| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2011 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Two unknowns from two coefficient conditions |
| Difficulty | Standard +0.3 This is a standard C4 binomial expansion question with a straightforward two-equation system to solve. Part (a) is routine application of the formula, while parts (b)-(c) require multiplying the expansion by (a+bx) and matching coefficients—mechanical algebra with no conceptual challenges beyond the standard syllabus. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \((2-3x)^{-2} = 2^{-2}\left(1-\frac{3}{2}x\right)^{-2}\) | B1 | |
| \(\left(1-\frac{3}{2}x\right)^{-2} = 1+(-2)\left(-\frac{3}{2}x\right)+\frac{-2\cdot-3}{1\cdot2}\left(-\frac{3}{2}x\right)^2+\frac{-2\cdot-3\cdot-4}{1\cdot2\cdot3}\left(-\frac{3}{2}x\right)^3+\ldots\) | M1 A1 | |
| \(= 1+3x+\frac{27}{4}x^2+\frac{27}{2}x^3+\ldots\) | ||
| \((2-3x)^{-2} = \frac{1}{4}+\frac{3}{4}x+\frac{27}{16}x^2+\frac{27}{8}x^3+\ldots\) | M1 A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(f(x)=(a+bx)\left(\frac{1}{4}+\frac{3}{4}x+\frac{27}{16}x^2+\frac{27}{8}x^3+\ldots\right)\) | ||
| Coefficient of \(x\): \(\frac{3a}{4}+\frac{b}{4}=0 \Rightarrow (3a+b=0)\) | M1 | |
| Coefficient of \(x^2\): \(\frac{27a}{16}+\frac{3b}{4}=\frac{9}{16} \Rightarrow (9a+4b=3)\) | M1 A1 | A1 either correct |
| Leading to \(a=-1,\ b=3\) | M1 A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| Coefficient of \(x^3\) is \(\frac{27a}{8}+\frac{27b}{16}=\frac{27}{8}\times(-1)+\frac{27}{16}\times3\) | M1 A1ft | |
| \(=\frac{27}{16}\) | A1 | cao, (3) |
## Question 5:
### Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| $(2-3x)^{-2} = 2^{-2}\left(1-\frac{3}{2}x\right)^{-2}$ | B1 | |
| $\left(1-\frac{3}{2}x\right)^{-2} = 1+(-2)\left(-\frac{3}{2}x\right)+\frac{-2\cdot-3}{1\cdot2}\left(-\frac{3}{2}x\right)^2+\frac{-2\cdot-3\cdot-4}{1\cdot2\cdot3}\left(-\frac{3}{2}x\right)^3+\ldots$ | M1 A1 | |
| $= 1+3x+\frac{27}{4}x^2+\frac{27}{2}x^3+\ldots$ | | |
| $(2-3x)^{-2} = \frac{1}{4}+\frac{3}{4}x+\frac{27}{16}x^2+\frac{27}{8}x^3+\ldots$ | M1 A1 | (5) |
### Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| $f(x)=(a+bx)\left(\frac{1}{4}+\frac{3}{4}x+\frac{27}{16}x^2+\frac{27}{8}x^3+\ldots\right)$ | | |
| Coefficient of $x$: $\frac{3a}{4}+\frac{b}{4}=0 \Rightarrow (3a+b=0)$ | M1 | |
| Coefficient of $x^2$: $\frac{27a}{16}+\frac{3b}{4}=\frac{9}{16} \Rightarrow (9a+4b=3)$ | M1 A1 | A1 either correct |
| Leading to $a=-1,\ b=3$ | M1 A1 | (5) |
### Part (c):
| Working | Marks | Notes |
|---------|-------|-------|
| Coefficient of $x^3$ is $\frac{27a}{8}+\frac{27b}{16}=\frac{27}{8}\times(-1)+\frac{27}{16}\times3$ | M1 A1ft | |
| $=\frac{27}{16}$ | A1 | cao, (3) |
---\begin{enumerate}
\item (a) Use the binomial theorem to expand
\end{enumerate}
$$( 2 - 3 x ) ^ { - 2 } , \quad | x | < \frac { 2 } { 3 }$$
in ascending powers of $x$, up to and including the term in $x ^ { 3 }$. Give each coefficient as a simplified fraction.
$$\mathrm { f } ( x ) = \frac { a + b x } { ( 2 - 3 x ) ^ { 2 } } , \quad | x | < \frac { 2 } { 3 } , \quad \text { where } a \text { and } b \text { are constants. }$$
In the binomial expansion of $\mathrm { f } ( x )$, in ascending powers of $x$, the coefficient of $x$ is 0 and the coefficient of $x ^ { 2 }$ is $\frac { 9 } { 16 }$. Find\\
(b) the value of $a$ and the value of $b$,\\
(c) the coefficient of $x ^ { 3 }$, giving your answer as a simplified fraction.\\
\hfill \mbox{\textit{Edexcel C4 2011 Q5 [13]}}