| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Combined linear and quadratic inequalities |
| Difficulty | Moderate -0.8 This is a straightforward C1 question testing routine manipulation of linear and quadratic inequalities. Part (a) requires simple rearrangement, part (b) involves factorizing a quadratic and identifying the solution region, and part (c) combines the results—all standard techniques with no problem-solving insight required. Easier than average for A-level. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable |
| Answer | Marks |
|---|---|
| (a) \(3x - 7 > 3 - x\) and \(4x > 10\) | M1 |
| \(x > 2.5\), \(x > \frac{5}{2}\), \(\frac{5}{2} < x\) o.e. | A1 |
| (2 marks) | |
| (b) Obtain \(x^2 - 9x - 36\) and attempt to solve \(x^2 - 9x - 36 = 0\) e.g. \((x - 12)(x + 3) = 0\) so \(x = \quad\) or \(x = \frac{9 \pm \sqrt{81 + 144}}{2}\) | M1 |
| \(12, -3\) | A1 |
| \(-3 \le x \le 12\) | M1A1 |
| (4 marks) | |
| (c) \(2.5 < x \le 12\) | A1cso |
| (1 mark) | |
| Notes: | |
| (a) | M1: Reaching \(px > q\) with one or both of \(p\) or \(q\) correct. Also give for \(-4x < -10\) |
| A1: Cao \(x > 2.5\) o.e. Accept alternatives to \(2.5\) like \(2\frac{1}{2}\) and \(\frac{5}{2}\) even allow \(\frac{10}{4}\) and allow \(\frac{5}{2} < x\) o.e. This answer must occur and be credited as part (a). A correct answer implies M1A1 | |
| Mark parts (b) and (c) together. | |
| (b) | M1: Rearrange \(3TQ \le 0\) or \(3TQ = 0\) or even \(3TQ > 0\) Do not worry about the inequality at this stage AND attempt to solve by factorising, formula or completion of the square with the usual rules (see notes) |
| A1: \(12\) and \(-3\) seen as critical values | |
| M1: Inside region for their critical values – must be stated – not just a table or a graph | |
| A1: \(-3 \le x \le 12\) Accept \(x \ge -3\) and \(x \le 12\) or \([-3, 12]\) For the A mark: Do not accept \(x \ge -3\) or \(x \le 12\) nor \(-3 < x < 12\) nor \((-3, 12)\) nor \(x \ge -3, x \le 12\) However allow recovery if they follow these statements by a correct statement, either in (b) or as they start part (c). N.B. \(-3 \le 0 \le 12\) and \(x \ge -3, x \le 12\) are poor notation and get M1A0 here. | |
| (c) | A1 cso: \(2.5 < x \le 12\) Accept \(x > 2.5\) and \(x \le 12\) Allow \(\frac{10}{4}\). Do not accept \(x > 2.5\) or \(x \le 12\). Accept \(x > 2.5\) and \(x \le 12\) Allow \(\frac{10}{4}\). Do not accept \(x > 2.5\) or \(x \le 12\). Accept \((2.5, 12]\) A graph or table is not sufficient. Must follow correct earlier work – except for special case. |
| Special case: | \((c) x > 2.5, x < 12; 2.5 < 0 \le 12\) are poor notation – but if this poor notation has been penalised in (b) then allow A1 here. Any other errors are penalised in both (b) and (c). |
**(a)** $3x - 7 > 3 - x$ and $4x > 10$ | M1 |
$x > 2.5$, $x > \frac{5}{2}$, $\frac{5}{2} < x$ o.e. | A1 |
| | **(2 marks)** |
**(b)** Obtain $x^2 - 9x - 36$ and attempt to solve $x^2 - 9x - 36 = 0$ e.g. $(x - 12)(x + 3) = 0$ so $x = \quad$ or $x = \frac{9 \pm \sqrt{81 + 144}}{2}$ | M1 |
$12, -3$ | A1 |
$-3 \le x \le 12$ | M1A1 |
| | **(4 marks)** |
**(c)** $2.5 < x \le 12$ | A1cso |
| | **(1 mark)** |
| **Notes:** | |
|---|---|
| **(a)** | M1: Reaching $px > q$ with one or both of $p$ or $q$ correct. Also give for $-4x < -10$ |
| | A1: Cao $x > 2.5$ o.e. Accept alternatives to $2.5$ like $2\frac{1}{2}$ and $\frac{5}{2}$ even allow $\frac{10}{4}$ and allow $\frac{5}{2} < x$ o.e. This answer must occur and be credited as part (a). A correct answer implies M1A1 |
| **Mark parts (b) and (c) together.** | |
| **(b)** | M1: Rearrange $3TQ \le 0$ or $3TQ = 0$ or even $3TQ > 0$ Do not worry about the inequality at this stage AND attempt to solve by factorising, formula or completion of the square with the usual rules (see notes) |
| | A1: $12$ and $-3$ seen as critical values |
| | M1: Inside region for their critical values – must be stated – not just a table or a graph |
| | A1: $-3 \le x \le 12$ Accept $x \ge -3$ and $x \le 12$ or $[-3, 12]$ For the A mark: Do not accept $x \ge -3$ or $x \le 12$ nor $-3 < x < 12$ nor $(-3, 12)$ nor $x \ge -3, x \le 12$ However allow recovery if they follow these statements by a correct statement, either in (b) or as they start part (c). N.B. $-3 \le 0 \le 12$ and $x \ge -3, x \le 12$ are poor notation and get M1A0 here. |
| **(c)** | A1 cso: $2.5 < x \le 12$ Accept $x > 2.5$ and $x \le 12$ Allow $\frac{10}{4}$. Do not accept $x > 2.5$ or $x \le 12$. Accept $x > 2.5$ and $x \le 12$ Allow $\frac{10}{4}$. Do not accept $x > 2.5$ **or** $x \le 12$. Accept $(2.5, 12]$ A graph or table is not sufficient. **Must follow correct earlier work** – except for special case. |
| **Special case:** | $(c) x > 2.5, x < 12; 2.5 < 0 \le 12$ are poor notation – but if this poor notation has been penalised in (b) then allow A1 here. Any other errors are penalised in both (b) and (c). |
---3. Find the set of values of $x$ for which
\begin{enumerate}[label=(\alph*)]
\item $3 x - 7 > 3 - x$
\item $x ^ { 2 } - 9 x \leqslant 36$
\item both $3 x - 7 > 3 - x$ and $x ^ { 2 } - 9 x \leqslant 36$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2014 Q3 [7]}}