| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential growth/decay model setup |
| Difficulty | Moderate -0.3 Part (a) requires straightforward differentiation of an exponential function and substitution, which is standard C3 content. Part (b) involves setting up an exponential growth model from given conditions and solving for the constant, then evaluating at a future time. Both parts are routine applications of exponential models with no novel problem-solving required, making this slightly easier than average but still requiring multiple techniques. |
| Spec | 1.06b Gradient of e^(kx): derivative and exponential model1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Differentiate to produce \(ke^{-0.33t}\) | M1 | where constant \(k\) is different from 58; method must involve differentiation |
| Obtain \(-19.14e^{-0.33t}\) or \(19.14e^{-0.33t}\) | A1 | or unsimplified equiv |
| Obtain \(-5.1\) or \(5.1\) | A1 | whatever they claim value represents; accept \(5.11\) but not greater accuracy |
| Total | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Either: State or imply formula \(42e^{6k}\) or \(42a^t\) | B1 | \(42e^{-kt}\), \(42e^{-kx}\), etc. also acceptable |
| Attempt to find \(k\) from \(42e^{6k} = 51.8\) or \(a\) from \(42a^6 = 51.8\) | M1 | using sound process involving logarithms at least as far as \(6k = \ldots\) or \(a = \ldots\) |
| Obtain \(k = 0.035\) or \(a = 1.0356\) | A1 | or greater accuracy \(0.03495\ldots\) or exact equiv \(\frac{1}{6}\ln\frac{37}{30}\) |
| Substitute 24 to obtain value between \(97.1\) and \(97.3\) inclusive | A1 | allow greater accuracy than 3 s.f. |
| Or: Use ratio \(\frac{51.8}{42}\) in calculation | B1 | |
| Attempt calculation of form \(42 \times r^n\) | M1 | |
| Obtain \(42 \times \left(\frac{51.8}{42}\right)^4\) or \(51.8 \times \left(\frac{51.8}{42}\right)^3\) | A1 | |
| Obtain value between \(97.1\) and \(97.3\) inclusive | A1 | allow greater accuracy than 3 s.f. |
| Total | [4] |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Differentiate to produce $ke^{-0.33t}$ | M1 | where constant $k$ is different from 58; method must involve differentiation |
| Obtain $-19.14e^{-0.33t}$ or $19.14e^{-0.33t}$ | A1 | or unsimplified equiv |
| Obtain $-5.1$ or $5.1$ | A1 | whatever they claim value represents; accept $5.11$ but not greater accuracy |
| **Total** | **[3]** | |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Either: State or imply formula $42e^{6k}$ or $42a^t$ | B1 | $42e^{-kt}$, $42e^{-kx}$, etc. also acceptable |
| Attempt to find $k$ from $42e^{6k} = 51.8$ or $a$ from $42a^6 = 51.8$ | M1 | using sound process involving logarithms at least as far as $6k = \ldots$ or $a = \ldots$ |
| Obtain $k = 0.035$ or $a = 1.0356$ | A1 | or greater accuracy $0.03495\ldots$ or exact equiv $\frac{1}{6}\ln\frac{37}{30}$ |
| Substitute 24 to obtain value between $97.1$ and $97.3$ inclusive | A1 | allow greater accuracy than 3 s.f. |
| Or: Use ratio $\frac{51.8}{42}$ in calculation | B1 | |
| Attempt calculation of form $42 \times r^n$ | M1 | |
| Obtain $42 \times \left(\frac{51.8}{42}\right)^4$ or $51.8 \times \left(\frac{51.8}{42}\right)^3$ | A1 | |
| Obtain value between $97.1$ and $97.3$ inclusive | A1 | allow greater accuracy than 3 s.f. |
| **Total** | **[4]** | |
---5
\begin{enumerate}[label=(\alph*)]
\item The mass, $M$ grams, of a substance at time $t$ years is given by
$$M = 58 \mathrm { e } ^ { - 0.33 t }$$
Find the rate at which the mass is decreasing at the instant when $t = 4$. Give your answer correct to 2 significant figures.
\item The mass of a second substance is increasing exponentially. The initial mass is 42.0 grams and, 6 years later, the mass is 51.8 grams. Find the mass at a time 24 years after the initial value.
\end{enumerate}
\hfill \mbox{\textit{OCR C3 2014 Q5 [7]}}