| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find stationary points and nature |
| Difficulty | Standard +0.3 This is a straightforward application of the product rule to find stationary points. Part (a) requires differentiating (x-2)²e^(3x), setting f'(x)=0, and solving a simple linear equation after factoring. Part (b) is a standard interpretation of the graph. Slightly easier than average due to the clean factorization and simple algebra involved. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(x)=2(x-2)\times e^{3x}+(x-2)^2\times 3e^{3x}\) | M1 A1 | Product rule of form \(\alpha(x-2)e^{3x}+(x-2)^2\beta e^{3x}\); correct derivative (need not be simplified) |
| \(f'(x)=0 \Rightarrow e^{3x}(x-2)\bigl(2+3(x-2)\bigr)=0 \Rightarrow x=...\) | M1 | Sets derivative to zero; cancels/factorises exponential and \((x-2)\) term; or expand to 3-term quadratic \(3x^2-10x+8\) and solve |
| \(x=\frac{4}{3}\) | A1 | Correct \(x\) value |
| \(y=\frac{4}{9}e^4\), so \(A\) is \(\left(\frac{4}{3},\frac{4}{9}e^4\right)\) | A1 | Correct simplified \(y\) value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(k>0\) or \(k\leq \frac{4}{9}e^4\) with \(...\) or \(<\) respectively | M1 | Identifies one correct boundary for \(k\); direction of inequality must be correct |
| \(0 < k \leq \frac{4}{9}e^4\) | A1ft | Correct range following their positive \(y\)-coordinate in (a); accept separate inequalities or set/interval notation; awrt 24.3 accepted for upper bound |
# Question 3:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x)=2(x-2)\times e^{3x}+(x-2)^2\times 3e^{3x}$ | M1 A1 | Product rule of form $\alpha(x-2)e^{3x}+(x-2)^2\beta e^{3x}$; correct derivative (need not be simplified) |
| $f'(x)=0 \Rightarrow e^{3x}(x-2)\bigl(2+3(x-2)\bigr)=0 \Rightarrow x=...$ | M1 | Sets derivative to zero; cancels/factorises exponential and $(x-2)$ term; or expand to 3-term quadratic $3x^2-10x+8$ and solve |
| $x=\frac{4}{3}$ | A1 | Correct $x$ value |
| $y=\frac{4}{9}e^4$, so $A$ is $\left(\frac{4}{3},\frac{4}{9}e^4\right)$ | A1 | Correct simplified $y$ value |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k>0$ or $k\leq \frac{4}{9}e^4$ with $...$ or $<$ respectively | M1 | Identifies one correct boundary for $k$; direction of inequality must be correct |
| $0 < k \leq \frac{4}{9}e^4$ | A1ft | Correct range following their positive $y$-coordinate in (a); accept separate inequalities or set/interval notation; awrt 24.3 accepted for upper bound |
---3. In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{83e12fa4-1abb-4bea-bff4-8d36757bd9c3-08_535_839_402_555}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of part of the curve with equation $y = \mathrm { f } ( x )$ where
$$\mathrm { f } ( x ) = ( x - 2 ) ^ { 2 } \mathrm { e } ^ { 3 x } \quad x \in \mathbb { R }$$
The curve has a maximum turning point at $A$ and a minimum turning point at $( 2,0 )$
\begin{enumerate}[label=(\alph*)]
\item Use calculus to find the exact coordinates of $A$.
Given that the equation $\mathrm { f } ( x ) = k$, where $k$ is a constant, has at least two distinct roots,
\item state the range of possible values for $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2022 Q3 [7]}}