| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Solve quadratic inequality |
| Difficulty | Moderate -0.8 Part (i) is a one-step linear inequality requiring basic rearrangement. Part (ii) involves expanding a quadratic, rearranging to standard form, factoring, and applying sign analysis—standard C1 technique but more steps than typical. Overall easier than average A-level questions due to routine methods and no conceptual challenges. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02h Express solutions: using 'and', 'or', set and interval notation |
| Answer | Marks |
|---|---|
| Marks: B1 | B1 [2] |
| Guidance: soi, allow \(-8x > 1\) but not just \(8x + 1 < 0\) | Correct working only, allow \(-\frac{1}{8} > x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Marks: M1* | DM1* | A1 |
| Guidance: Expand brackets and rearrange to collect all terms on one side | Correct method to find roots of resulting quadratic | 0, 5 seen as roots – could be on sketch graph |
## (i)
**Answer:** $8x < -1$; $x < -\frac{1}{8}$
**Marks:** B1 | B1 [2]
**Guidance:** soi, allow $-8x > 1$ but not just $8x + 1 < 0$ | Correct working only, allow $-\frac{1}{8} > x$
**Additional notes:** Allow $\leq$ or $\geq$ for first mark; Do not ISW if contradictory; Do not allow $\leq$ or $\geq$; Do not allow $-\frac{1}{8}$
## (ii)
**Answer:** $2x^2 - 10x \leq 0$; $2x(x-5) \leq 0$; $0 \leq x \leq 5$
**Marks:** M1* | DM1* | A1 | DM1* | A1 [5]
**Guidance:** Expand brackets and rearrange to collect all terms on one side | Correct method to find roots of resulting quadratic | 0, 5 seen as roots – could be on sketch graph | Chooses "inside region" for their roots of their resulting quadratic (not the original) | Do not accept strict inequalities for final mark
**Additional notes:** Dependent on first M1 only; Allow $(2x + 0)(x - 5)$; Do not allow $(2x - 4)(x - 3)$, this is the original expression. No more than one incorrect term
---Solve the inequalities
\begin{enumerate}[label=(\roman*)]
\item $3 - 8x > 4$, [2]
\item $(2x - 4)(x - 3) \leq 12$. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR C1 2013 Q7 [7]}}