| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2006 |
| Session | January |
| Marks | 11 |
| 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 routine twist (showing B=0). Part (a) uses cover-up method or substitution, requiring algebraic manipulation but no novel insight. Part (b) applies binomial expansion to two terms—straightforward but involves careful arithmetic across multiple terms. Slightly easier than average due to the simplification when B=0. |
| Spec | 1.02y Partial fractions: decompose rational functions1.04c Extend binomial expansion: rational n, |x|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Considers \(3x^2+16 = A(2+x)^2 + B(1-3x)(2+x) + C(1-3x)\) and substitutes \(x=-2\) or \(x=\frac{1}{3}\), or compares coefficients | M1 | |
| \(A = 3\) and \(C = 4\) | A1, A1 | |
| Compares coefficients or uses simultaneous equation to show \(B = 0\) | B1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Writes \(3(1-3x)^{-1} + 4(2+x)^{-2}\) | M1 | |
| \(= 3(1 + 3x + 9x^2 + 27x^3 + \ldots) +\) | (M1, A1) | |
| \(\frac{4}{4}\left(1 + \frac{(-2)}{1}\left(\frac{x}{2}\right) + \frac{(-2)(-3)}{1.2}\left(\frac{x}{2}\right)^2 + \frac{(-2)(-3)(-4)}{1.2.3}\left(\frac{x}{2}\right)^3 + \ldots\right)\) | (M1, A1) | |
| \(= 4 + 8x + 27\tfrac{3}{4}x^2 + 80\tfrac{1}{2}x^3 + \ldots\) | A1, A1 | (7) [11] |
| Or uses \((3x^2+16)(1-3x)^{-1}(2+x)^{-2}\): \((3x^2+16)(1+3x+9x^2+27x^3+\ldots) \times \frac{1}{4}\left(1+\frac{(-2)}{1}\left(\frac{x}{2}\right)+\ldots\right)\) | M1, (M1A1)\(\times\), (M1A1) | |
| \(= 4 + 8x + 27\tfrac{3}{4}x^2 + 80\tfrac{1}{2}x^3 + \ldots\) | A1, A1 | (7) |
## Question 5:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Considers $3x^2+16 = A(2+x)^2 + B(1-3x)(2+x) + C(1-3x)$ and substitutes $x=-2$ or $x=\frac{1}{3}$, or compares coefficients | M1 | |
| $A = 3$ and $C = 4$ | A1, A1 | |
| Compares coefficients or uses simultaneous equation to show $B = 0$ | B1 | **(4)** |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Writes $3(1-3x)^{-1} + 4(2+x)^{-2}$ | M1 | |
| $= 3(1 + 3x + 9x^2 + 27x^3 + \ldots) +$ | (M1, A1) | |
| $\frac{4}{4}\left(1 + \frac{(-2)}{1}\left(\frac{x}{2}\right) + \frac{(-2)(-3)}{1.2}\left(\frac{x}{2}\right)^2 + \frac{(-2)(-3)(-4)}{1.2.3}\left(\frac{x}{2}\right)^3 + \ldots\right)$ | (M1, A1) | |
| $= 4 + 8x + 27\tfrac{3}{4}x^2 + 80\tfrac{1}{2}x^3 + \ldots$ | A1, A1 | **(7) [11]** |
| **Or** uses $(3x^2+16)(1-3x)^{-1}(2+x)^{-2}$: $(3x^2+16)(1+3x+9x^2+27x^3+\ldots) \times \frac{1}{4}\left(1+\frac{(-2)}{1}\left(\frac{x}{2}\right)+\ldots\right)$ | M1, (M1A1)$\times$, (M1A1) | |
| $= 4 + 8x + 27\tfrac{3}{4}x^2 + 80\tfrac{1}{2}x^3 + \ldots$ | A1, A1 | **(7)** |
---5.
$$f ( x ) = \frac { 3 x ^ { 2 } + 16 } { ( 1 - 3 x ) ( 2 + x ) ^ { 2 } } = \frac { A } { ( 1 - 3 x ) } + \frac { B } { ( 2 + x ) } + \frac { C } { ( 2 + x ) ^ { 2 } } , \quad | x | < \frac { 1 } { 3 } .$$
\begin{enumerate}[label=(\alph*)]
\item Find the values of $A$ and $C$ and show that $B = 0$.
\item Hence, or otherwise, find the series expansion of $\mathrm { f } ( x )$, in ascending powers of $x$, up to and including the term in $x ^ { 3 }$. Simplify each term.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2006 Q5 [11]}}