| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Direct single expansion substitution |
| Difficulty | Moderate -0.8 This is a straightforward application of the binomial expansion formula for fractional powers with standard substitution. Part (a) requires routine use of (a+bx)^n expansion with n=1/2, part (b)(i) is direct substitution, and part (b)(ii) tests basic understanding of validity condition |x|<1. All steps are mechanical with no problem-solving insight required, making it easier than average. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((4+5x)^{\frac{1}{2}} = (4)^{\frac{1}{2}}\left(1+\frac{5x}{4}\right)^{\frac{1}{2}} = 2\left(1+\frac{5x}{4}\right)^{\frac{1}{2}}\) | B1 | Manipulates \((4+5x)^{\frac{1}{2}}\) by taking out factor of \((4)^{\frac{1}{2}}\) or 2 |
| \(=\{2\}\left[1+\left(\frac{1}{2}\right)\left(\frac{5x}{4}\right)+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}\left(\frac{5x}{4}\right)^2+\ldots\right]\) | M1 | Expands \((\ldots+\lambda x)^{\frac{1}{2}}\) to give at least 2 terms, where \(\lambda \neq 1\) |
| Correct simplified or unsimplified expansion with consistent \((\lambda x)\) | A1ft | Follow through on consistent \(\lambda x\) |
| \(= 2+\frac{5}{4}x - \frac{25}{64}x^2+\ldots\) | A1 | Fully correct; \(k=-\frac{25}{64}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=\frac{1}{10} \Rightarrow (4+5(0.1))^{\frac{1}{2}}\) | M1 | Attempts to substitute \(x=\frac{1}{10}\) or \(0.1\) into \((4+5x)^{\frac{1}{2}}\) |
| \(=\sqrt{4.5}=\frac{3}{2}\sqrt{2}\) or \(\frac{3}{\sqrt{2}}\) | ||
| \(\frac{3}{2}\sqrt{2}\) or \(1.5\sqrt{2}\) or \(\frac{3}{\sqrt{2}} = 2+\frac{5}{4}\left(\frac{1}{10}\right)-\frac{25}{64}\left(\frac{1}{10}\right)^2+\ldots\ \{=2.121\ldots\}\) \(\Rightarrow \frac{3}{2}\sqrt{2}=\frac{543}{256}\) or \(\frac{3}{\sqrt{2}}=\frac{543}{256} \Rightarrow \sqrt{2}=\ldots\) | M1 | Complete method of finding approximate value for \(\sqrt{2}\) by substituting and equating to form \(\alpha\sqrt{2}\) or \(\frac{\beta}{\sqrt{2}}\), then rearranging |
| \(\sqrt{2}=\frac{181}{128}\) or \(\sqrt{2}=\frac{256}{181}\) | A1 | Also allow equivalent fractions e.g. \(\frac{362}{256}\), \(\frac{543}{384}\), \(\frac{256}{181}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=\frac{1}{10}\) satisfies \( | x | <\frac{4}{5}\), so the approximation is valid |
## Question 2:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(4+5x)^{\frac{1}{2}} = (4)^{\frac{1}{2}}\left(1+\frac{5x}{4}\right)^{\frac{1}{2}} = 2\left(1+\frac{5x}{4}\right)^{\frac{1}{2}}$ | B1 | Manipulates $(4+5x)^{\frac{1}{2}}$ by taking out factor of $(4)^{\frac{1}{2}}$ or 2 |
| $=\{2\}\left[1+\left(\frac{1}{2}\right)\left(\frac{5x}{4}\right)+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}\left(\frac{5x}{4}\right)^2+\ldots\right]$ | M1 | Expands $(\ldots+\lambda x)^{\frac{1}{2}}$ to give at least 2 terms, where $\lambda \neq 1$ |
| Correct simplified or unsimplified expansion with consistent $(\lambda x)$ | A1ft | Follow through on consistent $\lambda x$ |
| $= 2+\frac{5}{4}x - \frac{25}{64}x^2+\ldots$ | A1 | Fully correct; $k=-\frac{25}{64}$ |
### Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=\frac{1}{10} \Rightarrow (4+5(0.1))^{\frac{1}{2}}$ | M1 | Attempts to substitute $x=\frac{1}{10}$ or $0.1$ into $(4+5x)^{\frac{1}{2}}$ |
| $=\sqrt{4.5}=\frac{3}{2}\sqrt{2}$ or $\frac{3}{\sqrt{2}}$ | | |
| $\frac{3}{2}\sqrt{2}$ or $1.5\sqrt{2}$ or $\frac{3}{\sqrt{2}} = 2+\frac{5}{4}\left(\frac{1}{10}\right)-\frac{25}{64}\left(\frac{1}{10}\right)^2+\ldots\ \{=2.121\ldots\}$ $\Rightarrow \frac{3}{2}\sqrt{2}=\frac{543}{256}$ or $\frac{3}{\sqrt{2}}=\frac{543}{256} \Rightarrow \sqrt{2}=\ldots$ | M1 | Complete method of finding approximate value for $\sqrt{2}$ by substituting and equating to form $\alpha\sqrt{2}$ or $\frac{\beta}{\sqrt{2}}$, then rearranging |
| $\sqrt{2}=\frac{181}{128}$ or $\sqrt{2}=\frac{256}{181}$ | A1 | Also allow equivalent fractions e.g. $\frac{362}{256}$, $\frac{543}{384}$, $\frac{256}{181}$ |
### Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=\frac{1}{10}$ satisfies $|x|<\frac{4}{5}$, so the approximation is valid | B1 | Must explain that $x=\frac{1}{10}$ satisfies $|x|<\frac{4}{5}$ |
---\begin{enumerate}
\item (a) Show that the binomial expansion of
\end{enumerate}
$$( 4 + 5 x ) ^ { \frac { 1 } { 2 } }$$
in ascending powers of $x$, up to and including the term in $x ^ { 2 }$ is
$$2 + \frac { 5 } { 4 } x + k x ^ { 2 }$$
giving the value of the constant $k$ as a simplified fraction.\\
(b) (i) Use the expansion from part (a), with $x = \frac { 1 } { 10 }$, to find an approximate value for $\sqrt { 2 }$ Give your answer in the form $\frac { p } { q }$ where $p$ and $q$ are integers.\\
(ii) Explain why substituting $x = \frac { 1 } { 10 }$ into this binomial expansion leads to a valid approximation.
\hfill \mbox{\textit{Edexcel Paper 1 Q2 [8]}}