| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2020 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Graph y=a|bx+c|+d given: solve equation or inequality |
| Difficulty | Standard +0.3 This is a straightforward modulus function question requiring standard techniques: finding the vertex by setting the inside equal to zero (part a), solving by considering two cases where the expression is positive/negative (part b), and using geometric reasoning about when a line intersects the V-shape twice (part c). All parts follow predictable methods taught in P3 with no novel insight required, making it slightly easier than average. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02q Use intersection points: of graphs to solve equations1.02s Modulus graphs: sketch graph of |ax+b|1.02t Solve modulus equations: graphically with modulus function |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \((2.5, 3)\) oe | B1 B1 | (2) |
| (b) Attempts one solution usually \(4x - 10 + 3 = 3x - 2 \Rightarrow x = 5\) | M1 A1 | |
| Attempts both solutions \(-4x + 10 + 3 = 3x - 2 \Rightarrow x = \frac{15}{7}\) | dM1 A1 | (4) |
| (c) Attempts to solve \(y = kx + 2\) with \(x = 2.5, y = 3\) or states that \(k < 4\) | M1 | |
| \(k = ...\frac{2}{5}\) | A1 | |
| States \(\frac{2}{5} < k < 4\) | A1 | (3) |
**(a)** $(2.5, 3)$ oe | B1 B1 | (2)
**(b)** Attempts one solution usually $4x - 10 + 3 = 3x - 2 \Rightarrow x = 5$ | M1 A1 |
Attempts both solutions $-4x + 10 + 3 = 3x - 2 \Rightarrow x = \frac{15}{7}$ | dM1 A1 | (4)
**(c)** Attempts to solve $y = kx + 2$ with $x = 2.5, y = 3$ or states that $k < 4$ | M1 |
$k = ...\frac{2}{5}$ | A1 |
States $\frac{2}{5} < k < 4$ | A1 | (3)
**Total: 9 marks**
---6.
\begin{figure}[h]
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\includegraphics[alt={},max width=\textwidth]{1c700103-ecab-4a08-b411-3f445ed88885-18_736_1102_258_427}
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\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows part of the graph with equation $y = \mathrm { f } ( x )$, where
$$\mathrm { f } ( x ) = 2 | 2 x - 5 | + 3 \quad x \geqslant 0$$
The vertex of the graph is at point $P$ as shown.
\begin{enumerate}[label=(\alph*)]
\item State the coordinates of $P$.
\item Solve the equation $\mathrm { f } ( x ) = 3 x - 2$
Given that the equation
$$f ( x ) = k x + 2$$
where $k$ is a constant, has exactly two roots,
\item find the range of values of $k$.\\
\begin{center}
\end{center}
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2020 Q6 [9]}}