| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Show equation reduces to polynomial |
| Difficulty | Standard +0.3 This is a structured multi-part question requiring graphical work, algebraic manipulation to form a quadratic, solving using the quadratic formula, and applying the discriminant condition for tangency. While it involves several techniques, each step is clearly signposted and follows standard C1 procedures without requiring novel insight or complex problem-solving. |
| 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 equations |
\includegraphics{figure_12}
Fig. 12 shows the graph of $y = \frac{1}{x - 2}$.
\begin{enumerate}[label=(\roman*)]
\item Draw accurately the graph of $y = 2x + 3$ 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 $y = 2x + 3$. [3]
\item Show algebraically that the $x$-coordinates of the points of intersection of $y = \frac{1}{x - 2}$ and $y = 2x + 3$ satisfy the equation $2x^2 - x - 7 = 0$. Hence find the exact values of the $x$-coordinates of the points of intersection. [5]
\item 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]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C1 2013 Q12 [12]}}