AQA FP1 2012 June — Question 5 11 marks

Exam BoardAQA
ModuleFP1 (Further Pure Mathematics 1)
Year2012
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPartial Fractions
TypeRational curve sketching with asymptotes and inequalities
DifficultyStandard +0.3 This is a straightforward FP1 curve sketching question requiring identification of vertical asymptotes from denominator zeros, finding a horizontal asymptote (y=0 by degree comparison), solving a quadratic equation from the intersection, and using the sketch to write inequality solutions. All techniques are standard with no novel insight required, making it slightly easier than average.
Spec1.02g Inequalities: linear and quadratic in single variable1.02k Simplify rational expressions: factorising, cancelling, algebraic division

5 The curve \(C\) has equation \(y = \frac { x } { ( x + 1 ) ( x - 2 ) }\).
The line \(L\) has equation \(y = - \frac { 1 } { 2 }\).
  1. Write down the equations of the asymptotes of \(C\).
  2. The line \(L\) intersects the curve \(C\) at two points. Find the \(x\)-coordinates of these two points.
  3. Sketch \(C\) and \(L\) on the same axes.
    (You are given that the curve \(C\) has no stationary points.)
  4. Solve the inequality $$\frac { x } { ( x + 1 ) ( x - 2 ) } \leqslant - \frac { 1 } { 2 }$$

AnswerMarks Guidance
5(a)Asymptotes: \(x = -1\); \(x = 2\); \(y = 0\) B1, B1, B1
5(b)\(-\frac{1}{2} - \frac{x}{x^2 - x - 2} \Rightarrow x^2 - x - 2 = -2x\) M1
5(b)\(x^2 + x - 2 = 0 \Rightarrow x = 1, x = -2\) A1
5(c)(Graph showing) Three branches shown on sketch of C with either middle branch or outer two branches correct in shape M1
5(c) A1
5(d)\(-2 \leq x < -1\) B1
5(d)\(1 \leq x < 2\) B1
5(d)\(-2 \leq x < -1, 1 \leq x < 2\) B1
5(a) | Asymptotes: $x = -1$; $x = 2$; $y = 0$ | B1, B1, B1 | $x = -1$ OE; $x = 2$ OE; $y = 0$
5(b) | $-\frac{1}{2} - \frac{x}{x^2 - x - 2} \Rightarrow x^2 - x - 2 = -2x$ | M1 | Correctly removing brackets and fractions to reach $x^2 - x - 2 = -2x$ OE
5(b) | $x^2 + x - 2 = 0 \Rightarrow x = 1, x = -2$ | A1 | Correct two values for x-coordinates. NMS 2 or 0 marks
5(c) | (Graph showing) Three branches shown on sketch of C with either middle branch or outer two branches correct in shape | M1 | Three branches shown on sketch of C with either middle branch or outer two branches correct in shape. All three branches, correct shape and positions and approaching correct asymptotes in a correct manner. If middle branch does *clearly* not go through the origin, then A0
5(c) | | A1 |
5(d) | $-2 \leq x < -1$ | B1 | Condone < for ≤ or vice versa
5(d) | $1 \leq x < 2$ | B1 | Condone < for ≤ or vice versa
5(d) | $-2 \leq x < -1, 1 \leq x < 2$ | B1 | All complete and correct
5 The curve $C$ has equation $y = \frac { x } { ( x + 1 ) ( x - 2 ) }$.\\
The line $L$ has equation $y = - \frac { 1 } { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Write down the equations of the asymptotes of $C$.
\item The line $L$ intersects the curve $C$ at two points. Find the $x$-coordinates of these two points.
\item Sketch $C$ and $L$ on the same axes.\\
(You are given that the curve $C$ has no stationary points.)
\item Solve the inequality

$$\frac { x } { ( x + 1 ) ( x - 2 ) } \leqslant - \frac { 1 } { 2 }$$
\end{enumerate}

\hfill \mbox{\textit{AQA FP1 2012 Q5 [11]}}