| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| 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, requiring decomposition into A/(1+2x) + B/(2-x) + C/(2-x)², followed by routine integration. While it involves multiple steps and careful algebra, it's a textbook exercise testing well-practiced techniques without requiring novel insight or problem-solving beyond the standard method. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State or imply the form \(\dfrac{A}{1+2x} + \dfrac{B}{2-x} + \dfrac{C}{(2-x)^2}\) | B1 | Alternative form: \(\dfrac{A}{1+2x} + \dfrac{Dx+E}{(2-x)^2}\) |
| Use a correct method for finding a coefficient | M1 | e.g. \(A(2-x)^2 + B(1+2x)(2-x) + C(1+2x) = 2x^2+17x-17\) and compare coefficients or substitute for \(x\). \(A(2-x)^3 + B(1+2x)(2-x)^2 + C(1+2x)(2-x) = 2x^2+17x-17\) scores M0. |
| Obtain one of \(A=-4,\, B=-3\) and \(C=5\) | A1 | |
| Obtain a second value | A1 | |
| Obtain the third value | A1 | Extra term in partial fractions, then B0 unless recover at end. Allow marks for any constants found correctly. Missing terms in partial fractions: B0 but M1A1 available for correct method obtaining at least one correct constant (e.g. cover-up rule). Max 2/5. Ignore any substitution back into original expression. If alternative form used: \(A=-4,\, D=3\) and \(E=-1\). |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Integrate and obtain terms \(-2\ln(1+2x) + 3\ln(2-x) + \frac{5}{2-x}\) | B1FT | OE; FT is on correct use of their \(A\), \(B\), \(C\); or on \(A\), \(D\) and \(E\) |
| (second B1FT term) | B1FT | If using \(A\), \(D\), \(E\) form then B1 for the \(A\) term, but no further marks until partial fractions are used to split the second term or they use integration by parts to obtain \(\frac{Dx+E}{2-x} - \int\frac{D}{2-x}dx\) for 2nd B1 and 3rd B1 for correct completion |
| (third B1FT term) | B1FT | B0FT, B0FT, B0FT if they place their \(A\), \(B\), \(C\) with incorrect denominators |
| Substitute limits correctly in integral of form \(a\ln(1+2x) + b\ln(2-x) + \frac{c}{2-x}\), where \(abc \neq 0\) | M1 | Condone minor slips in substitution. Exact substitution required |
| Obtain answer \(\frac{5}{2} - \ln 72\) after full and correct working | A1 | AG – evidence of some correct work to combine or simplify logs required e.g. allow from \(-\ln 9 + \ln\frac{1}{8}\) or \(-\ln 2^3 - \ln 3^2\) |
## Question 9(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply the form $\dfrac{A}{1+2x} + \dfrac{B}{2-x} + \dfrac{C}{(2-x)^2}$ | B1 | Alternative form: $\dfrac{A}{1+2x} + \dfrac{Dx+E}{(2-x)^2}$ |
| Use a correct method for finding a coefficient | M1 | e.g. $A(2-x)^2 + B(1+2x)(2-x) + C(1+2x) = 2x^2+17x-17$ and compare coefficients or substitute for $x$. $A(2-x)^3 + B(1+2x)(2-x)^2 + C(1+2x)(2-x) = 2x^2+17x-17$ scores M0. |
| Obtain one of $A=-4,\, B=-3$ and $C=5$ | A1 | |
| Obtain a second value | A1 | |
| Obtain the third value | A1 | Extra term in partial fractions, then B0 unless recover at end. Allow marks for any constants found correctly. Missing terms in partial fractions: B0 but M1A1 available for correct method obtaining at least one correct constant (e.g. cover-up rule). Max 2/5. Ignore any substitution back into original expression. If alternative form used: $A=-4,\, D=3$ and $E=-1$. |
| | 5 | |
## Question 9(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Integrate and obtain terms $-2\ln(1+2x) + 3\ln(2-x) + \frac{5}{2-x}$ | B1FT | OE; FT is on correct use of their $A$, $B$, $C$; or on $A$, $D$ and $E$ |
| (second B1FT term) | B1FT | If using $A$, $D$, $E$ form then B1 for the $A$ term, but no further marks until partial fractions are used to split the second term or they use integration by parts to obtain $\frac{Dx+E}{2-x} - \int\frac{D}{2-x}dx$ for 2nd B1 and 3rd B1 for correct completion |
| (third B1FT term) | B1FT | B0FT, B0FT, B0FT if they place their $A$, $B$, $C$ with incorrect denominators |
| Substitute limits correctly in integral of form $a\ln(1+2x) + b\ln(2-x) + \frac{c}{2-x}$, where $abc \neq 0$ | M1 | Condone minor slips in substitution. Exact substitution required |
| Obtain answer $\frac{5}{2} - \ln 72$ after full and correct working | A1 | AG – evidence of some correct work to combine or simplify logs required e.g. allow from $-\ln 9 + \ln\frac{1}{8}$ or $-\ln 2^3 - \ln 3^2$ |
---9 Let $\mathrm { f } ( x ) = \frac { 2 x ^ { 2 } + 17 x - 17 } { ( 1 + 2 x ) ( 2 - x ) ^ { 2 } }$.
\begin{enumerate}[label=(\alph*)]
\item Express $\mathrm { f } ( x )$ in partial fractions.
\item Hence show that $\int _ { 0 } ^ { 1 } \mathrm { f } ( x ) \mathrm { d } x = \frac { 5 } { 2 } - \ln 72$.\\
\includegraphics[max width=\textwidth, alt={}, center]{60bb482b-fa41-42ea-a112-62851e5a19aa-16_524_725_269_696}
The diagram shows the curve $y = ( x + 5 ) \sqrt { 3 - 2 x }$ and its maximum point $M$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2023 Q9 [10]}}