| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Partial fractions with repeated linear factor |
| Difficulty | Standard +0.3 This is a standard partial fractions question with a repeated linear factor followed by routine integration. The algebraic manipulation requires setting up A/(2x+1) + B/(2x+3) + C/(2x+3)², solving for constants, then integrating standard forms. While it involves multiple steps and careful algebra, it follows a well-practiced procedure with no novel insight required, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State or imply the form \(\dfrac{A}{2x+1} + \dfrac{B}{2x+3} + \dfrac{C}{(2x+3)^2}\) | B1 | |
| Use a correct method to find a constant | M1 | |
| Obtain \(A=1,\, B=-1,\, C=3\) | A1 A1 A1 | Full marks for three correct constants — do not need to see partial fractions. [Mark form \(\dfrac{A}{2x+1}+\dfrac{Dx+E}{(2x+3)^2}\) where \(A=1, D=-2, E=0\), B1M1A1A1A1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Integrate and obtain \(\tfrac{1}{2}\ln(2x+1) - \tfrac{1}{2}\ln(2x+3) - \dfrac{3}{2(2x+3)}\) | B1 B1 B1 | FT on \(A\), \(B\) and \(C\). [Correct integration of \(A,D,E\) form gives \(\tfrac{1}{2}\ln(2x+1)+\tfrac{x}{2x+3}-\tfrac{1}{2}\ln(2x+3)\) if integration by parts used] |
| Substitute limits correctly in integral with terms \(a\ln(2x+1)\), \(b\ln(2x+3)\) and \(c/(2x+3)\), \(abc\neq 0\) | M1 | Value for upper limit minus value for lower limit; 1 slip substituting can still score M1; condone omission of \(\ln(1)\) |
| Obtain the given answer following full and correct working | A1 | Need to see at least one interim step of valid log work; AG |
## Question 8(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply the form $\dfrac{A}{2x+1} + \dfrac{B}{2x+3} + \dfrac{C}{(2x+3)^2}$ | B1 | |
| Use a correct method to find a constant | M1 | |
| Obtain $A=1,\, B=-1,\, C=3$ | A1 A1 A1 | Full marks for three correct constants — do not need to see partial fractions. [Mark form $\dfrac{A}{2x+1}+\dfrac{Dx+E}{(2x+3)^2}$ where $A=1, D=-2, E=0$, B1M1A1A1A1] |
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## Question 8(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Integrate and obtain $\tfrac{1}{2}\ln(2x+1) - \tfrac{1}{2}\ln(2x+3) - \dfrac{3}{2(2x+3)}$ | B1 B1 B1 | FT on $A$, $B$ and $C$. [Correct integration of $A,D,E$ form gives $\tfrac{1}{2}\ln(2x+1)+\tfrac{x}{2x+3}-\tfrac{1}{2}\ln(2x+3)$ if integration by parts used] |
| Substitute limits correctly in integral with terms $a\ln(2x+1)$, $b\ln(2x+3)$ and $c/(2x+3)$, $abc\neq 0$ | M1 | Value for upper limit minus value for lower limit; 1 slip substituting can still score M1; condone omission of $\ln(1)$ |
| Obtain the **given answer** following full and correct working | A1 | Need to see at least one interim step of valid log work; AG |
---8 Let $\mathrm { f } ( x ) = \frac { 10 x + 9 } { ( 2 x + 1 ) ( 2 x + 3 ) ^ { 2 } }$.\\
(i) Express $\mathrm { f } ( x )$ in partial fractions.\\
(ii) Hence show that $\int _ { 0 } ^ { 1 } \mathrm { f } ( x ) \mathrm { d } x = \frac { 1 } { 2 } \ln \frac { 9 } { 5 } + \frac { 1 } { 5 }$.\\
\hfill \mbox{\textit{CAIE P3 2019 Q8 [10]}}