| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simultaneous equations |
| Type | Line intersecting quadratic curve |
| Difficulty | Easy -1.3 This is a routine surds question testing basic recall and standard techniques. Part (a) is trivial application of index laws, while part (b) requires rationalizing the denominator—a standard textbook exercise with no problem-solving element. Significantly easier than average A-level questions. |
| Spec | 1.02b Surds: manipulation and rationalising denominators |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(20\) | B1 | Sight of 20. (\(4 \times 5\) is not sufficient) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{\sqrt{2}}{2\sqrt{5}-3\sqrt{2}} \times \frac{2\sqrt{5}+3\sqrt{2}}{2\sqrt{5}+3\sqrt{2}}\) | M1 | Multiplies top and bottom by a correct expression. NB \(2\sqrt{5}+3\sqrt{2} \equiv \sqrt{20}+\sqrt{18}\). Allow multiply by \(k(2\sqrt{5}+3\sqrt{2})\) |
| Denominator \(= 2\) | A1 | Obtains denominator of 2, or sight of \((2\sqrt{5}-3\sqrt{2})(2\sqrt{5}+3\sqrt{2})=2\) with no errors. May be implied by \(\frac{\cdots}{2k}\) |
| Numerator \(= \sqrt{2}(2\sqrt{5}\pm3\sqrt{2})=2\sqrt{10}\pm6\) | M1 | Attempt to multiply numerator by \(\pm(2\sqrt{5}\pm3\sqrt{2})\) and obtain form \(p+q\sqrt{10}\) where \(p\) and \(q\) are integers |
| \(\frac{\sqrt{2}}{2\sqrt{5}-3\sqrt{2}}=\frac{2\sqrt{10}+6}{2}=3+\sqrt{10}\) | A1 | Cso. Answer written as \(\sqrt{10}+3\) or statement \(a=3\), \(b=10\). Allow \(1\sqrt{10}\) for \(\sqrt{10}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{\sqrt{2}}{2\sqrt{5}-3\sqrt{2}}=\frac{1}{\sqrt{10}-3}\) or \(\frac{2}{2\sqrt{10}-6}\) | M1A1 | M1: Divides or multiplies top and bottom by \(\sqrt{2}\). A1: \(\frac{k}{k(\sqrt{10}-3)}\) |
| \(=\frac{1}{\sqrt{10}-3}\times\frac{\sqrt{10}+3}{\sqrt{10}+3}\) | M1 | Multiplies top and bottom by \(\sqrt{10}+3\) |
| \(=3+\sqrt{10}\) | A1 |
# Question 1:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $20$ | B1 | Sight of 20. ($4 \times 5$ is not sufficient) |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{\sqrt{2}}{2\sqrt{5}-3\sqrt{2}} \times \frac{2\sqrt{5}+3\sqrt{2}}{2\sqrt{5}+3\sqrt{2}}$ | M1 | Multiplies top and bottom by a **correct** expression. NB $2\sqrt{5}+3\sqrt{2} \equiv \sqrt{20}+\sqrt{18}$. Allow multiply by $k(2\sqrt{5}+3\sqrt{2})$ |
| Denominator $= 2$ | A1 | Obtains denominator of 2, or sight of $(2\sqrt{5}-3\sqrt{2})(2\sqrt{5}+3\sqrt{2})=2$ with no errors. May be implied by $\frac{\cdots}{2k}$ |
| Numerator $= \sqrt{2}(2\sqrt{5}\pm3\sqrt{2})=2\sqrt{10}\pm6$ | M1 | Attempt to multiply numerator by $\pm(2\sqrt{5}\pm3\sqrt{2})$ and obtain form $p+q\sqrt{10}$ where $p$ and $q$ are integers |
| $\frac{\sqrt{2}}{2\sqrt{5}-3\sqrt{2}}=\frac{2\sqrt{10}+6}{2}=3+\sqrt{10}$ | A1 | Cso. Answer written as $\sqrt{10}+3$ or statement $a=3$, $b=10$. Allow $1\sqrt{10}$ for $\sqrt{10}$ |
**Alternative for (b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{\sqrt{2}}{2\sqrt{5}-3\sqrt{2}}=\frac{1}{\sqrt{10}-3}$ or $\frac{2}{2\sqrt{10}-6}$ | M1A1 | M1: Divides or multiplies top and bottom by $\sqrt{2}$. A1: $\frac{k}{k(\sqrt{10}-3)}$ |
| $=\frac{1}{\sqrt{10}-3}\times\frac{\sqrt{10}+3}{\sqrt{10}+3}$ | M1 | Multiplies top and bottom by $\sqrt{10}+3$ |
| $=3+\sqrt{10}$ | A1 | |
---Simplify
\begin{enumerate}[label=(\alph*)]
\item $( 2 \sqrt { } 5 ) ^ { 2 }$
\item $\frac { \sqrt { } 2 } { 2 \sqrt { } 5 - 3 \sqrt { } 2 }$ giving your answer in the form $a + \sqrt { } b$, where $a$ and $b$ are integers.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2015 Q1 [5]}}