| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Graph y=a|bx+c|+d with unknown constants: find constants then solve |
| Difficulty | Standard +0.3 This is a straightforward modulus function question requiring standard techniques: finding intercepts by substitution, locating the vertex of a V-shaped graph, solving a modulus inequality, and finding conditions for intersection. All steps are routine applications of well-practiced methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02s Modulus graphs: sketch graph of |ax+b|1.02t Solve modulus equations: graphically with modulus function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(7\) | B1 | Condone \((0,7)\) but not \((7,0)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{9}{k}\) | B1 | Condone \(\left(\frac{9}{k}, -2\right)\) but not \(\left(-2, \frac{9}{k}\right)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(kx-9-2=0\) and \(-kx+9-2=0 \Rightarrow \frac{7}{k}, \frac{11}{k}\) | M1 | Correct attempt to find two critical values; forms two correct equations/inequalities with modulus removed |
| CVs \(\frac{7}{k}, \frac{11}{k}\) | A1 | Correct critical values, cao |
| \(\frac{7}{k} < x < \frac{11}{k}\) | A1 | Must be in terms of \(x\); do not accept \(x>\frac{7}{k}\), \(x<\frac{11}{k}\) as separate inequalities; do not accept \(\left[\frac{7}{k}, \frac{11}{k}\right]\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{"}{-2}\text{"} = -2 \times \text{"}\frac{9}{k}\text{"} + 3 \Rightarrow k = \ldots\) | M1 | Uses correct method to find value of \(k\), e.g. substitutes \(\left(\frac{9}{k}, -2\right)\) into \(y=3-2x\) |
| \((k=)\ 3.6\) | A1 | |
| \(k > 3.6\) | A1 |
# Question 5:
## Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $7$ | B1 | Condone $(0,7)$ but not $(7,0)$ |
## Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{9}{k}$ | B1 | Condone $\left(\frac{9}{k}, -2\right)$ but not $\left(-2, \frac{9}{k}\right)$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $kx-9-2=0$ and $-kx+9-2=0 \Rightarrow \frac{7}{k}, \frac{11}{k}$ | M1 | Correct attempt to find two critical values; forms two correct equations/inequalities with modulus removed |
| CVs $\frac{7}{k}, \frac{11}{k}$ | A1 | Correct critical values, cao |
| $\frac{7}{k} < x < \frac{11}{k}$ | A1 | Must be in terms of $x$; do not accept $x>\frac{7}{k}$, $x<\frac{11}{k}$ as separate inequalities; do not accept $\left[\frac{7}{k}, \frac{11}{k}\right]$ |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{"}{-2}\text{"} = -2 \times \text{"}\frac{9}{k}\text{"} + 3 \Rightarrow k = \ldots$ | M1 | Uses correct method to find value of $k$, e.g. substitutes $\left(\frac{9}{k}, -2\right)$ into $y=3-2x$ |
| $(k=)\ 3.6$ | A1 | |
| $k > 3.6$ | A1 | |5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{44035bf8-f54c-472a-b969-b4fa4fa3d203-14_668_812_258_566}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows part of the graph with equation $y = \mathrm { f } ( x )$, where
$$\mathrm { f } ( x ) = | k x - 9 | - 2 \quad x \in \mathbb { R }$$
and $k$ is a positive constant.
The graph intersects the $y$-axis at the point $A$ and has a minimum point at $B$ as shown.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the $y$ coordinate of $A$
\item Find, in terms of $k$, the $x$ coordinate of $B$
\end{enumerate}\item Find, in terms of $k$, the range of values of $x$ that satisfy the inequality
$$| k x - 9 | - 2 < 0$$
Given that the line $y = 3 - 2 x$ intersects the graph $y = \mathrm { f } ( x )$ at two distinct points,
\item find the range of possible values of $k$
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2022 Q5 [8]}}