| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2023 |
| Session | January |
| Marks | 9 |
| 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 P4/FP2 question combining routine partial fractions with binomial expansion. Part (a) is straightforward decomposition, part (b)(i) requires expanding two simple binomial terms and collecting coefficients, and part (b)(ii) involves finding the intersection of two standard validity conditions. All steps are textbook procedures 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|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{5x+10}{(1-x)(2+3x)} \equiv \frac{A}{1-x} + \frac{B}{2+3x}\), attempts value for \(A\) or \(B\) | M1 | Correct form identified, achieves at least one constant |
| One correct value: \(A=3\) or \(B=4\) | A1 | |
| Correct PF form: \(\frac{3}{1-x} + \frac{4}{2+3x}\) | A1 | May be awarded if seen in (b); accept \(3(1-x)^{-1}+4(2+3x)^{-1}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{A}{1-x} = A(1-x)^{-1} = A\left(1+x+x^2+...\right)\) | B1 | May be unsimplified; allow with their \(A\) or \(A=1\) |
| \(\left\{\frac{B}{2}\right\}\left(1+\frac{3x}{2}\right)^{-1} = \left\{\frac{B}{2}\right\}\left(1+(-1)\frac{3x}{2}+\frac{(-1)(-2)}{2}\left(\frac{3x}{2}\right)^2+...\right) = \frac{B}{2}\left(1-\frac{3x}{2}+\frac{9x^2}{4}+...\right)\) | M1; A1 | M1: attempts \((2+3x)^{-1}\) binomially taking out factor 2 first or directly; look for \((1+kx)^{-1}\) form where \(k\neq 1\); A1: correct expansion with their \(B\) |
| \(f(x) = 3\times\left(1+x+x^2+...\right) + \frac{4}{2}\left(1-\frac{3x}{2}+\frac{9x^2}{4}+...\right)\) | M1 | Uses coefficients and attempts to add both series |
| \(= 5 + \frac{15}{2}x^2 + ...\) | A1 | Condone additional higher order terms |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\left | x\right | < \frac{2}{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \((1-x)^{-1} = 1+x+x^2+...\) | B1 | May be unsimplified |
| \(\left\{\frac{1}{2}\right\}\left(1+\frac{3x}{2}\right)^{-1} = \frac{1}{2}\left(1-\frac{3x}{2}+\frac{9}{4}x^2+...\right)\) | M1; A1 | Same as main scheme; \(B=1\) allowed |
| \(\frac{5x+10}{(1-x)(2+3x)} = (5x+10)\left(1+x+x^2+...\right)\times\frac{1}{2}\left(1-\frac{3x}{2}+\frac{9x^2}{4}+...\right) = 5+...x+...x^2\) | M1 | Attempts to expand all three brackets, achieving correct constant term |
| \(= 5+\frac{15}{2}x^2+...\) | A1 | Condone additional higher order terms |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{5x+10}{(1-x)(2+3x)} = (5x+10)\left(2+\left(x-3x^2\right)\right)^{-1} = \frac{1}{2}(5x+10)\left(1+\frac{1}{2}\left(x-3x^2\right)\right)^{-1}\) | B1 | Writes \(f(x)\) in this form with \((x-3x^2)\) clear |
| \((1+p(x))^{-1} = \left(1\pm p(x)+\frac{(-1)(-2)}{2}(p(x))^2+...\right)\); \(\frac{1}{2}\left(1-\frac{1}{2}(x-3x^2)+\frac{1}{4}(x-3x^2)^2+...\right)\) | M1; A1 | M1: attempts binomial expansion on \((1+p(x))^{-1}\) or \((2+p(x))^{-1}\) for \(p(x)\) of form \(ax+bx^2\); A1: correct expansion |
| \((10+5x)\left(\frac{1}{2}-\frac{1}{4}x+\frac{3}{4}x^2+\frac{1}{8}x^2+...\right) = 5-\frac{5}{2}x+\frac{35}{4}x^2+\frac{5}{2}x-\frac{5}{4}x^2+...\) | M1 | Expands brackets achieving at least correct constant term |
| \(= 5+\frac{15}{2}x^2+...\) | A1 | Condone additional higher order terms |
