| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find stationary points coordinates |
| Difficulty | Standard +0.8 This question requires applying the quotient rule to a function involving both trigonometric and exponential components, solving the resulting equation involving tan, and then applying transformations. While the quotient rule application is standard, the algebraic manipulation to reach tan 2x = √2 requires care, and part (b) tests understanding of function transformations on stationary points, which is conceptually more demanding than routine differentiation. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Attempt to differentiate using quotient rule | M1 | Can use product rule with \(u=4\sin 2x\), \(v=e^{1-\sqrt{2}x}\) |
| \(f'(x) = \frac{e^{\sqrt{2}x-1}\times 8\cos 2x - 4\sin 2x \times \sqrt{2}e^{\sqrt{2}x-1}}{(e^{\sqrt{2}x-1})^2}\) | A1 | For product rule: \(f'(x)=e^{1-\sqrt{2}x}\times 8\cos 2x + 4\sin 2x \times -\sqrt{2}e^{1-\sqrt{2}x}\) |
| Sets \(f'(x)=0\) and cancels/factorises \(e^{\sqrt{2}x-1}\) terms | M1 | Look for equation in \(\cos 2x\) and \(\sin 2x\) only |
| \(\frac{\sin 2x}{\cos 2x} = \frac{8}{4\sqrt{2}} \Rightarrow \tan 2x = \sqrt{2}\) | A1* | Given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Solves \(\tan 4x = \sqrt{2}\), finds 2nd solution; look for \(x = \frac{\pi + \arctan\sqrt{2}}{4}\) | M1 | Alternatively finds 2nd solution of \(\tan 2x = \sqrt{2}\) and divides by 2 |
| \(x = 1.02\) | A1 | Correct answer with no incorrect working scores both marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Solves \(\tan 2x = \sqrt{2}\), finds 1st solution; look for \(x = \frac{\arctan\sqrt{2}}{2}\) | M1 | |
| \(x = 0.478\) | A1 | Correct answer with no incorrect working scores both marks |
## Question 15:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Attempt to differentiate using quotient rule | M1 | Can use product rule with $u=4\sin 2x$, $v=e^{1-\sqrt{2}x}$ |
| $f'(x) = \frac{e^{\sqrt{2}x-1}\times 8\cos 2x - 4\sin 2x \times \sqrt{2}e^{\sqrt{2}x-1}}{(e^{\sqrt{2}x-1})^2}$ | A1 | For product rule: $f'(x)=e^{1-\sqrt{2}x}\times 8\cos 2x + 4\sin 2x \times -\sqrt{2}e^{1-\sqrt{2}x}$ |
| Sets $f'(x)=0$ and cancels/factorises $e^{\sqrt{2}x-1}$ terms | M1 | Look for equation in $\cos 2x$ and $\sin 2x$ only |
| $\frac{\sin 2x}{\cos 2x} = \frac{8}{4\sqrt{2}} \Rightarrow \tan 2x = \sqrt{2}$ | A1* | Given answer |
### Part (b)(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Solves $\tan 4x = \sqrt{2}$, finds 2nd solution; look for $x = \frac{\pi + \arctan\sqrt{2}}{4}$ | M1 | Alternatively finds 2nd solution of $\tan 2x = \sqrt{2}$ and divides by 2 |
| $x = 1.02$ | A1 | Correct answer with no incorrect working scores both marks |
### Part (b)(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Solves $\tan 2x = \sqrt{2}$, finds 1st solution; look for $x = \frac{\arctan\sqrt{2}}{2}$ | M1 | |
| $x = 0.478$ | A1 | Correct answer with no incorrect working scores both marks |15.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f7994129-07ee-4f6d-9531-08a15a38b794-30_551_1026_219_523}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
Figure 5 shows a sketch of the curve with equation $y = \mathrm { f } ( x )$, where
$$\mathrm { f } ( x ) = \frac { 4 \sin 2 x } { \mathrm { e } ^ { \sqrt { 2 } x - 1 } } , \quad 0 \leqslant x \leqslant \pi$$
The curve has a maximum turning point at $P$ and a minimum turning point at $Q$ as shown in Figure 5.
\begin{enumerate}[label=(\alph*)]
\item Show that the $x$ coordinates of point $P$ and point $Q$ are solutions of the equation
$$\tan 2 x = \sqrt { 2 }$$
\item Using your answer to part (a), find the $x$-coordinate of the minimum turning point on the curve with equation
\begin{enumerate}[label=(\roman*)]
\item $y = \mathrm { f } ( 2 x )$.
\item $y = 3 - 2 \mathrm { f } ( x )$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 Q15 [8]}}