| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Substitute expression for variable |
| Difficulty | Standard +0.3 This is a standard C4 binomial expansion question with routine substitutions. Part (a) requires factoring out constants and applying the binomial theorem with n=-1/2, which is textbook procedure. Parts (b) and (c) involve simple substitutions (x→-x and x→2x) into the result from part (a), requiring minimal additional work beyond pattern recognition. The question is slightly easier than average due to its mechanical nature and clear scaffolding. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x) = \ldots(\ldots - \ldots x)^{-\frac{1}{2}}\) | M1 | |
| \(= 6 \times 9^{-\frac{1}{2}}(\ldots)\) | B1 | \(\frac{6}{9^{\frac{1}{2}}}, \frac{6}{3}, 2\) or equivalent |
| \(= \ldots\left(1 + \left(-\tfrac{1}{2}\right)(kx) + \frac{(-\frac{1}{2})(-\frac{3}{2})}{2}(kx)^2 + \frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})}{3!}(kx)^3 + \ldots\right)\) | M1; A1ft | |
| \(= 2\left(1 + \frac{2}{9}x + \ldots\right)\) or \(2 + \frac{4}{9}x\) | A1 | |
| \(= 2 + \frac{4}{9}x + \frac{4}{27}x^2 + \frac{40}{729}x^3 + \ldots\) | A1 (6) |
| Answer | Marks |
|---|---|
| \(g(x) = 2 - \frac{4}{9}x + \frac{4}{27}x^2 - \frac{40}{729}x^3 + \ldots\) | B1ft (1) |
| Answer | Marks |
|---|---|
| \(h(x) = 2 + \frac{4}{9}(2x) + \frac{4}{27}(2x)^2 + \frac{40}{729}(2x)^3 + \ldots\) | M1 A1 (2) |
| \(= 2 + \frac{8}{9}x + \frac{16}{27}x^2 + \frac{320}{729}x^3 + \ldots\) |
## Question 3:
### Part (a)
| $f(x) = \ldots(\ldots - \ldots x)^{-\frac{1}{2}}$ | M1 | |
| $= 6 \times 9^{-\frac{1}{2}}(\ldots)$ | B1 | $\frac{6}{9^{\frac{1}{2}}}, \frac{6}{3}, 2$ or equivalent |
| $= \ldots\left(1 + \left(-\tfrac{1}{2}\right)(kx) + \frac{(-\frac{1}{2})(-\frac{3}{2})}{2}(kx)^2 + \frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})}{3!}(kx)^3 + \ldots\right)$ | M1; A1ft | |
| $= 2\left(1 + \frac{2}{9}x + \ldots\right)$ or $2 + \frac{4}{9}x$ | A1 | |
| $= 2 + \frac{4}{9}x + \frac{4}{27}x^2 + \frac{40}{729}x^3 + \ldots$ | A1 (6) | |
### Part (b)
| $g(x) = 2 - \frac{4}{9}x + \frac{4}{27}x^2 - \frac{40}{729}x^3 + \ldots$ | B1ft (1) | |
### Part (c)
| $h(x) = 2 + \frac{4}{9}(2x) + \frac{4}{27}(2x)^2 + \frac{40}{729}(2x)^3 + \ldots$ | M1 A1 (2) | |
| $= 2 + \frac{8}{9}x + \frac{16}{27}x^2 + \frac{320}{729}x^3 + \ldots$ | | |
---3.
$$f ( x ) = \frac { 6 } { \sqrt { ( 9 - 4 x ) } } , \quad | x | < \frac { 9 } { 4 }$$
\begin{enumerate}[label=(\alph*)]
\item Find the binomial expansion of $\mathrm { f } ( x )$ in ascending powers of $x$, up to and including the term in $x ^ { 3 }$. Give each coefficient in its simplest form.\\
(6)
Use your answer to part (a) to find the binomial expansion in ascending powers of $x$, up to and including the term in $x ^ { 3 }$, of
\item $\quad \mathrm { g } ( x ) = \frac { 6 } { \sqrt { } ( 9 + 4 x ) } , \quad | x | < \frac { 9 } { 4 }$
\item $\mathrm { h } ( x ) = \frac { 6 } { \sqrt { } ( 9 - 8 x ) } , \quad | x | < \frac { 9 } { 8 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2012 Q3 [9]}}