# Question 1:
## Part (a)
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{5x+10}{(1-x)(2+3x)} \equiv \frac{A}{1-x} + \frac{B}{2+3x}$, attempts value for $A$ or $B$ | M1 | Correct form identified, achieves at least one constant |
| One correct value: $A=3$ or $B=4$ | A1 | |
| Correct PF form: $\frac{3}{1-x} + \frac{4}{2+3x}$ | A1 | May be awarded if seen in (b); accept $3(1-x)^{-1}+4(2+3x)^{-1}$ |
## Part (b)(i)
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{A}{1-x} = A(1-x)^{-1} = A\left(1+x+x^2+...\right)$ | B1 | May be unsimplified; allow with their $A$ or $A=1$ |
| $\left\{\frac{B}{2}\right\}\left(1+\frac{3x}{2}\right)^{-1} = \left\{\frac{B}{2}\right\}\left(1+(-1)\frac{3x}{2}+\frac{(-1)(-2)}{2}\left(\frac{3x}{2}\right)^2+...\right) = \frac{B}{2}\left(1-\frac{3x}{2}+\frac{9x^2}{4}+...\right)$ | M1; A1 | M1: attempts $(2+3x)^{-1}$ binomially taking out factor 2 first or directly; look for $(1+kx)^{-1}$ form where $k\neq 1$; A1: correct expansion with their $B$ |
| $f(x) = 3\times\left(1+x+x^2+...\right) + \frac{4}{2}\left(1-\frac{3x}{2}+\frac{9x^2}{4}+...\right)$ | M1 | Uses coefficients and attempts to add both series |
| $= 5 + \frac{15}{2}x^2 + ...$ | A1 | Condone additional higher order terms |
## Part (b)(ii)
| Working | Mark | Guidance |
|---------|------|----------|
| $\left|x\right| < \frac{2}{3}$ | B1 | Must be clearly identified as the answer; B0 if both ranges given with no choice; accept formal set notation with $\cap$ |
---
## Part (b)(i) Alt 1
| Working | Mark | Guidance |
|---------|------|----------|
| $(1-x)^{-1} = 1+x+x^2+...$ | B1 | May be unsimplified |
| $\left\{\frac{1}{2}\right\}\left(1+\frac{3x}{2}\right)^{-1} = \frac{1}{2}\left(1-\frac{3x}{2}+\frac{9}{4}x^2+...\right)$ | M1; A1 | Same as main scheme; $B=1$ allowed |
| $\frac{5x+10}{(1-x)(2+3x)} = (5x+10)\left(1+x+x^2+...\right)\times\frac{1}{2}\left(1-\frac{3x}{2}+\frac{9x^2}{4}+...\right) = 5+...x+...x^2$ | M1 | Attempts to expand all three brackets, achieving correct constant term |
| $= 5+\frac{15}{2}x^2+...$ | A1 | Condone additional higher order terms |
## Part (b)(i) Alt 2
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{5x+10}{(1-x)(2+3x)} = (5x+10)\left(2+\left(x-3x^2\right)\right)^{-1} = \frac{1}{2}(5x+10)\left(1+\frac{1}{2}\left(x-3x^2\right)\right)^{-1}$ | B1 | Writes $f(x)$ in this form with $(x-3x^2)$ clear |
| $(1+p(x))^{-1} = \left(1\pm p(x)+\frac{(-1)(-2)}{2}(p(x))^2+...\right)$; $\frac{1}{2}\left(1-\frac{1}{2}(x-3x^2)+\frac{1}{4}(x-3x^2)^2+...\right)$ | M1; A1 | M1: attempts binomial expansion on $(1+p(x))^{-1}$ or $(2+p(x))^{-1}$ for $p(x)$ of form $ax+bx^2$; A1: correct expansion |
| $(10+5x)\left(\frac{1}{2}-\frac{1}{4}x+\frac{3}{4}x^2+\frac{1}{8}x^2+...\right) = 5-\frac{5}{2}x+\frac{35}{4}x^2+\frac{5}{2}x-\frac{5}{4}x^2+...$ | M1 | Expands brackets achieving at least correct constant term |
| $= 5+\frac{15}{2}x^2+...$ | A1 | Condone additional higher order terms |
---1.
$$f ( x ) = \frac { 5 x + 10 } { ( 1 - x ) ( 2 + 3 x ) }$$
\begin{enumerate}[label=(\alph*)]
\item Write $\mathrm { f } ( x )$ in partial fraction form.
\item \begin{enumerate}[label=(\roman*)]
\item Hence find, in ascending powers of $x$ up to and including the terms in $x ^ { 2 }$, the binomial series expansion of $\mathrm { f } ( x )$. Give each coefficient as a simplified fraction.
\item Find the range of values of $x$ for which this expansion is valid.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2023 Q1 [9]}}