| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| 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 identification of the vertex (direct substitution), solving an equation with modulus (split into two cases), and finding intersection conditions (comparing gradients at the vertex). All techniques are standard A-level procedures with no novel insight 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 |
| \(x=2\) or \(y=5\) | B1 | One correct coordinate seen |
| \(P(2,5)\) | B1 | Deduces \((2,5)\); condone \(2, 5\) without brackets |
| Total | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(16-4x = 3(x-2)+5 \Rightarrow x = \ldots\) | M1 | Attempts to solve correct equation without modulus signs; must reach a value for \(x\) |
| \(x = \frac{17}{7}\) | A1 | Exact answer only; no other values; if other values found they must be rejected |
| Total | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(k_{\max} = 3\) or \(k_{\min} = \frac{"5"-4}{"2"}\) | M1 | Correct method to find either critical value following through on their \(P\); allow \(m=\) in place of \(k=\) |
| \(\frac{1}{2} < k < 3\) | A1 | Correct range in terms of \(k\); allow "and" or "\(\cap\)" to join regions but not "or", ",", or "\(\cup\)"; do not accept \(\frac{1}{2} < x < 3\) |
| Total | (2) |
# Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=2$ or $y=5$ | B1 | One correct coordinate seen |
| $P(2,5)$ | B1 | Deduces $(2,5)$; condone $2, 5$ without brackets |
| **Total** | **(2)** | |
---
# Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $16-4x = 3(x-2)+5 \Rightarrow x = \ldots$ | M1 | Attempts to solve correct equation without modulus signs; must reach a value for $x$ |
| $x = \frac{17}{7}$ | A1 | Exact answer only; no other values; if other values found they must be rejected |
| **Total** | **(2)** | |
---
# Question 6(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k_{\max} = 3$ or $k_{\min} = \frac{"5"-4}{"2"}$ | M1 | Correct method to find either critical value following through on their $P$; allow $m=$ in place of $k=$ |
| $\frac{1}{2} < k < 3$ | A1 | Correct range in terms of $k$; allow "and" or "$\cap$" to join regions but not "or", ",", or "$\cup$"; do not accept $\frac{1}{2} < x < 3$ |
| **Total** | **(2)** | |6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e116a86f-63e0-4e80-b49c-d9f3c819ce15-12_680_677_246_696}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of the graph with equation
$$y = 3 | x - 2 | + 5$$
The vertex of the graph is at the point $P$, shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of $P$.
\item Solve the equation
$$16 - 4 x = 3 | x - 2 | + 5$$
A line $l$ has equation $y = k x + 4$ where $k$ is a constant.\\
Given that $l$ intersects $y = 3 | x - 2 | + 5$ at 2 distinct points,
\item find the range of values of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 2024 Q6 [6]}}