| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Solve |quadratic| compared to linear: algebraic inequality |
| Difficulty | Standard +0.8 This FP2 question requires solving an equation involving a modulus in the denominator by considering two cases (x > -2 and x < -2), then solving resulting quadratics and checking validity of solutions against case conditions. Part (b) adds complexity by requiring inequality analysis across different regions. While systematic, it demands careful case-work, algebraic manipulation, and understanding of how the modulus affects the inequality direction—more demanding than standard C1-C3 questions but routine for Further Maths students who have practiced modulus techniques. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02p Interpret algebraic solutions: graphically1.02q Use intersection points: of graphs to solve equations |
| Answer | Marks | Guidance |
|---|---|---|
| (a) For \(x \geq -2\): Attempt to solve \(x^2 - 1 = 3(1-x)(x+2)\) | M1 | |
| \([4x^2 + 3x - 7 = 0]\) | ||
| \(x = 1\) or \(x = \frac{4}{7}\) | B1, A1 | |
| For \(x < -2\): Attempt to solve \(x^2 - 1 = -3(1-x)(x+2)\) | M1 | |
| Solving \(x + 1 = 3x + 6\) \((2x^2 + 3x - 5 = 0)\) | M1dep | |
| \(x = -\frac{5}{2}\) | A16 | |
| "Squaring" route: If candidates do not notice the factor of \((x-1)^2\) they have quartic to solve | ||
| Squaring and finding quartic = 0: \([8x^4 + 18x^3 - 25x^2 - 36x + 35 = 0]\) | ||
| Finding one factor and factorising: \((x-1)(8x^3 + 26x^2 + x - 35) = 0\) | M1 | |
| Finding one other factor and reducing other factor to quadratic, likely to be \((x-1)^2(8x^2 + 34x + 35) = 0\) | M1 | |
| Complete factorisation: \((x-1)^2(2x+5)(4x+7) = 0\) | M1 | |
| [Second M1 implies the first, if candidate starts there or cancels \((x-1)^2\)] | ||
| \(x = 1\), B1, \(x = -\frac{7}{4}\) A1, \(x = -\frac{5}{2}\) | A1 | |
| \(x = 1\) allowed anywhere, no penalty in (b) | ||
| (b) \(-\frac{7}{4} < x < 1\) | M1 | One part |
| Both correct and enclosed | A1 | |
| \(x < -\frac{5}{2}\) | B1ft3 | {Must be for \(x < -2\) and only one value} |
| Correct answers seen with no working is independent of (a) (graphical calculator) mark as scheme. Only allow the accuracy mark if no other interval, in both parts. \(\leq\) used penalise first time used |
**(a)** For $x \geq -2$: Attempt to solve $x^2 - 1 = 3(1-x)(x+2)$ | M1 |
$[4x^2 + 3x - 7 = 0]$ | |
$x = 1$ or $x = \frac{4}{7}$ | B1, A1 |
For $x < -2$: Attempt to solve $x^2 - 1 = -3(1-x)(x+2)$ | M1 |
Solving $x + 1 = 3x + 6$ $(2x^2 + 3x - 5 = 0)$ | M1dep |
$x = -\frac{5}{2}$ | A16 |
"Squaring" route: If candidates do not notice the factor of $(x-1)^2$ they have quartic to solve | |
Squaring and finding quartic = 0: $[8x^4 + 18x^3 - 25x^2 - 36x + 35 = 0]$ | |
Finding one factor and factorising: $(x-1)(8x^3 + 26x^2 + x - 35) = 0$ | M1 |
Finding one other factor and reducing other factor to quadratic, likely to be $(x-1)^2(8x^2 + 34x + 35) = 0$ | M1 |
Complete factorisation: $(x-1)^2(2x+5)(4x+7) = 0$ | M1 |
[Second M1 implies the first, if candidate starts there or cancels $(x-1)^2$] | |
$x = 1$, B1, $x = -\frac{7}{4}$ A1, $x = -\frac{5}{2}$ | A1 |
$x = 1$ allowed anywhere, no penalty in (b) | |
**(b)** $-\frac{7}{4} < x < 1$ | M1 | One part
| | Both correct and enclosed | A1 |
$x < -\frac{5}{2}$ | B1ft3 | {Must be for $x < -2$ and only one value} |
Correct answers seen with no working is independent of (a) (graphical calculator) mark as scheme. Only allow the accuracy mark if no other interval, in both parts. $\leq$ used penalise first time used | |
**Total: [9]**
---2.\\
\includegraphics[max width=\textwidth, alt={}, center]{d6befd60-de40-41b6-8ae5-48656dbca40c-1_734_1228_888_479}
The diagram above shows a sketch of the curve with equation
$$y = \frac { x ^ { 2 } - 1 } { | x + 2 | } , \quad x \neq - 2$$
The curve crosses the $x$-axis at $x = 1$ and $x = - 1$ and the line $x = - 2$ is an asymptote of the curve.
\begin{enumerate}[label=(\alph*)]
\item Use algebra to solve the equation $\frac { x ^ { 2 } - 1 } { | x + 2 | } = 3 ( 1 - x )$.
\item Hence, or otherwise, find the set of values of $x$ for which
$$\frac { x ^ { 2 } - 1 } { | x + 2 | } < 3 ( 1 - x )$$
(Total 9 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2007 Q2 [9]}}