| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | June |
| Marks | 12 |
| 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 structured, multi-part question that guides students through standard C4 techniques: basic binomial expansion with negative indices, partial fractions decomposition, and validity conditions. While it requires multiple steps, each part is routine and scaffolded, with part (a) directly supporting part (c). The validity range requires understanding |x| < 1 and |3x| < 1, which is standard bookwork. Slightly easier than average due to the scaffolding and lack of novel problem-solving. |
| Spec | 1.02y Partial fractions: decompose rational functions1.04c Extend binomial expansion: rational n, |x|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| \((1+x)^{-1} = 1 + (-1)x + px^2 + qx^3 = 1 - x + x^2 - x^3\) | M1, A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \((1+3x)^{-1} = 1 - 3x + (3x)^2 - (3x)^3 = 1 - 3x + 9x^2 - 27x^3\) | M1 | |
| A1 | 2 marks | CAO. SC \(x^3\)-term: \(1 - 3x + \frac{9}{2}x^2\) 1/2 |
| Alt (starting again): | ||
| \((1+3x)^{-1} = 1 - (3x) + \frac{(-1)(-2)(3x)^2}{2!} + \frac{(-1)(-2)(-3)(3x)^3}{3!}\) | M1 | |
| \(= 1 - 3x + 9x^2 - 27x^3\) | A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1+4x}{(1+x)(1+3x)} = \frac{A}{1+x} + \frac{B}{1+3x}\) | M1 | |
| \(1 + 4x = A(1+3x) + B(1+x)\) | ||
| \(x = -1, x = -\frac{1}{3}\) | m1 | |
| \(A = \frac{3}{2}, B = -\frac{1}{2}\) | A1 | 3 marks |
| Alt: | ||
| \(\frac{1+4x}{(1+x)(1+3x)} = \frac{A}{1+x} + \frac{B}{1+3x}\) | M1 | |
| \(1 + 4x = A(1+3x) + B(1+x)\) | ||
| \(A + B = 1, 3A + B = 4\) | m1 | |
| \(A = \frac{3}{2}, B = -\frac{1}{2}\) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1+4x}{(1+x)(1+3x)} = \frac{3}{2(1+x)} - \frac{1}{2(1+3x)}\) | M1 | |
| \(= \frac{3}{2}(1-x+x^2-x^3) - \frac{1}{2}(1-3x+9x^2-27x^3)\) | m1 | |
| \(= 1 - 3x^2 + 12x^3\) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\ | x\ | < 1 \text{ and } \ |
| \(\ | x\ | < \frac{1}{3}\) (0.33) |
**2(a)(i)**
| $(1+x)^{-1} = 1 + (-1)x + px^2 + qx^3 = 1 - x + x^2 - x^3$ | M1, A1 | 2 marks | $p \neq 0, q \neq 0$; SC 1/2 for $= 1 - x + px^2$ |
**2(a)(ii)**
| $(1+3x)^{-1} = 1 - 3x + (3x)^2 - (3x)^3 = 1 - 3x + 9x^2 - 27x^3$ | M1 | | x replaced by 3x in candidate's (a)(i); condone missing brackets |
| | A1 | 2 marks | CAO. SC $x^3$-term: $1 - 3x + \frac{9}{2}x^2$ 1/2 |
| **Alt (starting again):** | | | |
| $(1+3x)^{-1} = 1 - (3x) + \frac{(-1)(-2)(3x)^2}{2!} + \frac{(-1)(-2)(-3)(3x)^3}{3!}$ | M1 | | condone missing brackets |
| $= 1 - 3x + 9x^2 - 27x^3$ | A1 | 2 marks | CAO |
**2(b)**
| $\frac{1+4x}{(1+x)(1+3x)} = \frac{A}{1+x} + \frac{B}{1+3x}$ | M1 | | correct partial fractions form, multiplication by denominator |
| $1 + 4x = A(1+3x) + B(1+x)$ | | | |
| $x = -1, x = -\frac{1}{3}$ | m1 | | Use (any) two values of $x$ to find $A$ and $B$ |
| $A = \frac{3}{2}, B = -\frac{1}{2}$ | A1 | 3 marks | $A$ and $B$ both correct |
| **Alt:** | | | |
| $\frac{1+4x}{(1+x)(1+3x)} = \frac{A}{1+x} + \frac{B}{1+3x}$ | M1 | | correct partial fractions form, multiplication by denominator |
| $1 + 4x = A(1+3x) + B(1+x)$ | | | |
| $A + B = 1, 3A + B = 4$ | m1 | | Set up and solve |
| $A = \frac{3}{2}, B = -\frac{1}{2}$ | A1 | 3 marks | $A$ and $B$ both correct |
**2(c)(i)**
| $\frac{1+4x}{(1+x)(1+3x)} = \frac{3}{2(1+x)} - \frac{1}{2(1+3x)}$ | M1 | | correct partial fractions form, multiplication by denominator |
| $= \frac{3}{2}(1-x+x^2-x^3) - \frac{1}{2}(1-3x+9x^2-27x^3)$ | m1 | | multiply candidate's expansions by $A$ and $B$, and expand and simplify |
| $= 1 - 3x^2 + 12x^3$ | A1 | 3 marks | CAO; SC $A$ and $B$ interchanged, treat as miscopy. $(1 - 4x + 13x^2 - 40x^3)$ |
**2(c)(ii)**
| $\|x\| < 1 \text{ and } \|3x\| < 1$ | M1 | | OE and nothing else incorrect |
| $\|x\| < \frac{1}{3}$ (0.33) | A1 | 2 marks | OE. Condone $\leq$ |2
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the binomial expansion of $( 1 + x ) ^ { - 1 }$ up to the term in $x ^ { 3 }$.
\item Hence, or otherwise, obtain the binomial expansion of $\frac { 1 } { 1 + 3 x }$ up to the term in $x ^ { 3 }$.
\end{enumerate}\item Express $\frac { 1 + 4 x } { ( 1 + x ) ( 1 + 3 x ) }$ in partial fractions.
\item \begin{enumerate}[label=(\roman*)]
\item Find the binomial expansion of $\frac { 1 + 4 x } { ( 1 + x ) ( 1 + 3 x ) }$ up to the term in $x ^ { 3 }$.
\item Find the range of values of $x$ for which the binomial expansion of $\frac { 1 + 4 x } { ( 1 + x ) ( 1 + 3 x ) }$ is valid.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C4 2007 Q2 [12]}}