| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Rational inequality algebraically |
| Difficulty | Standard +0.8 This FP1 rational inequality requires factorizing the denominator, finding a common denominator, combining fractions, solving the resulting inequality with sign analysis across multiple critical points, and carefully handling discontinuities. It's more demanding than standard A-level inequalities but routine for FP1 students who have practiced the technique. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable |
| V349 SIHI NI IMIMM ION OC | VJYV SIHIL NI LIIIM ION OO | VJYV SIHIL NI JIIYM ION OC |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{x(x+3)-(x^2-2x-3)}{(x^2-2x-3)(x+3)} \leq 0\) or \(x(x-3)(x+1)(x+3)^2-(x-3)^2(x+1)^2(x+3) \leq 0\) or \(x(x^2-2x-3)(x+3)^2-(x^2-2x-3)^2(x+3) \leq 0\) | M1 | 2.1 — Gathers terms on one side with common denominator, or multiplies by \((x+1)^2(x-3)^2(x+3)^2\) and gathers terms onto one side |
| \(\frac{5x+3}{(x-3)(x+3)(x+1)}\ \{\leq 0\}\) or \((x-3)(x+1)(x+3)(5x+3)\ \{\leq 0\}\) | M1 | Expands/simplifies numerator fully or factors out \((x-3)(x+1)(x+3)\) |
| Same expression as above confirmed correct | A1 | 1.1b |
| All three critical values \(-3,\ 3,\ -1\) | B1 | 1.1b — Correct critical values, can be implied by inequalities |
| Critical value \(-\dfrac{3}{5}\) | B1ft | 1.1b — Follow through on their fourth factor of form \((ax+b)\), giving C.V. \(= -\frac{b}{a}\) |
| \(\left\{x \in \mathbb{R}: -3 < x < -1\right\} \cup \left\{x \in \mathbb{R}: -\dfrac{3}{5} \leq x < 3\right\}\) | M1 | 2.2a — Deduces 2 "inside" inequalities with critical values in ascending order |
| Same final answer confirmed | A1 | 2.5 — Exactly 2 correct intervals; condone omission of union symbol |
| Total: 7 marks | Also accept: \(-3 |
## Question 3:
$$\frac{x}{x^2-2x-3} \leq \frac{1}{x+3}$$
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{x(x+3)-(x^2-2x-3)}{(x^2-2x-3)(x+3)} \leq 0$ **or** $x(x-3)(x+1)(x+3)^2-(x-3)^2(x+1)^2(x+3) \leq 0$ **or** $x(x^2-2x-3)(x+3)^2-(x^2-2x-3)^2(x+3) \leq 0$ | M1 | 2.1 — Gathers terms on one side with common denominator, or multiplies by $(x+1)^2(x-3)^2(x+3)^2$ and gathers terms onto one side |
| $\frac{5x+3}{(x-3)(x+3)(x+1)}\ \{\leq 0\}$ **or** $(x-3)(x+1)(x+3)(5x+3)\ \{\leq 0\}$ | M1 | Expands/simplifies numerator fully or factors out $(x-3)(x+1)(x+3)$ |
| Same expression as above confirmed correct | A1 | 1.1b |
| All three critical values $-3,\ 3,\ -1$ | B1 | 1.1b — Correct critical values, can be implied by inequalities |
| Critical value $-\dfrac{3}{5}$ | B1ft | 1.1b — Follow through on their fourth factor of form $(ax+b)$, giving C.V. $= -\frac{b}{a}$ |
| $\left\{x \in \mathbb{R}: -3 < x < -1\right\} \cup \left\{x \in \mathbb{R}: -\dfrac{3}{5} \leq x < 3\right\}$ | M1 | 2.2a — Deduces 2 "inside" inequalities with critical values in ascending order |
| Same final answer confirmed | A1 | 2.5 — Exactly 2 correct intervals; condone omission of union symbol |
| **Total: 7 marks** | | Also accept: $-3<x<-1,\ -\frac{3}{5}\leq x<3$; or $(-3,-1),\ \left[-\frac{3}{5},3\right)$; or $-1>x>-3,\ 3>x\geq-\frac{3}{5}$ |
---\begin{enumerate}
\item Use algebra to find the values of $x$ for which
\end{enumerate}
$$\frac { x } { x ^ { 2 } - 2 x - 3 } \leqslant \frac { 1 } { x + 3 }$$
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V349 SIHI NI IMIMM ION OC & VJYV SIHIL NI LIIIM ION OO & VJYV SIHIL NI JIIYM ION OC \\
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\hfill \mbox{\textit{Edexcel FP1 AS 2018 Q3 [7]}}