| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Show derivative equals given algebraic form |
| Difficulty | Moderate -0.3 This is a straightforward product rule application with exponential function, followed by routine stationary point finding. Part (a) is guided ('show that'), requiring standard differentiation with some algebraic manipulation. Parts (b) and (c) involve solving simple equations and function transformations. The question is slightly easier than average due to its structured, step-by-step nature and use of standard techniques without requiring novel insight. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f'(x) = 6(2x+1)^2 e^{-4x} - 4(2x+1)^3 e^{-4x}\) | M1 A1 | Attempts product rule to achieve \(P(2x+1)^2 e^{-4x} \pm Q(2x+1)^3 e^{-4x}\) |
| \(= 2(2x+1)^2 e^{-4x}\{3-2(2x+1)\}\) | dM1 | Correctly takes out common factor of \((2x+1)^2 e^{-4x}\); allow minor slips in factor |
| \(= 2(2x+1)^2(1-4x)e^{-4x}\) | A1 | Achieves \(2(2x+1)^2(1-4x)e^{-4x}\) with no incorrect algebra |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Sets \(f'(x)=0 \Rightarrow x = -\frac{1}{2}, \frac{1}{4}\) | B1 | Both values required |
| Either \(f\!\left(-\frac{1}{2}\right)=\ldots\) or \(f\!\left(\frac{1}{4}\right)=\ldots\) | M1 | Attempts to substitute one of \(x=\pm\frac{1}{2}, \pm\frac{1}{4}\) into \(f(x)\) |
| Both \(\left(-\frac{1}{2}, 0\right)\) and \(\left(\frac{1}{4}, \frac{27}{8e}\right)\) | A1 | Must be exact; both coordinates correctly paired |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left(\frac{9}{4}, \frac{27}{e}\right)\) | B1ft B1ft | First B1ft: one correct aspect (e.g. \(x\)-coord \(+2\), or \(y\)-coord \(\times 8\)); Second B1ft: full correct point only; B0 if another point given; accept awrt 9.93 for \(y\)-coord |
## Question 8:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f'(x) = 6(2x+1)^2 e^{-4x} - 4(2x+1)^3 e^{-4x}$ | M1 A1 | Attempts product rule to achieve $P(2x+1)^2 e^{-4x} \pm Q(2x+1)^3 e^{-4x}$ |
| $= 2(2x+1)^2 e^{-4x}\{3-2(2x+1)\}$ | dM1 | Correctly takes out common factor of $(2x+1)^2 e^{-4x}$; allow minor slips in factor |
| $= 2(2x+1)^2(1-4x)e^{-4x}$ | A1 | Achieves $2(2x+1)^2(1-4x)e^{-4x}$ with no incorrect algebra |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sets $f'(x)=0 \Rightarrow x = -\frac{1}{2}, \frac{1}{4}$ | B1 | Both values required |
| Either $f\!\left(-\frac{1}{2}\right)=\ldots$ or $f\!\left(\frac{1}{4}\right)=\ldots$ | M1 | Attempts to substitute one of $x=\pm\frac{1}{2}, \pm\frac{1}{4}$ into $f(x)$ |
| Both $\left(-\frac{1}{2}, 0\right)$ and $\left(\frac{1}{4}, \frac{27}{8e}\right)$ | A1 | Must be exact; both coordinates correctly paired |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(\frac{9}{4}, \frac{27}{e}\right)$ | B1ft B1ft | First B1ft: one correct aspect (e.g. $x$-coord $+2$, or $y$-coord $\times 8$); Second B1ft: full correct point only; B0 if another point given; accept awrt 9.93 for $y$-coord |
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bef290fb-fbac-4c9c-981e-5e323ac7182e-22_687_698_255_685}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of the curve $C$ with equation $y = \mathrm { f } ( x )$, where
$$f ( x ) = ( 2 x + 1 ) ^ { 3 } e ^ { - 4 x }$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\mathrm { f } ^ { \prime } ( x ) = A ( 2 x + 1 ) ^ { 2 } ( 1 - 4 x ) \mathrm { e } ^ { - 4 x }$$
where $A$ is a constant to be found.
\item Hence find the exact coordinates of the two stationary points on $C$.
The function g is defined by
$$g ( x ) = 8 f ( x - 2 )$$
\item Find the coordinates of the maximum stationary point on the curve with equation $y = g ( x )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2023 Q8 [9]}}