CAIE FP1 2009 June — Question 10 11 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2009
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCurve Sketching
TypeRational functions with parameters: analysis depending on parameter sign/range
DifficultyStandard +0.8 This Further Maths question requires finding asymptotes (vertical and oblique via polynomial division), locating stationary points using the quotient rule, and sketching two cases with different topologies. It demands multiple techniques, careful algebraic manipulation, and understanding how the parameter affects curve behavior—more demanding than standard C3/C4 curve sketching but routine for FP1.
Spec1.02n Sketch curves: simple equations including polynomials1.07n Stationary points: find maxima, minima using derivatives

10 The curve \(C\) has equation $$y = \frac { x ^ { 2 } } { x + \lambda }$$ where \(\lambda\) is a non-zero constant. Obtain the equation of each of the asymptotes of \(C\). In separate diagrams, sketch \(C\) for the cases \(\lambda > 0\) and \(\lambda < 0\). In both cases the coordinates of the turning points must be indicated.

AnswerMarks
One asymptote is \(x = -\lambda\)B1
\(y = x - \lambda + \frac{\lambda^2}{(x + \lambda)}\)M1
Other asymptote is \(y = x - \lambda\)A1
or for previous 2 marks:
\(mx + c = \frac{x^2}{(x + \lambda)} \Rightar(m-1)x^2 + (c + \lambda m)x + \lambda c = 0\)
AnswerMarks
\(m - 1 = 0, c + \lambda m = 0\) (both)M1
\(m = 1, c = -\lambda\) (both)A1
Differentiates and \(y' = 0\)M1
Identifies turning point at \((-2\lambda, -4\lambda)\) and \((0, 0)\)A1A1
(Award D1 if correct turning points stated with no working)
First graph:
AnswerMarks
Axes and both asymptotesB1
BranchesB1B1
Deduct at most 1, overall, for bad forms at infinity
Second graph:
AnswerMarks
Axes and both asymptotes + 1 branchB1
Other branchB1
Deduct at most 1, overall, for bad forms at infinity
Choose which is 1st graph and which is 2nd to the benefit of candidates, i.e. best = 1st
For differentiating a special caseM1A0A0
Graphs of special casesB2 (max) B1 (max)
If stated in text, turning points need not be labelled on graph
One asymptote is $x = -\lambda$ | B1 |

$y = x - \lambda + \frac{\lambda^2}{(x + \lambda)}$ | M1 |

Other asymptote is $y = x - \lambda$ | A1 |

**or for previous 2 marks:**

$mx + c = \frac{x^2}{(x + \lambda)} \Rightar(m-1)x^2 + (c + \lambda m)x + \lambda c = 0$

$m - 1 = 0, c + \lambda m = 0$ (both) | M1 |

$m = 1, c = -\lambda$ (both) | A1 |

Differentiates and $y' = 0$ | M1 |

Identifies turning point at $(-2\lambda, -4\lambda)$ and $(0, 0)$ | A1A1 |

(Award D1 if correct turning points stated with no working) |

**First graph:**

Axes and both asymptotes | B1 |

Branches | B1B1 |

Deduct at most 1, overall, for bad forms at infinity |

**Second graph:**

Axes and both asymptotes + 1 branch | B1 |

Other branch | B1 |

Deduct at most 1, overall, for bad forms at infinity |

Choose which is 1st graph and which is 2nd to the benefit of candidates, i.e. best = 1st |

**For differentiating a special case** | M1A0A0 |

**Graphs of special cases** | B2 (max) B1 (max) |

If stated in text, turning points need not be labelled on graph |
10 The curve $C$ has equation

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

where $\lambda$ is a non-zero constant. Obtain the equation of each of the asymptotes of $C$.

In separate diagrams, sketch $C$ for the cases $\lambda > 0$ and $\lambda < 0$. In both cases the coordinates of the turning points must be indicated.

\hfill \mbox{\textit{CAIE FP1 2009 Q10 [11]}}