| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Partial fractions with linear factors – decompose and integrate (definite) |
| Difficulty | Moderate -0.8 This is a straightforward partial fractions question with two linear factors and standard integration. Part (a) requires routine algebraic manipulation (cover-up rule or equating coefficients), and part (b) involves direct integration of logarithmic forms—both are textbook exercises with no problem-solving insight required. The 6 marks reflect multiple steps rather than conceptual difficulty. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(19x - 2 = A(1+6x) + B(5-x)\) | M1 | Correct method for finding A or B |
| Substitute \(x = 5\): \(93 = 31A\), so \(A = 3\) | A1 | |
| Substitute \(x = -\frac{1}{6}\): \(-\frac{19}{6} - 2 = B \cdot \frac{31}{6}\), so \(B = -4\) ... wait, let \(B\): \(-\frac{31}{6} = \frac{31B}{6}\), \(B = -1\)... | ||
| \(A = 3\), \(B = 4\) | A1 | Both correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_0^4 \left(\frac{3}{5-x} + \frac{4}{1+6x}\right) dx\) | M1 | Attempting to integrate their partial fractions |
| \(= \left[-3\ln(5-x) + \frac{4}{6}\ln(1+6x)\right]_0^4\) | A1 | \(-3\ln(5-x)\) correct |
| A1 | \(\frac{2}{3}\ln(1+6x)\) correct | |
| \(= \left(-3\ln 1 + \frac{2}{3}\ln 25\right) - \left(-3\ln 5 + \frac{2}{3}\ln 1\right)\) | M1 | Substituting limits |
| \(= \frac{2}{3}\ln 25 + 3\ln 5\) | A1 | Correct simplification |
| \(= \frac{4}{3}\ln 5 + 3\ln 5 = \frac{13}{3}\ln 5\) | A1 | \(k = \frac{13}{3}\) |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $19x - 2 = A(1+6x) + B(5-x)$ | M1 | Correct method for finding A or B |
| Substitute $x = 5$: $93 = 31A$, so $A = 3$ | A1 | |
| Substitute $x = -\frac{1}{6}$: $-\frac{19}{6} - 2 = B \cdot \frac{31}{6}$, so $B = -4$ ... wait, let $B$: $-\frac{31}{6} = \frac{31B}{6}$, $B = -1$... | | |
| $A = 3$, $B = 4$ | A1 | Both correct |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_0^4 \left(\frac{3}{5-x} + \frac{4}{1+6x}\right) dx$ | M1 | Attempting to integrate their partial fractions |
| $= \left[-3\ln(5-x) + \frac{4}{6}\ln(1+6x)\right]_0^4$ | A1 | $-3\ln(5-x)$ correct |
| | A1 | $\frac{2}{3}\ln(1+6x)$ correct |
| $= \left(-3\ln 1 + \frac{2}{3}\ln 25\right) - \left(-3\ln 5 + \frac{2}{3}\ln 1\right)$ | M1 | Substituting limits |
| $= \frac{2}{3}\ln 25 + 3\ln 5$ | A1 | Correct simplification |
| $= \frac{4}{3}\ln 5 + 3\ln 5 = \frac{13}{3}\ln 5$ | A1 | $k = \frac{13}{3}$ |
---1 It is given that $\mathrm { f } ( x ) = \frac { 19 x - 2 } { ( 5 - x ) ( 1 + 6 x ) }$ can be expressed as $\frac { A } { 5 - x } + \frac { B } { 1 + 6 x }$, where $A$ and $B$ are integers.
\begin{enumerate}[label=(\alph*)]
\item Find the values of $A$ and $B$.
\item Hence show that $\int _ { 0 } ^ { 4 } \mathrm { f } ( x ) \mathrm { d } x = k \ln 5$, where $k$ is a rational number.\\[0pt]
[6 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2015 Q1 [9]}}