| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Linear transformation to find constants |
| Difficulty | Moderate -0.3 This is a structured multi-part question on exponential models and logarithms that guides students through standard techniques: writing a linear equation from gradient/intercept, converting between exponential and logarithmic forms, and interpreting parameters. While it requires understanding of logs and exponentials across multiple steps (13 marks total), each individual part follows routine AS-level procedures with clear scaffolding. The most challenging aspect is part (b) requiring log manipulation, but this is standard bookwork. Slightly easier than average due to the step-by-step guidance. |
| Spec | 1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| 14(a) | log P = mt + c | |
| 10 | M1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| 200 | A1 | 1.1b |
| Answer | Marks |
|---|---|
| (b) | Way 1: |
| Answer | Marks |
|---|---|
| 10 10 10 | Way 2: |
| Answer | Marks | Guidance |
|---|---|---|
| P(cid:32)10(cid:169)200 (cid:185) (cid:32)10510(cid:169)200(cid:185) | M1 | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 200 | (cid:167) 1 (cid:183) | |
| a (cid:32)105or b(cid:32)10 (cid:168) (cid:169)200 (cid:184) (cid:185) | M1 | 1.1b |
| So a = 100 000 or b = 1.0116 | A1 | 1.1b |
| Both a = 100 000 and b = 1.0116 (awrt 1.01) | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| (c)(i) | The initial population | B1 |
| (c)(ii) | The proportional increase of population each year | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| (d)(i) | 300000 to nearest hundred thousand | B1 |
| (d)(ii) | Uses 200000(cid:32)abt with their values of a and b or |
| Answer | Marks | Guidance |
|---|---|---|
| 200 | M1 | 3.4 |
| 60.2 years to 3sf | A1ft | 1.1b |
| Answer | Marks |
|---|---|
| (e) | Any two valid reasons- e.g. |
| Answer | Marks | Guidance |
|---|---|---|
| (cid:120) The model predicts unlimited growth | B2 | 3.5b |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Scheme | Marks |
Question 14:
--- 14(a) ---
14(a) | log P = mt + c
10 | M1 | 1.1b
1
log P(cid:32) t(cid:14)5
10
200 | A1 | 1.1b
(2)
(b) | Way 1:
As P(cid:32)abt then
log P(cid:32)tlog b(cid:14)log a
10 10 10 | Way 2:
t
As log P (cid:32) (cid:14)5 then
10
200
(cid:167) t (cid:183) (cid:167) t (cid:183)
(cid:168) (cid:14)5(cid:184) (cid:168) (cid:184)
P(cid:32)10(cid:169)200 (cid:185) (cid:32)10510(cid:169)200(cid:185) | M1 | 2.1
1
log b(cid:32) or log a(cid:32)5
10 10
200 | (cid:167) 1 (cid:183)
a (cid:32)105or b(cid:32)10 (cid:168) (cid:169)200 (cid:184) (cid:185) | M1 | 1.1b
So a = 100 000 or b = 1.0116 | A1 | 1.1b
Both a = 100 000 and b = 1.0116 (awrt 1.01) | A1 | 1.1b
(4)
(c)(i) | The initial population | B1 | 3.4
(c)(ii) | The proportional increase of population each year | B1 | 3.4
(2)
(d)(i) | 300000 to nearest hundred thousand | B1 | 3.4
(d)(ii) | Uses 200000(cid:32)abt with their values of a and b or
1
log 200000(cid:32) t(cid:14)5 and rearranges to give t =
10
200 | M1 | 3.4
60.2 years to 3sf | A1ft | 1.1b
(3)
(e) | Any two valid reasons- e.g.
(cid:120) 100 years is a long time and population may be affected by
wars and disease
(cid:120) Inaccuracies in measuring gradient may result in widely
different estimates
(cid:120) Population growth may not be proportional to population
size
(cid:120) The model predicts unlimited growth | B2 | 3.5b
(2)
Question 14 continued
Notes:
(a)
M1: Uses a linear equation to relate log P and t
A1: Correct use of gradient and intercept to give a correct line equation
(b)
M1: Way 1: Uses logs correctly to give log equation; Way 2: Uses powers correctly to “undo”
log equation and expresses as product of two powers
M1: Way 1: Identifies log b or log a or both; Way 2: Identifies a or b as powers of 10
A1: Correct value for a or b
A1: Correct values for both
(c)(i)
B1: Accept equivalent answers e.g. The population at t = 0
(c)(ii)
B1: So accept rate at which the population is increasing each year or scale factor 1.01 or
increase of 1% per year
(d)(i)
B1: cao
(d)(ii)
M1: As in the scheme
A1ft: On their values of a and b with correct log work
(e)
B2: As given in the scheme – any two valid reasons
Question | Scheme | Marks | AOs\includegraphics{figure_2}
A town's population, $P$, is modelled by the equation $P = ab^t$, where $a$ and $b$ are constants and $t$ is the number of years since the population was first recorded.
The line $l$ shown in Figure 2 illustrates the linear relationship between $t$ and $\log_{10} P$ for the population over a period of 100 years.
The line $l$ meets the vertical axis at $(0, 5)$ as shown. The gradient of $l$ is $\frac{1}{200}$.
\begin{enumerate}[label=(\alph*)]
\item Write down an equation for $l$.
[2]
\item Find the value of $a$ and the value of $b$.
[4]
\item With reference to the model interpret
\begin{enumerate}[label=(\roman*)]
\item the value of the constant $a$,
\item the value of the constant $b$.
\end{enumerate}
[2]
\item Find
\begin{enumerate}[label=(\roman*)]
\item the population predicted by the model when $t = 100$, giving your answer to the nearest hundred thousand,
\item the number of years it takes the population to reach $200\,000$, according to the model.
\end{enumerate}
[3]
\item State two reasons why this may not be a realistic population model.
[2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 Q14 [13]}}