| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simultaneous equations |
| Type | Two curves intersecting |
| Difficulty | Moderate -0.3 This is a straightforward simultaneous equations problem requiring equating curves, simplifying to a cubic, factoring out a known root (x-1), then solving the resulting quadratic. While it involves multiple steps and a cubic equation, the process is mechanical and follows standard AS-level techniques with no novel insight required. Slightly easier than average due to being given one intersection point to verify. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts both \(y=8-10\times1+6\times1^2-1^3\) and \(y=1^2-12\times1+14\) | M1 | 1.1b — must be seen in (a) |
| Achieves \(y=3\) for both equations and gives minimal conclusion e.g. \((1,3)\) lies on both curves so they intersect at \(x=1\) | A1 | 1.1b — requires both correct calculations with minimal conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Curves intersect when \(x^2-12x+14=8-10x+6x^2-x^3 \Rightarrow x^3-5x^2-2x+6=0\) | M1 | 1.1b — sets equal and proceeds to cubic \(= 0\) |
| Key step: dividing by \((x-1)\): \(x^3-5x^2-2x+6=(x-1)(x^2+px\pm6)\) | dM1 | 3.1a — for realising \((x-1)\) is a factor and dividing |
| \(x^3-5x^2-2x+6=(x-1)(x^2-4x-6)\) | A1 | 1.1b — or just \(x^2-4x-6\) as quadratic factor |
| Solves \(x^2-4x-6=0\): \((x-2)^2=10 \Rightarrow x=\ldots\) | ddM1 | 1.1b — must be 3TQ solved by completing square or quadratic formula; quadratic must not factorise |
| \(x=2-\sqrt{10}\) only | A1 | 1.1b — must discard \(2+\sqrt{10}\); exact equivalent e.g. \(\frac{4-\sqrt{40}}{2}\) acceptable |
## Question 15(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts both $y=8-10\times1+6\times1^2-1^3$ and $y=1^2-12\times1+14$ | M1 | 1.1b — must be seen in (a) |
| Achieves $y=3$ for both equations and gives minimal conclusion e.g. $(1,3)$ lies on both curves so they intersect at $x=1$ | A1 | 1.1b — requires both correct calculations with minimal conclusion |
**(2)**
## Question 15(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Curves intersect when $x^2-12x+14=8-10x+6x^2-x^3 \Rightarrow x^3-5x^2-2x+6=0$ | M1 | 1.1b — sets equal and proceeds to cubic $= 0$ |
| Key step: dividing by $(x-1)$: $x^3-5x^2-2x+6=(x-1)(x^2+px\pm6)$ | dM1 | 3.1a — for realising $(x-1)$ is a factor and dividing |
| $x^3-5x^2-2x+6=(x-1)(x^2-4x-6)$ | A1 | 1.1b — or just $x^2-4x-6$ as quadratic factor |
| Solves $x^2-4x-6=0$: $(x-2)^2=10 \Rightarrow x=\ldots$ | ddM1 | 1.1b — must be 3TQ solved by completing square or quadratic formula; quadratic must not factorise |
| $x=2-\sqrt{10}$ only | A1 | 1.1b — must discard $2+\sqrt{10}$; exact equivalent e.g. $\frac{4-\sqrt{40}}{2}$ acceptable |
**(5)**
**(7 marks total)**\begin{enumerate}
\item In this question you must show detailed reasoning.
\end{enumerate}
\section*{Solutions relying on calculator technology are not acceptable.}
The curve $C _ { 1 }$ has equation $y = 8 - 10 x + 6 x ^ { 2 } - x ^ { 3 }$\\
The curve $C _ { 2 }$ has equation $y = x ^ { 2 } - 12 x + 14$\\
(a) Verify that when $x = 1$ the curves $C _ { 1 }$ and $C _ { 2 }$ intersect.
The curves also intersect when $x = k$.\\
Given that $k < 0$\\
(b) use algebra to find the exact value of $k$.
\hfill \mbox{\textit{Edexcel AS Paper 1 2023 Q15 [7]}}