| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2012 |
| Session | June |
| 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 to reach a given answer. The question is slightly easier than average because it's a textbook-style exercise with no conceptual challenges, though the logarithm simplification in part (ii) requires some care. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08j Integration using partial fractions |
**Part (i)**
- Attempt to eliminate fractions by choosing suitable $x$ values or sim eqns: M1
- Obtain $2x + 11 = A(x+3) + B(2x+1)$ OR $A + 2B = 2$ and $3A + B = 11$: A1
- Obtain $A = 4$: A1
- $B = -1$: A1 **[4]**
**Part (ii)**
- Attempt integration to obtain at least one ln term, either $P\ln(2x+1)$ or $Q\ln(x+3)$: M1
- Obtain $2\ln(2x+1) - \ln(x+3)$: A1
- Use limits of 2 and 0 in correct order in any function: M1
- Attempt use of any log law once on their exact expression: M1
- Obtain $\ln 15$ NIS: A1 **[5]**
**[Total: 9]**8 (i) Given that $\frac { 2 x + 11 } { ( 2 x + 1 ) ( x + 3 ) } \equiv \frac { A } { 2 x + 1 } + \frac { B } { x + 3 }$, find the values of the constants $A$ and $B$.\\
(ii) Hence show that $\int _ { 0 } ^ { 2 } \frac { 2 x + 11 } { ( 2 x + 1 ) ( x + 3 ) } \mathrm { d } x = \ln 15$.
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2012 Q8 [9]}}