| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem and Partial Fractions |
| Type | Partial fractions showing coefficient is zero |
| Difficulty | Standard +0.3 This is a standard two-part question combining partial fractions with binomial expansion. Part (a) uses routine cover-up/substitution methods to find constants, with the 'show B=0' adding minimal difficulty. Part (b) requires expanding two binomial terms and collecting coefficients—straightforward application of techniques with no novel insight required. Slightly easier than average due to the helpful hint that B=0 and only requiring terms up to x². |
| Spec | 1.02y Partial fractions: decompose rational functions1.04c Extend binomial expansion: rational n, |x|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4(x^2+6) = A(2+x)^2 + B(1-2x)(2+x) + C(1-2x)\) | M1 | Uses correct form; allow sign errors |
| Let \(x = -2 \Rightarrow 40 = 5C \Rightarrow C = 8\) | dM1 | Uses correct method (substitution or equating coefficients) to find at least one constant |
| Let \(x = \frac{1}{2} \Rightarrow 25 = 6.25A \Rightarrow A = 4\); so \(A=4, C=8\) | A1 | Both \(A=4, C=8\). Correct answers with no working imply M1M1A1 |
| Compare constants/terms in \(x\) or substitute another value; conclude \(B = 0\), e.g. \(24 = 4A + 2B + C \Rightarrow B = 0\)* | A1* | Needs a method as answer is given; must conclude \(B=0\) |
| Way 2: Compare \(x^2\): \(4 = A - 2B\); \(x\): \(0 = 4A - 3B - 2C\); constants: \(24 = 4A + 2B + C\); so \(A=4, C=8, B=0\)* | M1, dM1, A1, A1 | As Way 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{4(x^2+6)}{(1-2x)(2+x)^2} = 4(1-2x)^{-1} + 8(2+x)^{-2} = 4(1-2x)^{-1} + 8 \times \frac{1}{2^2}\left(1+\frac{x}{2}\right)^{-2}\) | B1 ft | Writes in form \(A(1-2x)^{-1} + C \times \frac{1}{2^2}\left(1+\frac{x}{2}\right)^{-2}\) ft on \(A\) and \(C\); no \(B\) term |
| \(\left(1+(-1)(-2x)+\frac{(-1)(-2)(-2x)^2}{2!}+\cdots\right)\) or \(\left(1+(-2)\left(\frac{x}{2}\right)+\frac{(-2)(-3)}{2!}\left(\frac{x}{2}\right)^2+\cdots\right)\) | M1 | Uses binomial expansion correctly for one expansion; power of 2 outside second bracket ignored. Allow missing brackets |
| Both expansions in brackets completely correct and unsimplified | A1 | Can be awarded without mention of \(A\) and/or \(C\) or with power of 2 outside second bracket ignored |
| \(= 4(1+2x+4x^2+\cdots) + 2\left(1-x+\frac{3}{4}x^2\right) = 6 + 6x + \frac{35}{2}x^2\) | dM1, A1 | dM1: Multiplies out and collects terms (allow sign slips). A1: \(6+6x+\frac{35}{2}x^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{4(x^2+6)}{(1-2x)(2+x)^2} = 4(x^2+6)\times(1-2x)^{-1}\times\frac{1}{4}\left(1+\frac{x}{2}\right)^{-2}\) | B1 | May be awarded for writing \(4(x^2+6)\times(1-2x)^{-1}\times(2+x)^{-2}\) and writing separately \((2+x)^{-2} = \frac{1}{2^2}\left(1+\frac{x}{2}\right)^{-2}\) |
| Both expansions in brackets correct and unsimplified | M1, A1 | Follow Way 1 notes |
| \(= 4(x^2+6)(1+2x+4x^2+\cdots)\times\frac{1}{4}\left(1-x+\frac{3}{4}x^2\right) = 6+6x+\frac{35}{2}x^2\) | dM1, A1 | dM1: Must multiply out three brackets. A1: \(6+6x+\frac{35}{2}x^2\) |
