| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem and Partial Fractions |
| Type | Partial fractions with validity range |
| Difficulty | Standard +0.3 This is a standard three-part partial fractions question requiring decomposition with a repeated linear factor, binomial expansions of three terms, and identification of convergence conditions. While it involves multiple steps, each component is routine A-level technique with no novel insight required, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.04c Extend binomial expansion: rational n, |x|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State or imply the form \(\frac{A}{1-2x} + \frac{B}{2+x} + \frac{C}{(2+x)^2}\) | B1 | |
| Use a correct method for finding a coefficient | M1 | \(A(2+x)^2 + B(1-2x)(2+x) + C(1-2x) = 24x+13\) |
| Obtain one of \(A=4\), \(B=2\) and \(C=-7\) | A1 | If errors in equating still allow A marks for \(A\) and \(C\). |
| Obtain a second value | A1 | |
| Obtain the third value | A1 | Mark the form \(\frac{A}{1-2x} + \frac{Dx+E}{(2+x)^2}\), where \(A=4\), \(D=2\) and \(E=-3\), B1 M1 A1 A1 A1 as above. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use a correct method to find the first two terms of the expansion of \((1-2x)^{-1}\), \((2+x)^{-1}\), \((2+x)^{-2}\), \(\left(1+\frac{x}{2}\right)^{-1}\) or \(\left(1+\frac{x}{2}\right)^{-2}\) | M1 | Symbolic coefficients are not sufficient for the M1 |
| \(A\left(1+(-1)(-2x)+\frac{(-1)(-2)}{2}(-2x)^2+..\right)\) where \(A=4\) | A1 FT | |
| \(\frac{B}{2}\left(1+(-1)\left(\frac{x}{2}\right)+\frac{(-1)(-2)}{2}\left(\frac{x}{2}\right)^2+...\right)\) where \(B=2\) | A1 FT | |
| \(\frac{C}{4}\left(1+(-2)\left(\frac{x}{2}\right)+\frac{(-2)(-3)}{2}\left(\frac{x}{2}\right)^2+...\right)\) where \(C=-7\) | A1 FT | \(= 4(1+2x+4x^2)+2/2(1-x/2+x^2/4)-7/4(1-x+3x^2/4)\) \(=(4+1-7/4)+(8-1/2+7/4)x+(16+1/4-21/16)x^2\) The FT is on \(A\), \(B\), \(C\) |
| Obtain final answer \(\frac{13}{4}+\frac{37}{4}x+\frac{239}{16}x^2\) | A1 | OE; \((Dx+E)/4\left[1+(-2)(x/2)+(-2)(-3)(x/2)^2/2...\right]\), \(D=2\), \(E=-3\); FT is on \(A\), \(D\), \(E\). Maclaurin's: \(f(0)=13/4\) B1, \(f'(0)=37/4\) B1, \(f''(0)=239/8\) B1; \(\frac{13}{4}+\frac{37}{4}x+\frac{239}{8}x^2/2\) or equivalent M1 A1 |
| Total: 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\ | x\ | <\frac{1}{2}\) |
| Total: 1 mark |
## Question 10(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply the form $\frac{A}{1-2x} + \frac{B}{2+x} + \frac{C}{(2+x)^2}$ | **B1** | |
| Use a correct method for finding a coefficient | **M1** | $A(2+x)^2 + B(1-2x)(2+x) + C(1-2x) = 24x+13$ |
| Obtain one of $A=4$, $B=2$ and $C=-7$ | **A1** | If errors in equating still allow A marks for $A$ and $C$. |
| Obtain a second value | **A1** | |
| Obtain the third value | **A1** | Mark the form $\frac{A}{1-2x} + \frac{Dx+E}{(2+x)^2}$, where $A=4$, $D=2$ and $E=-3$, B1 M1 A1 A1 A1 as above. |
## Question 10(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use a correct method to find the first two terms of the expansion of $(1-2x)^{-1}$, $(2+x)^{-1}$, $(2+x)^{-2}$, $\left(1+\frac{x}{2}\right)^{-1}$ or $\left(1+\frac{x}{2}\right)^{-2}$ | M1 | Symbolic coefficients are not sufficient for the M1 |
| $A\left(1+(-1)(-2x)+\frac{(-1)(-2)}{2}(-2x)^2+..\right)$ where $A=4$ | A1 FT | |
| $\frac{B}{2}\left(1+(-1)\left(\frac{x}{2}\right)+\frac{(-1)(-2)}{2}\left(\frac{x}{2}\right)^2+...\right)$ where $B=2$ | A1 FT | |
| $\frac{C}{4}\left(1+(-2)\left(\frac{x}{2}\right)+\frac{(-2)(-3)}{2}\left(\frac{x}{2}\right)^2+...\right)$ where $C=-7$ | A1 FT | $= 4(1+2x+4x^2)+2/2(1-x/2+x^2/4)-7/4(1-x+3x^2/4)$ $=(4+1-7/4)+(8-1/2+7/4)x+(16+1/4-21/16)x^2$ The FT is on $A$, $B$, $C$ |
| Obtain final answer $\frac{13}{4}+\frac{37}{4}x+\frac{239}{16}x^2$ | A1 | OE; $(Dx+E)/4\left[1+(-2)(x/2)+(-2)(-3)(x/2)^2/2...\right]$, $D=2$, $E=-3$; FT is on $A$, $D$, $E$. Maclaurin's: $f(0)=13/4$ B1, $f'(0)=37/4$ B1, $f''(0)=239/8$ B1; $\frac{13}{4}+\frac{37}{4}x+\frac{239}{8}x^2/2$ or equivalent M1 A1 |
| **Total: 5 marks** | | |
## Question 10(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\|x\|<\frac{1}{2}$ | B1 | OE |
| **Total: 1 mark** | | |10 Let $\mathrm { f } ( x ) = \frac { 24 x + 13 } { ( 1 - 2 x ) ( 2 + x ) ^ { 2 } }$.
\begin{enumerate}[label=(\alph*)]
\item Express $\mathrm { f } ( x )$ in partial fractions.
\item Hence obtain the expansion of $\mathrm { f } ( x )$ in ascending powers of $x$, up to and including the term in $x ^ { 2 }$.
\item State the set of values of $x$ for which the expansion in (b) is valid.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2023 Q10 [11]}}