AQA C4 2011 January — Question 3 12 marks

Exam BoardAQA
ModuleC4 (Core Mathematics 4)
Year2011
SessionJanuary
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeneralised Binomial Theorem and Partial Fractions
TypePartial fractions with validity range
DifficultyStandard +0.3 This is a standard C4 partial fractions question with routine binomial expansion. Part (a) is mechanical algebra, part (b) applies standard binomial theorem to two simple terms, and part (c) requires finding the intersection of two convergence conditions |x|<1 and |5x/3|<1. Slightly above average due to the multi-step nature and the validity range requiring careful consideration of both fractions, but all techniques are standard textbook exercises with no novel insight required.
Spec1.02y Partial fractions: decompose rational functions1.04c Extend binomial expansion: rational n, |x|<1

3
  1. Express \(\frac { 3 + 9 x } { ( 1 + x ) ( 3 + 5 x ) }\) in the form \(\frac { A } { 1 + x } + \frac { B } { 3 + 5 x }\), where \(A\) and \(B\) are integers.
  2. Hence, or otherwise, find the binomial expansion of \(\frac { 3 + 9 x } { ( 1 + x ) ( 3 + 5 x ) }\) up to and including the term in \(x ^ { 2 }\).
  3. Find the range of values of \(x\) for which the binomial expansion of \(\frac { 3 + 9 x } { ( 1 + x ) ( 3 + 5 x ) }\) is valid.
    (2 marks)

Question 3:
Part (a)
AnswerMarks
\(3+9x \equiv A(3+5x) + B(1+x)\)M1
\(x = -1\): \(-6 = -2A \Rightarrow A = 3\)A1
\(x = -\frac{3}{5}\): \(\frac{-12}{5} = \frac{2B}{5} \Rightarrow B = -6\)A1
Part (b)
AnswerMarks
\(\frac{3}{1+x} = 3(1+x)^{-1} = 3\left(1 - x + x^2 - \cdots\right)\)M1 A1
\(\frac{-6}{3+5x} = \frac{-6}{3}\left(1+\frac{5x}{3}\right)^{-1} = -2\left(1 - \frac{5x}{3} + \frac{25x^2}{9} - \cdots\right)\)M1 A1
Combining: \(1 + \frac{1}{3}x - \frac{23}{9}x^2\)M1 A1 A1
Part (c)
AnswerMarks Guidance
Valid for \(x < 1\) and \(\left
\(-\frac{3}{5} < x < \frac{3}{5}\)A1
# Question 3:

## Part (a)
| $3+9x \equiv A(3+5x) + B(1+x)$ | M1 | |
|---|---|---|
| $x = -1$: $-6 = -2A \Rightarrow A = 3$ | A1 | |
| $x = -\frac{3}{5}$: $\frac{-12}{5} = \frac{2B}{5} \Rightarrow B = -6$ | A1 | |

## Part (b)
| $\frac{3}{1+x} = 3(1+x)^{-1} = 3\left(1 - x + x^2 - \cdots\right)$ | M1 A1 | |
|---|---|---|
| $\frac{-6}{3+5x} = \frac{-6}{3}\left(1+\frac{5x}{3}\right)^{-1} = -2\left(1 - \frac{5x}{3} + \frac{25x^2}{9} - \cdots\right)$ | M1 A1 | |
| Combining: $1 + \frac{1}{3}x - \frac{23}{9}x^2$ | M1 A1 A1 | |

## Part (c)
| Valid for $|x| < 1$ and $\left|\frac{5x}{3}\right| < 1 \Rightarrow |x| < \frac{3}{5}$ | M1 | |
|---|---|---|
| $-\frac{3}{5} < x < \frac{3}{5}$ | A1 | |

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3
\begin{enumerate}[label=(\alph*)]
\item Express $\frac { 3 + 9 x } { ( 1 + x ) ( 3 + 5 x ) }$ in the form $\frac { A } { 1 + x } + \frac { B } { 3 + 5 x }$, where $A$ and $B$ are integers.
\item Hence, or otherwise, find the binomial expansion of $\frac { 3 + 9 x } { ( 1 + x ) ( 3 + 5 x ) }$ up to and including the term in $x ^ { 2 }$.
\item Find the range of values of $x$ for which the binomial expansion of $\frac { 3 + 9 x } { ( 1 + x ) ( 3 + 5 x ) }$ is valid.\\
(2 marks)
\end{enumerate}

\hfill \mbox{\textit{AQA C4 2011 Q3 [12]}}