| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Time to reach target in exponential model |
| Difficulty | Moderate -0.3 This is a standard exponential decay question testing geometric sequences and logarithms. Parts (i)-(iii) involve routine calculations with percentage decrease and solving exponential inequalities using logarithms—core C2 skills with clear scaffolding. Part (iv) requires converting between exponential forms, which is slightly more demanding but still follows a standard pattern. The multi-part structure and 11 total marks indicate moderate length, but each step is procedural without requiring novel insight. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<11.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| \(65 \times 0.983^n < 3\) or \(\log_{10}(65 \times 0.983^n) < \log_{10}3\) oe; \(\log_{10}65 + \log_{10}0.983^n < \log_{10}3\) www; \([\log_{10}65 + n \log_{10}0.983 < \log_{10}3]\); \(n \log_{10}0.983 < \log_{10}3 - \log_{10}65\) and; completion to \(n > \frac{\log_{10}3 - \log_{10}65}{\log_{10}0.983}\) AG www; \(n = 180\) cao | M1*, M1dep, A1, B1 | condone omission of base 10 throughout; may be implied by eg \(\log_{10}65 + n \log_{10}0.983 < \log_{10}3\) even if \(<\) is replaced by eg \(=\) or \(>\) with no prior incorrect log moves; or \([\log_{10}0.983^n < \log_{10}3 - \log_{10}65]\); inequality signs must be correct throughout; reason for change of inequality sign not required; B0 for \(n > 180\); watch for correct inequality sign at each step |
| Answer | Marks | Guidance |
|---|---|---|
| \(63.895 = 65 \times 10^k\) soi; \(\log_{10}(\text{their } 63.895) = \log_{10}65 - k\); or \(-k = \log_{10}(\text{their } 0.983)\); \([ k ] = 17.4 \times 10^{-3}\) to \(7.45 \times 10^{-3}\); \([ d ] = 142.1...\) to \(42.123\) [°C] isw | B1, M1, A1, A1 | accept 63.895 rot to 3 or 4 sf; B1 may be awarded for substitution of \(t = 1\) after manipulation; their 63.895 must be from attempt to reduce 65 by 1.7% at least once; M1A1A1 may be awarded if other value of \(t\) with correct \(d\) is used; NB B1M1A0A1 is possible; unsupported answers for \(k\) and/or \(d\) do not score |
### Part (iii)
$65 \times 0.983^n < 3$ or $\log_{10}(65 \times 0.983^n) < \log_{10}3$ oe; $\log_{10}65 + \log_{10}0.983^n < \log_{10}3$ www; $[\log_{10}65 + n \log_{10}0.983 < \log_{10}3]$; $n \log_{10}0.983 < \log_{10}3 - \log_{10}65$ and; completion to $n > \frac{\log_{10}3 - \log_{10}65}{\log_{10}0.983}$ AG www; $n = 180$ cao | M1*, M1dep, A1, B1 | condone omission of base 10 throughout; may be implied by eg $\log_{10}65 + n \log_{10}0.983 < \log_{10}3$ even if $<$ is replaced by eg $=$ or $>$ with no prior incorrect log moves; or $[\log_{10}0.983^n < \log_{10}3 - \log_{10}65]$; inequality signs must be correct throughout; reason for change of inequality sign not required; B0 for $n > 180$; watch for correct inequality sign at each step
### Part (iv)
$63.895 = 65 \times 10^k$ soi; $\log_{10}(\text{their } 63.895) = \log_{10}65 - k$; or $-k = \log_{10}(\text{their } 0.983)$; $[ k ] = 17.4 \times 10^{-3}$ to $7.45 \times 10^{-3}$; $[ d ] = 142.1...$ to $42.123$ [°C] isw | B1, M1, A1, A1 | accept 63.895 rot to 3 or 4 sf; B1 may be awarded for substitution of $t = 1$ after manipulation; their 63.895 must be from attempt to reduce 65 by 1.7% at least once; M1A1A1 may be awarded if other value of $t$ with correct $d$ is used; NB B1M1A0A1 is possible; unsupported answers for $k$ and/or $d$ do not scoreA hot drink when first made has a temperature which is $65°C$ higher than room temperature. The temperature difference, $d °C$, between the drink and its surroundings decreases by $1.7\%$ each minute.
\begin{enumerate}[label=(\roman*)]
\item Show that 3 minutes after the drink is made, $d = 61.7$ to 3 significant figures. [2]
\item Write down an expression for the value of $d$ at time $n$ minutes after the drink is made, where $n$ is an integer. [1]
\item Show that when $d < 3$, $n$ must satisfy the inequality
$$n > \frac{\log_{10} 3 - \log_{10} 65}{\log_{10} 0.983}.$$
Hence find the least integer value of $n$ for which $d < 3$. [4]
\item The temperature difference at any time $t$ minutes after the drink is made can also be expressed as $d = 65 \times 10^{-kt}$, for some constant $k$. Use the value of $d$ for 1 minute after the drink is made to calculate the value of $k$. Hence find the temperature difference 25.3 minutes after the drink is made. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 2013 Q11 [11]}}