| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Rational function curve sketching |
| Difficulty | Standard +0.8 This Further Maths question requires finding asymptotes, proving a range exclusion using algebraic manipulation, finding turning points via quotient rule differentiation, and synthesizing all information into a sketch. The range exclusion proof (part ii) requires rearranging to form a quadratic and analyzing discriminant conditions, which demands more sophisticated reasoning than standard curve sketching. While systematic, the multi-step nature and proof element elevate this above typical A-level questions. |
| 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}| 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]{68e31138-756a-433a-bf42-0fdfadad091e-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]}}