| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2015 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential model with shifted asymptote |
| Difficulty | Standard +0.3 This is a straightforward exponential modeling question requiring: (a) reading the horizontal asymptote from the formula (V ≥ 1000), (b) differentiating and substituting t=10, and (c) solving an exponential equation. All techniques are standard C3/C4 material with no novel insight required, making it slightly easier than average. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06b Gradient of e^(kx): derivative and exponential model1.06g Equations with exponentials: solve a^x = b1.07j Differentiate exponentials: e^(kx) and a^(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1000 < V \leq 23000\) | B1, B1 | Accept \(V < 23000\) or \(V \leq 23000\) for upper; \(V > 1000\) or \(V \geq 1000\) for lower. \(V \geq 23000\) is B0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dV}{dt} = 18000 \times -0.2e^{-0.2t} + 4000 \times -0.1e^{-0.1t}\) | M1 | Score for \(\frac{dV}{dt} = Ae^{-0.2t} + Be^{-0.1t}\) where \(A \neq 18000, B \neq 4000\) |
| \(\frac{dV}{dt}\bigg | _{t=10} = 18000 \times -0.2e^{-2} + 4000 \times -0.1e^{-1} = \text{awrt}(-)634\) | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(15000 = 18000e^{-0.2t} + 4000e^{-0.1t} + 1000\) leading to \(0 = 9e^{-0.2t} + 2e^{-0.1t} - 7\) | M1A1 | Setting up 3TQ in \(e^{\pm 0.1t}\) AND correct attempt to factorise or solve by formula; \(e^{\pm 0.2t}\) term must be the \(x^2\) term |
| \(0 = (9e^{-0.1t}-7)(e^{-0.1t}+1)\) | Correct factors \((9e^{-0.1t}-7)(e^{-0.1t}+1)\) or root \(e^{-0.1t} = \frac{7}{9}\) | |
| \(9e^{-0.1t} = 7 \Rightarrow t = 10\ln\left(\frac{9}{7}\right)\) | dM1A1 | Setting \(ae^{\pm 0.1t} - b = 0\) and using correct ln work to \(t=\ldots\); accept \(t = \frac{1}{0.1}\ln\left(\frac{9}{7}\right), \frac{1}{-0.1}\ln\left(\frac{7}{9}\right), -10\ln\left(\frac{7}{9}\right)\). Extra solutions: withhold mark |
# Question 8:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1000 < V \leq 23000$ | B1, B1 | Accept $V < 23000$ or $V \leq 23000$ for upper; $V > 1000$ or $V \geq 1000$ for lower. $V \geq 23000$ is B0 |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dV}{dt} = 18000 \times -0.2e^{-0.2t} + 4000 \times -0.1e^{-0.1t}$ | M1 | Score for $\frac{dV}{dt} = Ae^{-0.2t} + Be^{-0.1t}$ where $A \neq 18000, B \neq 4000$ |
| $\frac{dV}{dt}\bigg|_{t=10} = 18000 \times -0.2e^{-2} + 4000 \times -0.1e^{-1} = \text{awrt}(-)634$ | M1A1 | Condone substitution of $t=10$ into $\frac{dV}{dt}$ of form $Ae^{-0.2t}+Be^{-0.1t}+1000$. Students who sub $t=10$ into $V$ first then differentiate score 0,0,0 |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $15000 = 18000e^{-0.2t} + 4000e^{-0.1t} + 1000$ leading to $0 = 9e^{-0.2t} + 2e^{-0.1t} - 7$ | M1A1 | Setting up 3TQ in $e^{\pm 0.1t}$ AND correct attempt to factorise or solve by formula; $e^{\pm 0.2t}$ term must be the $x^2$ term |
| $0 = (9e^{-0.1t}-7)(e^{-0.1t}+1)$ | | Correct factors $(9e^{-0.1t}-7)(e^{-0.1t}+1)$ or root $e^{-0.1t} = \frac{7}{9}$ |
| $9e^{-0.1t} = 7 \Rightarrow t = 10\ln\left(\frac{9}{7}\right)$ | dM1A1 | Setting $ae^{\pm 0.1t} - b = 0$ and using correct ln work to $t=\ldots$; accept $t = \frac{1}{0.1}\ln\left(\frac{9}{7}\right), \frac{1}{-0.1}\ln\left(\frac{7}{9}\right), -10\ln\left(\frac{7}{9}\right)$. Extra solutions: withhold mark |
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{03548211-79cb-4629-b6ca-aa9dfcc77a33-13_743_1198_219_372}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The value of Lin's car is modelled by the formula
$$V = 18000 \mathrm { e } ^ { - 0.2 t } + 4000 \mathrm { e } ^ { - 0.1 t } + 1000 , \quad t \geqslant 0$$
where the value of the car is $V$ pounds when the age of the car is $t$ years.\\
A sketch of $t$ against $V$ is shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item State the range of $V$.
According to this model,
\item find the rate at which the value of the car is decreasing when $t = 10$
Give your answer in pounds per year.
\item Calculate the exact value of $t$ when $V = 15000$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2015 Q8 [9]}}