| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2005 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Partial fractions with linear factors – decompose and integrate (indefinite) |
| Difficulty | Moderate -0.8 This is a straightforward partial fractions question with two linear factors followed by direct integration. Part (a) requires standard algebraic manipulation to find constants A and B, and part (b) is immediate application of logarithm integration rules. No problem-solving insight needed—purely routine technique that students practice extensively. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(3x - 5 = A(2x - 1) + B(x + 3)\) | M1 | |
| \(x = -3\), \(A = 2\); \(x = \frac{1}{2}\), \(B = -1\) | m1A1 | ml: sub in 2 values or set up simultaneous equations |
| (b) \(\int\left(\frac{2}{x+3} - \frac{1}{2x-1}\right)dx\) | M1 | |
| \(= 2\ln(x+3) - \frac{1}{2}\ln(2x-1)(+c)\) | m1, A1 | Use of lns for either integral ft \(A\) and \(B\) |
**(a)** $3x - 5 = A(2x - 1) + B(x + 3)$ | M1 |
$x = -3$, $A = 2$; $x = \frac{1}{2}$, $B = -1$ | m1A1 | ml: sub in 2 values or set up simultaneous equations
**(b)** $\int\left(\frac{2}{x+3} - \frac{1}{2x-1}\right)dx$ | M1 |
$= 2\ln(x+3) - \frac{1}{2}\ln(2x-1)(+c)$ | m1, A1 | Use of lns for either integral ft $A$ and $B$
**Total: 6 marks**
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\begin{enumerate}[label=(\alph*)]
\item Express $\frac { 3 x - 5 } { ( x + 3 ) ( 2 x - 1 ) }$ in the form $\frac { A } { x + 3 } + \frac { B } { 2 x - 1 }$.
\item Hence find $\int \frac { 3 x - 5 } { ( x + 3 ) ( 2 x - 1 ) } \mathrm { d } x$.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2005 Q2 [6]}}