| 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 showing coefficient is zero |
| Difficulty | Standard +0.3 This is a standard C4 partial fractions question with a straightforward twist (showing A=0). Part (i) uses routine cover-up/substitution methods, while part (ii) applies standard binomial expansions with negative indices—both are textbook techniques requiring minimal problem-solving insight. Slightly above average difficulty only due to the two-part structure and algebraic manipulation 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 |
| \(\frac{3+2x^2}{(1+x)^2(1-4x)} = \frac{A}{1+x} + \frac{B}{(1+x)^2} + \frac{C}{1-4x}\) \(\Rightarrow 3+2x^2 = A(1+x)(1-4x)+B(1-4x)+C(1+x)^2\) | M1 | Clearing fractions (or any 2 correct equations) |
| \(x=-1 \Rightarrow 5 = 5B \Rightarrow B=1\) | B1 | \(B=1\) www |
| \(x=\frac{1}{4} \Rightarrow \frac{3\cdot 1}{8} = \frac{25}{16}C \Rightarrow C=2\) | B1 | \(C=2\) www |
| Coefficient of \(x^2\): \(2 = -4A + C \Rightarrow A = 0\) | E1 | \(A=0\) needs justification |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((1+x)^{-2} = 1+(-2)x+(-2)(-3)\frac{x^2}{2!}+\ldots\) \(= 1-2x+3x^2+\ldots\) | M1 | Binomial series (coefficients unsimplified) for either |
| A1 | \(1-2x+3x^2+\ldots\) | |
| \((1-4x)^{-1} = 1+(-1)(-4x)+(-1)(-2)\frac{(-4x)^2}{2!}+\ldots\) \(= 1+4x+16x^2+\ldots\) | A1 | \(1+4x+16x^2+\ldots\) |
| \(\frac{3+2x^2}{(1+x)^2(1-4x)} = (1+x)^{-2} + 2(1-4x)^{-1}\) \(\approx 1-2x+3x^2+2(1+4x+16x^2) = 3+6x+35x^2\) | A1ft | Their \(A,B,C\) and their expansions |
# Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{3+2x^2}{(1+x)^2(1-4x)} = \frac{A}{1+x} + \frac{B}{(1+x)^2} + \frac{C}{1-4x}$ $\Rightarrow 3+2x^2 = A(1+x)(1-4x)+B(1-4x)+C(1+x)^2$ | M1 | Clearing fractions (or any 2 correct equations) |
| $x=-1 \Rightarrow 5 = 5B \Rightarrow B=1$ | B1 | $B=1$ www |
| $x=\frac{1}{4} \Rightarrow \frac{3\cdot 1}{8} = \frac{25}{16}C \Rightarrow C=2$ | B1 | $C=2$ www |
| Coefficient of $x^2$: $2 = -4A + C \Rightarrow A = 0$ | E1 | $A=0$ needs justification |
**[4 marks total]**
# Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(1+x)^{-2} = 1+(-2)x+(-2)(-3)\frac{x^2}{2!}+\ldots$ $= 1-2x+3x^2+\ldots$ | M1 | Binomial series (coefficients unsimplified) for either |
| | A1 | $1-2x+3x^2+\ldots$ |
| $(1-4x)^{-1} = 1+(-1)(-4x)+(-1)(-2)\frac{(-4x)^2}{2!}+\ldots$ $= 1+4x+16x^2+\ldots$ | A1 | $1+4x+16x^2+\ldots$ |
| $\frac{3+2x^2}{(1+x)^2(1-4x)} = (1+x)^{-2} + 2(1-4x)^{-1}$ $\approx 1-2x+3x^2+2(1+4x+16x^2) = 3+6x+35x^2$ | A1ft | Their $A,B,C$ and their expansions |
**[4 marks total]**6 (i) Given that
$$\frac { 3 + 2 x ^ { 2 } } { ( 1 + x ) ^ { 2 } ( 1 - 4 x ) } = \frac { A } { 1 + x } + \frac { B } { ( 1 + x ) ^ { 2 } } + \frac { C } { 1 - 4 x }$$
where $A , B$ and $C$ are constants, find $B$ and $C$, and show that $A = 0$.\\
(ii) Given that $x$ is sufficiently small, find the first three terms of the binomial expansions of $( 1 + x ) ^ { - 2 }$ and $( 1 - 4 x ) ^ { - 1 }$.
Hence find the first three terms of the expansion of $\frac { 3 + 2 x ^ { 2 } } { ( 1 + x ) ^ { 2 } ( 1 - 4 x ) }$.
\hfill \mbox{\textit{OCR MEI C4 Q6 [8]}}