| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem and Partial Fractions |
| Type | Partial fractions with validity range |
| Difficulty | Moderate -0.3 This is a standard C4 question combining routine partial fractions with binomial expansions. Part (a) is trivial recall, part (b)(i) uses the cover-up method with simple linear factors, and part (b)(ii) requires combining two standard binomial expansions. Part (c) tests understanding of validity ranges by finding the intersection of |x| < 1 and |3x/2| < 1. 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 |
|---|---|---|
| Working | Marks | Guidance |
| \((1-x)^{-1} = 1 + (-1)(-x) + \frac{1}{2}(-1\cdot-2)(-x)^2\) | M1 | \(1 \pm x + kx^2\) |
| \(= 1 + x + x^2\) | A1 | Fully simplified |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(3x-1 = A(2-3x) + B(1-x)\) | M1 | |
| \(x=1\), \(x=\frac{2}{3}\) | m1 | Use 2 values of \(x\) or equate coefficients and solve \(-3A-B=3\), \(2A+B=-1\) |
| \(A=-2\), \(B=3\) | A1 | Both values; NMS 3/3 if both correct, 1/3 if one correct |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\frac{-2}{1-x} = -2 - 2x - 2x^2\) | B1F | F on \((1-x)^{-1}\) and \(A\) |
| \(\frac{1}{2-3x} = \frac{1}{2}\left(1-\frac{3}{2}x\right)^{-1}\) | B1 | |
| \(= (p)\left(1 + kx + (kx)^2\right)\) | M1 | \(p\), \(k\) = candidate's \(\frac{1}{2}\), \(\frac{3}{2}\), \(k \neq \pm 1\) |
| \(= (p)\left(1 + \frac{3}{2}x + \frac{9}{4}x^2\right)\) | A1 | Use (a) or start binomial again; condone missing brackets and one sign error |
| \(\frac{3x-1}{(1-x)(2-3x)} = -2(1-x)^{-1} + 3(2-3x)^{-1}\) | M1 | Valid combination of both expansions |
| \(= -\frac{1}{2} + \frac{1}{4}x + \frac{11}{8}x^2\) | A1 | CSO |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \((2-3x)^{-1} = \frac{1}{2}\left(1-\frac{3}{2}x\right)^{-1}\) | (B1) | \(k =\) candidate's \(\frac{3}{2}\), \(k \neq \pm 1\) |
| \((1-kx)^{-1} = 1 + kx + (kx)^2\) | (M1) | Use (a) or start binomial again |
| \(= 1 + \frac{3}{2}x + \frac{9}{4}x^2\) | (A1) | |
| \(\frac{3x-1}{(1-x)(2-3x)} = (3x-1)(1-x)^{-1}(2-3x)^{-1}\) | (M1) | \((3x-1) \times\) both expansions |
| \(\frac{3x-1}{(1-x)(2-3x)} = -\frac{1}{2}+\frac{1}{4}x+\frac{11}{8}x^2\) | (m1)(A1) | Multiply out; collect terms; CSO |
# Question 3:
## Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| $(1-x)^{-1} = 1 + (-1)(-x) + \frac{1}{2}(-1\cdot-2)(-x)^2$ | M1 | $1 \pm x + kx^2$ |
| $= 1 + x + x^2$ | A1 | Fully simplified |
**Total: 2 marks**
## Part (b)(i)
| Working | Marks | Guidance |
|---------|-------|----------|
| $3x-1 = A(2-3x) + B(1-x)$ | M1 | |
| $x=1$, $x=\frac{2}{3}$ | m1 | Use 2 values of $x$ or equate coefficients and solve $-3A-B=3$, $2A+B=-1$ |
| $A=-2$, $B=3$ | A1 | Both values; NMS 3/3 if both correct, 1/3 if one correct |
**Total: 3 marks**
## Part (b)(ii)
| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{-2}{1-x} = -2 - 2x - 2x^2$ | B1F | F on $(1-x)^{-1}$ and $A$ |
| $\frac{1}{2-3x} = \frac{1}{2}\left(1-\frac{3}{2}x\right)^{-1}$ | B1 | |
| $= (p)\left(1 + kx + (kx)^2\right)$ | M1 | $p$, $k$ = candidate's $\frac{1}{2}$, $\frac{3}{2}$, $k \neq \pm 1$ |
| $= (p)\left(1 + \frac{3}{2}x + \frac{9}{4}x^2\right)$ | A1 | Use (a) or start binomial again; condone missing brackets and one sign error |
| $\frac{3x-1}{(1-x)(2-3x)} = -2(1-x)^{-1} + 3(2-3x)^{-1}$ | M1 | Valid combination of both expansions |
| $= -\frac{1}{2} + \frac{1}{4}x + \frac{11}{8}x^2$ | A1 | CSO |
**Alternative:**
| Working | Marks | Guidance |
|---------|-------|----------|
| $(2-3x)^{-1} = \frac{1}{2}\left(1-\frac{3}{2}x\right)^{-1}$ | (B1) | $k =$ candidate's $\frac{3}{2}$, $k \neq \pm 1$ |
| $(1-kx)^{-1} = 1 + kx + (kx)^2$ | (M1) | Use (a) or start binomial again |
| $= 1 + \frac{3}{2}x + \frac{9}{4}x^2$ | (A1) | |
| $\frac{3x-1}{(1-x)(2-3x)} = (3x-1)(1-x)^{-1}(2-3x)^{-1}$ | (M1) | $(3x-1) \times$ both expansions |
| $\frac{3x-1}{(1-x)(2-3x)} = -\frac{1}{2}+\frac{1}{4}x+\frac{11}{8}x^2$ | (m1)(A1) | Multiply out; collect terms; CSO |
**Total: 6 marks**3
\begin{enumerate}[label=(\alph*)]
\item Find the binomial expansion of $( 1 - x ) ^ { - 1 }$ up to and including the term in $x ^ { 2 }$.
\item \begin{enumerate}[label=(\roman*)]
\item Express $\frac { 3 x - 1 } { ( 1 - x ) ( 2 - 3 x ) }$ in the form $\frac { A } { 1 - x } + \frac { B } { 2 - 3 x }$, where $A$ and $B$ are integers.
\item Find the binomial expansion of $\frac { 3 x - 1 } { ( 1 - x ) ( 2 - 3 x ) }$ up to and including the term in $x ^ { 2 }$.
\end{enumerate}\item Find the range of values of $x$ for which the binomial expansion of $\frac { 3 x - 1 } { ( 1 - x ) ( 2 - 3 x ) }$ is valid.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2009 Q3 [13]}}