| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Completing the square and sketching |
| Type | Line-curve intersection points |
| Difficulty | Moderate -0.8 This is a straightforward C1 question testing routine completing the square (standard technique with simple coefficients), identifying a minimum point, basic sketching, and solving simultaneous equations by substitution. All parts are textbook exercises requiring only standard procedures with no problem-solving insight or challenging algebra. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02q Use intersection points: of graphs to solve equations |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(x^2 + 2x + 4 = (x+1)^2 - 1 + 4 = (x+1)^2 + 3\) | M1, A1, B2 | Completing the square; minimum: \((-1, 3)\) |
| (ii) Graph showing parabola C with vertex near \((-1, 3)\), crossing positive y-axis, and line \(l\) | B2, B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\therefore (-4, 12)\) and \((1, 7)\) | M1, A1, M1, A1 | (11 marks) |
**(i)** $x^2 + 2x + 4 = (x+1)^2 - 1 + 4 = (x+1)^2 + 3$ | M1, A1, B2 | Completing the square; minimum: $(-1, 3)$
**(ii)** Graph showing parabola C with vertex near $(-1, 3)$, crossing positive y-axis, and line $l$ | B2, B1 |
**(iii)** $x^2 + 2x + 4 = 8 - x$
$x^2 + 3x - 4 = 0$
$(x+4)(x-1) = 0$
$x = -4, 1$
$\therefore (-4, 12)$ and $(1, 7)$ | M1, A1, M1, A1 | (11 marks)
---The curve $C$ has the equation $y = x^2 + 2x + 4$.
\begin{enumerate}[label=(\roman*)]
\item Express $x^2 + 2x + 4$ in the form $(x + p)^2 + q$ and hence state the coordinates of the minimum point of $C$. [4]
\end{enumerate}
The straight line $l$ has the equation $x + y = 8$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Sketch $l$ and $C$ on the same set of axes. [3]
\item Find the coordinates of the points where $l$ and $C$ intersect. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR C1 Q7 [11]}}