| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - partial fractions |
| Difficulty | Standard +0.3 This is a separable differential equation requiring standard integration techniques (partial fractions for the right side), followed by applying initial conditions and straightforward algebraic manipulation. Part (b) involves routine substitution and limit evaluation. While it requires multiple steps and careful algebra, all techniques are standard A-level methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)1.08l Interpret differential equation solutions: in context |
| Answer | Marks | Guidance |
|---|---|---|
| 14 | (a) | dm dt |
| Answer | Marks |
|---|---|
| (cid:11)1 (cid:14) 2t(cid:12) | M1 |
| Answer | Marks |
|---|---|
| [8] | 1.1a |
| Answer | Marks |
|---|---|
| 2.1 | separating variables |
| Answer | Marks | Guidance |
|---|---|---|
| 14 | (b) | (i) |
| Answer | Marks |
|---|---|
| (cid:159) t = 1.25 ÷ 0.5 = 2.5 minutes | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1a |
| Answer | Marks | Guidance |
|---|---|---|
| 14 | (b) | (ii) |
| Answer | Marks |
|---|---|
| (cid:111) 1.5 [grams] | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1b |
| 2.2a | or substituting a large value for t |
Question 14:
14 | (a) | dm dt
(cid:179) (cid:32)(cid:179)
m t(1(cid:14)2t)
1 A B
(cid:123) (cid:14)
t(1(cid:14)2t) t 1(cid:14)2t
(cid:159) 1 (cid:123) A(1 + 2t) + Bt
t = 0 (cid:159) A = 1
t = −½ (cid:159) 1 = − ½ B (cid:159) B = −2
dm (cid:167)1 2 (cid:183)
(cid:159) (cid:179) (cid:32)(cid:179) (cid:168) (cid:16) (cid:184) dt
m (cid:169)t 1(cid:14)2t(cid:185)
(cid:159) ln m = ln t – ln(1 + 2t) + c
t = 1, m = 1 (cid:159) c = ln 3
(cid:167) 3t (cid:183)
(cid:159)lnm(cid:32) ln
(cid:168) (cid:184)
(cid:169)1(cid:14)2t(cid:185)
e
3t
(cid:159)m(cid:32)
(cid:11)1 (cid:14) 2t(cid:12) | M1
M1
M1
A1A1
B1FT
M1
cE1
[8] | 1.1a
3.1b
1.1
1.1
1.1
2.1
m
i1.1
2.1 | separating variables
using partial fractions
substituting values, equating coeffs
n
or cover up
A = 1, B = −2
e
FT their A, B, condone no c
evaluating constant of integration
AG
14 | (b) | (i) | p
3t
1.25 (cid:32) S
(cid:11)1(cid:14)2t(cid:12)
(cid:159) 1.25 + 2.5t = 3t
(cid:159) t = 1.25 ÷ 0.5 = 2.5 minutes | M1
A1
[2] | 1.1a
1.1
14 | (b) | (ii) | 3
m(cid:32)
(cid:167)1 (cid:183)
(cid:168) (cid:14) 2(cid:184)
(cid:169)t (cid:185)
(cid:111) 1.5 [grams] | M1
A1
[2] | 3.1b
2.2a | or substituting a large value for tIn a chemical reaction, the mass $m$ grams of a chemical at time $t$ minutes is modelled by the differential equation
$$\frac{dm}{dt} = \frac{m}{t(1 + 2t)}.$$
At time 1 minute, the mass of the chemical is 1 gram.
\begin{enumerate}[label=(\alph*)]
\item Solve the differential equation to show that $m = \frac{3t}{(1 + 2t)}$. [8]
\item Hence
\begin{enumerate}[label=(\roman*)]
\item find the time when the mass is 1.25 grams, [2]
\item show what happens to the mass of the chemical as $t$ becomes large. [2]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 2 Q14 [12]}}