| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Solve quadratic inequality |
| Difficulty | Moderate -0.8 Part (a) is a straightforward linear inequality requiring basic expansion and collection of terms. Part (b) is a standard quadratic inequality requiring rearrangement, factorization, and identification of the solution region—routine C1 content with no conceptual challenges beyond applying the standard method. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable |
| Answer | Marks | Guidance |
|---|---|---|
| \(3 - 6x - 15x - 10 > 0\) | M1 | Expanding correctly |
| \(-21x - 7 > 0 \Rightarrow x < -\frac{1}{3}\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(6x^2 - x - 12 \leq 0\) | M1 | Rearranging to \(= 0\) |
| \((2x - 3)(3x + 4) = 0\) | M1 | Factorising |
| \(x = \frac{3}{2}\), \(x = -\frac{4}{3}\) | A1 | Both critical values |
| \(-\frac{4}{3} \leq x \leq \frac{3}{2}\) | A1 | Correct inequality, correct notation |
# Question 8:
## Part (a):
| $3 - 6x - 15x - 10 > 0$ | M1 | Expanding correctly |
|---|---|---|
| $-21x - 7 > 0 \Rightarrow x < -\frac{1}{3}$ | A1 | Correct answer |
## Part (b):
| $6x^2 - x - 12 \leq 0$ | M1 | Rearranging to $= 0$ |
|---|---|---|
| $(2x - 3)(3x + 4) = 0$ | M1 | Factorising |
| $x = \frac{3}{2}$, $x = -\frac{4}{3}$ | A1 | Both critical values |
| $-\frac{4}{3} \leq x \leq \frac{3}{2}$ | A1 | Correct inequality, correct notation |8 Solve the following inequalities:
\begin{enumerate}[label=(\alph*)]
\item $\quad 3 ( 1 - 2 x ) - 5 ( 3 x + 2 ) > 0$
\item $\quad 6 x ^ { 2 } \leqslant x + 12$\\[0pt]
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2014 Q8 [6]}}