| 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 coefficients. The validity range requires understanding convergence conditions for binomial series, which is standard C4 content. Slightly above average due to the multi-step nature and need to combine two topics, but no novel insight required. |
| Spec | 1.02y Partial fractions: decompose rational functions1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| 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 |
| \(x = \frac{1}{2} \Rightarrow \frac{1}{2} = B(1+\frac{1}{2}) \Rightarrow B = \frac{1}{3}\) | A1 | www cao |
| \(x = -1 \Rightarrow -1 = 3A \Rightarrow A = -\frac{1}{3}\) | A1 | www cao. Accept \(A/(1+x) + B/(1-2x)\), \(A=-\frac{1}{3}\), \(B=\frac{1}{3}\) as sufficient for full marks without needing to reassemble fractions with numerical numerators |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{x}{(1+x)(1-2x)} = \frac{-1/3}{1+x} + \frac{1/3}{1-2x}\) \(= \frac{1}{3}\left[(1-2x)^{-1} - (1+x)^{-1}\right]\) | ||
| \(= \frac{1}{3}\left[1+(-1)(-2x)+\frac{(-1)(-2)}{2}(-2x)^2+\ldots - (1+(-1)x + \frac{(-1)(-2)}{2}x^2+\ldots)\right]\) | M1 | Correct binomial coefficients throughout for first three terms of either \((1-2x)^{-1}\) or \((1+x)^{-1}\), ie \(1,(-1),\frac{(-1)(-2)}{2}\), not nCr form. Or correct simplified coefficients seen |
| \(= \frac{1}{3}[1+2x+4x^2+\ldots-(1-x+x^2+\ldots)]\) | A1 | \(1+2x+4x^2\) |
| A1 | \(1-x+x^2\) (or \(\frac{1}{3}/-\frac{1}{3}\) of each expression, ft their \(A/B\)) | |
| \(= \frac{1}{3}(3x+3x^2+\ldots) = x + x^2 + \ldots\) so \(a=1\) and \(b=1\) | A1 | www cao |
| OR: \(x(1-x-2x^2) = x(1-(x+2x^2))\) \(= x(1+x+2x^2+(-1)(-2)(x+2x^2)^2/2+\ldots)\) \(= x(1+x+2x^2+x^2\ldots)\) \(= x+x^2\ldots\) so \(a=1\) and \(b=1\) | M1, A2, A1 | Correct binomial coefficients for \((1-(x+2x^2))\), ie \(1,(-1)\). \(x(1+x)\) www. ww cao |
| Valid for \(-\frac{1}{2} < x < \frac{1}{2}\) or \(\ | x\ | < \frac{1}{2}\) |
# Question 3(i):
| 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 |
| $x = \frac{1}{2} \Rightarrow \frac{1}{2} = B(1+\frac{1}{2}) \Rightarrow B = \frac{1}{3}$ | A1 | www cao |
| $x = -1 \Rightarrow -1 = 3A \Rightarrow A = -\frac{1}{3}$ | A1 | www cao. Accept $A/(1+x) + B/(1-2x)$, $A=-\frac{1}{3}$, $B=\frac{1}{3}$ as sufficient for full marks without needing to reassemble fractions with numerical numerators |
**[3 marks total]**
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# Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{x}{(1+x)(1-2x)} = \frac{-1/3}{1+x} + \frac{1/3}{1-2x}$ $= \frac{1}{3}\left[(1-2x)^{-1} - (1+x)^{-1}\right]$ | | |
| $= \frac{1}{3}\left[1+(-1)(-2x)+\frac{(-1)(-2)}{2}(-2x)^2+\ldots - (1+(-1)x + \frac{(-1)(-2)}{2}x^2+\ldots)\right]$ | M1 | Correct binomial coefficients throughout for first three terms of either $(1-2x)^{-1}$ or $(1+x)^{-1}$, ie $1,(-1),\frac{(-1)(-2)}{2}$, not nCr form. Or correct simplified coefficients seen |
| $= \frac{1}{3}[1+2x+4x^2+\ldots-(1-x+x^2+\ldots)]$ | A1 | $1+2x+4x^2$ |
| | A1 | $1-x+x^2$ (or $\frac{1}{3}/-\frac{1}{3}$ of each expression, ft their $A/B$) |
| $= \frac{1}{3}(3x+3x^2+\ldots) = x + x^2 + \ldots$ so $a=1$ and $b=1$ | A1 | www cao |
| **OR:** $x(1-x-2x^2) = x(1-(x+2x^2))$ $= x(1+x+2x^2+(-1)(-2)(x+2x^2)^2/2+\ldots)$ $= x(1+x+2x^2+x^2\ldots)$ $= x+x^2\ldots$ so $a=1$ and $b=1$ | M1, A2, A1 | Correct binomial coefficients for $(1-(x+2x^2))$, ie $1,(-1)$. $x(1+x)$ www. ww 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. Condone $\leq$. Condone $-\frac{1}{2} < \|x\| < \frac{1}{2}$ but not $x < \frac{1}{2}$ or $-1 < 2x < 1$ or $-\frac{1}{2} > x > \frac{1}{2}$ |
**[5 marks total]**
---3 (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. constants to be determined.
State the set of values of $x$ for which the expansion is valid.
\hfill \mbox{\textit{OCR MEI C4 Q3 [8]}}