| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indices and Surds |
| Type | Rationalize denominator simple |
| Difficulty | Easy -1.2 This is a straightforward rationalizing the denominator question requiring multiplication by the conjugate and simplification. It's a standard C1 exercise with a single technique and minimal steps, making it easier than average but not trivial since students must correctly expand and collect terms. |
| Spec | 1.02b Surds: manipulation and rationalising denominators |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{7+\sqrt{5}}{\sqrt{5}-1} \times \frac{(\sqrt{5}+1)}{(\sqrt{5}+1)}\) | M1 | Multiplies top and bottom by a correct expression. This statement is sufficient. (Allow multiply by \(k(\sqrt{5}+1)\)) |
| \(= \frac{\cdots}{4}\) | A1cso | Obtains a denominator of 4 or sight of \((\sqrt{5}-1)(\sqrt{5}+1)=4\) |
| \((7+\sqrt{5})(\sqrt{5}+1) = 7\sqrt{5}+5+7+\sqrt{5}\) | M1 | An attempt to multiply the numerator by \((\pm\sqrt{5}\pm1)\) and get 4 terms with at least 2 correct for their \((\pm\sqrt{5}\pm1)\). (May be implied) |
| \(3+2\sqrt{5}\) | A1cso | Answer as written or \(a=3\) and \(b=2\). (Allow \(2\sqrt{5}+3\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{7+\sqrt{5}}{\sqrt{5}-1} \times \frac{(-\sqrt{5}-1)}{(-\sqrt{5}-1)}\) | M1 | Multiplies top and bottom by a correct expression. (Allow \(k(-\sqrt{5}-1)\)) |
| \(= \frac{\cdots}{-4}\) | A1cso | Obtains a denominator of \(-4\) |
| \((7+\sqrt{5})(-\sqrt{5}-1) = -7\sqrt{5}-5-7-\sqrt{5}\) | M1 | An attempt to multiply numerator by \((\pm\sqrt{5}\pm1)\), get 4 terms with at least 2 correct. (May be implied) |
| \(3+2\sqrt{5}\) | A1cso | Answer as written or \(a=3\) and \(b=2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{(7+\sqrt{5})}{\sqrt{5}-1} = a+b\sqrt{5} \Rightarrow 7+\sqrt{5}=(a-b)\sqrt{5}+5b-a\) | M1 | Multiplies and collects rational and irrational parts |
| \(a-b=1,\quad 5b-a=7\) | A1 | Correct equations |
| \(a=3,\quad b=2\) | — | M1 for attempt to solve simultaneous equations, A1 both answers correct |
## Question 1:
**Way 1:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{7+\sqrt{5}}{\sqrt{5}-1} \times \frac{(\sqrt{5}+1)}{(\sqrt{5}+1)}$ | M1 | Multiplies top and bottom by a correct expression. This statement is sufficient. (Allow multiply by $k(\sqrt{5}+1)$) |
| $= \frac{\cdots}{4}$ | A1cso | Obtains a denominator of 4 or sight of $(\sqrt{5}-1)(\sqrt{5}+1)=4$ |
| $(7+\sqrt{5})(\sqrt{5}+1) = 7\sqrt{5}+5+7+\sqrt{5}$ | M1 | An attempt to multiply the numerator by $(\pm\sqrt{5}\pm1)$ and get 4 terms with at least 2 correct for their $(\pm\sqrt{5}\pm1)$. (May be implied) |
| $3+2\sqrt{5}$ | A1cso | Answer as written or $a=3$ and $b=2$. (Allow $2\sqrt{5}+3$) |
**Way 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{7+\sqrt{5}}{\sqrt{5}-1} \times \frac{(-\sqrt{5}-1)}{(-\sqrt{5}-1)}$ | M1 | Multiplies top and bottom by a correct expression. (Allow $k(-\sqrt{5}-1)$) |
| $= \frac{\cdots}{-4}$ | A1cso | Obtains a denominator of $-4$ |
| $(7+\sqrt{5})(-\sqrt{5}-1) = -7\sqrt{5}-5-7-\sqrt{5}$ | M1 | An attempt to multiply numerator by $(\pm\sqrt{5}\pm1)$, get 4 terms with at least 2 correct. (May be implied) |
| $3+2\sqrt{5}$ | A1cso | Answer as written or $a=3$ and $b=2$ |
**Alternative (Simultaneous Equations):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{(7+\sqrt{5})}{\sqrt{5}-1} = a+b\sqrt{5} \Rightarrow 7+\sqrt{5}=(a-b)\sqrt{5}+5b-a$ | M1 | Multiplies and collects rational and irrational parts |
| $a-b=1,\quad 5b-a=7$ | A1 | Correct equations |
| $a=3,\quad b=2$ | — | M1 for attempt to solve simultaneous equations, A1 both answers correct |
**Total: [4]**\begin{enumerate}
\item Simplify
\end{enumerate}
$$\frac { 7 + \sqrt { 5 } } { \sqrt { 5 } - 1 }$$
giving your answer in the form $a + b \sqrt { 5 }$, where $a$ and $b$ are integers.\\
\hfill \mbox{\textit{Edexcel C1 2013 Q1 [4]}}