| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Show line is tangent, verify |
| Difficulty | Moderate -0.3 This is a multi-part question covering standard C1 techniques: completing the square, finding intersections, sketching curves, and solving simultaneous equations. While it requires multiple steps and combines several skills, each individual technique is routine and well-practiced at this level. The tangent verification in part (iii) is slightly more sophisticated than pure recall, bringing it slightly below average difficulty overall. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02e Complete the square: quadratic polynomials and turning points1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02n Sketch curves: simple equations including polynomials1.02q Use intersection points: of graphs to solve equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x^2 - 5x + 6 = 2 - x\) | M1 | For attempt to equate or subtract equations, or attempt at rearrangement and elimination of \(x\). Accept calculus approach: \(y' = 2x - 5\) |
| \(x^2 - 4x + 4 [= 0]\) | M1 | For rearrangement to zero and collection of terms; condone one error. If using completing the square, need to get as far as \((x-k)^2 = c\), with at most one error \([(x-2)^2 = 0\) if correct]. Use of \(y' = -1\) M1 |
| \(x = 2, [y = 0]\) | A1 | Condone omission of \(y = 0\) since already found in (ii). If they have eliminated \(x\), \(y = 0\) is not sufficient for A1 — need to get \(x = 2\). A0 for \(x = 2\) and another root. \(x = 2\) A1 |
| 'double root at \(x = 2\) so tangent' oe; www | A1 | e.g. 'only one point of contact, so tangent'; or showing \(b^2 - 4ac = 0\) and concluding 'so tangent'; www. tgt is \(y[-0] = -(x-2)\) and obtaining given line A1 |
| [4] |
## Question 1(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^2 - 5x + 6 = 2 - x$ | M1 | For attempt to equate or subtract equations, or attempt at rearrangement and elimination of $x$. Accept calculus approach: $y' = 2x - 5$ |
| $x^2 - 4x + 4 [= 0]$ | M1 | For rearrangement to zero and collection of terms; condone one error. If using completing the square, need to get as far as $(x-k)^2 = c$, with at most one error $[(x-2)^2 = 0$ if correct]. Use of $y' = -1$ M1 |
| $x = 2, [y = 0]$ | A1 | Condone omission of $y = 0$ since already found in (ii). If they have eliminated $x$, $y = 0$ is not sufficient for A1 — need to get $x = 2$. A0 for $x = 2$ and another root. $x = 2$ A1 |
| 'double root at $x = 2$ so tangent' oe; www | A1 | e.g. 'only one point of contact, so tangent'; or showing $b^2 - 4ac = 0$ and concluding 'so tangent'; www. tgt is $y[-0] = -(x-2)$ and obtaining given line A1 |
| | **[4]** | |
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\begin{enumerate}[label=(\roman*)]
\item Express $x ^ { 2 } - 5 x + 6$ in the form $( x - a ) ^ { 2 } - b$. Hence state the coordinates of the turning point of the curve $y = x ^ { 2 } - 5 x + 6$.
\item Find the coordinates of the intersections of the curve $y = x ^ { 2 } - 5 x + 6$ with the axes and sketch this curve.
\item Solve the simultaneous equations $y = x ^ { 2 } - 5 x + 6$ and $x + y = 2$. Hence show that the line $x + y = 2$ is a tangent to the curve $y = x ^ { 2 } - 5 x + 6$ at one of the points where the curve intersects the axes. [4]
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{973ad9eb-33f2-432e-9449-e54c1728008b-1_1292_1401_887_359}
\captionsetup{labelformat=empty}
\caption{Fig. 12}
\end{center}
\end{figure}
Fig. 12 shows the graph of $y = \frac { 1 } { x - 3 }$.
\item Draw accurately, on the copy of Fig. 12, the graph of $y = x ^ { 2 } - 4 x + 1$ for $- 1 \leqslant x \leqslant 5$. Use your graph to estimate the coordinates of the intersections of $y = \frac { 1 } { x - 3 }$ and $y = x ^ { 2 } - 4 x + 1$.
\item Show algebraically that, where the curves intersect, $x ^ { 3 } - 7 x ^ { 2 } + 13 x - 4 = 0$.
\item Use the fact that $x = 4$ is a root of $x ^ { 3 } - 7 x ^ { 2 } + 13 x - 4 = 0$ to find a quadratic factor of $x ^ { 3 } - 7 x ^ { 2 } + 13 x - 4$. Hence find the exact values of the other two roots of this equation. [5]
\item Find algebraically the coordinates of the points of intersection of the curve $y = 4 x ^ { 2 } + 24 x + 31$ and the line $x + y = 10$.
\item Express $4 x ^ { 2 } + 24 x + 31$ in the form $a ( x + b ) ^ { 2 } + c$.
\item For the curve $y = 4 x ^ { 2 } + 24 x + 31$,\\
(A) write down the equation of the line of symmetry,\\
(B) write down the minimum $y$-value on the curve.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C1 Q1 [12]}}