| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Applied differentiation |
| Type | Velocity-time graph modelling |
| Difficulty | Standard +0.3 This is a structured multi-part question on differentiation applications with clear scaffolding. Part (a) requires setting v=0 and solving, part (b) is a 'show that' requiring product rule differentiation and algebraic manipulation, and part (c) involves straightforward iteration. While it combines several techniques (logarithms, product rule, iteration), each step is routine and well-signposted, making it slightly easier than average for A-level. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.07q Product and quotient rules: differentiation1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| 25 | B1 | Accept 25 but condone "25 seconds". If another value given (apart from 0) it is B0 |
| Total: (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempts product rule differentiation | M1 | Look for \((10-0.4t)\times\frac{1}{(t+1)}\pm k\ln(t+1)\), where \(k\) is constant |
| \(\frac{dv}{dt} = \ln(t+1)\times-0.4+\frac{(10-0.4t)}{t+1}\) | A1 | Correct differentiation; condone missing lhs or seen as \(v'\), \(\frac{dy}{dx}\) or \(=0\) |
| Sets \(\frac{dv}{dt}=0\Rightarrow\frac{(10-0.4t)}{(t+1)}=0.4\ln(t+1)\) and makes progress towards making \(t\) the subject | dM1 | Set allowable derivative \(=0\); cross multiply/divide and rearrange so \(t\) occurs only once |
| \(t=\frac{25-\ln(t+1)}{1+\ln(t+1)}\) or \(t=\frac{26}{1+\ln(t+1)}-1\) * | A1* | Must show key steps including correct \(\frac{dv}{dt}\) and a correct preceding line such as \(t=\frac{25-\ln(t+1)}{1+\ln(t+1)}\) |
| Total: (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempts \(t_2=\frac{26}{1+\ln 8}-1\) | M1 | Attempts iteration formula at least once; implied by awrt 7.44 |
| awrt 7.298 | A1 | This value alone scores both marks as iteration is implied; ISW after sight of this value |
| Total for (i): |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| awrt 7.33 seconds | A1 | Requires units and evidence of repeated iteration; allow awrt 7.33 s |
| Total: (3) |
## Question 8:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| 25 | B1 | Accept 25 but condone "25 seconds". If another value given (apart from 0) it is B0 |
| **Total: (1)** | | |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempts product rule differentiation | M1 | Look for $(10-0.4t)\times\frac{1}{(t+1)}\pm k\ln(t+1)$, where $k$ is constant |
| $\frac{dv}{dt} = \ln(t+1)\times-0.4+\frac{(10-0.4t)}{t+1}$ | A1 | Correct differentiation; condone missing lhs or seen as $v'$, $\frac{dy}{dx}$ or $=0$ |
| Sets $\frac{dv}{dt}=0\Rightarrow\frac{(10-0.4t)}{(t+1)}=0.4\ln(t+1)$ and makes progress towards making $t$ the subject | dM1 | Set allowable derivative $=0$; cross multiply/divide and rearrange so $t$ occurs only once |
| $t=\frac{25-\ln(t+1)}{1+\ln(t+1)}$ or $t=\frac{26}{1+\ln(t+1)}-1$ * | A1* | Must show key steps including correct $\frac{dv}{dt}$ and a correct preceding line such as $t=\frac{25-\ln(t+1)}{1+\ln(t+1)}$ |
| **Total: (4)** | | |
### Part (c)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempts $t_2=\frac{26}{1+\ln 8}-1$ | M1 | Attempts iteration formula at least once; implied by awrt 7.44 |
| awrt 7.298 | A1 | This value alone scores both marks as iteration is implied; ISW after sight of this value |
| **Total for (i):** | | |
### Part (c)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| awrt 7.33 seconds | A1 | Requires units and evidence of repeated iteration; allow awrt 7.33 s |
| **Total: (3)** | | |
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{129adfbb-98fa-4e88-b636-7b4d111f3349-16_522_673_248_696}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A car stops at two sets of traffic lights.\\
Figure 2 shows a graph of the speed of the car, $v \mathrm {~ms} ^ { - 1 }$, as it travels between the two sets of traffic lights.
The car takes $T$ seconds to travel between the two sets of traffic lights.\\
The speed of the car is modelled by the equation
$$v = ( 10 - 0.4 t ) \ln ( t + 1 ) \quad 0 \leqslant t \leqslant T$$
where $t$ seconds is the time after the car leaves the first set of traffic lights.\\
According to the model,
\begin{enumerate}[label=(\alph*)]
\item find the value of $T$
\item show that the maximum speed of the car occurs when
$$t = \frac { 26 } { 1 + \ln ( t + 1 ) } - 1$$
Using the iteration formula
$$t _ { n + 1 } = \frac { 26 } { 1 + \ln \left( t _ { n } + 1 \right) } - 1$$
with $t _ { 1 } = 7$
\item \begin{enumerate}[label=(\roman*)]
\item find the value of $t _ { 3 }$ to 3 decimal places,
\item find, by repeated iteration, the time taken for the car to reach maximum speed.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 2022 Q8 [8]}}