| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Sketch y=|f(x)| or y=f(|x|) for non-linear f(x) and solve |
| Difficulty | Standard +0.3 This is a straightforward modulus function question requiring standard techniques: finding the vertex by setting the inner expression to zero, solving a modulus inequality by considering cases, and sketching |f(x)| by reflecting negative portions. All are routine P3/C3 procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02l Modulus function: notation, relations, equations and inequalities1.02s Modulus graphs: sketch graph of |ax+b|1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(-\frac{7}{2}, -10\right)\) | B1 B1 | First B1 for one correct coordinate; second B1 for both correct. Allow \(x = -\frac{7}{2}, y = -10\). Allow exact equivalents e.g. \(x = -\frac{14}{4}\). Must be answered in (a), cannot imply from (c) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempts to solve \(\frac{1}{2}(2x+7)-10 \geq \frac{1}{3}x+1\) or \(-\frac{1}{2}(2x+7)-10 \geq \frac{1}{3}x+1\) | M1 | Must proceed as far as \(x...\); condone slips attempting to make \(\ |
| Both critical values \(x = -\frac{87}{8}, \frac{45}{4}\) | A1 | May be part of an incorrect inequality |
| Selects outside region for their critical values | dM1 | Dependent on M1; allow written as one inequality; allow strict inequalities |
| \(x \leq -\frac{87}{8},\ x \geq \frac{45}{4}\) | A1 | Allow forms such as \(x \in \left(-\infty, -\frac{87}{8}\right] \cup \left[\frac{45}{4}, \infty\right)\); do not allow \(\frac{45}{4} \leq x \leq -\frac{87}{8}\) or intersection form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| W shape, any position | B1 | Condone asymmetric arms |
| Maximum point at \(\left(-\frac{7}{2}, 10\right)\) | B1ft | Follow through on coordinates from part (a) |
| One correct minimum shown | B1 | Must be minimum point, not just x-axis crossing; condone \((0, 6.5)\) for \((6.5, 0)\) if on correct axis |
| Fully correct: \(x = -\frac{27}{2}\) and \(x = \frac{13}{2}\) marked, all points correct | B1 | Allow exact equivalents; condone asymmetric arms; there must be a sketch — graph takes precedence over text if they contradict |
# Question 7:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(-\frac{7}{2}, -10\right)$ | B1 B1 | First B1 for one correct coordinate; second B1 for both correct. Allow $x = -\frac{7}{2}, y = -10$. Allow exact equivalents e.g. $x = -\frac{14}{4}$. Must be answered in (a), cannot imply from (c) |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempts to solve $\frac{1}{2}(2x+7)-10 \geq \frac{1}{3}x+1$ or $-\frac{1}{2}(2x+7)-10 \geq \frac{1}{3}x+1$ | M1 | Must proceed as far as $x...$; condone slips attempting to make $\|2x+7\|$ the subject |
| Both critical values $x = -\frac{87}{8}, \frac{45}{4}$ | A1 | May be part of an incorrect inequality |
| Selects outside region for their critical values | dM1 | Dependent on M1; allow written as one inequality; allow strict inequalities |
| $x \leq -\frac{87}{8},\ x \geq \frac{45}{4}$ | A1 | Allow forms such as $x \in \left(-\infty, -\frac{87}{8}\right] \cup \left[\frac{45}{4}, \infty\right)$; do not allow $\frac{45}{4} \leq x \leq -\frac{87}{8}$ or intersection form |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| W shape, any position | B1 | Condone asymmetric arms |
| Maximum point at $\left(-\frac{7}{2}, 10\right)$ | B1ft | Follow through on coordinates from part (a) |
| One correct minimum shown | B1 | Must be minimum point, not just x-axis crossing; condone $(0, 6.5)$ for $(6.5, 0)$ if on correct axis |
| Fully correct: $x = -\frac{27}{2}$ and $x = \frac{13}{2}$ marked, all points correct | B1 | Allow exact equivalents; condone asymmetric arms; there must be a sketch — graph takes precedence over text if they contradict |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f3272b4c-d8dc-4f32-add9-153de90f4d0a-18_720_746_210_591}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of part of the graph with equation $y = \mathrm { f } ( x )$, where
$$f ( x ) = \frac { 1 } { 2 } | 2 x + 7 | - 10$$
\begin{enumerate}[label=(\alph*)]
\item State the coordinates of the vertex, V, of the graph.
\item Solve, using algebra,
$$\frac { 1 } { 2 } | 2 x + 7 | - 10 \geqslant \frac { 1 } { 3 } x + 1$$
\item Sketch the graph with equation
$$y = | \mathrm { f } ( x ) |$$
stating the coordinates of the local maximum point and each local minimum point.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2022 Q7 [10]}}