| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Expand and state validity |
| Difficulty | Moderate -0.3 Part (a) is a standard partial fractions decomposition with linear factors (routine C4 technique). Part (b)(i) requires binomial expansions of two terms and combining them, which is more involved but still follows a well-practiced algorithm. Part (b)(ii) is straightforward recall of validity conditions. This is a typical textbook-style C4 question testing standard techniques without requiring novel insight, making it slightly easier than average overall. |
| 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 |
|---|---|---|
| \(19x - 3 = A(3 - 4x) + B(1 + 2x)\) Correct equation and 'attempt' to find \(A\) or \(B\) | M1 | e.g. Using \(x = \frac{3}{4}\) or \(x = -\frac{1}{2}\) or simultaneous equation such as \(19 = -4A + 2B\) and \(-3 = 3A + B\) |
| \(A = -\frac{5}{2}\) | A1 | |
| \(B = \frac{9}{2}\) | A1 | Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \((1 + 2x)^{-1} = 1 - 2x + kx^2\) | M1 | provided \(k \neq 0\) |
| \(= 1 - 2x + 4x^2\) | A1 | Accept \((+2x)^2\) |
| \((3 - 4x)^{-1} = 3^{-1}\left(1 - \frac{4}{3}x\right)^{-1}\) | B1 | ACF for \(3^{-1}\) eg \(\frac{1}{3}\) or 0.33 ...or 0.3 |
| \(\left(1 - \frac{4}{3}x\right)^{-1}\) | ||
| \(= 1 + (-1)\left(-\frac{4}{3}x\right) + \frac{(-1)(-2)}{2!}\left(-\frac{4}{3}x\right)^2\) | M1 | Condone poor use of or missing brackets. |
| \(= \left(1 + \frac{4}{3}x + \frac{16}{9}x^2\right)\) | A1 | PI by later work |
| Answer | Marks | Guidance |
|---|---|---|
| \(-\frac{5}{2}(1 - 2x + 4x^2) + \frac{9}{2}\cdot\frac{1}{3}\left(1 + \frac{4}{3}x + \frac{16}{9}x^2\right)\) | M1 | PI by correct answer ft on candidate's \(A\) and \(B\) with their relevant series. |
| \(= -1 + 7x - \frac{22}{3}x^2\) | A1 | Total: 7 |
| Alt. \((3 - 4x)^{-1} = 3^{-1} + (-1)3^{-2}(-4x) + \frac{(-1)(-2)}{2!}3^{-3}(-4x)^2\) | M1 | \(= \frac{1}{3} + \frac{4}{9}x + \frac{16}{27}x^2\) |
| Answer | Marks |
|---|---|
| Attempt to multiply any two of their three brackets together as far as the term in \(x^2\) | M1 then A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(-\frac{1}{2} < x < \frac{1}{2}\) only | B1 | Total: 1 |
| Do NOT accept \(-1 < 2x < 1\) or \( | 2x | < 1\) or inclusion of the equality sign. |
## Part (a)
$19x - 3 = A(3 - 4x) + B(1 + 2x)$ Correct equation and 'attempt' to find $A$ or $B$ | M1 | e.g. Using $x = \frac{3}{4}$ or $x = -\frac{1}{2}$ or simultaneous equation such as $19 = -4A + 2B$ and $-3 = 3A + B$
$A = -\frac{5}{2}$ | A1 |
$B = \frac{9}{2}$ | A1 | Total: 3
NMS or cover up rule scores SC2 for $A = -\frac{5}{2}$ or $B = \frac{9}{2}$ or SC3 for both $A = -\frac{5}{2}$ and $B = \frac{9}{2}$
## Part (b)(i)
$(1 + 2x)^{-1} = 1 - 2x + kx^2$ | M1 | provided $k \neq 0$
$= 1 - 2x + 4x^2$ | A1 | Accept $(+2x)^2$
$(3 - 4x)^{-1} = 3^{-1}\left(1 - \frac{4}{3}x\right)^{-1}$ | B1 | ACF for $3^{-1}$ eg $\frac{1}{3}$ or 0.33 ...or 0.3
$\left(1 - \frac{4}{3}x\right)^{-1}$ | | |
$= 1 + (-1)\left(-\frac{4}{3}x\right) + \frac{(-1)(-2)}{2!}\left(-\frac{4}{3}x\right)^2$ | M1 | Condone poor use of or missing brackets.
$= \left(1 + \frac{4}{3}x + \frac{16}{9}x^2\right)$ | A1 | PI by later work
$\frac{19x - 3}{(1 + 2x)(3 - 4x)} =$
$-\frac{5}{2}(1 - 2x + 4x^2) + \frac{9}{2}\cdot\frac{1}{3}\left(1 + \frac{4}{3}x + \frac{16}{9}x^2\right)$ | M1 | PI by correct answer ft on candidate's $A$ and $B$ with their relevant series.
$= -1 + 7x - \frac{22}{3}x^2$ | A1 | Total: 7 | Must have $\frac{22}{3}, 7\frac{1}{3}$ or 7.3
Alt. $(3 - 4x)^{-1} = 3^{-1} + (-1)3^{-2}(-4x) + \frac{(-1)(-2)}{2!}3^{-3}(-4x)^2$ | M1 | $= \frac{1}{3} + \frac{4}{9}x + \frac{16}{27}x^2$ | A2
If $A$ and $B$ are correct full expansions are $-\frac{5}{2} + 5x - 10x^2$ and $\frac{3}{2} + 2x + \frac{8}{3}x^2$
Alt. for combined expansions without using PI's.
$\frac{19x-3}{(1+2x)(3-4x)} = (19x - 3)(1 + 2x)^{-1}(3 - 4x)^{-1} = (19x - 3)\left(1 - 2x + 4x^2\right)\left(\frac{1}{3} + \frac{4}{9}x + \frac{16}{27}x^2\right)$
Attempt to multiply any two of their three brackets together as far as the term in $x^2$ | M1 then A1
Condone unsimplified fractions in the binomial expansion(s), but final answer must be fully simplified.
$A$ and $B$ could be included in the series – e.g. $\frac{9}{2(3-4x)} = \frac{9}{6-8x} = \frac{9}{6}\left(1 - \frac{8x}{6}\right)^{-1}$ could still score B1 M1 A1.
## Part (b)(ii)
$-\frac{1}{2} < x < \frac{1}{2}$ only | B1 | Total: 1 | Strictly $<$ : Accept $|x| < \frac{1}{2}$
Do NOT accept $-1 < 2x < 1$ or $|2x| < 1$ or inclusion of the equality sign.
If $-\frac{3}{4} < x < \frac{3}{4}$ is also seen, it must be clear that $-\frac{1}{2} < x < \frac{1}{2}$ is the answer.
---\begin{enumerate}[label=(\alph*)]
\item Express $\frac{19x - 3}{(1 + 2x)(3 - 4x)}$ in the form $\frac{A}{1 + 2x} + \frac{B}{3 - 4x}$.
[3 marks]
\item \begin{enumerate}[label=(\roman*)]
\item Find the binomial expansion of $\frac{19x - 3}{(1 + 2x)(3 - 4x)}$ up to and including the term in $x^2$.
[7 marks]
\item State the range of values of $x$ for which this expansion is valid.
[1 mark]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2016 Q1 [11]}}