| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2021 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simultaneous equations |
| Type | Line intersecting reciprocal curve |
| Difficulty | Moderate -0.8 This is a straightforward simultaneous equations question requiring substitution of y = 2x + 8 into y = 12/x, leading to a quadratic equation. The sketching in part (a) is routine, and solving 2x² + 8x - 12 = 0 uses standard factorization or quadratic formula techniques. While it requires multiple steps and 'detailed reasoning,' the methods are entirely standard with no conceptual challenges beyond basic AS-level algebra. |
| Spec | 1.02o Sketch reciprocal curves: y=a/x and y=a/x^21.02q Use intersection points: of graphs to solve equations1.03a Straight lines: equation forms y=mx+c, ax+by+c=0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| General shape for curve with axes as asymptotes | B1 | — |
| One point labelled, e.g. \((1, 12)\), \((2, 6)\), \((3, 4)\) etc | B1dep [2] | One correct point clearly seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = 2x + 8\) drawn, line with positive gradient and positive intercept | B1 | Allow marks for the line even if the curve is wrong or missing |
| \((-4, 0)\) and \((0, 8)\) clearly labelled | B1 [2] | Both intercepts correctly labelled |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2x + 8 = \frac{12}{x}\) | M1 | Attempt to eliminate one variable. Allow SC1 for \((-5.16, -2.32)\) and \((1.16, 10.3)\) BC |
| \(2x^2 + 8x - 12 = 0\) | M1 | Attempt to rewrite into the form \(ax^2 + bx + c = 0\) and attempt to solve |
| \(x = -2 \pm \sqrt{10}\) | A1 | Both roots seen (allow BC if both M marks awarded) |
| When \(x = -2 + \sqrt{10}\), \(y = 2(-2+\sqrt{10})+8 = 4 + 2\sqrt{10}\); \(x = -2-\sqrt{10}\), \(y = 4 - 2\sqrt{10}\) | M1 | Substituting their roots into either equation. If substitution into the curve used, allow \(\frac{12}{-2\pm\sqrt{10}}\) for M1A0 if denominator not rationalised |
| Coordinates are \(\left(-2+\sqrt{10},\ 4+2\sqrt{10}\right)\) and \(\left(-2-\sqrt{10},\ 4-2\sqrt{10}\right)\) | A1 [5] | Both coordinates correct and exact |
# Question 9(a)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| General shape for curve with axes as asymptotes | B1 | — |
| One point labelled, e.g. $(1, 12)$, $(2, 6)$, $(3, 4)$ etc | B1dep [2] | One correct point clearly seen |
---
# Question 9(a)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = 2x + 8$ drawn, line with positive gradient and positive intercept | B1 | Allow marks for the line even if the curve is wrong or missing |
| $(-4, 0)$ and $(0, 8)$ clearly labelled | B1 [2] | Both intercepts correctly labelled |
---
# Question 9(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2x + 8 = \frac{12}{x}$ | M1 | Attempt to eliminate one variable. Allow SC1 for $(-5.16, -2.32)$ and $(1.16, 10.3)$ BC |
| $2x^2 + 8x - 12 = 0$ | M1 | Attempt to rewrite into the form $ax^2 + bx + c = 0$ and attempt to solve |
| $x = -2 \pm \sqrt{10}$ | A1 | Both roots seen (allow BC if both M marks awarded) |
| When $x = -2 + \sqrt{10}$, $y = 2(-2+\sqrt{10})+8 = 4 + 2\sqrt{10}$; $x = -2-\sqrt{10}$, $y = 4 - 2\sqrt{10}$ | M1 | Substituting their roots into either equation. If substitution into the curve used, allow $\frac{12}{-2\pm\sqrt{10}}$ for M1A0 if denominator not rationalised |
| Coordinates are $\left(-2+\sqrt{10},\ 4+2\sqrt{10}\right)$ and $\left(-2-\sqrt{10},\ 4-2\sqrt{10}\right)$ | A1 [5] | Both coordinates correct and exact |
---9
\begin{enumerate}[label=(\alph*)]
\item Sketch both of the following on the axes provided in the Printed Answer Booklet.
\begin{enumerate}[label=(\roman*)]
\item The curve $\mathrm { y } = \frac { 12 } { \mathrm { x } }$, stating the coordinates of at least one point on the curve.
\item The line $y = 2 x + 8$, stating the coordinates of the points at which the line crosses the axes.
\end{enumerate}\item In this question you must show detailed reasoning.
Determine the exact coordinates of the points of intersection of the curve and the line.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2021 Q9 [9]}}