| Exam Board | WJEC |
|---|---|
| Module | Unit 3 (Unit 3) |
| Session | Specimen |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Direct single expansion substitution |
| Difficulty | Moderate -0.8 This is a straightforward binomial expansion question requiring standard application of the formula for negative/fractional powers, followed by simple substitution and arithmetic. Both parts are routine textbook exercises with no problem-solving or insight required—easier than average A-level questions. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| \((1-x)^{-\frac{1}{2}} = 1 + \frac{x}{2} + \left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\frac{x^2}{2} + ...\) | B1 | |
| \(= 1 + \frac{x}{2} + \frac{3x^2}{8} + ...\) | B1 | |
| Valid for \( | x | < 1\) |
| When \(x = \frac{1}{10}\): \(\left(\frac{9}{10}\right)^{\frac{1}{2}} \approx 1 + \frac{1}{20} + \frac{3}{800} = \frac{843}{800}\) | B1 | |
| So that \((10)^{\frac{1}{2}} = 3 \times \frac{843}{800} = \frac{2529}{800}\) | B1 |
$(1-x)^{-\frac{1}{2}} = 1 + \frac{x}{2} + \left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\frac{x^2}{2} + ...$ | B1 |
$= 1 + \frac{x}{2} + \frac{3x^2}{8} + ...$ | B1 |
Valid for $|x| < 1$ | B1 |
When $x = \frac{1}{10}$: $\left(\frac{9}{10}\right)^{\frac{1}{2}} \approx 1 + \frac{1}{20} + \frac{3}{800} = \frac{843}{800}$ | B1 |
So that $(10)^{\frac{1}{2}} = 3 \times \frac{843}{800} = \frac{2529}{800}$ | B1 |
**Total: [4]**\begin{enumerate}[label=(\alph*)]
\item Expand $(1-x)^{-\frac{1}{2}}$ in ascending power of $x$ as far as the term in $x^2$. State the range of $x$ for which the expansion is valid. [2]
\item By taking $x = \frac{1}{10}$, find an approximation for $\sqrt{10}$ in the form $\frac{a}{b}$, where $a$ and $b$ are to be determined. [2]
\end{enumerate}
\hfill \mbox{\textit{WJEC Unit 3 Q4 [4]}}