| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simultaneous equations |
| Type | Line intersecting reciprocal curve |
| Difficulty | Standard +0.3 This is a standard C1 simultaneous equations question combining graphical and algebraic methods. Part (i) is routine graph sketching, part (ii) is straightforward algebraic manipulation (equating expressions, clearing fractions, solving quadratic), and part (iii) applies the same technique with the tangency condition (discriminant = 0). While multi-part, each step uses well-practiced techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02f Solve quadratic equations: including in a function of unknown1.02o 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 | Mark | Guidance |
| \(y = 2x + 3\) drawn accurately | M1 | At least as far as intersecting curve twice; ruled straight line and within 2mm of \((2, 7)\) and \((-1, 1)\) |
| \((-1.6\) to \(-1.7, -0.2\) to \(-0.3)\) | B1 | Intersections may be in form \(x = ..., y = ...\) |
| \((2.1\) to \(2.2, 7.2\) to \(7.4)\) | B1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{x-2} = 2x + 3\) | M1 | Or attempt at elimination of \(x\) by rearrangement and substitution; may be seen in (i) — allow marks |
| \(1 = (2x+3)(x-2)\) | M1 | Condone lack of brackets; implies first M1 if that step not seen |
| \(1 = 2x^2 - x - 6\) oe | A1 | For correct expansion; need not be simplified; NB A0 for \(2x^2 - x - 7 = 0\) without expansion seen [given answer]; implies second M1 if that step not seen after \(\frac{1}{x-2} = 2x+3\) seen |
| \(\frac{1 \pm \sqrt{1^2 - 4 \times 2 \times -7}}{2 \times 2}\) oe | M1 | Use of formula or completing the square on given equation, with at most one error; completing square attempt must reach at least \([2](x-a)^2 = b\) or \((2x-c)^2 = d\) stage oe with at most one error |
| \(\frac{1 \pm \sqrt{57}}{4}\) isw | A1 | e.g. coordinates; after completing square, accept \(\frac{1}{4} \pm \sqrt{\frac{57}{16}}\) or better |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{x-2} = -x + k\) and attempt at rearrangement | M1 | |
| \(x^2 - (k+2)x + 2k + 1 [= 0]\) | M1 | For simplifying and rearranging to zero; condone one error; collection of \(x\) terms with bracket not required; e.g. M1 bod for \(x^2 - (k+2)x + 2k\) or M1 for \(x^2 - 2kx + 2k + 1 [= 0]\) |
| \(b^2 - 4ac = 0\) oe seen or used | M1 | \(= 0\) may not be seen, but may be implied by their final values of \(k\) |
| \([k =]\ 0\) or \(4\) as final answer, both required | A1 | SC1 for 0 and 4 found if 3rd M1 not earned (may or may not have earned first two Ms); e.g. obtained graphically or using calculus and/or final answer given as a range |
| [4] |
## Question 3(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = 2x + 3$ drawn accurately | M1 | At least as far as intersecting curve twice; ruled straight line and within 2mm of $(2, 7)$ and $(-1, 1)$ |
| $(-1.6$ to $-1.7, -0.2$ to $-0.3)$ | B1 | Intersections may be in form $x = ..., y = ...$ |
| $(2.1$ to $2.2, 7.2$ to $7.4)$ | B1 | |
| **[3]** | | |
---
## Question 3(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{x-2} = 2x + 3$ | M1 | Or attempt at elimination of $x$ by rearrangement and substitution; may be seen in (i) — allow marks |
| $1 = (2x+3)(x-2)$ | M1 | Condone lack of brackets; implies first M1 if that step not seen |
| $1 = 2x^2 - x - 6$ oe | A1 | For correct expansion; need not be simplified; NB A0 for $2x^2 - x - 7 = 0$ without expansion seen [given answer]; implies second M1 if that step not seen after $\frac{1}{x-2} = 2x+3$ seen |
| $\frac{1 \pm \sqrt{1^2 - 4 \times 2 \times -7}}{2 \times 2}$ oe | M1 | Use of formula or completing the square on given equation, with at most one error; completing square attempt must reach at least $[2](x-a)^2 = b$ or $(2x-c)^2 = d$ stage oe with at most one error |
| $\frac{1 \pm \sqrt{57}}{4}$ isw | A1 | e.g. coordinates; after completing square, accept $\frac{1}{4} \pm \sqrt{\frac{57}{16}}$ or better |
| **[5]** | | |
---
## Question 3(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{x-2} = -x + k$ and attempt at rearrangement | M1 | |
| $x^2 - (k+2)x + 2k + 1 [= 0]$ | M1 | For simplifying and rearranging to zero; condone one error; collection of $x$ terms with bracket not required; e.g. M1 bod for $x^2 - (k+2)x + 2k$ or M1 for $x^2 - 2kx + 2k + 1 [= 0]$ |
| $b^2 - 4ac = 0$ oe seen or used | M1 | $= 0$ may not be seen, but may be implied by their final values of $k$ |
| $[k =]\ 0$ or $4$ as final answer, both required | A1 | SC1 for 0 and 4 found if 3rd M1 not earned (may or may not have earned first two Ms); e.g. obtained graphically or using calculus and/or final answer given as a range |
| **[4]** | | |
---3
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e93e3c51-ae2b-420b-abb8-bf0c483caff8-3_1270_1219_326_463}
\captionsetup{labelformat=empty}
\caption{Fig. 12}
\end{center}
\end{figure}
Fig. 12 shows the graph of $y = \frac { 1 } { x - 2 }$.\\
(i) Draw accurately the graph of $y = 2 x + 3$ on the copy of Fig. 12 and use it to estimate the coordinates of the points of intersection of $y = \frac { 1 } { x - 2 }$ and $y = 2 x + 3$.\\
(ii) Show algebraically that the $x$-coordinates of the points of intersection of $y = \frac { 1 } { x - 2 }$ and $y = 2 x + 3$ satisfy the equation $2 x ^ { 2 } - x - 7 = 0$. Hence find the exact values of the $x$-coordinates of the points of intersection.\\
(iii) Find the quadratic equation satisfied by the $x$-coordinates of the points of intersection of $y = \frac { 1 } { x - 2 }$ and $y = - x + k$. Hence find the exact values of $k$ for which $y = - x + k$ is a tangent to $y = \frac { 1 } { x - 2 }$. [4]
\hfill \mbox{\textit{OCR MEI C1 Q3 [4]}}