| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Factoring out constants before expansion |
| Difficulty | Moderate -0.3 This is a standard C4 binomial expansion question requiring routine application of the formula (1+x)^n with fractional index. Part (a) involves straightforward substitution into the binomial formula, part (b) is simple arithmetic simplification, and part (c) connects the expansion to an approximation. While it requires careful algebraic manipulation and understanding of validity conditions, it follows a well-practiced template with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02b Surds: manipulation and rationalising denominators1.04c Extend binomial expansion: rational n, |x|<1 |
| Answer | Marks |
|---|---|
| \((4)^{\frac{1}{2}}\) or \(2\) | B1 |
| Expands \((\ldots + kx)^{\frac{1}{2}}\) to give any 2 terms out of 3 terms simplified or un-simplified, e.g: \(1+\left(\frac{1}{2}\right)(kx)\) or \(\left(\frac{1}{2}\right)(kx)+\frac{(\frac{1}{2})(-\frac{1}{2})}{2!}(kx)^2\) or \(1 + \ldots + \frac{(\frac{1}{2})(-\frac{1}{2})}{2!}(kx)^2\) where \(k\) is a numerical value and where \(k \neq 1\) | M1 A1 ft |
| A correct simplified or un-simplified \(1+\left(\frac{1}{2}\right)(kx)+\frac{(\frac{1}{2})(-\frac{1}{2})}{2!}(kx)^2\) expansion with consistent \((kx)\) | A1 |
| Award B1M1A0 for \(2\left[1+\left(\frac{1}{2}\right)\left(\frac{5x}{4}\right)+\frac{(\frac{1}{2})(-\frac{1}{2})}{2!}\left(\frac{5x}{4}\right)^2+\ldots\right]\) because \((kx)\) is not consistent | Note |
| Incorrect bracketing: \(2\left[1+\left(\frac{1}{2}\right)\frac{5x}{4}+\frac{(\frac{1}{2})(-\frac{1}{2})}{2!}\frac{5x}{4}+\ldots\right]\) is B1M1A0 unless recovered | Note |
| \(2+\frac{5}{4}x\) (simplified fractions) or allow \(2+1.25x\) or \(2+1\frac{1}{4}x\) | A1 |
| Accept only \(-\frac{25}{64}x^2\) or \(-0.390625x^2\) | A1 |
| If a candidate would otherwise score \(2^{\text{A0}}, 3^{\text{A0}}\) then allow Special Case \(2^{\text{M1}}\) A1 for either | SC |
| \(2 + \frac{5}{8}x; \ldots\) or SC: \(2 + 1 + \ldots - \frac{25}{128}x^2 + \ldots\) or SC: \(\lambda\left[1+\frac{5}{8}x - \frac{25}{128}x^2+\ldots\right]\) or SC: \(\lambda + \frac{52}{8}x - \frac{25\lambda}{128}x^2+\ldots\) (where \(\lambda\) can be 1 or omitted), where each term in the \([\ldots]\) is a simplified fraction or a decimal | |
| OR SC: for \(2 + \frac{10}{8}x - \frac{50}{128}x^2 + \ldots\) (i.e. for not simplifying their correct coefficients.) | |
| Candidates who write \(2\left[1+\left(\frac{1}{2}\right)\left(\frac{5x}{4}\right)+\frac{(\frac{1}{2})(-\frac{1}{2})}{2!}\left(\frac{5x}{4}\right)^2+\ldots\right]\) where \(k=-\frac{5}{4}\) and not \(\frac{5}{4}\) and achieve \(2-\frac{5}{4}x-\frac{25}{64}x^2+\ldots\) will get B1M1A1A0A1 | Note |
| Ignore extra terms beyond the term in \(x^2\) | Note |
| You can ignore subsequent working following a correct answer | Note |
| Answer | Marks |
|---|---|
| \(\frac{3}{2}\sqrt{2}\) or \(1.5\sqrt{2}\) or \(k = \frac{3}{2}\) or \(1.5\) o.e. (Ignore how \(k = \frac{3}{2}\) is found.) | B1 |
| \(\frac{3}{2}\sqrt{2}\) or \(1.5\sqrt{2}\) or \(\frac{3}{\sqrt{2}}\) | B1 |
| \(= 2 + \frac{5}{4}\left(\frac{1}{10}\right) - \frac{25}{64}\left(\frac{1}{10}\right)^2 + \ldots (\text{or} = 2.121\ldots)\) | M1 |
| \(\frac{3}{2}\sqrt{2}\) or \(1.5\sqrt{2}\) or \(\frac{3}{\sqrt{2}} = 2 + \frac{5}{4}\left(\frac{1}{10}\right) - \frac{25}{64}\left(\frac{1}{10}\right)^2 + \ldots\) | |
| So, \(\frac{3}{2}\sqrt{2} = \frac{543}{256}\) or \(\frac{3}{\sqrt{2}} = \frac{543}{256}\) | |
