| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2016 |
| Session | Specimen |
| Marks | 9 |
| Topic | Partial Fractions |
| Type | Partial fractions with linear factors – decompose and integrate (definite) |
| Difficulty | Moderate -0.3 This is a straightforward partial fractions question with simple linear factors and standard integration. Part (i) requires routine algebraic manipulation to find constants A and B, while part (ii) involves integrating logarithmic terms and simplifying using log laws to reach a given answer. The question is slightly easier than average because it's a textbook application with no conceptual challenges, though the definite integral requires careful arithmetic with logarithms. |
| Spec | 1.02y Partial fractions: decompose rational functions1.06f Laws of logarithms: addition, subtraction, power rules1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| - Obtain \(P\ln | 2x-1 | + Q\ln |
**Question 7(i)**
- Attempt to eliminate fractions — M1
- Obtain $8x - 1 = A(x+1) + B(2x-1)$ — A1
- Obtain $A = 2$ — B1
- Obtain $B = 3$ — B1
**Question 7(ii)**
- Attempt integration to obtain at least one ln term — M1
- Obtain $P\ln|2x-1| + Q\ln|x+1|$ — A1
- Use limits in correct order — M1
- Attempt use of log laws — DM1
- Obtain $\ln 24$ AG — A1
**Total: 9 marks**7 (i) Express $\frac { 8 x - 1 } { ( 2 x - 1 ) ( x + 1 ) }$ in the form $\frac { A } { 2 x - 1 } + \frac { B } { x + 1 }$ where $A$ and $B$ are constants.\\
(ii) Hence show that $\int _ { 2 } ^ { 5 } \frac { 8 x - 1 } { ( 2 x - 1 ) ( x + 1 ) } \mathrm { d } x = \ln 24$.
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2016 Q7 [9]}}