| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2005 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem and Partial Fractions |
| Type | Partial fractions with validity range |
| Difficulty | Standard +0.3 This is a standard C4 partial fractions question with a repeated linear factor, followed by routine binomial expansion of each term. Part (i) uses the cover-up method and substitution (textbook technique), part (ii) requires expanding three simple binomial terms to x², and part (iii) asks for the standard radius of convergence. All steps are algorithmic with no problem-solving 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 |
|---|---|---|
| (i) \(3x + 4 \equiv A(2+x)^2 + B(2+x)(1+x) + C(1+x)\) | M1, A/B1 | Accept \(≡\) or \(=\); If identity used, award 'A' mark, if cover-up rule used, award 'B' mark. Any correct eqn for B from identity |
| \(A = 1\), \(C = 2\) | A1 | |
| \(A + B = 0\) or \(4A + 3B + C = 3\) or \(4A + 2B + C = 4\) | A1 5 | |
| \(B = -1\) | ||
| (ii) \(1 - x + x^2\) | B1, B1, B1 | Expansion of \((1 + x)^{-1}\); Expansion of \((1 + \frac{1}{2}x)^{-1}\); First 2 terms of \((1 + \frac{1}{2}x)^{-2}\) |
| \(1 - \frac{1}{2}x + \frac{1}{4}x^2\) | B1 | |
| \(1 - x + \frac{3}{4}x^2\) | B1 5 | |
| \(1 - \frac{5}{4}x + \frac{5}{4}x^2\) | ||
| Other inequalities to be discarded. | 11 | |
| (iii) \(-1 < x < 1\) | AEF |
(i) $3x + 4 \equiv A(2+x)^2 + B(2+x)(1+x) + C(1+x)$ | M1, A/B1 | Accept $≡$ or $=$; If identity used, award 'A' mark, if cover-up rule used, award 'B' mark. Any correct eqn for B from identity
$A = 1$, $C = 2$ | | A1
$A + B = 0$ or $4A + 3B + C = 3$ or $4A + 2B + C = 4$ | | A1 5 |
$B = -1$ | |
(ii) $1 - x + x^2$ | B1, B1, B1 | Expansion of $(1 + x)^{-1}$; Expansion of $(1 + \frac{1}{2}x)^{-1}$; First 2 terms of $(1 + \frac{1}{2}x)^{-2}$
$1 - \frac{1}{2}x + \frac{1}{4}x^2$ | | B1 | Third term of $(1 + \frac{1}{2}x)^{-2}$
$1 - x + \frac{3}{4}x^2$ | | B1 5 | Complete correct expansion
$1 - \frac{5}{4}x + \frac{5}{4}x^2$ | | | **If partial fractions not used:** Award B1 for expansion of $(1+x)^{-1}$; B1+B1 for expansion of $(1 + \frac{1}{2}x)^{-2}$, and B1 for $1 - \frac{5}{4}x...$ & B1 for..., $+\frac{5}{4}x^2$. **Or** if denom expanded to give $a+bx+cx^2$ with $a=4,b=-8,c=5$, award B1. Expansion of $[1+(b/a)x+(c/a)x^2]^{-1} = 1 - (b/a)x + ... , (-c/a + b^2/a^2)x^2$ B1+B1. Final ans $= (1 - \frac{5}{4}x ... + \frac{5}{4}x^2)B1+B1$
| | | Other inequalities to be discarded. | **11**
(iii) $-1 < x < 1$ | | **AEF**\begin{enumerate}[label=(\roman*)]
\item Given that $\frac{3x + 4}{(1 + x)(2 + x)^2} \equiv \frac{A}{1 + x} + \frac{B}{2 + x} + \frac{C}{(2 + x)^2}$, find $A$, $B$ and $C$. [5]
\item Hence or otherwise expand $\frac{3x + 4}{(1 + x)(2 + x)^2}$ in ascending powers of $x$, up to and including the term in $x^2$. [5]
\item State the set of values of $x$ for which the expansion in part (ii) is valid. [1]
\end{enumerate}
\hfill \mbox{\textit{OCR C4 2005 Q8 [11]}}