| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Solve equation involving derivatives |
| Difficulty | Standard +0.3 This is a straightforward product rule differentiation question with standard techniques. Part (a) requires differentiating e^(-0.25x)sin(x) using product rule and setting equal to zero, leading to tan(x)=4 through basic algebraic manipulation. Parts (b-d) involve sketching absolute value transformations and interpreting the model, which are routine A-level tasks. The question is slightly above average difficulty due to the multi-part nature and modulus function, but all techniques are standard. |
| Spec | 1.02z Models in context: use functions in modelling1.05o Trigonometric equations: solve in given intervals1.06b Gradient of e^(kx): derivative and exponential model1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow f'(x) = -2.5e^{-0.25x}\sin x + 10e^{-0.25x}\cos x\) | M1, A1 | 1.1b, 1.1b |
| \(f'(x) = 0 \Rightarrow -2.5e^{-0.25x}\sin x + 10e^{-0.25x}\cos x = 0\) | M1 | 2.1 |
| \(\dfrac{\sin x}{\cos x} = \dfrac{10}{2.5} \Rightarrow \tan x = 4\)* | A1* | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Correct shape for at least 2 loops (height of second loop lower than first) | M1 | 1.1b |
| Fully correct with at least 4 loops with decreasing heights and no rounding at cusps | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Solves \(\tan x = 4\) and substitutes into \(H(t)\) | M1 | 3.1a |
| \(H(4.47) = \left | 10e^{-0.25\times4.47}\sin 4.47\right | \) |
| awrt 3.18 (metres) | A1 | 3.2a |
| Answer | Marks | Guidance |
|---|---|---|
| The times between each bounce should not stay the same when the heights of each bounce is getting smaller | B1 | 3.5b |
# Question 12:
**Part (a):**
$f(x) = 10e^{-0.25x}\sin x$
$\Rightarrow f'(x) = -2.5e^{-0.25x}\sin x + 10e^{-0.25x}\cos x$ | M1, A1 | 1.1b, 1.1b
$f'(x) = 0 \Rightarrow -2.5e^{-0.25x}\sin x + 10e^{-0.25x}\cos x = 0$ | M1 | 2.1
$\dfrac{\sin x}{\cos x} = \dfrac{10}{2.5} \Rightarrow \tan x = 4$* | A1* | 1.1b
**Total: (4)**
---
**Part (b):**
Correct shape for at least 2 loops (height of second loop lower than first) | M1 | 1.1b
Fully correct with at least 4 loops with decreasing heights and no rounding at cusps | A1 | 1.1b
**Total: (2)**
---
**Part (c):**
Solves $\tan x = 4$ and substitutes into $H(t)$ | M1 | 3.1a
$H(4.47) = \left|10e^{-0.25\times4.47}\sin 4.47\right|$ | M1 | 1.1b
awrt 3.18 (metres) | A1 | 3.2a
**Total: (3)**
---
**Part (d):**
The times between each bounce should not stay the same when the heights of each bounce is getting smaller | B1 | 3.5b
**Total: (1)**
**Overall Total: (10 marks)**12.
$$\mathrm { f } ( x ) = 10 \mathrm { e } ^ { - 0.25 x } \sin x , \quad x \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Show that the $x$ coordinates of the turning points of the curve with equation $y = \mathrm { f } ( x )$ satisfy the equation $\tan x = 4$
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{91a2f26a-add2-4b58-997d-2ae229548217-34_687_1029_495_518}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of part of the curve with equation $y = \mathrm { f } ( x )$.
\item Sketch the graph of $H$ against $t$ where
$$\mathrm { H } ( t ) = \left| 10 \mathrm { e } ^ { - 0.25 t } \sin t \right| \quad t \geqslant 0$$
showing the long-term behaviour of this curve.
The function $\mathrm { H } ( t )$ is used to model the height, in metres, of a ball above the ground $t$ seconds after it has been kicked.
Using this model, find
\item the maximum height of the ball above the ground between the first and second bounce.
\item Explain why this model should not be used to predict the time of each bounce.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 2019 Q12 [10]}}