| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Graph y = a|bx+c| + d: identify vertex and intercepts |
| Difficulty | Standard +0.3 This is a standard modulus equation problem requiring sketching two graphs and finding intersections. Part (a) involves routine sketching of modulus functions with axis intercepts. Part (b) uses the graphical approach to identify regions and solve algebraically, but the structure is predictable and the algebra straightforward. Slightly easier than average due to the guided 'hence' structure and standard techniques. |
| 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 |
| Upside down V shape with maximum on \(y\)-axis | M1 | Shape of \(y = k - 2\ |
| Intercepts at \(k\) on \(y\)-axis and \(\pm\frac{k}{2}\) on \(x\)-axis | A1 | Allow as values or coordinates either way round; graph must cross axes at these points |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| V shape in quadrants 1 and 2 with minimum on \(x\)-axis | M1 | Shape of \(y = \ |
| Intercepts at \(\frac{k}{3}\) on \(y\)-axis and \(\frac{k}{6}\) on \(x\)-axis | A1 | Graph must cross \(y\)-axis and touch \(x\)-axis |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts to solve either \(k + 2x = -2x + \frac{k}{3}\) or \(k - 2x = 2x - \frac{k}{3}\) | M1 | One correct equation not involving moduli |
| Achieves either \(x = -\frac{1}{6}k\) or \(x = \frac{1}{3}k\) | A1 | Allow unsimplified e.g. \(-\frac{2k}{12}\) |
| Attempts to solve both \(k + 2x = -2x + \frac{k}{3}\) and \(k - 2x = 2x - \frac{k}{3}\) | M1 | Two correct equations not involving moduli |
| \(x = -\frac{1}{6}k\) and \(x = \frac{1}{3}k\) | A1 | Two correct simplified solutions, no other solutions given |
## Question 6:
### Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Upside down V shape with maximum on $y$-axis | M1 | Shape of $y = k - 2\|x\|$ |
| Intercepts at $k$ on $y$-axis and $\pm\frac{k}{2}$ on $x$-axis | A1 | Allow as values or coordinates either way round; graph must cross axes at these points |
### Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| V shape in quadrants 1 and 2 with minimum on $x$-axis | M1 | Shape of $y = \|2x - \frac{k}{3}\|$ |
| Intercepts at $\frac{k}{3}$ on $y$-axis and $\frac{k}{6}$ on $x$-axis | A1 | Graph must cross $y$-axis and touch $x$-axis |
**(4 marks)**
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to solve either $k + 2x = -2x + \frac{k}{3}$ **or** $k - 2x = 2x - \frac{k}{3}$ | M1 | One correct equation not involving moduli |
| Achieves either $x = -\frac{1}{6}k$ **or** $x = \frac{1}{3}k$ | A1 | Allow unsimplified e.g. $-\frac{2k}{12}$ |
| Attempts to solve both $k + 2x = -2x + \frac{k}{3}$ **and** $k - 2x = 2x - \frac{k}{3}$ | M1 | Two correct equations not involving moduli |
| $x = -\frac{1}{6}k$ **and** $x = \frac{1}{3}k$ | A1 | Two correct simplified solutions, no other solutions given |
**Note (squaring method):** $\left(2x - \frac{k}{3}\right)^2 = (k-2x)^2 \Rightarrow x = \frac{k}{3}$; Score M1 for squaring to obtain at least 3 terms on both sides and solving; A1 for $x = \frac{k}{3}$. Candidates unlikely to find other root this way.
**(4 marks total, 8 marks)**
---6. Given that $k$ is a positive constant,
\begin{enumerate}[label=(\alph*)]
\item on separate diagrams, sketch the graph with equation
\begin{enumerate}[label=(\roman*)]
\item $y = k - 2 | x |$
\item $y = \left| 2 x - \frac { k } { 3 } \right|$
Show on each sketch the coordinates, in terms of $k$, of each point where the graph meets or cuts the axes.
\end{enumerate}\item Hence find, in terms of $k$, the values of $x$ for which
$$\left| 2 x - \frac { k } { 3 } \right| = k - 2 | x |$$
giving your answers in simplest form.
\includegraphics[max width=\textwidth, alt={}, center]{76205772-5395-4ab2-96f9-ad9803b8388c-23_2647_1840_118_111}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2021 Q6 [8]}}