| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2010 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Iterative or numerical methods after integration |
| Difficulty | Standard +0.3 This is a standard differential equations question requiring partial fractions (routine), separation of variables, and integration—all core C4 techniques. The algebraic manipulation is straightforward, and part (ii)(b) involves substituting t=2 into the derived formula. While multi-step, it follows a well-practiced procedure without requiring novel insight or particularly challenging algebra. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks |
|---|---|
| \(\frac{\frac{1}{3}}{3-x}\) \(......\) \(-\frac{\frac{1}{3}}{6-x}\) | B1+1 |
| Answer | Marks | Guidance |
|---|---|---|
| Separate variables \(\int \frac{1}{(3-x)(6-x)}\,dx = \int k\,dt\) | M1 | or invert both sides |
| Change \(\frac{1}{(3-x)(6-x)}\) into partial fractions from (i) | \(\sqrt{}\)B1 | |
| \(\int\frac{A}{3-x}\,dx = \left(-A \text{ or } -\frac{1}{A}\right)\ln(3-x)\) | B1 | or \(\int\frac{B}{6-x}\,dx = \left(-B \text{ or } -\frac{1}{B}\right)\ln(6-x)\) |
| \(-\frac{1}{3}\ln(3-x) + \frac{1}{3}\ln(6-x) = kt\,(+c)\) | \(\sqrt{}\)A1 | f.t. from wrong multiples in (i) |
| Subst \((x=0, t=0)\) & \((x=1, t=1)\) into eqn with \(c\) | M1 | and solve for \(k\) |
| Use \(\ln a + \ln b = \ln ab\) or \(\ln a - \ln b = \ln\frac{a}{b}\) | M1 | |
| Obtain \(k = \frac{1}{3}\ln\frac{5}{4}\) with sufficient working & WWW | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Substitute \(k = \frac{1}{3}\ln\frac{5}{4}\), \(t=2\) & their value of \(c\) | *M1 | |
| Reduce to an eqn of form \(\frac{6-x}{3-x} = \lambda\) | dep*M1 | where \(\lambda\) is a const |
| Obtain \(x = \frac{27}{17}\) or \(1.6\) or better \((1.5882353...)\) | A2 | S.R. A1\(\sqrt{}\) for \(x = \frac{3\lambda-6}{\lambda-1}\) |
# Question 10:
## Part (i):
| $\frac{\frac{1}{3}}{3-x}$ $......$ $-\frac{\frac{1}{3}}{6-x}$ | B1+1 | |
|---|---|---|
## Part (ii)(a):
| Separate variables $\int \frac{1}{(3-x)(6-x)}\,dx = \int k\,dt$ | M1 | or invert both sides |
|---|---|---|
| Change $\frac{1}{(3-x)(6-x)}$ into partial fractions from (i) | $\sqrt{}$B1 | |
| $\int\frac{A}{3-x}\,dx = \left(-A \text{ or } -\frac{1}{A}\right)\ln(3-x)$ | B1 | or $\int\frac{B}{6-x}\,dx = \left(-B \text{ or } -\frac{1}{B}\right)\ln(6-x)$ |
| $-\frac{1}{3}\ln(3-x) + \frac{1}{3}\ln(6-x) = kt\,(+c)$ | $\sqrt{}$A1 | f.t. from wrong multiples in (i) |
| Subst $(x=0, t=0)$ & $(x=1, t=1)$ into eqn with $c$ | M1 | and solve for $k$ |
| Use $\ln a + \ln b = \ln ab$ or $\ln a - \ln b = \ln\frac{a}{b}$ | M1 | |
| Obtain $k = \frac{1}{3}\ln\frac{5}{4}$ with sufficient working & WWW | A1 | AG |
## Part (ii)(b):
| Substitute $k = \frac{1}{3}\ln\frac{5}{4}$, $t=2$ & their value of $c$ | *M1 | |
|---|---|---|
| Reduce to an eqn of form $\frac{6-x}{3-x} = \lambda$ | dep*M1 | where $\lambda$ is a const |
| Obtain $x = \frac{27}{17}$ or $1.6$ or better $(1.5882353...)$ | A2 | S.R. A1$\sqrt{}$ for $x = \frac{3\lambda-6}{\lambda-1}$ |10 (i) Express $\frac { 1 } { ( 3 - x ) ( 6 - x ) }$ in partial fractions.\\
(ii) In a chemical reaction, the amount $x$ grams of a substance at time $t$ seconds is related to the rate at which $x$ is changing by the equation
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = k ( 3 - x ) ( 6 - x )$$
where $k$ is a constant. When $t = 0 , x = 0$ and when $t = 1 , x = 1$.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac { 1 } { 3 } \ln \frac { 5 } { 4 }$.
\item Find the value of $x$ when $t = 2$.
\end{enumerate}
\hfill \mbox{\textit{OCR C4 2010 Q10 [13]}}