| yields, \(\sqrt{2} = \frac{181}{128}\) or \(\sqrt{2} = \frac{256}{181}\) | A1 oe |
## Part (a)
$(4)^{\frac{1}{2}}$ or $2$ | B1 |
Expands $(\ldots + kx)^{\frac{1}{2}}$ to give any 2 terms out of 3 terms simplified or un-simplified, e.g: $1+\left(\frac{1}{2}\right)(kx)$ or $\left(\frac{1}{2}\right)(kx)+\frac{(\frac{1}{2})(-\frac{1}{2})}{2!}(kx)^2$ or $1 + \ldots + \frac{(\frac{1}{2})(-\frac{1}{2})}{2!}(kx)^2$ where $k$ is a numerical value and where $k \neq 1$ | M1 A1 ft |
A correct simplified or un-simplified $1+\left(\frac{1}{2}\right)(kx)+\frac{(\frac{1}{2})(-\frac{1}{2})}{2!}(kx)^2$ expansion with **consistent** $(kx)$ | A1 |
Award B1M1A0 for $2\left[1+\left(\frac{1}{2}\right)\left(\frac{5x}{4}\right)+\frac{(\frac{1}{2})(-\frac{1}{2})}{2!}\left(\frac{5x}{4}\right)^2+\ldots\right]$ because $(kx)$ is not consistent | Note |
Incorrect bracketing: $2\left[1+\left(\frac{1}{2}\right)\frac{5x}{4}+\frac{(\frac{1}{2})(-\frac{1}{2})}{2!}\frac{5x}{4}+\ldots\right]$ is B1M1A0 unless recovered | Note |
$2+\frac{5}{4}x$ (simplified fractions) or allow $2+1.25x$ or $2+1\frac{1}{4}x$ | A1 |
Accept only $-\frac{25}{64}x^2$ or $-0.390625x^2$ | A1 |
If a candidate **would otherwise score** $2^{\text{A0}}, 3^{\text{A0}}$ then allow Special Case $2^{\text{M1}}$ A1 for either | SC |
$2 + \frac{5}{8}x; \ldots$ or SC: $2 + 1 + \ldots - \frac{25}{128}x^2 + \ldots$ or SC: $\lambda\left[1+\frac{5}{8}x - \frac{25}{128}x^2+\ldots\right]$ or SC: $\lambda + \frac{52}{8}x - \frac{25\lambda}{128}x^2+\ldots$ (where $\lambda$ can be 1 or omitted), where each term in the $[\ldots]$ is a simplified fraction or a decimal | |
OR SC: for $2 + \frac{10}{8}x - \frac{50}{128}x^2 + \ldots$ (i.e. for not simplifying their correct coefficients.) | |
Candidates who write $2\left[1+\left(\frac{1}{2}\right)\left(\frac{5x}{4}\right)+\frac{(\frac{1}{2})(-\frac{1}{2})}{2!}\left(\frac{5x}{4}\right)^2+\ldots\right]$ where $k=-\frac{5}{4}$ and not $\frac{5}{4}$ and achieve $2-\frac{5}{4}x-\frac{25}{64}x^2+\ldots$ will get B1M1A1A0A1 | Note |
Ignore extra terms beyond the term in $x^2$ | Note |
You can ignore subsequent working following a correct answer | Note |
## Part (b)
$\frac{3}{2}\sqrt{2}$ or $1.5\sqrt{2}$ or $k = \frac{3}{2}$ or $1.5$ o.e. (Ignore how $k = \frac{3}{2}$ is found.) | B1 |
$\frac{3}{2}\sqrt{2}$ or $1.5\sqrt{2}$ or $\frac{3}{\sqrt{2}}$ | B1 |
$= 2 + \frac{5}{4}\left(\frac{1}{10}\right) - \frac{25}{64}\left(\frac{1}{10}\right)^2 + \ldots (\text{or} = 2.121\ldots)$ | M1 |
$\frac{3}{2}\sqrt{2}$ or $1.5\sqrt{2}$ or $\frac{3}{\sqrt{2}} = 2 + \frac{5}{4}\left(\frac{1}{10}\right) - \frac{25}{64}\left(\frac{1}{10}\right)^2 + \ldots$ | |
So, $\frac{3}{2}\sqrt{2} = \frac{543}{256}$ or $\frac{3}{\sqrt{2}} = \frac{543}{256}$ | |
yields, $\sqrt{2} = \frac{181}{128}$ or $\sqrt{2} = \frac{256}{181}$ | A1 oe |
---\begin{enumerate}[label=(\alph*)]
\item Find the binomial expansion of
$$(4 + 5x)^{\frac{1}{2}}, \quad |x| < \frac{4}{5}$$
in ascending powers of $x$, up to and including the term in $x^2$.
Give each coefficient in its simplest form.
[5]
\item Find the exact value of $(4 + 5x)^{\frac{1}{2}}$ when $x = \frac{1}{10}$
Give your answer in the form $k\sqrt{2}$, where $k$ is a constant to be determined.
[1]
\item Substitute $x = \frac{1}{10}$ into your binomial expansion from part (a) and hence find an approximate value for $\sqrt{2}$
Give your answer in the form $\frac{p}{q}$ where $p$ and $q$ are integers.
[2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2015 Q1 [8]}}