| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2019 |
| 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 FP1 rational inequality question with a pedagogical twist—students must identify algebraic errors in a given solution before solving correctly. While it requires careful sign analysis and understanding of critical values, the error-spotting format and straightforward factorization make it easier than a typical problem-solving question. The main challenge is recognizing the squaring error and excluded values, which are routine FP1 skills. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Line 3 error — allow any of: bracketing error; \(-24\) should be \(24\) in square brackets; \(x(x+11)-(x-24)\) should be \(x(x+11)-(x-24)\); \(x(x+11)-x-24\) should be \(x(x+11)-x+24\) | B1 | AO2.3. Give B0 for contradictory reasons |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Line 7 error — allow any of: should be \(\{x \in \mathbb{R}: x < -12\) or \(-11 < x < 2\) or \(x > 24\}\); they have found regions where inequality is \(< 0\); they have reversed the inequality | B1 | AO2.3. Give B0 for contradictory reasons. Allow "Should be \(x < -12,\ -11 < x < 2,\ x > 24\)" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((x-24)(x+11)[x(x+11)-(x-24)] > 0\) \((x-24)(x+11)[x^2+10x+24] > 0\) \((x-24)(x+11)(x+6)(x+4) > 0\) Critical values \(x = -11, -6, -4, 24\) | M1, A1 | Uses brackets to correct error, forms 3TQ and uses correct method to solve. AO1.1b |
| \(\{x \in \mathbb{R}: x < -11\} \cup \{x \in \mathbb{R}: -6 < x < -4\} \cup \{x \in \mathbb{R}: x > 24\}\) | M1, A1 | Deduces 2 "outsides" and "middle interval". Exactly 3 correct intervals in set notation. AO2.2a, AO2.5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{x}{x-24} > \frac{1}{x+11} \Rightarrow \frac{x}{x-24} - \frac{1}{x+11} > 0 \Rightarrow \frac{x(x+11)-(x-24)}{(x-24)(x+11)} > 0\) \(\Rightarrow \frac{x^2+10x+24}{(x-24)(x+11)} > 0 \Rightarrow \frac{(x+6)(x+4)}{(x-24)(x+11)} > 0\) Critical values \(x = -11, -6, -4, 24\) | M1, A1 | Gathers terms, common denominator, simplifies numerator, forms 3TQ and solves correctly. AO1.1b |
| \(\{x \in \mathbb{R}: x < -11\} \cup \{x \in \mathbb{R}: -6 < x < -4\} \cup \{x \in \mathbb{R}: x > 24\}\) | M1, A1 | See Way 1. AO2.2a, AO2.5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Considers each of intervals \(x < -11\), \(-11 < x < 24\), \(x > 24\) separately and evaluates which parts satisfy original inequality | M1 | AO1.1b |
| Obtains correct inequality statement for each of intervals \(x < -11\), \(-11 < x < 24\), \(x > 24\) | A1 | AO1.1b |
| \(\{x \in \mathbb{R}: x < -11\} \cup \{x \in \mathbb{R}: -6 < x < -4\} \cup \{x \in \mathbb{R}: x > 24\}\) | M1, A1 | See Way 1 |
# Question 2:
## Part (a)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Line 3 error — allow any of: bracketing error; $-24$ should be $24$ in square brackets; $x(x+11)-(x-24)$ should be $x(x+11)-(x-24)$; $x(x+11)-x-24$ should be $x(x+11)-x+24$ | B1 | AO2.3. Give B0 for contradictory reasons |
## Part (a)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Line 7 error — allow any of: should be $\{x \in \mathbb{R}: x < -12$ or $-11 < x < 2$ or $x > 24\}$; they have found regions where inequality is $< 0$; they have reversed the inequality | B1 | AO2.3. Give B0 for contradictory reasons. Allow "Should be $x < -12,\ -11 < x < 2,\ x > 24$" |
## Part (b) — Way 1
| Answer | Mark | Guidance |
|--------|------|----------|
| $(x-24)(x+11)[x(x+11)-(x-24)] > 0$ $(x-24)(x+11)[x^2+10x+24] > 0$ $(x-24)(x+11)(x+6)(x+4) > 0$ Critical values $x = -11, -6, -4, 24$ | M1, A1 | Uses brackets to correct error, forms 3TQ and uses correct method to solve. AO1.1b |
| $\{x \in \mathbb{R}: x < -11\} \cup \{x \in \mathbb{R}: -6 < x < -4\} \cup \{x \in \mathbb{R}: x > 24\}$ | M1, A1 | Deduces 2 "outsides" and "middle interval". Exactly 3 correct intervals in set notation. AO2.2a, AO2.5 |
## Part (b) — Way 2
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{x}{x-24} > \frac{1}{x+11} \Rightarrow \frac{x}{x-24} - \frac{1}{x+11} > 0 \Rightarrow \frac{x(x+11)-(x-24)}{(x-24)(x+11)} > 0$ $\Rightarrow \frac{x^2+10x+24}{(x-24)(x+11)} > 0 \Rightarrow \frac{(x+6)(x+4)}{(x-24)(x+11)} > 0$ Critical values $x = -11, -6, -4, 24$ | M1, A1 | Gathers terms, common denominator, simplifies numerator, forms 3TQ and solves correctly. AO1.1b |
| $\{x \in \mathbb{R}: x < -11\} \cup \{x \in \mathbb{R}: -6 < x < -4\} \cup \{x \in \mathbb{R}: x > 24\}$ | M1, A1 | See Way 1. AO2.2a, AO2.5 |
## Part (b) — Way 3
| Answer | Mark | Guidance |
|--------|------|----------|
| Considers each of intervals $x < -11$, $-11 < x < 24$, $x > 24$ separately and evaluates which parts satisfy original inequality | M1 | AO1.1b |
| Obtains correct inequality statement for each of intervals $x < -11$, $-11 < x < 24$, $x > 24$ | A1 | AO1.1b |
| $\{x \in \mathbb{R}: x < -11\} \cup \{x \in \mathbb{R}: -6 < x < -4\} \cup \{x \in \mathbb{R}: x > 24\}$ | M1, A1 | See Way 1 |
---\begin{enumerate}
\item A student was set the following problem.
\end{enumerate}
Use algebra to find the set of values of $x$ for which
$$\frac { x } { x - 24 } > \frac { 1 } { x + 11 }$$
The student's attempt at a solution is written below.
$$\begin{gathered}
x ( x - 24 ) ( x + 11 ) ^ { 2 } > ( x + 11 ) ( x - 24 ) ^ { 2 } \\
x ( x - 24 ) ( x + 11 ) ^ { 2 } - ( x + 11 ) ( x - 24 ) ^ { 2 } > 0 \\
( x - 24 ) ( x + 11 ) [ x ( x + 11 ) - x - 24 ] > 0 \\
( x - 24 ) ( x + 11 ) \left[ x ^ { 2 } + 10 x - 24 \right] > 0 \\
( x - 24 ) ( x + 11 ) ( x + 12 ) ( x - 2 ) > 0 \\
x = 24 , x = - 11 , x = - 12 , x = 2 \\
\{ x \in \mathbb { R } : - 12 < x < - 11 \} \cup \{ x \in \mathbb { R } : 2 < x < 24 \}
\end{gathered}$$
Line 3
There are errors in the student's solution.\\
(a) Identify the error made\\
(i) in line 3\\
(ii) in line 7\\
(b) Find a correct solution to this problem.
\hfill \mbox{\textit{Edexcel FP1 AS 2019 Q2 [6]}}