Edexcel C4 2006 June — Question 2 9 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Year2006
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeneralised Binomial Theorem and Partial Fractions
TypePartial fractions then binomial expansion
DifficultyStandard +0.3 This is a straightforward C4 question combining routine partial fractions with repeated linear factors and standard binomial expansion. Part (a) uses the cover-up method or substitution (very mechanical), and part (b) applies the binomial theorem to two simple terms. The algebra is clean, the techniques are standard textbook exercises, and no problem-solving insight is required—slightly easier than average for A-level.
Spec1.02y Partial fractions: decompose rational functions1.04c Extend binomial expansion: rational n, |x|<1

2. $$f ( x ) = \frac { 3 x - 1 } { ( 1 - 2 x ) ^ { 2 } } , \quad | x | < \frac { 1 } { 2 }$$ Given that, for \(x \neq \frac { 1 } { 2 } , \quad \frac { 3 x - 1 } { ( 1 - 2 x ) ^ { 2 } } = \frac { A } { ( 1 - 2 x ) } + \frac { B } { ( 1 - 2 x ) ^ { 2 } } , \quad\) where \(A\) and \(B\) are constants,
  1. find the values of \(A\) and \(B\).
  2. Hence, or otherwise, find the series expansion of \(\mathrm { f } ( x )\), in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\), simplifying each term.
    (6)

2.

$$f ( x ) = \frac { 3 x - 1 } { ( 1 - 2 x ) ^ { 2 } } , \quad | x | < \frac { 1 } { 2 }$$

Given that, for $x \neq \frac { 1 } { 2 } , \quad \frac { 3 x - 1 } { ( 1 - 2 x ) ^ { 2 } } = \frac { A } { ( 1 - 2 x ) } + \frac { B } { ( 1 - 2 x ) ^ { 2 } } , \quad$ where $A$ and $B$ are constants,
\begin{enumerate}[label=(\alph*)]
\item find the values of $A$ and $B$.
\item Hence, or otherwise, find the series expansion of $\mathrm { f } ( x )$, in ascending powers of $x$, up to and including the term in $x ^ { 3 }$, simplifying each term.\\
(6)
\end{enumerate}

\hfill \mbox{\textit{Edexcel C4 2006 Q2 [9]}}