| Exam Board | WJEC |
|---|---|
| Module | Unit 1 (Unit 1) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Express result in specific form |
| Difficulty | Standard +0.8 This question requires systematic application of the binomial theorem with surds (5 terms, careful coefficient calculation), algebraic manipulation to separate √3 and √2 terms, then a non-routine insight to rearrange the approximation to isolate √6. The second part demands creative algebraic manipulation beyond standard textbook exercises, though the individual techniques are A-level standard. |
| Spec | 1.02b Surds: manipulation and rationalising denominators1.04a Binomial expansion: (a+b)^n for positive integer n |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \((\sqrt{3} – \sqrt{2})^5 = (\sqrt{3})^5 + 5(\sqrt{3})^4(– \sqrt{2}) + 10(\sqrt{3})^4(– \sqrt{2})^2 + 10(\sqrt{3})(– \sqrt{2})^3 + 5(\sqrt{3})(– \sqrt{2})^4 + (– \sqrt{2})^5\) | B2 | AO1 (five or six terms correct) |
| \((\sqrt{3} – \sqrt{2})^5 = 9\sqrt{3} – 45\sqrt{2} + 60\sqrt{3} – 60\sqrt{2} + 20\sqrt{3} – 4\sqrt{2}\) | B2 | AO1 (six terms correct) |
| Answer | Marks | Guidance |
|---|---|---|
| \((\sqrt{3} – \sqrt{2})^5 = 89\sqrt{3} – 109\sqrt{2}\) | B1 | AO1 (f.t. one error) |
| (b) Since \((\sqrt{3} – \sqrt{2})^5 \approx 0\), we may assume that \(89\sqrt{3} \approx 109\sqrt{2}\) | M1 | AO3 (f.t. candidate's answer to part (a) provided one coefficient is negative) |
| Either: \(89\sqrt{3} \times \sqrt{3} \approx 109\sqrt{2} \times \sqrt{3}\) | m1 | AO3 (f.t. candidate's answer to part (a) provided one coefficient is negative) |
| \(\sqrt{6} \approx \frac{267}{109}\) | A1 | AO3 (c.a.o.) |
| Or: \(89\sqrt{3} \times \sqrt{2} \approx 109\sqrt{2} \times \sqrt{2}\) | (m1) | (AO3) (f.t. candidate's answer to part (a) provided one coefficient is negative) |
| \(\sqrt{6} \approx \frac{218}{89}\) | (A1) | (AO3) (c.a.o.) |
**(a)** $(\sqrt{3} – \sqrt{2})^5 = (\sqrt{3})^5 + 5(\sqrt{3})^4(– \sqrt{2}) + 10(\sqrt{3})^4(– \sqrt{2})^2 + 10(\sqrt{3})(– \sqrt{2})^3 + 5(\sqrt{3})(– \sqrt{2})^4 + (– \sqrt{2})^5$ | B2 | AO1 (five or six terms correct)
$(\sqrt{3} – \sqrt{2})^5 = 9\sqrt{3} – 45\sqrt{2} + 60\sqrt{3} – 60\sqrt{2} + 20\sqrt{3} – 4\sqrt{2}$ | B2 | AO1 (six terms correct)
If B2 not awarded, award B1 for three, four or five correct terms
$(\sqrt{3} – \sqrt{2})^5 = 89\sqrt{3} – 109\sqrt{2}$ | B1 | AO1 (f.t. one error)
**(b)** Since $(\sqrt{3} – \sqrt{2})^5 \approx 0$, we may assume that $89\sqrt{3} \approx 109\sqrt{2}$ | M1 | AO3 (f.t. candidate's answer to part (a) provided one coefficient is negative)
Either: $89\sqrt{3} \times \sqrt{3} \approx 109\sqrt{2} \times \sqrt{3}$ | m1 | AO3 (f.t. candidate's answer to part (a) provided one coefficient is negative)
$\sqrt{6} \approx \frac{267}{109}$ | A1 | AO3 (c.a.o.)
Or: $89\sqrt{3} \times \sqrt{2} \approx 109\sqrt{2} \times \sqrt{2}$ | (m1) | (AO3) (f.t. candidate's answer to part (a) provided one coefficient is negative)
$\sqrt{6} \approx \frac{218}{89}$ | (A1) | (AO3) (c.a.o.)
**Total: [8]**
---\begin{enumerate}[label=(\alph*)]
\item Use the binomial theorem to express $\left(\sqrt{3} - \sqrt{2}\right)^5$ in the form $a\sqrt{3} + b\sqrt{2}$, where $a$, $b$ are integers whose values are to be found. [5]
\item Given that $\left(\sqrt{3} - \sqrt{2}\right)^5 \approx 0$, use your answer to part (a) to find an approximate value for $\sqrt{6}$ in the form $\frac{c}{d}$, where $c$ and $d$ are positive integers whose values are to be found. [3]
\end{enumerate}
\hfill \mbox{\textit{WJEC Unit 1 Q10 [8]}}