| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2014 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indices and Surds |
| Type | Show surd expression equals value |
| Difficulty | Moderate -0.8 This is a routine surds manipulation question requiring rationalizing denominators and simplifying surds using standard techniques (multiplying by conjugate, simplifying √27 = 3√3, etc.). Both parts are straightforward applications of Core 1 surd rules with no problem-solving insight needed, making it easier than average but not trivial since it requires careful algebraic manipulation. |
| Spec | 1.02b Surds: manipulation and rationalising denominators |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{4(2\sqrt{2}+\sqrt{6})}{(2\sqrt{2}-\sqrt{6})(2\sqrt{2}+\sqrt{6})}\) | M1 | M1: Multiplies numerator and denominator by \(\pm(2\sqrt{2}+\sqrt{6})\) |
| \((2\sqrt{2}-\sqrt{6})(2\sqrt{2}+\sqrt{6}) = 8 - 6 = 2\) | B1 | B1: correct treatment of denominator to give 2 |
| \(\sqrt{6} = \sqrt{2}\sqrt{3}\) used in numerator | B1 | B1: may be implied by correct factorisation of numerator. B0 for \(2\sqrt{6}=2\sqrt{2}(\sqrt{3}\ldots)\) |
| Concludes \(\frac{4(2\sqrt{2}+\sqrt{6})}{2} = 2\sqrt{2}(2+\sqrt{3})\) | A1 | A1: cao, no errors seen |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sqrt{27} = 3\sqrt{3}\) and \(\sqrt{21} \times \sqrt{7} = 7\sqrt{3}\) | B1 | B1: both first two terms correct as multiples of \(\sqrt{3}\) |
| \(2\sqrt{3}\) or \(\frac{6\sqrt{3}}{3}\) (3rd term) | B1 | B1: rationalises denominator in second term |
| \(3\sqrt{3} + 7\sqrt{3} - 2\sqrt{3} = 8\sqrt{3}\) or \(3\sqrt{3}+7\sqrt{3}-\frac{6\sqrt{3}}{3}=8\sqrt{3}\) | B1 | B1: correct conclusion. N.B. \(3\sqrt{3}+7\sqrt{3}-\frac{6}{\sqrt{3}}=8\sqrt{3}\) is B1B0B0 |
| [3] |
# Question 4:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{4(2\sqrt{2}+\sqrt{6})}{(2\sqrt{2}-\sqrt{6})(2\sqrt{2}+\sqrt{6})}$ | M1 | M1: Multiplies numerator and denominator by $\pm(2\sqrt{2}+\sqrt{6})$ |
| $(2\sqrt{2}-\sqrt{6})(2\sqrt{2}+\sqrt{6}) = 8 - 6 = 2$ | B1 | B1: correct treatment of denominator to give 2 |
| $\sqrt{6} = \sqrt{2}\sqrt{3}$ used in numerator | B1 | B1: may be implied by correct factorisation of numerator. B0 for $2\sqrt{6}=2\sqrt{2}(\sqrt{3}\ldots)$ |
| Concludes $\frac{4(2\sqrt{2}+\sqrt{6})}{2} = 2\sqrt{2}(2+\sqrt{3})$ | A1 | A1: cao, no errors seen |
| | [4] | |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sqrt{27} = 3\sqrt{3}$ and $\sqrt{21} \times \sqrt{7} = 7\sqrt{3}$ | B1 | B1: both first two terms correct as multiples of $\sqrt{3}$ |
| $2\sqrt{3}$ or $\frac{6\sqrt{3}}{3}$ (3rd term) | B1 | B1: rationalises denominator in second term |
| $3\sqrt{3} + 7\sqrt{3} - 2\sqrt{3} = 8\sqrt{3}$ or $3\sqrt{3}+7\sqrt{3}-\frac{6\sqrt{3}}{3}=8\sqrt{3}$ | B1 | B1: correct conclusion. N.B. $3\sqrt{3}+7\sqrt{3}-\frac{6}{\sqrt{3}}=8\sqrt{3}$ is B1B0B0 |
| | [3] | |
---4. Answer this question without the use of a calculator and show all your working.\\
(i) Show that
$$\frac { 4 } { 2 \sqrt { 2 } - \sqrt { 6 } } = 2 \sqrt { 2 } ( 2 + \sqrt { 3 } )$$
(ii) Show that
$$\sqrt { 27 } + \sqrt { 21 } \times \sqrt { 7 } - \frac { 6 } { \sqrt { 3 } } = 8 \sqrt { 3 }$$
\hfill \mbox{\textit{Edexcel C12 2014 Q4 [7]}}