| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Theorem (positive integer n) |
| Type | Express result in specific form |
| Difficulty | Moderate -0.8 Part (a) is a direct application of binomial expansion with n=4, requiring only Pascal's triangle or the binomial formula. Parts (b)(i) and (b)(ii) are straightforward substitution and logarithm manipulation following from part (a), with all steps clearly signposted by 'hence'. This is easier than average as it's highly structured with minimal problem-solving required. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n1.06f Laws of logarithms: addition, subtraction, power rules |
| Answer | Marks | Guidance |
|---|---|---|
| 6(a) | \((1 + x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4\) | M1, A2,1 |
| 6(b)(i) | \((1 + \sqrt{5})^4 = 1 + 4\sqrt{5} + 6(\sqrt{5})^2 + 4(\sqrt{5})^3 + (\sqrt{5})^4\) = \(1 + 4\sqrt{5} + 6(5) + 4(5\sqrt{5}) + (25)\) = ............ = 56 + 24\(\sqrt{5}\) | M1, A1 ft, A1 |
| 6(b)(ii) | \(\log_s(1 + \sqrt{5})^4 = \log_s[8(7 + 3\sqrt{5})]\) = \(\log_s 8 + \log_s(7 + 3\sqrt{5})\) = \(3 + \log_s(7 + 3\sqrt{5})\) | M1, m1, A1 |
6(a) | $(1 + x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4$ | M1, A2,1 | Full method; A1 if four terms correct or just one slip |
6(b)(i) | $(1 + \sqrt{5})^4 = 1 + 4\sqrt{5} + 6(\sqrt{5})^2 + 4(\sqrt{5})^3 + (\sqrt{5})^4$ = $1 + 4\sqrt{5} + 6(5) + 4(5\sqrt{5}) + (25)$ = ............ = 56 + 24$\sqrt{5}$ | M1, A1 ft, A1 | Substitute. $\sqrt{5}$ for x.; Two of 3 terms shown in brackets; AG CSO Be convinced |
6(b)(ii) | $\log_s(1 + \sqrt{5})^4 = \log_s[8(7 + 3\sqrt{5})]$ = $\log_s 8 + \log_s(7 + 3\sqrt{5})$ = $3 + \log_s(7 + 3\sqrt{5})$ | M1, m1, A1 | OE seen; CSO; SC B1 Change to base 10 and verify |6
\begin{enumerate}[label=(\alph*)]
\item Using the binomial expansion, or otherwise, express $( 1 + x ) ^ { 4 }$ in ascending powers of $x$.
\item \begin{enumerate}[label=(\roman*)]
\item Hence show that $( 1 + \sqrt { 5 } ) ^ { 4 } = 56 + 24 \sqrt { 5 }$.
\item Hence show that $\log _ { 2 } ( 1 + \sqrt { 5 } ) ^ { 4 } = k + \log _ { 2 } ( 7 + 3 \sqrt { 5 } )$, where $k$ is an integer.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2005 Q6 [9]}}