| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2024 |
| Session | March |
| Marks | 6 |
| 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 of the line equation into the curve equation, then applying the discriminant condition (b²-4ac=0) for a repeated root. The algebra is straightforward and the technique is commonly practiced in P1, making it slightly easier than average but still requiring proper method application. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02d Quadratic functions: graphs and discriminant conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt substitution for \(y\) in quadratic equation | \*M1 | Or substitution for \(x\)… |
| Obtain \(5x^2+30x+75-k[=0]\) or \(5y^2-20y+50-k[=0]\) | A1 | OE e.g. \(x^2+6x+15-\frac{k}{5}\) (all terms gathered together). |
| Use \(b^2-4ac=0\) with *their* \(a\), \(b\) and \(c\) | DM1 | \('=0'\) may be implied in subsequent working or the answer. |
| Obtain \(900-20(75-k)=0\) or equivalent and hence \(k=30\) | A1 | …obtaining \(400-20(50-k)=0\) and \(k=30\) |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substitute *their* value of \(k\) in equation from part (a) and attempt solution | M1 | Expect \(5x^2+30x+45[=0]\) or \(5y^2-20y+20[=0]\) |
| Obtain coordinates \((-3, 2)\) | A1 | SC B1 only \((-3, 2)\) without attempt at quadratic solution. |
| Total: 2 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt substitution for $y$ in quadratic equation | \*M1 | Or substitution for $x$… |
| Obtain $5x^2+30x+75-k[=0]$ or $5y^2-20y+50-k[=0]$ | A1 | OE e.g. $x^2+6x+15-\frac{k}{5}$ (all terms gathered together). |
| Use $b^2-4ac=0$ with *their* $a$, $b$ and $c$ | DM1 | $'=0'$ may be implied in subsequent working or the answer. |
| Obtain $900-20(75-k)=0$ or equivalent and hence $k=30$ | A1 | …obtaining $400-20(50-k)=0$ and $k=30$ |
| **Total: 4** | | |
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## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute *their* value of $k$ in equation from part **(a)** and attempt solution | M1 | Expect $5x^2+30x+45[=0]$ or $5y^2-20y+20[=0]$ |
| Obtain coordinates $(-3, 2)$ | A1 | **SC B1** only $(-3, 2)$ without attempt at quadratic solution. |
| **Total: 2** | | |
---7 The straight line $\mathrm { y } = \mathrm { x } + 5$ meets the curve $2 \mathrm { x } ^ { 2 } + 3 \mathrm { y } ^ { 2 } = \mathrm { k }$ at a single point $P$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of the constant $k$.
\item Find the coordinates of $P$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2024 Q7 [6]}}