CAIE FP1 2011 November — Question 7 11 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2011
SessionNovember
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolynomial Division & Manipulation
TypeSketching Rational Functions with Oblique Asymptote
DifficultyStandard +0.8 This is a multi-part Further Maths question requiring: (1) identifying vertical and oblique asymptotes via polynomial division, (2) finding conditions for two distinct turning points by differentiating a rational function using quotient rule and analyzing a discriminant condition, and (3) curve sketching with specific parameter value. While each technique is standard for FP1, the combination of algebraic manipulation, calculus, and inequality solving with parameters makes this moderately challenging, above typical A-level but not requiring exceptional insight.
Spec1.02n Sketch curves: simple equations including polynomials1.02q Use intersection points: of graphs to solve equations
7 The curve \(C\) has equation \(y = \frac { x ^ { 2 } + p x + 1 } { x - 2 }\), where \(p\) is a constant. Given that \(C\) has two asymptotes, find the equation of each asymptote. Find the set of values of \(p\) for which \(C\) has two distinct turning points. Sketch \(C\) in the case \(p = - 1\). Your sketch should indicate the coordinates of any intersections with the axes, but need not show the coordinates of any turning points.
Question 7:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(x = 2\)B1 Vertical asymptote
\(y = x + p + 2 + \frac{2p+5}{x-2}\)M1 Divides by \((x-2)\)
\(y = x + p + 2\)A1 Oblique asymptote
\(\frac{dy}{dx} = \frac{x^2 - 4x + 4 - 2p - 5}{(x-2)^2}\)M1A1 Differentiates
\(y' = 0 \Rightarrow x^2 - 4x - (2p+1) = 0\)M1
\(B^2 - 4AC > 0 \Rightarrow 16 + 4(2p+1) > 0\)M1
\(\Rightarrow p > -\frac{5}{2}\)A1
Axes and \((0, -0.5)\) marked; Upper branch with minimum; Lower branch with maximum below \(x\)-axis; (Deduct at most 1 for poor forms at infinity)B1, B1, B1 Sketches graph; working to show either \(x^2 - x + 1 = 0\) has no real roots, or maximum value 
# Question 7:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = 2$ | B1 | Vertical asymptote | **Part total: 3** |
| $y = x + p + 2 + \frac{2p+5}{x-2}$ | M1 | Divides by $(x-2)$ |
| $y = x + p + 2$ | A1 | Oblique asymptote | **Part total: 3** |
| $\frac{dy}{dx} = \frac{x^2 - 4x + 4 - 2p - 5}{(x-2)^2}$ | M1A1 | Differentiates |
| $y' = 0 \Rightarrow x^2 - 4x - (2p+1) = 0$ | M1 | |
| $B^2 - 4AC > 0 \Rightarrow 16 + 4(2p+1) > 0$ | M1 | |
| $\Rightarrow p > -\frac{5}{2}$ | A1 | | **Part total: 5** |
| Axes and $(0, -0.5)$ marked; Upper branch with minimum; Lower branch with maximum below $x$-axis; (Deduct at most 1 for poor forms at infinity) | B1, B1, B1 | Sketches graph; working to show either $x^2 - x + 1 = 0$ has no real roots, or maximum value | **Part total: 3** |

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7 The curve $C$ has equation $y = \frac { x ^ { 2 } + p x + 1 } { x - 2 }$, where $p$ is a constant. Given that $C$ has two asymptotes, find the equation of each asymptote.

Find the set of values of $p$ for which $C$ has two distinct turning points.

Sketch $C$ in the case $p = - 1$. Your sketch should indicate the coordinates of any intersections with the axes, but need not show the coordinates of any turning points.

\hfill \mbox{\textit{CAIE FP1 2011 Q7 [11]}}
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Q1 5 Q2 5 Q3 7 Q4 7 Q5 7 Q6 8 Q7 11 Q8 11 Q9 12 Q10 13 Q11 EITHER Q11 OR