| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simultaneous equations |
| Type | Tangency condition for line and curve |
| Difficulty | Moderate -0.3 This is a standard tangency condition problem requiring substitution to form a quadratic, then applying the discriminant condition (b²-4ac=0) for one solution. While it involves multiple steps and the discriminant concept, it's a routine textbook exercise with a clear algorithmic approach that P1 students practice extensively. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02f Solve quadratic equations: including in a function of unknown1.02q Use intersection points: of graphs to solve equations |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Sets \(4x + c = x(x-3)\) and attempts to write as a 3TQ | M1 | All terms don't need to be on same side; condone slips |
| Uses \(b^2 = 4ac\) for their \(x^2 - 7x - c = 0\) | dM1 | May use \(b^2 - 4ac = 0\); condone \(\pm 49 \pm 4c = 0\) |
| Correct equation \(49 = -4c\) or \(49 + 4c = 0\) | A1 | |
| \(c = -12.25\) oe | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Attempts to solve \(x^2 - 7x - c = 0\) with their \(c\) \((c \neq 0)\) | M1 | Allow usual methods; calculator answer must be correct for their 3TQ |
| Attempts to find \(y\) coordinate for their \(x\) coordinate | dM1 | Dependent on previous M; if two \(x\) values, only need \(y\) for one |
| \(\left(\frac{7}{2}, \frac{7}{4}\right)\) oe | A1 | Accept written separately: \(x = 3.5, y = 1.75\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{d}{dx}(x^2 - 3x) = 4 \Rightarrow 2x - 3 = 4 \Rightarrow x = \ldots\) | M1 | |
| Attempts \(y\) coordinate by substituting into \(y = x(x-3)\) | dM1 | |
| Correct equation in \(c\): e.g. \(\frac{7}{4} = 4 \times \frac{7}{2} + c\) | A1 | |
| \(c = -12.25\) oe | A1 |
## Question 6:
### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| Sets $4x + c = x(x-3)$ and attempts to write as a 3TQ | M1 | All terms don't need to be on same side; condone slips |
| Uses $b^2 = 4ac$ for their $x^2 - 7x - c = 0$ | dM1 | May use $b^2 - 4ac = 0$; condone $\pm 49 \pm 4c = 0$ |
| Correct equation $49 = -4c$ or $49 + 4c = 0$ | A1 | |
| $c = -12.25$ oe | A1 | |
### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| Attempts to solve $x^2 - 7x - c = 0$ with their $c$ $(c \neq 0)$ | M1 | Allow usual methods; calculator answer must be correct for their 3TQ |
| Attempts to find $y$ coordinate for their $x$ coordinate | dM1 | Dependent on previous M; if two $x$ values, only need $y$ for one |
| $\left(\frac{7}{2}, \frac{7}{4}\right)$ oe | A1 | Accept written separately: $x = 3.5, y = 1.75$ |
**Alternative via gradients:**
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{d}{dx}(x^2 - 3x) = 4 \Rightarrow 2x - 3 = 4 \Rightarrow x = \ldots$ | M1 | |
| Attempts $y$ coordinate by substituting into $y = x(x-3)$ | dM1 | |
| Correct equation in $c$: e.g. $\frac{7}{4} = 4 \times \frac{7}{2} + c$ | A1 | |
| $c = -12.25$ oe | A1 | |
---\begin{enumerate}
\item The line with equation $y = 4 x + c$, where $c$ is a constant, meets the curve with equation $y = x ( x - 3 )$ at only one point.\\
(a) Find the value of $c$.\\
(b) Hence find the coordinates of the point of intersection.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2019 Q6 [7]}}