| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Rational inequality algebraically |
| Difficulty | Moderate -0.3 This is a standard rational inequality requiring sign analysis and critical point identification. While it's Further Maths (FP1), the technique is straightforward: rearrange to standard form, find critical points x=2 and x=24/7, test intervals. Part (b) is trivial once (a) is solved. Slightly easier than average due to being a routine textbook exercise with a clear method. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02h Express solutions: using 'and', 'or', set and interval notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| E.g. \((x \neq 2 \Rightarrow) 5x = 12(x-2) \Rightarrow x = ...\) or \(\frac{5x}{x-2} \geqslant 12 \Rightarrow 5x(x-2) \geqslant 12(x-2)^2 \Rightarrow (x-2)(5x-12)(x-2) \geqslant 0 \Rightarrow x = ...\) or \(\frac{5x}{x-2} \geqslant 12 \Rightarrow \frac{5x - 12(x-2)}{x-2} \geqslant 0 \Rightarrow x = ...\) o.e. | M1 | Algebraic method to find critical value aside from \(x=2\). May set equal using \(x \neq 2\), or multiply through by \((x-2)^2\) and solve quadratic, or gather terms over common denominator. Allow with any inequality or equality for first two marks. |
| \(x - 2 \cdot (-7x + 24) \geqslant 0\) or \(x-2 \cdot 7x - 24 \leqslant 0 \Rightarrow x = ..., ...\) or \(7x^2 - 38x + 48 \leqslant 0\) or \(-7x^2 + 38x - 48 \geqslant 0 \Rightarrow x = ..., ...\) or \(\frac{24-7x}{x-2} \geqslant 0 \Rightarrow x = ..., ...\) | dM1 | Finds both critical values by correct algebraic method (allowing slips rearranging). \(x=2\) may be stated or used as boundary value. Dependent on previous M mark. |
| Critical values \(x = 2, \dfrac{24}{7}\) | A1 | Correct critical values stated or used. Allow awrt 3.43 for \(\frac{24}{7}\). A0 if other values also used. |
| \(2 < x \leqslant \dfrac{24}{7}\) | A1 | Must be exact. Accept alternative notations such as \(\left(2, \frac{24}{7}\right]\) but formal set notation not required. Accept as separate inequalities. |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 3\) | B1 | B1cao: Deduces correct value for \(x\) and no other values, as long as 3 is in their solution set from (a). Allow if endpoint 2 was included in answer to (a) as long as it is not given as a solution for (b). |
| (1) |
## Question 1:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| E.g. $(x \neq 2 \Rightarrow) 5x = 12(x-2) \Rightarrow x = ...$ or $\frac{5x}{x-2} \geqslant 12 \Rightarrow 5x(x-2) \geqslant 12(x-2)^2 \Rightarrow (x-2)(5x-12)(x-2) \geqslant 0 \Rightarrow x = ...$ or $\frac{5x}{x-2} \geqslant 12 \Rightarrow \frac{5x - 12(x-2)}{x-2} \geqslant 0 \Rightarrow x = ...$ o.e. | M1 | Algebraic method to find critical value aside from $x=2$. May set equal using $x \neq 2$, or multiply through by $(x-2)^2$ and solve quadratic, or gather terms over common denominator. Allow with any inequality or equality for first two marks. |
| $x - 2 \cdot (-7x + 24) \geqslant 0$ or $x-2 \cdot 7x - 24 \leqslant 0 \Rightarrow x = ..., ...$ or $7x^2 - 38x + 48 \leqslant 0$ or $-7x^2 + 38x - 48 \geqslant 0 \Rightarrow x = ..., ...$ or $\frac{24-7x}{x-2} \geqslant 0 \Rightarrow x = ..., ...$ | dM1 | Finds **both** critical values by correct algebraic method (allowing slips rearranging). $x=2$ may be stated or used as boundary value. Dependent on previous M mark. |
| Critical values $x = 2, \dfrac{24}{7}$ | A1 | Correct critical values stated or used. Allow awrt 3.43 for $\frac{24}{7}$. A0 if other values also used. |
| $2 < x \leqslant \dfrac{24}{7}$ | A1 | Must be exact. Accept alternative notations such as $\left(2, \frac{24}{7}\right]$ but formal set notation not required. Accept as separate inequalities. |
| | **(4)** | |
---
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 3$ | B1 | **B1cao**: Deduces correct value for $x$ and no other values, as long as 3 is in their solution set from (a). Allow if endpoint 2 was included in answer to (a) as long as it is not given as a solution for (b). |
| | **(1)** | |
**Total: (5 marks)**\begin{enumerate}
\item (a) Use algebra to determine the values of $x$ for which
\end{enumerate}
$$\frac { 5 x } { x - 2 } \geqslant 12$$
(b) Hence, given that $x$ is an integer, deduce the value of $x$.
\hfill \mbox{\textit{Edexcel FP1 AS 2023 Q1 [5]}}