CAIE FP1 2017 November — Question 9 12 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2017
SessionNovember
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPartial Fractions
TypeRational curve analysis with turning points and range restrictions
DifficultyStandard +0.3 This is a standard Further Maths curve sketching question requiring asymptote identification, range analysis via partial fractions/algebra, differentiation for turning points, and sketching. While it involves multiple techniques, each step follows routine procedures with no novel insight required, making it slightly easier than average for FP1.
Spec1.02k 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

9 The curve \(C\) has equation $$y = \frac { 3 x - 9 } { ( x - 2 ) ( x + 1 ) }$$
  1. Find the equations of the asymptotes of \(C\). \includegraphics[max width=\textwidth, alt={}, center]{a0987277-06e9-451b-ae18-bb7de9e7661c-14_61_1566_513_328}
  2. Show that there is no point on \(C\) for which \(\frac { 1 } { 3 } < y < 3\).
  3. Find the coordinates of the turning points of \(C\).
  4. Sketch \(C\).

Question 9(i):
AnswerMarks Guidance
AnswerMarks Guidance
Degree of numerator < degree of denominator \(\Rightarrow y = 0\) is horizontal asymptoteB1
\((x+1)(x-2) = 0 \Rightarrow x = -1\) and \(x = 2\) are vertical asymptotesB1
Total: 2
Question 9(ii):
AnswerMarks Guidance
AnswerMarks 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):
AnswerMarks Guidance
AnswerMarks 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):
AnswerMarks Guidance
AnswerMarks Guidance
Axes, asymptotes and points on axes \((0, 4.5)\), \((3, 0)\)B1
RH branch; Other two branchesB1B1
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]{a0987277-06e9-451b-ae18-bb7de9e7661c-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]}}
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