| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Show line is tangent, verify |
| Difficulty | Moderate -0.8 This is a straightforward C1 question testing completing the square (routine algebraic manipulation), sketching basic graphs, and finding intersection points by substitution. All techniques are standard with no problem-solving insight required, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02e Complete the square: quadratic polynomials and turning points |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(x^2 + 2x + 4 = (x + 1)^2 - 1 + 4\) | M1 | |
| \(= (x + 1)^2 + 3\) | A1 | |
| minimum: \((-1, 3)\) | A2 | |
| (b) Sketch showing parabola \(C\) and line \(l\) with no intersection points | B2 B1 | |
| (c) \(x^2 + 2x + 4 = 8 - x\) | ||
| \(x^2 + 3x - 4 = 0\) | ||
| \((x + 4)(x - 1) = 0\) | M1 | |
| \(x = -4, 1\) | A1 | |
| \(\therefore (-4, 12) \text{ and } (1, 7)\) | M1 A1 | (11) |
(a) $x^2 + 2x + 4 = (x + 1)^2 - 1 + 4$ | M1 |
$= (x + 1)^2 + 3$ | A1 |
minimum: $(-1, 3)$ | A2 |
(b) Sketch showing parabola $C$ and line $l$ with no intersection points | B2 B1 |
(c) $x^2 + 2x + 4 = 8 - x$ | |
$x^2 + 3x - 4 = 0$ | |
$(x + 4)(x - 1) = 0$ | M1 |
$x = -4, 1$ | A1 |
$\therefore (-4, 12) \text{ and } (1, 7)$ | M1 A1 | (11)9. The curve $C$ has the equation $y = x ^ { 2 } + 2 x + 4$.
\begin{enumerate}[label=(\alph*)]
\item Express $x ^ { 2 } + 2 x + 4$ in the form $a ( x + b ) ^ { 2 } + c$ and hence state the coordinates of the minimum point of $C$.
The straight line $l$ has the equation $x + y = 8$.
\item Sketch $l$ and $C$ on the same set of axes.
\item Find the coordinates of the points where $I$ and $C$ intersect.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q9 [11]}}