| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Rational inequality algebraically |
| Difficulty | Standard +0.8 This is a multi-part Further Maths inequality question requiring algebraic manipulation to standard form, understanding of sign analysis with rational functions, and recognition that multiplying inequalities by expressions of unknown sign is invalid. Part (a) requires careful rearrangement over a common denominator (4 marks suggests non-trivial algebra), part (b) tests conceptual understanding of a common error, and part (c) applies sign analysis. The combination of algebraic technique, conceptual reasoning about inequalities, and the Further Maths context places this moderately above average difficulty. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02k Simplify rational expressions: factorising, cancelling, algebraic division |
| Answer | Marks | Guidance |
|---|---|---|
| 14(a) | Writes both terms on one side of the inequality, with a common | |
| denominator. | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Correctly combines their two fractions into one fraction. | 1.1b | A1F |
| Obtains a single fraction in which the numerator is a three-term quadratic. | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| (𝑥𝑥+−2)(𝑥𝑥+3) | 2.1 | R1 |
| Answer | Marks |
|---|---|
| 14(b) | 0≤ 𝑥𝑥+1 |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑥𝑥 = −𝑟𝑟 | 2.4 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 14(c) | Obtains one correct region, FT their three critical values. | |
| Condone . | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| , | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Total | 7 | |
| Q | Marking instructions | AO |
Question 14:
--- 14(a) ---
14(a) | Writes both terms on one side of the inequality, with a common
denominator. | 1.1a | M1 | 2
(𝑥𝑥+1 ) 𝑥𝑥+7
0 ≤ −
𝑥𝑥+1 𝑥𝑥+1
2
𝑥𝑥 +2𝑥𝑥 +1−𝑥𝑥−7
0 ≤
𝑥𝑥+1
2
𝑥𝑥 + 𝑥𝑥−6
0 ≤
𝑥𝑥+1
(𝑥𝑥+3)(𝑥𝑥 −2)
≥ 0
𝑥𝑥+1
Correctly combines their two fractions into one fraction. | 1.1b | A1F
Obtains a single fraction in which the numerator is a three-term quadratic. | 1.1a | M1
Completes a rigorous argument to show the correct inequality in the
required form.
Accept .
(𝑥𝑥+−2)(𝑥𝑥+3) | 2.1 | R1
--- 14(b) ---
14(b) | 0≤ 𝑥𝑥+1
Explains that is a solution of the inequality on the RHS, but not
the one of the LHS.
𝑥𝑥 = −𝑟𝑟 | 2.4 | B1 | is a solution of the inequality on the right, but
not the one on the left.
𝑥𝑥 = −𝑟𝑟
--- 14(c) ---
14(c) | Obtains one correct region, FT their three critical values.
Condone . | 1.1a | M1 | ,
−3 ≤ 𝑥𝑥 < −1 𝑥𝑥 ≥ 2
−3≤ 𝑥𝑥 ≤ −1
Obtains both correct regions.
, | 1.1b | A1
−3 ≤ 𝑥𝑥 < −1 𝑥𝑥 ≥ 2
Total | 7
Q | Marking instructions | AO | Marks | Typical solution\begin{enumerate}[label=(\alph*)]
\item Given
$$\frac{x + 7}{x + 1} \leq x + 1$$
show that
$$\frac{(x + a)(x + b)}{x + c} \geq 0$$
where $a$, $b$, and $c$ are integers to be found.
[4 marks]
\item Briefly explain why this statement is incorrect.
$$\frac{(x + p)(x + q)}{x + r} \geq 0 \Leftrightarrow (x + p)(x + q)(x + r) \geq 0$$
[1 mark]
\item Solve
$$\frac{x + 7}{x + 1} \leq x + 1$$
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2020 Q14 [7]}}