| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simultaneous equations |
| Type | Tangency condition for two curves |
| Difficulty | Standard +0.3 This is a standard tangency condition problem requiring students to set equations equal, form a quadratic, and apply the discriminant condition (b²-4ac=0) for a single intersection point. The algebra is straightforward, and part (b) follows directly from solving the given equation. Slightly above average difficulty due to the multi-step algebraic manipulation, but this is a textbook exercise with no novel insight required. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02f Solve quadratic equations: including in a function of unknown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^2 + kx - 9 = -3x^2 - 5x + k \Rightarrow 4x^2 + kx + 5x - 9 - k (= 0)\) | M1 | Equates both curves and collects terms; "= 0" may be implied by discriminant attempt |
| \(b^2 - 4ac = 0 \Rightarrow (k+5)^2 - 4 \times 4(-9-k) = 0\) | M1 | Attempts discriminant of 3TQ; requires \(b\) and \(c\) both of form \(pk+q\), \(p,q \neq 0\); wrong formula e.g. \(b^2+4ac\) scores M0 |
| \(k^2 + 26k + 169 = 0\)* | A1* | Achieves printed answer with no errors and sufficient working; "= 0" must appear at least once before printed answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(k^2 + 26k + 169 = 0 \Rightarrow k = -13\) | B1 | Condone \(x = -13\) if subsequently used as value for \(k\) |
| \(k = -13 \Rightarrow 4x^2 - 8x + 4 = 0 \Rightarrow x = \ldots\) | M1 | Either substitute \(k\) into equation from (a) and solve, OR substitute \(k\) into both given equations, equate and solve |
| \((1, -21)\) | A1 | Correct coordinates; allow e.g. \(x=1\), \(y=-21\) |
## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^2 + kx - 9 = -3x^2 - 5x + k \Rightarrow 4x^2 + kx + 5x - 9 - k (= 0)$ | M1 | Equates both curves and collects terms; "= 0" may be implied by discriminant attempt |
| $b^2 - 4ac = 0 \Rightarrow (k+5)^2 - 4 \times 4(-9-k) = 0$ | M1 | Attempts discriminant of 3TQ; requires $b$ and $c$ both of form $pk+q$, $p,q \neq 0$; wrong formula e.g. $b^2+4ac$ scores M0 |
| $k^2 + 26k + 169 = 0$* | A1* | Achieves printed answer with no errors and sufficient working; "= 0" must appear at least once before printed answer |
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## Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k^2 + 26k + 169 = 0 \Rightarrow k = -13$ | B1 | Condone $x = -13$ if subsequently used as value for $k$ |
| $k = -13 \Rightarrow 4x^2 - 8x + 4 = 0 \Rightarrow x = \ldots$ | M1 | Either substitute $k$ into equation from (a) and solve, OR substitute $k$ into both given equations, equate and solve |
| $(1, -21)$ | A1 | Correct coordinates; allow e.g. $x=1$, $y=-21$ |
---\begin{enumerate}
\item The curve $C _ { 1 }$ has equation
\end{enumerate}
$$y = x ^ { 2 } + k x - 9$$
and the curve $C _ { 2 }$ has equation
$$y = - 3 x ^ { 2 } - 5 x + k$$
where $k$ is a constant.\\
Given that $C _ { 1 }$ and $C _ { 2 }$ meet at a single point $P$\\
(a) show that
$$k ^ { 2 } + 26 k + 169 = 0$$
(b) Hence find the coordinates of $P$
\hfill \mbox{\textit{Edexcel P1 2024 Q4 [6]}}