| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem and Partial Fractions |
| Type | Partial fractions with percentage error |
| Difficulty | Standard +0.3 This is a standard C4 question combining partial fractions with binomial expansion. Part (a) uses routine cover-up/substitution methods, part (b) requires straightforward binomial expansions of (3x+2)^{-2} and (1-x)^{-1}, and part (c) is simple percentage error calculation. All techniques are textbook exercises 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 | Mark | Guidance |
| \(27x^2 + 32x + 16 \equiv A(3x+2)(1-x) + B(1-x) + C(3x+2)^2\) | M1 | Forming this identity |
| \(x^2\): \(27 = -3A + 9C\) (1); \(x\): \(32 = A - B + 12C\) (2); constants: \(16 = 2A + B + 4C\) (3) | M1 | Equates 3 terms |
| \((2)+(3)\): \(48 = 3A + 16C\) (4) | ||
| \((1)+(4)\): \(75 = 25C \Rightarrow C = 3\) | ||
| (1): \(27 = -3A + 27 \Rightarrow A = 0\) | ||
| (2): \(32 = -B + 36 \Rightarrow B = 4\) | A1 | Both \(B=4\) and \(C=3\) |
| \(A = 0\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(f(x) = 4(3x+2)^{-2} + 3(1-x)^{-1}\) moving to \(4(2+3x)^{-2} + 3(1-x)^{-1}\) | M1 | Moving powers to top on any one of the two expressions |
| \(4\left\{(2)^{-2}+(-2)(2)^{-3}(3x)+\frac{(-2)(-3)}{2!}(2)^{-4}(3x)^2+\cdots\right\}\) | dM1 | Either \((2)^{-2}\pm(-2)(2)^{-3}(3x)\) or \(1\pm(-1)(-x)\) from either first or second expansions respectively |
| \(+3\left\{1+(-1)(-x)+\frac{(-1)(-2)}{2!}(-x)^2+\cdots\right\}\) | A1 | Ignoring 1 and 3, any one correct \(\{\ldots\}\) expansion |
| Both \(\{\ldots\}\) correct | A1 | Both \(\{\ldots\}\) correct |
| \(= 4\left\{\frac{1}{4}-\frac{3}{4}x+\frac{27}{16}x^2+\cdots\right\}+3\left\{1+x+x^2+\cdots\right\}\) | — | — |
| \(= 4+(0x)+\frac{39}{4}x^2\) | A1; A1 | \(4+(0x)\); \(\frac{39}{4}x^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Actual \(= f(0.2) = \frac{1.08+6.4+16}{(6.76)(0.8)} = \frac{23.48}{5.408} = 4.341715976\ldots = \frac{2935}{676}\) | M1 | Attempt to find the actual value of \(f(0.2)\) |
| Estimate \(= f(0.2) = 4+\frac{39}{4}(0.2)^2 = 4+0.39 = 4.39\) | M1\(\checkmark\) | Attempt to find an estimate for \(f(0.2)\) using their answer to (b) |
| \(\%\text{age error} = \left | 100-\left(\frac{4.39}{4.341715976\ldots}\times100\right)\right | \) |
| \(= \ | -1.112095408\ldots\ | = 1.1\%\ (2\text{sf})\) |
# Question 3 (Appendix):
## Part (a) — Aliter Way 2:
| Answer | Mark | Guidance |
|--------|------|----------|
| $27x^2 + 32x + 16 \equiv A(3x+2)(1-x) + B(1-x) + C(3x+2)^2$ | M1 | Forming this identity |
| $x^2$: $27 = -3A + 9C$ (1); $x$: $32 = A - B + 12C$ (2); constants: $16 = 2A + B + 4C$ (3) | M1 | Equates 3 terms |
| $(2)+(3)$: $48 = 3A + 16C$ (4) | | |
| $(1)+(4)$: $75 = 25C \Rightarrow C = 3$ | | |
| (1): $27 = -3A + 27 \Rightarrow A = 0$ | | |
| (2): $32 = -B + 36 \Rightarrow B = 4$ | A1 | Both $B=4$ and $C=3$ |
| $A = 0$ | B1 | |
## Question 3(b) — Aliter Way 2:
| Working/Answer | Marks | Guidance |
|---|---|---|
| $f(x) = 4(3x+2)^{-2} + 3(1-x)^{-1}$ moving to $4(2+3x)^{-2} + 3(1-x)^{-1}$ | M1 | Moving powers to top on any one of the two expressions |
| $4\left\{(2)^{-2}+(-2)(2)^{-3}(3x)+\frac{(-2)(-3)}{2!}(2)^{-4}(3x)^2+\cdots\right\}$ | dM1 | Either $(2)^{-2}\pm(-2)(2)^{-3}(3x)$ or $1\pm(-1)(-x)$ from either first or second expansions respectively |
| $+3\left\{1+(-1)(-x)+\frac{(-1)(-2)}{2!}(-x)^2+\cdots\right\}$ | A1 | Ignoring 1 and 3, any one correct $\{\ldots\}$ expansion |
| Both $\{\ldots\}$ correct | A1 | Both $\{\ldots\}$ correct |
| $= 4\left\{\frac{1}{4}-\frac{3}{4}x+\frac{27}{16}x^2+\cdots\right\}+3\left\{1+x+x^2+\cdots\right\}$ | — | — |
| $= 4+(0x)+\frac{39}{4}x^2$ | A1; A1 | $4+(0x)$; $\frac{39}{4}x^2$ |
---
## Question 3(c) — Aliter Way 2:
| Working/Answer | Marks | Guidance |
|---|---|---|
| Actual $= f(0.2) = \frac{1.08+6.4+16}{(6.76)(0.8)} = \frac{23.48}{5.408} = 4.341715976\ldots = \frac{2935}{676}$ | M1 | Attempt to find the actual value of $f(0.2)$ |
| Estimate $= f(0.2) = 4+\frac{39}{4}(0.2)^2 = 4+0.39 = 4.39$ | M1$\checkmark$ | Attempt to find an estimate for $f(0.2)$ using their answer to (b) |
| $\%\text{age error} = \left|100-\left(\frac{4.39}{4.341715976\ldots}\times100\right)\right|$ | M1 | $\left|100-\left(\frac{\text{their estimate}}{\text{actual}}\times100\right)\right|$ |
| $= \|-1.112095408\ldots\| = 1.1\%\ (2\text{sf})$ | A1 cao | $1.1\%$ |
---3.
$$f ( x ) = \frac { 27 x ^ { 2 } + 32 x + 16 } { ( 3 x + 2 ) ^ { 2 } ( 1 - x ) } , \quad | x | < \frac { 2 } { 3 }$$
Given that $\mathrm { f } ( x )$ can be expressed in the form
$$f ( x ) = \frac { A } { ( 3 x + 2 ) } + \frac { B } { ( 3 x + 2 ) ^ { 2 } } + \frac { C } { ( 1 - x ) }$$
\begin{enumerate}[label=(\alph*)]
\item find the values of $B$ and $C$ and show that $A = 0$.
\item Hence, or otherwise, find the series expansion of $\mathrm { f } ( x )$, in ascending powers of $x$, up to and including the term in $x ^ { 2 }$. Simplify each term.
\item Find the percentage error made in using the series expansion in part (b) to estimate the value of $\mathrm { f } ( 0.2 )$. Give your answer to 2 significant figures.
\section*{LU}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2009 Q3 [14]}}