| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 8 |
| 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 straightforward two-part question combining standard partial fractions decomposition with binomial expansion. Part (i) is routine A-level technique, and part (ii) requires applying the binomial theorem to two simple terms and combining them—a standard C4 exercise. The validity range question is bookwork. Slightly above average difficulty only due to the multi-step nature and need to combine two expansions. |
| Spec | 1.02y Partial fractions: decompose rational functions1.04c Extend binomial expansion: rational n, |x|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{x}{(1+x)(1-2x)} = \frac{A}{1+x} + \frac{B}{1-2x}\) | ||
| \(\Rightarrow x = A(1-2x)+B(1+x)\) | M1 | expressing in partial fractions of correct form and attempting cover-up, substitution or equating coefficients. Condone a single sign error for M1 only. |
| \(x=\frac{1}{2} \Rightarrow \frac{1}{2} = B\cdot\frac{3}{2} \Rightarrow B=\frac{1}{3}\) | A1 | www cao |
| \(x=-1 \Rightarrow -1=3A \Rightarrow A=-\frac{1}{3}\) | A1 | www cao [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \frac{1}{3}\left[(1-2x)^{-1}-(1+x)^{-1}\right]\) | M1 | correct binomial coefficients throughout for first three terms of either \((1-2x)^{-1}\) or \((1+x)^{-1}\), i.e. \(1,(-1),(-1)(-2)/2\), not nCr form. Or correct simplified coefficients seen. |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \frac{1}{3}[1+2x+4x^2+\cdots]\) | A1 | \(1+2x+4x^2\) |
| \(-\frac{1}{3}[1-x+x^2+\cdots]\) | A1 | \(1-x+x^2\) (or \(\frac{1}{3}/-\frac{1}{3}\) of each expression, ft their \(A/B\)) |
| \(= \frac{1}{3}(3x+3x^2+\cdots) = x+x^2+\cdots\) so \(a=1\) and \(b=1\) | A1 | www cao |
| Valid for \(-\frac{1}{2} < x < \frac{1}{2}\) or \( | x | <\frac{1}{2}\) |
## Question 6(i):
$\frac{x}{(1+x)(1-2x)} = \frac{A}{1+x} + \frac{B}{1-2x}$ | |
$\Rightarrow x = A(1-2x)+B(1+x)$ | M1 | expressing in partial fractions of correct form and attempting cover-up, substitution or equating coefficients. Condone a single sign error for M1 only.
$x=\frac{1}{2} \Rightarrow \frac{1}{2} = B\cdot\frac{3}{2} \Rightarrow B=\frac{1}{3}$ | A1 | www cao
$x=-1 \Rightarrow -1=3A \Rightarrow A=-\frac{1}{3}$ | A1 | www cao [3] |
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## Question 6(ii):
$\frac{x}{(1+x)(1-2x)} = \frac{-1/3}{1+x} + \frac{1/3}{1-2x}$
$= \frac{1}{3}\left[(1-2x)^{-1}-(1+x)^{-1}\right]$ | M1 | correct binomial coefficients throughout for first three terms of either $(1-2x)^{-1}$ or $(1+x)^{-1}$, i.e. $1,(-1),(-1)(-2)/2$, not nCr form. Or correct simplified coefficients seen.
$= \frac{1}{3}\left[1+(-1)(-2x)+\frac{(-1)(-2)}{2}(-2x)^2+\cdots -\left(1+(-1)x+\frac{(-1)(-2)}{2}x^2+\cdots\right)\right]$
$= \frac{1}{3}[1+2x+4x^2+\cdots]$ | A1 | $1+2x+4x^2$
$-\frac{1}{3}[1-x+x^2+\cdots]$ | A1 | $1-x+x^2$ (or $\frac{1}{3}/-\frac{1}{3}$ of each expression, ft their $A/B$)
$= \frac{1}{3}(3x+3x^2+\cdots) = x+x^2+\cdots$ so $a=1$ and $b=1$ | A1 | www cao
Valid for $-\frac{1}{2} < x < \frac{1}{2}$ or $|x|<\frac{1}{2}$ | B1 | independent of expansion. Must combine as one overall range. [5] |
---6 (i) Express $\frac { x } { ( 1 + x ) ( 1 - 2 x ) }$ in partial fractions.\\
(ii) Hence use binomial expansions to show that $\frac { x } { ( 1 + x ) ( 1 - 2 x ) } = a x + b x ^ { 2 } + \ldots$, where $a$ and $b$ are\\
constants to be determined.
State the set of values of $x$ for which the expansion is valid.
\hfill \mbox{\textit{OCR MEI C4 Q6 [8]}}