| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Function Transformations |
| Type | Multiple separate transformations (sketch-based, standard transformations) |
| Difficulty | Moderate -0.3 This is a standard P1 transformation question requiring application of horizontal stretch (factor 1/2) and translation rules. Students must identify key points and apply transformations systematically, but the transformations are routine textbook exercises with no novel problem-solving required. The 7 total marks reflect straightforward application rather than conceptual difficulty. |
| Spec | 1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| 6(a) | y | |
| x | Shape through (0, 0) | B1 |
| (3, 0) | B1 | |
| (1.5, (cid:16)1) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | y | |
| x | Shape , not through (0, 0) | B1 |
| Minimum in 4th quadrant | B1 | |
| ((cid:16)p, 0) and (6 (cid:16)p, 0) | B1 | |
| (3 (cid:16)p, (cid:16)1) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Scheme | Marks |
Question 6:
--- 6(a) ---
6(a) | y
x | Shape through (0, 0) | B1
(3, 0) | B1
(1.5, (cid:16)1) | B1
(3)
(b) | y
x | Shape , not through (0, 0) | B1
Minimum in 4th quadrant | B1
((cid:16)p, 0) and (6 (cid:16)p, 0) | B1
(3 (cid:16)p, (cid:16)1) | B1
(4)
(7 marks)
Notes:
(a)
B1: U shaped parabola through origin.
B1: (3,0) stated or 3 labelled on x - axis (even (0,3) on x - axis).
B1: (1.5, -1) or equivalent e.g. (3/2, -1) labelled or stated and matching minimum point on the
graph.
(b)
B1: Is for any translated curve to left or right or up or down not through origin
B1: Is for minimum in 4th quadrant and x intercepts to left and right of y axis
(i.e. correct position).
B1: Coordinates stated or shown on x axis (Allow (0 – p, 0) instead of (-p, 0))
B1: Coordinates stated.
Note: If values are taken for p, then it is possible to give M1A1B0B0 even if there are
several attempts. (In this case none of the curves should go through the origin for M1 and
all minima should be in fourth quadrant and all x intercepts need to be to left and right of
y axis for A1
y
x
Question | Scheme | Marks\includegraphics{figure_1}
Figure 1 shows a sketch of the curve $C$ with equation $y = \text{f}(x)$
The curve $C$ passes through the origin and through $(6, 0)$
The curve $C$ has a minimum at the point $(3, -1)$
On separate diagrams, sketch the curve with equation
\begin{enumerate}[label=(\alph*)]
\item $y = \text{f}(2x)$ [3]
\item $y = \text{f}(x + p)$, where $p$ is a constant and $0 < p < 3$ [4]
\end{enumerate}
On each diagram show the coordinates of any points where the curve intersects the $x$-axis and of any minimum or maximum points.
\hfill \mbox{\textit{Edexcel P1 2018 Q6 [7]}}