CAIE FP1 2012 November — Question 9 12 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2012
SessionNovember
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolynomial Division & Manipulation
TypeSketching Rational Functions with Oblique Asymptote
DifficultyStandard +0.8 This FP1 question requires multiple sophisticated techniques: finding asymptotes of a rational function, proving a range restriction by rearranging to form a quadratic in x and analyzing the discriminant, finding stationary points by quotient rule differentiation, and synthesizing all information into a sketch. While each individual step is accessible, the combination of algebraic manipulation, discriminant analysis for range restrictions, and integration of multiple features into a coherent sketch makes this significantly above average difficulty for A-level.
Spec1.02z Models in context: use functions in modelling1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx
9 The curve \(C\) has equation \(y = \frac { x ^ { 2 } - 3 x + 3 } { x - 2 }\). Find the equations of the asymptotes of \(C\). Show that there are no points on \(C\) for which \(- 1 < y < 3\). Find the coordinates of the turning points of \(C\). Sketch \(C\).
Question 9:
AnswerMarks Guidance
Working/AnswerMarks Guidance
Vertical asymptote is \(x = 2\)B1
Oblique asymptote is \(y = x - 1\)M1, A1 Divides and states oblique asymptote
\(xy - 2y = x^2 - 3x + 3 \Rightarrow x^2 - (y+3)x + (3+2y) = 0\)B1 Rearranges as quadratic in \(x\)
For real \(x\): \(B^2 - 4AC \geq 0\)M1 Uses discriminant
\(\therefore (y+3)^2 - 4(3+2y) \geq 0\)
\(\Rightarrow \ldots \Rightarrow (y-3)(y+1) \geq 0\)A1
\(\Rightarrow y \leq -1\) or \(y \geq 3\)
\(\therefore\) no points for \(-1 < y < 3\) (AG)A1
\(y' = 1-(x-2)^{-2} = 0 \Rightarrow x = 1\) or \(3\)M1 Differentiates, puts \(= 0\)
Stationary points are \((1,-1)\) and \((3,3)\)A1A1
One mark for each branch correctly placedB1B1 Sketch 
## Question 9:

| Working/Answer | Marks | Guidance |
|---|---|---|
| Vertical asymptote is $x = 2$ | B1 | |
| Oblique asymptote is $y = x - 1$ | M1, A1 | Divides and states oblique asymptote |
| $xy - 2y = x^2 - 3x + 3 \Rightarrow x^2 - (y+3)x + (3+2y) = 0$ | B1 | Rearranges as quadratic in $x$ |
| For real $x$: $B^2 - 4AC \geq 0$ | M1 | Uses discriminant |
| $\therefore (y+3)^2 - 4(3+2y) \geq 0$ | | |
| $\Rightarrow \ldots \Rightarrow (y-3)(y+1) \geq 0$ | A1 | |
| $\Rightarrow y \leq -1$ or $y \geq 3$ | | |
| $\therefore$ no points for $-1 < y < 3$ (AG) | A1 | |
| $y' = 1-(x-2)^{-2} = 0 \Rightarrow x = 1$ or $3$ | M1 | Differentiates, puts $= 0$ |
| Stationary points are $(1,-1)$ and $(3,3)$ | A1A1 | |
| One mark for each branch correctly placed | B1B1 | Sketch |
9 The curve $C$ has equation $y = \frac { x ^ { 2 } - 3 x + 3 } { x - 2 }$. Find the equations of the asymptotes of $C$.

Show that there are no points on $C$ for which $- 1 < y < 3$.

Find the coordinates of the turning points of $C$.

Sketch $C$.

\hfill \mbox{\textit{CAIE FP1 2012 Q9 [12]}}
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