4 ) \geqslant 0$$
\item Given that there is only one stationary point on $C$, find the coordinates of this stationary point.\\
(No credit will be given for solutions based on differentiation.)
\end{enumerate}\item Sketch the curve $C$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2006 Q9 [16]}}