## Question 2:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4(x^2+6) = A(2+x)^2 + B(1-2x)(2+x) + C(1-2x)$ | M1 | Uses correct form; allow sign errors |
| Let $x = -2 \Rightarrow 40 = 5C \Rightarrow C = 8$ | dM1 | Uses correct method (substitution or equating coefficients) to find at least one constant |
| Let $x = \frac{1}{2} \Rightarrow 25 = 6.25A \Rightarrow A = 4$; so $A=4, C=8$ | A1 | Both $A=4, C=8$. Correct answers with no working imply M1M1A1 |
| Compare constants/terms in $x$ or substitute another value; conclude $B = 0$, e.g. $24 = 4A + 2B + C \Rightarrow B = 0$* | A1* | **Needs a method** as answer is given; must conclude $B=0$ |
| **Way 2:** Compare $x^2$: $4 = A - 2B$; $x$: $0 = 4A - 3B - 2C$; constants: $24 = 4A + 2B + C$; so $A=4, C=8, B=0$* | M1, dM1, A1, A1 | As Way 1 |
### Part (b) — Way 1
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{4(x^2+6)}{(1-2x)(2+x)^2} = 4(1-2x)^{-1} + 8(2+x)^{-2} = 4(1-2x)^{-1} + 8 \times \frac{1}{2^2}\left(1+\frac{x}{2}\right)^{-2}$ | B1 ft | Writes in form $A(1-2x)^{-1} + C \times \frac{1}{2^2}\left(1+\frac{x}{2}\right)^{-2}$ ft on $A$ and $C$; no $B$ term |
| $\left(1+(-1)(-2x)+\frac{(-1)(-2)(-2x)^2}{2!}+\cdots\right)$ or $\left(1+(-2)\left(\frac{x}{2}\right)+\frac{(-2)(-3)}{2!}\left(\frac{x}{2}\right)^2+\cdots\right)$ | M1 | Uses binomial expansion correctly for one expansion; power of 2 outside second bracket ignored. Allow missing brackets |
| Both expansions in brackets completely correct and unsimplified | A1 | Can be awarded without mention of $A$ and/or $C$ or with power of 2 outside second bracket ignored |
| $= 4(1+2x+4x^2+\cdots) + 2\left(1-x+\frac{3}{4}x^2\right) = 6 + 6x + \frac{35}{2}x^2$ | dM1, A1 | dM1: Multiplies out and collects terms (allow sign slips). A1: $6+6x+\frac{35}{2}x^2$ |
### Part (b) — Way 2
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{4(x^2+6)}{(1-2x)(2+x)^2} = 4(x^2+6)\times(1-2x)^{-1}\times\frac{1}{4}\left(1+\frac{x}{2}\right)^{-2}$ | B1 | May be awarded for writing $4(x^2+6)\times(1-2x)^{-1}\times(2+x)^{-2}$ and writing separately $(2+x)^{-2} = \frac{1}{2^2}\left(1+\frac{x}{2}\right)^{-2}$ |
| Both expansions in brackets correct and unsimplified | M1, A1 | Follow Way 1 notes |
| $= 4(x^2+6)(1+2x+4x^2+\cdots)\times\frac{1}{4}\left(1-x+\frac{3}{4}x^2\right) = 6+6x+\frac{35}{2}x^2$ | dM1, A1 | dM1: Must multiply out three brackets. A1: $6+6x+\frac{35}{2}x^2$ |2. Given that
$$\frac { 4 \left( x ^ { 2 } + 6 \right) } { ( 1 - 2 x ) ( 2 + x ) ^ { 2 } } \equiv \frac { A } { ( 1 - 2 x ) } + \frac { B } { ( 2 + x ) } + \frac { C } { ( 2 + x ) ^ { 2 } }$$
\begin{enumerate}[label=(\alph*)]
\item find the values of the constants $A$ and $C$ and show that $B = 0$\\
(4)
\item Hence, or otherwise, find the series expansion of
$$\frac { 4 \left( x ^ { 2 } + 6 \right) } { ( 1 - 2 x ) ( 2 + x ) ^ { 2 } } , \quad | x | < \frac { 1 } { 2 }$$
in ascending powers of $x$, up to and including the term in $x ^ { 2 }$, simplifying each term.\\
(5)
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2015 Q2 [9]}}