| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2018 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Factoring out constants before expansion |
| Difficulty | Standard +0.3 This is a straightforward application of the binomial expansion with fractional powers. Students must factor out 125^(2/3) = 25, expand (1 - x/25)^(2/3), and substitute x=5 for the approximation. While it requires careful algebraic manipulation and understanding of the factoring technique, it follows a standard template with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(f(x) = (125-5x)^{\frac{2}{3}} = 125^{\frac{2}{3}}\left(1 - \frac{1}{25}x\right)^{\frac{2}{3}}\); \(125^{\frac{2}{3}}\) or \(25\) | B1 | For taking out factor of \(125^{\frac{2}{3}}\) or \(25\) |
| \(= 25\times\left[1 + \left(\frac{2}{3}\right)\left(-\frac{1}{25}x\right) + \frac{\left(\frac{2}{3}\right)\left(\frac{2}{3}-1\right)}{2}\left(-\frac{1}{25}x\right)^2 + \ldots\right]\) | M1, A1 | M1: Binomial expansion with \(n=\frac{2}{3}\) and term of \(\left(\pm\frac{1}{25}x\right)\), \(\left(\pm\frac{5}{125}x\right)\) or \(\left(\pm\left(\frac{1}{5}\right)^2 x\right)\); A1: Any unsimplified correct form |
| \(= 25 - \frac{2}{3}x - \frac{1}{225}x^2\ldots\) | A1 | Must be simplified; ignore extra terms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| Let \(x = 1\) | B1 | Must state \(x=1\) or explicitly seen |
| Evaluate \(= 25 - \frac{2}{3} - \frac{1}{225} = 24.32889\) | M1, A1 | M1: Substitute value of \(x\) consistently into their series; A1: cao \(24.32889\) — DO NOT ACCEPT AWRT |
# Question 2(a):
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $f(x) = (125-5x)^{\frac{2}{3}} = 125^{\frac{2}{3}}\left(1 - \frac{1}{25}x\right)^{\frac{2}{3}}$; $125^{\frac{2}{3}}$ or $25$ | B1 | For taking out factor of $125^{\frac{2}{3}}$ or $25$ |
| $= 25\times\left[1 + \left(\frac{2}{3}\right)\left(-\frac{1}{25}x\right) + \frac{\left(\frac{2}{3}\right)\left(\frac{2}{3}-1\right)}{2}\left(-\frac{1}{25}x\right)^2 + \ldots\right]$ | M1, A1 | M1: Binomial expansion with $n=\frac{2}{3}$ and term of $\left(\pm\frac{1}{25}x\right)$, $\left(\pm\frac{5}{125}x\right)$ or $\left(\pm\left(\frac{1}{5}\right)^2 x\right)$; A1: Any unsimplified correct form |
| $= 25 - \frac{2}{3}x - \frac{1}{225}x^2\ldots$ | A1 | Must be simplified; ignore extra terms |
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# Question 2(b):
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| Let $x = 1$ | B1 | Must state $x=1$ or explicitly seen |
| Evaluate $= 25 - \frac{2}{3} - \frac{1}{225} = 24.32889$ | M1, A1 | M1: Substitute value of $x$ consistently into their series; A1: cao $24.32889$ — DO NOT ACCEPT AWRT |
---2.
$$f ( x ) = ( 125 - 5 x ) ^ { \frac { 2 } { 3 } } \quad | x | < 25$$
\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 $x ^ { 2 }$, giving the coefficient of $x$ and the coefficient of $x ^ { 2 }$ as simplified fractions.
\item Use your expansion to find an approximate value for $120 ^ { \frac { 2 } { 3 } }$, stating the value of $x$ which you have used and showing your working. Give your answer to 5 decimal places.
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\hfill \mbox{\textit{Edexcel C34 2018 Q2 [7]}}