| Exam Board | Edexcel |
|---|---|
| Module | PMT Mocks (PMT Mocks) |
| 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 question requiring substitution and solving a quartic that factors nicely. Part (a) is simple verification by substitution, and part (b) involves equating the curves, expanding, and factoring—all standard techniques with no novel insight required. Slightly easier than average due to the helpful verification in part (a) and the fact that the algebra, while involving a quartic, is designed to factor cleanly. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | Verifies that \(x = 3\) gives \(y = 9\) in both curves. E.g. \(y = (3)^3 - 2(3)^2 = 9\) and \(y = 9 - \frac{5}{2}(3-3)^2 = 9\) | B1 |
| (b) | \((1,-1)\) | A1 |
**(a)** | Verifies that $x = 3$ gives $y = 9$ in both curves. E.g. $y = (3)^3 - 2(3)^2 = 9$ and $y = 9 - \frac{5}{2}(3-3)^2 = 9$ | B1 | |
**(b)** | $(1,-1)$ | A1 | B1 Sets equations equal to each other and proceeds to $2x^3 + x^2 - 30x + 27 = 0$. E.g. $x^3 - 2x^2 = 9 - \frac{5}{2}(x-3)^2 \Rightarrow 2x^3 - 4x^2 = 18 - 5x^2 + 30x - 45 \Rightarrow 2x^3 + x^2 - 30x + 27 = 0$. M1 Divides by $(x-3)$ to form a quadratic factor. Allow any suitable algebraic method including division or inspection. If attempted via inspection look for correct first term and last terms. E.g. $2x^3 + x^2 - 30x + 27 = (x-3)(2x^2 + ax + 9)$ if cubic expression is correct. If attempted via division look for correct first and second terms. E.g. $\begin{matrix} x-3 \end{matrix} 2x^3 + x^2 - 30x + 27$. A1 $2x^3 + x^2 - 30x + 27 = (x-3)(2x^2 + 7x - 9)$. M1 Solves their quadratic equation $2x^2 + 7x - 9 = 0$ using a suitable method including calculator. E.g. $2x^2 + 7x - 9 = (2x + 9)(x - 1) = 0 \Rightarrow x = -4.5, x = 1$. A1 Gives $x = 1$ only. A1 Coordinates of $P = (1,-1)$ |
---11.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{48f9a252-61a2-491d-94d0-8470aee96942-16_1123_1031_280_511}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
The figure 5 shows part of the curves $C _ { 1 }$ and $C _ { 2 }$ with equations
$$\begin{array} { c c }
C _ { 1 } : y = x ^ { 3 } - 2 x ^ { 2 } & x > 0 \\
C _ { 2 } : y = 9 - \frac { 5 } { 2 } ( x - 3 ) ^ { 2 } & x > 0
\end{array}$$
The curves $C _ { 1 }$ and $C _ { 2 }$ intersect at the points $P$ and $Q$.\\
a. Verify that the point $Q$ has coordinates $( 3,9 )$\\
b. Use algebra to find the coordinates of the point $P$.\\
\hfill \mbox{\textit{Edexcel PMT Mocks Q11 [7]}}