| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential growth/decay model setup |
| Difficulty | Moderate -0.8 This is a straightforward linear modeling question requiring only the calculation of gradient from two points and forming y = mx + c, followed by a simple check of the y-intercept against given data. It involves basic algebraic manipulation with no conceptual challenges beyond GCSE-level straight line work. |
| Spec | 1.03c Straight line models: in variety of contexts |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Attempts \(H=mt+c\) with both \((3, 2.35)\) and \((6, 3.28)\) | M1 | Accept sight of \(2.35=3m+c\) and \(3.28=6m+c\). Allow attempt at gradient \(m=\frac{3.28-2.35}{6-3}(=0.31)\) |
| Method to find both \(m\) and \(c\) | dM1 | Full method to find both constants; dependent on M1 |
| \(H=0.31t+1.42\) oe | A1 | Allow equivalents such as \(H=\frac{31}{100}t+\frac{142}{100}\). Do not allow with units attached |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Uses model and states initial height is their \(b\) | B1ft | States/implies \(1.42\) (m) or \(142\) cm is original height (when \(t=0\)). Follow through on their \(c\) if \(c>0\) |
| Compares 140 cm with their 1.42 m and makes a valid comment | B1ft | For \(H=0.31t+1.42\): values are close/approximately the same, supporting linear model. Rule of thumb: "good model" for \(135\leq c\leq 145\) cm |
# Question 4:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Attempts $H=mt+c$ with both $(3, 2.35)$ and $(6, 3.28)$ | M1 | Accept sight of $2.35=3m+c$ and $3.28=6m+c$. Allow attempt at gradient $m=\frac{3.28-2.35}{6-3}(=0.31)$ |
| Method to find both $m$ and $c$ | dM1 | Full method to find both constants; dependent on M1 |
| $H=0.31t+1.42$ oe | A1 | Allow equivalents such as $H=\frac{31}{100}t+\frac{142}{100}$. Do not allow with units attached |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Uses model and states initial height is their $b$ | B1ft | States/implies $1.42$ (m) or $142$ cm is original height (when $t=0$). Follow through on their $c$ if $c>0$ |
| Compares 140 cm with their 1.42 m and makes a valid comment | B1ft | For $H=0.31t+1.42$: values are close/approximately the same, supporting linear model. Rule of thumb: "good model" for $135\leq c\leq 145$ cm |\begin{enumerate}
\item A tree was planted in the ground.
\end{enumerate}
Its height, $H$ metres, was measured $t$ years after planting.\\
Exactly 3 years after planting, the height of the tree was 2.35 metres.\\
Exactly 6 years after planting, the height of the tree was 3.28 metres.\\
Using a linear model,\\
(a) find an equation linking $H$ with $t$.
The height of the tree was approximately 140 cm when it was planted.\\
(b) Explain whether or not this fact supports the use of the linear model in part (a).
\hfill \mbox{\textit{Edexcel AS Paper 1 2019 Q4 [5]}}