| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Solve equation involving derivatives |
| Difficulty | Challenging +1.2 This is a structured multi-part question on differentiation of transcendental functions. Part (a) requires simple substitution to find where cos x = 0. Part (b) involves applying the product rule and chain rule to find f'(x), setting it to zero, and algebraic manipulation to reach the given form—this is methodical but requires careful execution. Part (c) uses the quadratic formula on a trigonometric equation. While it requires multiple techniques and careful algebra, each step follows standard procedures without requiring novel insight or particularly sophisticated reasoning. Slightly above average difficulty due to the algebraic manipulation required in part (b). |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\left(\frac{3\pi}{2}, 0\right)\) | B1 | Accept identification that \(x = \frac{3\pi}{2}\) at \(R\), e.g. \(\cos x = 0\) so \(x = \frac{3\pi}{2}\). Accept \(\frac{3\pi}{2}\) written on diagram at point \(R\). Must be exact. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f'(x) = (6\cos^2 x)e^{3\sin x} - (2\sin x)e^{3\sin x}\) | M1A1 | M1: Attempts product rule, form \(f'(x) = \pm\alpha\cos^2 x e^{3\sin x} \pm \beta\sin x e^{3\sin x}\). A1: Correct derivative in any form. |
| \(\Rightarrow \left(6(1-\sin^2 x) - 2\sin x\right)(=0)\) | dM1 | Attempts to use identity \(\pm\sin^2 x \pm \cos^2 x = \pm 1\) to produce 3-term quadratic in \(\sin x\) with \(e^{3\sin x}\) factored. Depends on previous M mark. |
| \(\Rightarrow 3\sin^2 x + \sin x - 3 = 0\) | A1 | Or any integer multiple. Must be "extracted". Allow if seen/used/implied in part (c). Note: \(f'(x) = (-6\cos^2 x)e^{3\sin x} + (2\sin x)e^{3\sin x} = 0\) scores M1A0dM1A0, but allow full recovery in (c). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\sin x = \frac{-1+\sqrt{37}}{6} \Rightarrow x = \ldots\) | M1 | Solves their 3-term quadratic in \(\sin x\) and finds value of \(x\) using \(\sin^{-1}\), where \(\lvert\sin x\rvert < 1\). |
| \(x = 2.131\) | A1 | awrt 2.131 only, no other values offered. Correct answer only scores no marks. |
## Question 6:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left(\frac{3\pi}{2}, 0\right)$ | B1 | Accept identification that $x = \frac{3\pi}{2}$ at $R$, e.g. $\cos x = 0$ so $x = \frac{3\pi}{2}$. Accept $\frac{3\pi}{2}$ written on diagram at point $R$. Must be exact. |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $f'(x) = (6\cos^2 x)e^{3\sin x} - (2\sin x)e^{3\sin x}$ | M1A1 | M1: Attempts product rule, form $f'(x) = \pm\alpha\cos^2 x e^{3\sin x} \pm \beta\sin x e^{3\sin x}$. A1: Correct derivative in any form. |
| $\Rightarrow \left(6(1-\sin^2 x) - 2\sin x\right)(=0)$ | dM1 | Attempts to use identity $\pm\sin^2 x \pm \cos^2 x = \pm 1$ to produce 3-term quadratic in $\sin x$ with $e^{3\sin x}$ factored. Depends on previous M mark. |
| $\Rightarrow 3\sin^2 x + \sin x - 3 = 0$ | A1 | Or any integer multiple. Must be "extracted". Allow if seen/used/implied in part (c). Note: $f'(x) = (-6\cos^2 x)e^{3\sin x} + (2\sin x)e^{3\sin x} = 0$ scores M1A0dM1A0, but allow full recovery in (c). |
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sin x = \frac{-1+\sqrt{37}}{6} \Rightarrow x = \ldots$ | M1 | Solves their 3-term quadratic in $\sin x$ and finds value of $x$ using $\sin^{-1}$, where $\lvert\sin x\rvert < 1$. |
| $x = 2.131$ | A1 | awrt 2.131 only, no other values offered. **Correct answer only scores no marks.** |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{76989f19-2624-4e86-a8ee-4978dd1014c2-14_741_844_258_612}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
Figure 1 shows a sketch of the curve with equation $y = \mathrm { f } ( x )$, where
$$f ( x ) = 2 e ^ { 3 \sin x } \cos x \quad 0 \leqslant x \leqslant 2 \pi$$
The curve intersects the $x$-axis at point $R$, as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item State the coordinates of $R$
The curve has two turning points, at point $P$ and point $Q$, also shown in Figure 1.
\item Show that, at points $P$ and $Q$,
$$a \sin ^ { 2 } x + b \sin x + c = 0$$
where $a$, $b$ and $c$ are integers to be found.
\item Hence find the $x$ coordinate of point $Q$, giving your answer to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2024 Q6 [7]}}