| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2013 |
| Session | June |
| Marks | 14 |
| Topic | Exponential Equations & Modelling |
| Type | log(y) vs x: convert and interpret |
| Difficulty | Moderate -0.3 This is a standard A-level exponential modelling question requiring log transformation to linearize data, plotting, and interpretation. Part (i) is routine algebra, parts (ii)-(iv) are textbook exercises in reading from a graph and substituting into formulas, and part (v) requires basic contextual commentary. While multi-part with 5 sections, each component is straightforward with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form1.06i Exponential growth/decay: in modelling context |
| Year | 2001 | 2002 | 2003 | 2004 | 2005 | 2006 | 2007 | 2008 |
| Number of breeding pairs | 8 | 10 | 16 | 24 | 33 | 40 | 47 | 69 |
| Answer | Marks | Guidance |
|---|---|---|
| \(t\) | 1 | 2 |
| \(\log N\) | 0.9 | 1 |
(i) Obtain $\log N = \log a + t \log b$ o.e. w.w.w. — M1; Compare with $y = mx + c$ — A1 **[2]**
(ii) Table of values:
| $t$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| $\log N$ | 0.9 | 1 | 1.2 | 1.38 | 1.52 | 1.6 | 1.67 | 1.84 |
M1, A1; Plot points (condone 1 error) — B1; Line of best fit — B1; Obtain $a$ between 5.5 and 6.5 — B1; $b$ between 1.32 and 1.42 — B1 **[6]**
SC M1A1 for $a$ and $b$ from data in the table only if no graph drawn.
(iii) Follow through their $a$ and $b$ given answers in these ranges:
- 2008: 50–95 — B1 ft
- 2020: 1400–5500 — B1 ft **[2]**
(iv) Use logs (or their expression from part (i)), or evaluate enough terms to get $N > 500$ — M1; Solve for $t$ and interpret as a year — M1; 2013–2017 — A1 ft **[3]**
(v) Any reasonable observation about the model, e.g.: It predicts unrestricted growth which is unrealistic; It predicts that the growth rate is not constant, but increases with population size, which is realistic; Extrapolation is not valid when breeding conditions may change, so not suitable. — B1 **[1]**
**Total: [14]**6 The table below gives the population of breeding pairs of red kites in Yorkshire from 2001 to 2008.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | }
\hline
Year & 2001 & 2002 & 2003 & 2004 & 2005 & 2006 & 2007 & 2008 \\
\hline
Number of breeding pairs & 8 & 10 & 16 & 24 & 33 & 40 & 47 & 69 \\
\hline
\end{tabular}
\end{center}
Source: \href{http://www.gigrin.co.uk}{www.gigrin.co.uk}\\
The following model for the population has been proposed:
$$N = a \times b ^ { t } ,$$
where $N$ is the number of breeding pairs $t$ years after the year 2000, and $a$ and $b$ are constants.\\
(i) Show that the model can be transformed to a linear relationship between $\log _ { 10 } N$ and $t$.\\
(ii) On graph paper, plot $\log _ { 10 } N$ against $t$ and draw by eye a line of best fit. Use your line to estimate the values of $a$ and $b$ in the equation for $N$ in terms of $t$.\\
(iii) What values of $N$ does the model give for the years 2008 and 2020?\\
(iv) In which year will the number of breeding pairs first exceed 500 according to the model?\\
(v) Comment on the suitability of the model to predict the population of breeding pairs of red kites in Yorkshire.
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2013 Q6 [14]}}