| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Show constant equals specific form |
| Difficulty | Moderate -0.8 This is a straightforward exponential modelling question requiring only routine substitution and logarithm manipulation. Part (a) is trivial substitution, part (b) involves substituting given values and rearranging to the required form (standard C3 technique), and part (c) is another substitution followed by taking logarithms. No problem-solving insight or novel approaches are needed—purely mechanical application of exponential/logarithm rules. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(25e\) or equivalent decimal, accept awrt 68 | B1 | For \(25e\) or numerical answer e.g. \(67.957\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(50=25e^{1-10k}\) | M1 | Uses \(t=10\), \(m=50\) in \(m=25e^{1-kt}\) |
| Way 1: \(e^{1-10k}=2 \Rightarrow 1-10k=\ln 2\) | A1 | Correct equation \(e^{f(k)}=B\) |
| Way 2: \(e^{-10k}=2/e\), \(-10k=\ln(2/e)\) | A1 | |
| Way 3: \(e^{10k}=e/2\), \(10k=\ln(e/2)\) | A1 | No intermediate step needed |
| Taking logs correctly to give \(f(k)=\ln B\) | M1 | |
| \(k=\frac{\ln e - \ln 2}{10}\) or \(k=\frac{-\ln(2/e)}{10}\) or \(k=\frac{\ln\left(\frac{1}{2}e\right)}{10}\) | A1* | cso – needs both M marks, everything correct and exact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses \(m=20\) and numerical \(k\): \(20=25e^{1-'k't} \Rightarrow e^{1-'k't}=0.8\) | M1 | NB \(e^{(1-'k')t}=0.8\) is M0 |
| \(\Rightarrow t=\frac{1-\ln 0.8}{`k`}\) or equivalent e.g. \(t=\frac{\ln\left(\frac{5e}{4}\right)}{`k`}\) | dM1 | Correct use of logs |
| \(t=\) awrt \(40\) (years) | A1 | Allow awrt 40 (may see 39.86 or 39.9); do not allow \(-40\) |
# Question 6:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $25e$ or equivalent decimal, accept awrt 68 | B1 | For $25e$ or numerical answer e.g. $67.957$ |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $50=25e^{1-10k}$ | M1 | Uses $t=10$, $m=50$ in $m=25e^{1-kt}$ |
| Way 1: $e^{1-10k}=2 \Rightarrow 1-10k=\ln 2$ | A1 | Correct equation $e^{f(k)}=B$ |
| Way 2: $e^{-10k}=2/e$, $-10k=\ln(2/e)$ | A1 | |
| Way 3: $e^{10k}=e/2$, $10k=\ln(e/2)$ | A1 | No intermediate step needed |
| Taking logs correctly to give $f(k)=\ln B$ | M1 | |
| $k=\frac{\ln e - \ln 2}{10}$ or $k=\frac{-\ln(2/e)}{10}$ or $k=\frac{\ln\left(\frac{1}{2}e\right)}{10}$ | A1* | cso – needs both M marks, everything correct and exact |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $m=20$ and numerical $k$: $20=25e^{1-'k't} \Rightarrow e^{1-'k't}=0.8$ | M1 | NB $e^{(1-'k')t}=0.8$ is M0 |
| $\Rightarrow t=\frac{1-\ln 0.8}{`k`}$ or equivalent e.g. $t=\frac{\ln\left(\frac{5e}{4}\right)}{`k`}$ | dM1 | Correct use of logs |
| $t=$ awrt $40$ (years) | A1 | Allow awrt 40 (may see 39.86 or 39.9); do not allow $-40$ |
---6. The mass, $m$ grams, of a radioactive substance $t$ years after first being observed, is modelled by the equation
$$m = 25 \mathrm { e } ^ { 1 - k t }$$
where $k$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item State the value of $m$ when the radioactive substance was first observed.
Given that the mass is 50 grams, 10 years after first being observed,
\item show that $k = \frac { 1 } { 10 } \ln \left( \frac { 1 } { 2 } \mathrm { e } \right)$
\item Find the value of $t$ when $m = 20$, giving your answer to the nearest year.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2015 Q6 [8]}}