| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2006 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem and Partial Fractions |
| Type | Partial fractions then binomial expansion |
| Difficulty | Standard +0.3 This is a standard two-part C4 question combining routine partial fractions decomposition with straightforward binomial expansions. Part (i) involves solving a system of equations for constants A, B, C using standard techniques. Part (ii) requires expanding (2-x)^{-1} and (1+x)^{-1}, (1+x)^{-2} using the binomial theorem and collecting terms—all textbook procedures with no novel insight required. Slightly above average difficulty due to the repeated factor requiring careful handling and the algebraic manipulation involved. |
| Spec | 1.02y Partial fractions: decompose rational functions1.04c Extend binomial expansion: rational n, |x|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A = 3\) | B1 | For correct value stated |
| \(C = 1\) | B1 | For correct value stated |
| \(11+8x \equiv A(1+x)^2 + B(2-x)(1+x) + C(2-x)\) | M1 | AEF; any suitable identity |
| e.g. \(A-B=0\), \(2A+B-C=8\), \(A+2B+2C=11\) | A1 | For any correct (f.t.) equation involving \(B\) |
| \(B = 3\) | A1 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((1-\frac{x}{2})^{-1} = 1+\frac{x}{2}+\frac{x^2}{4}+\ldots\) | B1 | s.o.i. |
| \((1+x)^{-1} = 1-x+x^2-\ldots\) | B1 | s.o.i. |
| \((1+x)^{-2} = 1-2x + 3x^2-\ldots\) | B1,B1 | s.o.i. |
| Expansion \(= \frac{11}{2}-\frac{17}{4}x+\frac{51}{8}x^2+\ldots\) | B1 | 5 CAO. No f.t. for wrong \(A\) and/or \(B\) and/or \(C\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((11+8x)(2-x)^{-1}(1+x)^{-2}\) attempted | ||
| B1 for \((1-\frac{x}{2})^{-1} = 1+\frac{x}{2}+\frac{x^2}{4}+\ldots\) | B1 | |
| B1,B1 for \((1+x)^{-2} = 1-2x+\ldots+3x^2+\ldots\) | B1,B1 | |
| B1,B1 for \(\frac{11}{2}-\frac{17}{4}x+\ldots+\frac{51}{8}x^2+\ldots\) | B1,B1 |
# Question 7:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A = 3$ | B1 | For correct value stated |
| $C = 1$ | B1 | For correct value stated |
| $11+8x \equiv A(1+x)^2 + B(2-x)(1+x) + C(2-x)$ | M1 | AEF; any suitable identity |
| e.g. $A-B=0$, $2A+B-C=8$, $A+2B+2C=11$ | A1 | For any correct (f.t.) equation involving $B$ |
| $B = 3$ | A1 | **5** |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(1-\frac{x}{2})^{-1} = 1+\frac{x}{2}+\frac{x^2}{4}+\ldots$ | B1 | s.o.i. |
| $(1+x)^{-1} = 1-x+x^2-\ldots$ | B1 | s.o.i. |
| $(1+x)^{-2} = 1-2x + 3x^2-\ldots$ | B1,B1 | s.o.i. |
| Expansion $= \frac{11}{2}-\frac{17}{4}x+\frac{51}{8}x^2+\ldots$ | B1 | **5** CAO. No f.t. for wrong $A$ and/or $B$ and/or $C$ |
**SR(1)** If partial fractions not used but product of **SR(2)** attempted:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(11+8x)(2-x)^{-1}(1+x)^{-2}$ attempted | | |
| B1 for $(1-\frac{x}{2})^{-1} = 1+\frac{x}{2}+\frac{x^2}{4}+\ldots$ | B1 | |
| B1,B1 for $(1+x)^{-2} = 1-2x+\ldots+3x^2+\ldots$ | B1,B1 | |
| B1,B1 for $\frac{11}{2}-\frac{17}{4}x+\ldots+\frac{51}{8}x^2+\ldots$ | B1,B1 | |
*N.B. In both SR, if final expansion given B0, allow SR B1 for $22-17x+51/2\, x^2$*7 The expression $\frac { 11 + 8 x } { ( 2 - x ) ( 1 + x ) ^ { 2 } }$ is denoted by $\mathrm { f } ( x )$.\\
(i) Express $\mathrm { f } ( x )$ in the form $\frac { A } { 2 - x } + \frac { B } { 1 + x } + \frac { C } { ( 1 + x ) ^ { 2 } }$, where $A , B$ and $C$ are constants.\\
(ii) Given that $| x | < 1$, find the first 3 terms in the expansion of $\mathrm { f } ( x )$ in ascending powers of $x$.
\hfill \mbox{\textit{OCR C4 2006 Q7 [10]}}