| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2020 |
| Session | Specimen |
| Marks | 4 |
| 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 a routine definite integral. Part (a) requires standard algebraic manipulation to find constants A and B, while part (b) 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 complications (distinct linear factors, clean numbers, answer provided to verify), though it does require competent execution of multiple techniques. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Obtain \(P\ln | 2x-1 | + Q\ln |
(a) Attempt to eliminate fractions — **M1**
Obtain $8x - 1 = A(x+1) + B(2x-1)$ — **A1**
Obtain $A = 2$ — **B1**
Obtain $B = 3$ — **B1**
**Total: 4**
(b) 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 — **M1** (DM1)
Obtain $\ln 24$ **AG** — **A1**
**Total: 5**7
\begin{enumerate}[label=(\alph*)]
\item 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.
\item Hence show that $\int _ { 2 } ^ { 5 } \frac { 8 x - 1 } { ( 2 x - 1 ) ( x + 1 ) } \mathrm { d } x = \ln 24$.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2020 Q7 [4]}}