| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | November |
| Marks | 12 |
| 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 is a multi-part Further Maths question requiring asymptote identification, algebraic proof of a range restriction, calculus for turning points, and curve sketching. While individual techniques are standard, the combination—especially proving the range gap using partial fractions or solving a quadratic inequality—requires solid algebraic manipulation and synthesis beyond typical A-level pure maths. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02n Sketch curves: simple equations including polynomials1.02y Partial fractions: decompose rational functions1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Degree of numerator < degree of denominator \(\Rightarrow y = 0\) is horizontal asymptote | B1 | |
| \((x+1)(x-2) = 0 \Rightarrow x = -1\) and \(x = 2\) are vertical asymptotes | B1 | |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(yx^2 - (y+3)x + 9 - 2y = 0\) | M1 | |
| No points on \(C\) if \((y+3)^2 - 4y(9-2y) < 0\) | M1 | |
| \(\Rightarrow 9y^2 - 30y + 9 < 0 \Rightarrow 3y^2 - 10y + 3 < 0\) | A1 | |
| \(\Rightarrow (3y-1)(y-3) < 0 \Rightarrow \frac{1}{3} < y < 3\) | A1 | AG |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = 0 \Rightarrow 3(x^2 - x - 2) - (3x-9)(2x-1) = 0\) | B1 | |
| \(\Rightarrow \ldots \Rightarrow (x-1)(x-5) = 0\) | B1 | |
| \(\Rightarrow\) Turning points are \((1,3)\) and \(\left(5, \frac{1}{3}\right)\) | B1 | |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Axes, asymptotes and points on axes \((0, 4.5)\), \((3,0)\) | B1 | |
| RH branch; Other two branches | B1B1 | |
| Total | 3 |
## Question 9(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Degree of numerator < degree of denominator $\Rightarrow y = 0$ is horizontal asymptote | B1 | |
| $(x+1)(x-2) = 0 \Rightarrow x = -1$ and $x = 2$ are vertical asymptotes | B1 | |
| **Total** | **2** | |
## Question 9(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $yx^2 - (y+3)x + 9 - 2y = 0$ | M1 | |
| No points on $C$ if $(y+3)^2 - 4y(9-2y) < 0$ | M1 | |
| $\Rightarrow 9y^2 - 30y + 9 < 0 \Rightarrow 3y^2 - 10y + 3 < 0$ | A1 | |
| $\Rightarrow (3y-1)(y-3) < 0 \Rightarrow \frac{1}{3} < y < 3$ | A1 | AG |
| **Total** | **4** | |
## Question 9(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = 0 \Rightarrow 3(x^2 - x - 2) - (3x-9)(2x-1) = 0$ | B1 | |
| $\Rightarrow \ldots \Rightarrow (x-1)(x-5) = 0$ | B1 | |
| $\Rightarrow$ Turning points are $(1,3)$ and $\left(5, \frac{1}{3}\right)$ | B1 | |
| **Total** | **3** | |
## Question 9(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Axes, asymptotes and points on axes $(0, 4.5)$, $(3,0)$ | B1 | |
| RH branch; Other two branches | B1B1 | |
| **Total** | **3** | |9 The curve $C$ has equation
$$y = \frac { 3 x - 9 } { ( x - 2 ) ( x + 1 ) }$$
(i) Find the equations of the asymptotes of $C$.\\
\includegraphics[max width=\textwidth, alt={}, center]{9221f480-4af6-44be-a535-d2ceb0f8b5d2-14_61_1566_513_328}\\
(ii) Show that there is no point on $C$ for which $\frac { 1 } { 3 } < y < 3$.\\
(iii) Find the coordinates of the turning points of $C$.\\
(iv) Sketch $C$.
\hfill \mbox{\textit{CAIE FP1 2017 Q9 [12]}}