| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Transformations of modulus graphs from given f(x) sketch |
| Difficulty | Moderate -0.3 This is a standard C3 question testing routine transformations of graphs (modulus and combined stretch/scale) and basic logarithm/differentiation skills. Part (a) requires applying well-practiced transformation rules, while parts (b)-(d) involve straightforward substitution, solving ln equations, and finding a tangent—all standard textbook exercises with no novel problem-solving required. Slightly easier than average due to the mechanical nature of all parts. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.06d Natural logarithm: ln(x) function and properties1.07l Derivative of ln(x): and related functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Shape correct | B1 | |
| Intercepts labelled \(p\) on \(x\)-axis and \(q\) on \(y\)-axis | B1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Shape correct | B1 | |
| \((2p, 0)\) on \(x\)-axis | B1 | |
| \((0, 3q)\) on \(y\)-axis | B1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(q = 3\ln 3\) | B1 | (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(\ln(2p+3) = 0 \Rightarrow 2p+3 = 1\); \(p = -1\) | M1 A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(\frac{dy}{dx} = \frac{6}{2x+3}\); evaluated at \(x = p\) | M1 A1 | |
| Equation: \(y = 6(x+1)\) any form | M1 A1ft | (4) |
## Question 5:
### Part (a)(i):
| Answer/Working | Marks | Notes |
|---|---|---|
| Shape correct | B1 | |
| Intercepts labelled $p$ on $x$-axis and $q$ on $y$-axis | B1 | **(2)** |
### Part (a)(ii):
| Answer/Working | Marks | Notes |
|---|---|---|
| Shape correct | B1 | |
| $(2p, 0)$ on $x$-axis | B1 | |
| $(0, 3q)$ on $y$-axis | B1 | **(3)** |
### Part (b):
| Answer/Working | Marks | Notes |
|---|---|---|
| $q = 3\ln 3$ | B1 | **(1)** |
### Part (c):
| Answer/Working | Marks | Notes |
|---|---|---|
| $\ln(2p+3) = 0 \Rightarrow 2p+3 = 1$; $p = -1$ | M1 A1 | **(2)** |
### Part (d):
| Answer/Working | Marks | Notes |
|---|---|---|
| $\frac{dy}{dx} = \frac{6}{2x+3}$; evaluated at $x = p$ | M1 A1 | |
| Equation: $y = 6(x+1)$ any form | M1 A1ft | **(4)** |
---5.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{937edb48-ef4c-4974-a571-60b0fded841b-3_394_680_982_680}
\end{center}
\end{figure}
Figure 1 shows part of the curve with equation $y = \mathrm { f } ( x ) , x \in \mathbb { R }$. The curve meets the $x$-axis at $P ( p , 0 )$ and meets the $y$-axis at $Q ( 0 , q )$.
\begin{enumerate}[label=(\alph*)]
\item On separate diagrams, sketch the curve with equation
\begin{enumerate}[label=(\roman*)]
\item $y = | \mathrm { f } ( x ) |$,
\item $y = 3 \mathrm { f } \left( \frac { 1 } { 2 } x \right)$.
In each case show, in terms of $p$ or $q$, the coordinates of points at which the curve meets the axes.
Given that $\mathrm { f } ( x ) = 3 \ln ( 2 x + 3 )$,
\end{enumerate}\item state the exact value of $q$,
\item find the value of $p$,
\item find an equation for the tangent to the curve at $P$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q5 [12]}}