Question 7 - A-Level Maths

CAIE FP1 2012 November — Question 7 9 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2012
Session	November
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Curve Sketching
Type	Rational functions with parameters: analysis depending on parameter sign/range
Difficulty	Standard +0.8 This FP1 question requires finding asymptotes of a rational function with parameter, proving a condition on turning points using calculus and inequalities, and sketching. The derivative involves quotient rule and analyzing a quadratic discriminant, requiring multi-step algebraic manipulation and conceptual understanding of how parameters affect curve properties—more demanding than standard C1/C2 curve sketching.
Spec	1.02k Simplify rational expressions: factorising, cancelling, algebraic division 1.02n Sketch curves: simple equations including polynomials 1.07n Stationary points: find maxima, minima using derivatives

7 The curve $C$ has equation $$y = \lambda x + \frac { x } { x - 2 }$$ where $\lambda$ is a non-zero constant. Find the equations of the asymptotes of $C$. Show that $C$ has no turning points if $\lambda < 0$. Sketch $C$ in the case $\lambda = - 1$, stating the coordinates of the intersections with the axes.

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Question 7:

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
Vertical asymptote $x = 2$	B1
$y = \lambda x + 1 + \dfrac{2}{x-2} \Rightarrow y = \lambda x + 1$ is oblique asymptote	M1A1	Part total 3
$y' = \lambda - 2(x-2)^{-2} = 0$ for turning points	M1A1	Differentiates and equates to 0
$\lambda = \dfrac{2}{(x-2)^2} > 0 \Rightarrow$ no turning points if $\lambda < 0$	A1	Part total 3
Or $y'=0 \Rightarrow \lambda x^2 - 4\lambda x + 4\lambda - 2 = 0$; uses discriminant to show $8\lambda < 0 \Rightarrow$ no T.P.s	(M1A1)(A1)	Alternative method
Axes and asymptotes; LH branch; RH branch (indicating intersections with axes at $(0,0)$ and $(3,0)$)	B1, B1B1	Part total 3; Total [9]

## Question 7:

| Working/Answer | Mark | Guidance |
|---|---|---|
| Vertical asymptote $x = 2$ | B1 | |
| $y = \lambda x + 1 + \dfrac{2}{x-2} \Rightarrow y = \lambda x + 1$ is oblique asymptote | M1A1 | Part total 3 |
| $y' = \lambda - 2(x-2)^{-2} = 0$ for turning points | M1A1 | Differentiates and equates to 0 |
| $\lambda = \dfrac{2}{(x-2)^2} > 0 \Rightarrow$ no turning points if $\lambda < 0$ | A1 | Part total 3 |
| Or $y'=0 \Rightarrow \lambda x^2 - 4\lambda x + 4\lambda - 2 = 0$; uses discriminant to show $8\lambda < 0 \Rightarrow$ no T.P.s | (M1A1)(A1) | Alternative method |
| Axes and asymptotes; LH branch; RH branch (indicating intersections with axes at $(0,0)$ and $(3,0)$) | B1, B1B1 | Part total 3; **Total [9]** |

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7 The curve $C$ has equation

$$y = \lambda x + \frac { x } { x - 2 }$$

where $\lambda$ is a non-zero constant. Find the equations of the asymptotes of $C$.

Show that $C$ has no turning points if $\lambda < 0$.

Sketch $C$ in the case $\lambda = - 1$, stating the coordinates of the intersections with the axes.

\hfill \mbox{\textit{CAIE FP1 2012 Q7 [9]}}

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