| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Solve inequality with reciprocal in modulus |
| Difficulty | Standard +0.3 Part (a) is a standard rational inequality requiring sign analysis or rearrangement to quadratic form. Part (b) extends this by introducing modulus, requiring consideration of two cases (x+3 positive/negative), but follows directly from part (a) with guided structure. This is routine Further Maths content with clear methodology, slightly above average difficulty due to the modulus extension but well within standard FP2 expectations. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02l Modulus function: notation, relations, equations and inequalities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((x+2)(x+3)^2 - 12(x+3) = 0\) OR \(\frac{(x+3)(x+2)-12}{(x+3)} > 0\) | M1 | Multiply through by \((x+3)^2\) and collect on one side, or other valid method (NOT calculator). Multiplying by \((x+3)\) is NOT valid unless two cases \(x>3\) and \(x<3\) considered separately or \(-3\) stated as cv |
| \((x+3)(x^2+5x-6) = 0 \Rightarrow (x+3)(x+6)(x-1)=0\) | ||
| CVs: \(-3, -6, 1\) | B1, A1, A1 | B1 for \(-3\) seen anywhere. A1A1 for other cvs (A1A0 if only one correct) |
| \(-6 < x < -3,\quad x > 1\) OR \(x \in (-6,-3) \cup (1,\infty)\) | dM1, A1 | dM1: obtaining inequalities using critical values and no other numbers. A1: correct inequalities and no extras. Use of \(\ldots\) or \(,,\) scores A0. May be written in set notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x > 1\) | B1 | Correct answer only shown. Allow \(x \ldots 1\) if already penalised in (a) |
# Question 1:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(x+2)(x+3)^2 - 12(x+3) = 0$ OR $\frac{(x+3)(x+2)-12}{(x+3)} > 0$ | M1 | Multiply through by $(x+3)^2$ and collect on one side, or other valid method (NOT calculator). Multiplying by $(x+3)$ is NOT valid unless two cases $x>3$ and $x<3$ considered separately or $-3$ stated as cv |
| $(x+3)(x^2+5x-6) = 0 \Rightarrow (x+3)(x+6)(x-1)=0$ | | |
| CVs: $-3, -6, 1$ | B1, A1, A1 | B1 for $-3$ seen anywhere. A1A1 for other cvs (A1A0 if only one correct) |
| $-6 < x < -3,\quad x > 1$ OR $x \in (-6,-3) \cup (1,\infty)$ | dM1, A1 | dM1: obtaining inequalities using critical values and no other numbers. A1: correct inequalities and no extras. Use of $\ldots$ or $,,$ scores A0. May be written in set notation |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x > 1$ | B1 | Correct answer only shown. Allow $x \ldots 1$ if already penalised in (a) |
---\begin{enumerate}
\item (a) Use algebra to find the set of values of $x$ for which
\end{enumerate}
$$x + 2 > \frac { 12 } { x + 3 }$$
(b) Hence, or otherwise, find the set of values of $x$ for which
$$x + 2 > \frac { 12 } { | x + 3 | }$$
\hfill \mbox{\textit{Edexcel FP2 2015 Q1 [7]}}