| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2008 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Function Transformations |
| Type | Transformation of specific function type |
| Difficulty | Moderate -0.8 This is a straightforward C1 question on horizontal translation of a reciprocal function. Part (i) requires basic transformation knowledge (shift right 2 units), part (ii) is simple equation solving, and part (iii) involves forming and solving a quadratic—all standard techniques with no novel insight required. Easier than average A-level questions. |
| Spec | 1.02o Sketch reciprocal curves: y=a/x and y=a/x^21.02q Use intersection points: of graphs to solve equations1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| i correct graph with clear asymptote \(x = 2\) (though need not be marked) | Marks: G2 | Guidance: G1 for one branch correct; condone (0, \(-\frac{1}{2}\)) not shown; SC1 for both sections of graph shifted two to left |
| ii \((0, -\frac{1}{2})\) shown; 11/5 or 2.2 o.e. isw | Marks: 2 | Guidance: M1 for correct first step |
| iii \(x = \frac{1}{x - 2}\); \(x(x - 2) = 1\) o.e.; \(x^2 - 2x - 1 = [0]\); ft their equiv eqn; attempt at quadratic formula; \(1 \pm \sqrt{2}\) (condone one error); position 1 of points shown | Marks: M1, M1, M1, M1, A1, B1 | Guidance: or equiv with ys; or \((x - 1)^2 - 1 = 1\) o.e.; or \((x - 1) = \pm\sqrt{2}\) (condone one error); on their curve with \(y = x\) (line drawn or \(y = x\) indicated by both chords); condone intent of diagonal line with gradient approx 1 through origin as \(y = x\) if unlabelled |
**i** correct graph with clear asymptote $x = 2$ (though need not be marked) | **Marks:** G2 | **Guidance:** G1 for one branch correct; condone (0, $-\frac{1}{2}$) not shown; SC1 for both sections of graph shifted two to left | **Total:** 3
**ii** $(0, -\frac{1}{2})$ shown; 11/5 or 2.2 o.e. isw | **Marks:** 2 | **Guidance:** M1 for correct first step | **Total:** 2
**iii** $x = \frac{1}{x - 2}$; $x(x - 2) = 1$ o.e.; $x^2 - 2x - 1 = [0]$; ft their equiv eqn; attempt at quadratic formula; $1 \pm \sqrt{2}$ (condone one error); position 1 of points shown | **Marks:** M1, M1, M1, M1, A1, B1 | **Guidance:** or equiv with ys; or $(x - 1)^2 - 1 = 1$ o.e.; or $(x - 1) = \pm\sqrt{2}$ (condone one error); on their curve with $y = x$ (line drawn or $y = x$ indicated by both chords); condone intent of diagonal line with gradient approx 1 through origin as $y = x$ if unlabelled | **Total:** 6
**Total for Question 10:** 1110 (i)
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{450c1c3a-9290-4afa-a051-112b60cf19c0-3_753_775_360_726}
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\caption{Fig. 10}
\end{center}
\end{figure}
Fig. 10 shows a sketch of the graph of $y = \frac { 1 } { x }$.\\
Sketch the graph of $y = \frac { 1 } { x - 2 }$, showing clearly the coordinates of any points where it crosses the axes.\\
(ii) Find the value of $x$ for which $\frac { 1 } { x - 2 } = 5$.\\
(iii) Find the $x$-coordinates of the points of intersection of the graphs of $y = x$ and $y = \frac { 1 } { x - 2 }$. Give your answers in the form $a \pm \sqrt { b }$.
Show the position of these points on your graph in part (i).
\hfill \mbox{\textit{OCR MEI C1 2008 Q10 [11]}}