| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Function Transformations |
| Type | Identify transformation from equations |
| Difficulty | Moderate -0.3 This is a multi-part question covering standard P1 content: sketching a reciprocal function, identifying a horizontal translation, and finding intersection points. Parts (a)-(b) are routine recall, while parts (c)-(d) require substitution and solving a quadratic, which are standard techniques. The question is slightly easier than average due to the straightforward nature of each component and clear scaffolding. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.02o Sketch reciprocal curves: y=a/x and y=a/x^21.02q Use intersection points: of graphs to solve equations1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Correct shape in quadrant 1 or 3 | M1 | Monotonically decreasing function in quadrant 1 or 3 with no incorrect asymptotes. Must not cross either axis |
| Fully correct shape and position | A1 | Correct shape and position with no incorrect asymptotes. Score best single attempt if both diagrams used |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Partial description implying at least one component e.g. "Translates 2 units left", "Shifts/moves 2 units right" | B1 | Partial description implying at least one of the two components but not fully correct |
| Translate 2 (units) to the right (parallel to the \(x\)-axis) | B1 | Requires (1): Translate/translation and (2): \(\begin{pmatrix}2\\0\end{pmatrix}\) or "2 (units to the) right" or "+2 in the \(x\) direction". Minimum e.g. "Translate \(+2\) on \(x\)" |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| When \(x = -4 \Rightarrow \frac{6}{x-2} = kx + 7 \Rightarrow \frac{6}{-6} = -4k + 7 \Rightarrow k = \ldots\) | M1 | |
| \(k = 2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{6}{x-2} =\) "2"\(x + 7 \Rightarrow 6 = 2x^2 + 3x - 14\) | M1 | |
| \(0 = 2x^2 + 3x - 20 \Rightarrow (2x-5)(x+4) = 0\) | dM1 | |
| \(Q = \left(\frac{5}{2}, 12\right)\) | A1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Substitutes \(x = \pm 4\) into \(\frac{6}{x-2} = kx + 7\) and solves for \(k\) | M1 | Allow equivalent work e.g. \(x = -4 \Rightarrow y = -1 \Rightarrow -1 = k(-4)+7\). Note rearrangement before substitution is fine. E.g. \(6 = (x-2)(kx+7) = kx^2+(7-2k)x-14\), \(6=24k-42\) |
| \(k = 2\) o.e. e.g. \(k = \frac{4}{2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Equates \(\frac{6}{x-2}\) with \(kx+7\) using their value for \(k\), cross multiplies to obtain a quadratic in \(x\) | M1 | Terms not necessarily collected. Rearrangement may have already been done in part (c) |
| Solves 3TQ by any acceptable method including via a calculator | dM1 | |
| \(x = \frac{5}{2}\). Condone \(Q = \frac{5}{2}\) | A1 | |
| \(Q = \left(\frac{5}{2}, 12\right)\) | A1 | Must be as coordinates or \(x = \ldots, y = \ldots\) |
## Question 7(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Correct shape in quadrant 1 or 3 | M1 | Monotonically decreasing function in quadrant 1 or 3 with no incorrect asymptotes. Must not cross either axis |
| Fully correct shape and position | A1 | Correct shape and position with no incorrect asymptotes. Score best single attempt if both diagrams used |
---
## Question 7(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Partial description implying at least one component e.g. "Translates 2 units left", "Shifts/moves 2 units right" | B1 | Partial description implying at least one of the two components but not fully correct |
| Translate 2 (units) to the right (parallel to the $x$-axis) | B1 | Requires (1): Translate/translation **and** (2): $\begin{pmatrix}2\\0\end{pmatrix}$ or "2 (units to the) right" or "+2 in the $x$ direction". Minimum e.g. "Translate $+2$ on $x$" |
---
## Question 7(c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| When $x = -4 \Rightarrow \frac{6}{x-2} = kx + 7 \Rightarrow \frac{6}{-6} = -4k + 7 \Rightarrow k = \ldots$ | M1 | |
| $k = 2$ | A1 | |
---
## Question 7(d):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{6}{x-2} =$ "2"$x + 7 \Rightarrow 6 = 2x^2 + 3x - 14$ | M1 | |
| $0 = 2x^2 + 3x - 20 \Rightarrow (2x-5)(x+4) = 0$ | dM1 | |
| $Q = \left(\frac{5}{2}, 12\right)$ | A1, A1 | |
# Question (c) [Previous question - tangent/curve intersection]:
| Answer | Mark | Guidance |
|--------|------|----------|
| Substitutes $x = \pm 4$ into $\frac{6}{x-2} = kx + 7$ and solves for $k$ | M1 | Allow equivalent work e.g. $x = -4 \Rightarrow y = -1 \Rightarrow -1 = k(-4)+7$. Note rearrangement before substitution is fine. E.g. $6 = (x-2)(kx+7) = kx^2+(7-2k)x-14$, $6=24k-42$ |
| $k = 2$ o.e. e.g. $k = \frac{4}{2}$ | A1 | |
# Question (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| Equates $\frac{6}{x-2}$ with $kx+7$ using their value for $k$, cross multiplies to obtain a quadratic in $x$ | M1 | Terms not necessarily collected. Rearrangement may have already been done in part (c) |
| Solves 3TQ by any acceptable method including via a calculator | dM1 | |
| $x = \frac{5}{2}$. Condone $Q = \frac{5}{2}$ | A1 | |
| $Q = \left(\frac{5}{2}, 12\right)$ | A1 | Must be as coordinates or $x = \ldots, y = \ldots$ |
---\begin{enumerate}
\item (a) On Diagram 1, sketch a graph of the curve $C$ with equation
\end{enumerate}
$$y = \frac { 6 } { x } \quad x \neq 0$$
The curve $C$ is transformed onto the curve with equation $y = \frac { 6 } { x - 2 } \quad x \neq 2$\\
(b) Fully describe this transformation.
The curve with equation
$$y = \frac { 6 } { x - 2 } \quad x \neq 2$$
and the line with equation
$$y = k x + 7 \quad \text { where } k \text { is a constant }$$
intersect at exactly two points, $P$ and $Q$.\\
Given that the $x$ coordinate of point $P$ is - 4\\
(c) find the value of $k$,\\
(d) find, using algebra, the coordinates of point $Q$.\\
(Solutions relying entirely on calculator technology are not acceptable.)
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{bb21001f-fe68-4776-992d-ede1aae233d7-17_710_743_248_662}
\end{center}
\section*{Diagram 1}
Only use this copy of Diagram 1 if you need to redraw your graph.\\
\includegraphics[max width=\textwidth, alt={}, center]{bb21001f-fe68-4776-992d-ede1aae233d7-19_709_739_1802_664}
Copy of Diagram 1\\
(Total for Question 7 is 10 marks)\\
\hfill \mbox{\textit{Edexcel P1 2023 Q7 [10]}}