| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Rational inequality algebraically |
| Difficulty | Standard +0.3 This is a standard rational inequality requiring sign analysis and critical point identification. While it's from FP1, the technique is straightforward: rearrange to get zero on one side, combine fractions, and analyze sign changes. More routine than average A-level questions but requires careful handling of the denominator restriction. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02h Express solutions: using 'and', 'or', set and interval notation |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(x = \frac{2x+15}{2x+3} \Rightarrow 2x^2+3x = 2x+15 \Rightarrow 2x^2+x-15=0 \Rightarrow x = \ldots\) Alt 1: \((2x+3)^2x \geqslant (2x+3)(2x+15) \Rightarrow (2x+3)(2x^2+3x-2x-15)\geqslant 0 \Rightarrow (2x+3)(x+3)(2x-5)\geqslant 0\) Alt 2: \(x - \frac{2x+15}{2x+3} \geqslant 0 \Rightarrow \frac{x(2x+3)-2x-15}{2x+3} \geqslant 0 \Rightarrow \frac{(x+3)(2x-5)}{2x+3} \geqslant 0\) | M1 | Complete method to find critical values other than \(-\frac{3}{2}\). Alt 1: Multiply by \((2x+3)^2\), collect terms, factorise into three brackets. Alt 2: Collect terms, combine into single fraction, factorise numerator. |
| \(\Rightarrow (x+3)(2x-5)=0 \Rightarrow\) CVs are \(-3,\ \frac{5}{2}\) | A1 | Correct critical values \(-3\) and \(\frac{5}{2}\) |
| Also \(2x+3=0 \Rightarrow x = -\frac{3}{2}\) a CV | B1 | For the critical value \(-\frac{3}{2}\) |
| Hence from graph (oe) the solution set is \(\left\{x \in \mathbb{R}: -3 \leqslant x < -\frac{3}{2},\ x \geqslant \frac{5}{2}\right\}\) | M1 | Selects correct regions for three CVs. Must include right-hand side open ended and another bounded region. CVs of \(a < b < c\) must be of form \(a \leqslant x \leqslant b,\ x \geqslant c\) or \(a < x < b,\ x > c\); direction of inequalities must be correct with or without strict inequalities. |
| A1 | At least one correct interval identified. Allow both intervals with correct endpoints but incorrect strict/inclusive inequalities. | |
| A1 | Fully correct solution as a set. Accept e.g. \(\left[-3, -\frac{3}{2}\right) \cup \left[\frac{5}{2}, \infty\right)\), but not just inequalities. Minimum use of set notation \(-3 \leqslant x < -\frac{3}{2} \cup x \geqslant \frac{5}{2}\) | |
| (6 marks) | Note: Correct answer with no working scores M0 A0 but can score B1 M1 A1 A1. No working shown to factorise a cubic e.g. \(4x^3+8x^2-27x-45=(x+3)(2x+3)(2x-5)\) is M0 A0 but can still score B1 M1 A1 A1. |
## Question 1:
| Working/Answer | Mark | Guidance |
|---|---|---|
| $x = \frac{2x+15}{2x+3} \Rightarrow 2x^2+3x = 2x+15 \Rightarrow 2x^2+x-15=0 \Rightarrow x = \ldots$ **Alt 1:** $(2x+3)^2x \geqslant (2x+3)(2x+15) \Rightarrow (2x+3)(2x^2+3x-2x-15)\geqslant 0 \Rightarrow (2x+3)(x+3)(2x-5)\geqslant 0$ **Alt 2:** $x - \frac{2x+15}{2x+3} \geqslant 0 \Rightarrow \frac{x(2x+3)-2x-15}{2x+3} \geqslant 0 \Rightarrow \frac{(x+3)(2x-5)}{2x+3} \geqslant 0$ | **M1** | Complete method to find critical values other than $-\frac{3}{2}$. Alt 1: Multiply by $(2x+3)^2$, collect terms, factorise into three brackets. Alt 2: Collect terms, combine into single fraction, factorise numerator. |
| $\Rightarrow (x+3)(2x-5)=0 \Rightarrow$ CVs are $-3,\ \frac{5}{2}$ | **A1** | Correct critical values $-3$ and $\frac{5}{2}$ |
| Also $2x+3=0 \Rightarrow x = -\frac{3}{2}$ a CV | **B1** | For the critical value $-\frac{3}{2}$ |
| Hence from graph (oe) the solution set is $\left\{x \in \mathbb{R}: -3 \leqslant x < -\frac{3}{2},\ x \geqslant \frac{5}{2}\right\}$ | **M1** | Selects correct regions for three CVs. Must include right-hand side open ended and another bounded region. CVs of $a < b < c$ must be of form $a \leqslant x \leqslant b,\ x \geqslant c$ or $a < x < b,\ x > c$; direction of inequalities must be correct with or without strict inequalities. |
| | **A1** | At least one correct interval identified. Allow both intervals with correct endpoints but incorrect strict/inclusive inequalities. |
| | **A1** | Fully correct solution as a set. Accept e.g. $\left[-3, -\frac{3}{2}\right) \cup \left[\frac{5}{2}, \infty\right)$, but not just inequalities. Minimum use of set notation $-3 \leqslant x < -\frac{3}{2} \cup x \geqslant \frac{5}{2}$ |
| **(6 marks)** | | **Note:** Correct answer with no working scores M0 A0 but can score B1 M1 A1 A1. No working shown to factorise a cubic e.g. $4x^3+8x^2-27x-45=(x+3)(2x+3)(2x-5)$ is M0 A0 but can still score B1 M1 A1 A1. |\begin{enumerate}
\item Use algebra to find the set of values of $x$ for which
\end{enumerate}
$$x \geqslant \frac { 2 x + 15 } { 2 x + 3 }$$
\hfill \mbox{\textit{Edexcel FP1 AS 2022 Q1 [6]}}