| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2015 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Multiply by polynomial |
| Difficulty | Standard +0.3 This is a standard binomial expansion question requiring routine application of the formula for (1-2x)^(-1/2), multiplication by a simple polynomial (2+3x), and a straightforward substitution. While it involves multiple steps, each step follows a well-practiced procedure with no novel insight 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 |
| \(\frac{1}{\sqrt{1-2x}} = (1-2x)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)(-2x) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}(-2x)^2\) | M1A1 | Uses correct binomial expansion with \(n = \pm\frac{1}{2}\) and '\(x\)' \(= \pm 2x\); fully correct unsimplified expression |
| \(= 1 + x + \frac{3}{2}x^2 + \ldots\) | Must show intermediate working for \(x^2\) term | |
| \(\frac{2+3x}{\sqrt{1-2x}} = (2+3x)(1+x+\frac{3}{2}x^2+\ldots)\) | M1 | Attempt to multiply quadratic binomial expansion by \((2+3x)\) |
| \(= 2 + 5x + 6x^2\) | A1* | Correct solution only; given answer, all aspects correct including bracketing |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Sub \(x = \frac{1}{20}\) into both sides of \(\frac{2+3x}{\sqrt{1-2x}} = 2+5x+6x^2\) | M1 | Condone missing brackets; accept equivalent e.g. \(\frac{2+3\times0.05}{\sqrt{1-2\times0.05}} = 2.265\) |
| \(\frac{43\sqrt{10}}{60} = \frac{453}{200}\) or \(\frac{43}{20} = \frac{453}{200} \times \frac{3}{\sqrt{10}}\) | dM1 | Attempt to simplify both sides resulting in expression involving \(\sqrt{10}\); sight of \(\frac{43\sqrt{10}}{60}\) and \(\frac{453}{200}\) |
| \(\sqrt{10} = \frac{1359}{430}\) (Accept \(\frac{4300}{1359}\)) | A1 | Accept \(\sqrt{10} = \frac{1359}{430} = 3\frac{69}{430}\) or rationalised form \(\sqrt{10} = \frac{4300}{1359} = 3\frac{223}{1359}\) |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{\sqrt{1-2x}} = (1-2x)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)(-2x) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}(-2x)^2$ | M1A1 | Uses correct binomial expansion with $n = \pm\frac{1}{2}$ and '$x$' $= \pm 2x$; fully correct unsimplified expression |
| $= 1 + x + \frac{3}{2}x^2 + \ldots$ | | Must show intermediate working for $x^2$ term |
| $\frac{2+3x}{\sqrt{1-2x}} = (2+3x)(1+x+\frac{3}{2}x^2+\ldots)$ | M1 | Attempt to multiply quadratic binomial expansion by $(2+3x)$ |
| $= 2 + 5x + 6x^2$ | A1* | Correct solution only; given answer, all aspects correct including bracketing |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sub $x = \frac{1}{20}$ into both sides of $\frac{2+3x}{\sqrt{1-2x}} = 2+5x+6x^2$ | M1 | Condone missing brackets; accept equivalent e.g. $\frac{2+3\times0.05}{\sqrt{1-2\times0.05}} = 2.265$ |
| $\frac{43\sqrt{10}}{60} = \frac{453}{200}$ or $\frac{43}{20} = \frac{453}{200} \times \frac{3}{\sqrt{10}}$ | dM1 | Attempt to simplify both sides resulting in expression involving $\sqrt{10}$; sight of $\frac{43\sqrt{10}}{60}$ and $\frac{453}{200}$ |
| $\sqrt{10} = \frac{1359}{430}$ (Accept $\frac{4300}{1359}$) | A1 | Accept $\sqrt{10} = \frac{1359}{430} = 3\frac{69}{430}$ or rationalised form $\sqrt{10} = \frac{4300}{1359} = 3\frac{223}{1359}$ |
---\begin{enumerate}
\item (a) Use the binomial expansion, in ascending powers of $x$, of $\frac { 1 } { \sqrt { } ( 1 - 2 x ) }$ to show that
\end{enumerate}
$$\frac { 2 + 3 x } { \sqrt { } ( 1 - 2 x ) } \approx 2 + 5 x + 6 x ^ { 2 } , \quad | x | < 0.5$$
(b) Substitute $x = \frac { 1 } { 20 }$ into
$$\frac { 2 + 3 x } { \sqrt { } ( 1 - 2 x ) } = 2 + 5 x + 6 x ^ { 2 }$$
to obtain an approximation to $\sqrt { 10 }$\\
Give your answer as a fraction in its simplest form.\\
\hfill \mbox{\textit{Edexcel C34 2015 Q5 [7]}}