Edexcel C1 2008 January — Question 3 4 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Year2008
SessionJanuary
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIndices and Surds
TypeRationalize denominator simple
DifficultyEasy -1.3 This is a straightforward rationalizing the denominator question requiring only 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 execute the algebraic manipulation correctly.
Spec1.02b Surds: manipulation and rationalising denominators

Simplify $$\frac { 5 - \sqrt { 3 } } { 2 + \sqrt { 3 } } ,$$ giving your answer in the form \(a + b \sqrt { } 3\), where \(a\) and \(b\) are integers.

AnswerMarks Guidance
\(\frac{(5-\sqrt{3})}{(2+\sqrt{3})} \times \frac{(2-\sqrt{3})}{(2-\sqrt{3})}\)M1 Multiplying top and bottom by \((2-\sqrt{3})\). (As shown above is sufficient.)
\(= \frac{10 - 2\sqrt{3} - 5\sqrt{3} + (\sqrt{3})^2}{...}\) (or \(= \frac{10 - 7\sqrt{3} + 3}{...}\))M1 Attempt to multiply out numerator \((5-\sqrt{3})(2-\sqrt{3})\). Must have at least 3 terms correct.
\(= 13 - 7\sqrt{3}\) (Allow \(\frac{13-7\sqrt{3}}{1}\))A1 Although 'denominator = 1' may be implied, the \(13 - 7\sqrt{3}\) must obviously be the final answer (not an intermediate step), to score full marks. (Also M0 M1 A1 A1 is not an option.)
\(a = 13\)A1 The A marks cannot be scored unless the 1st M mark has been scored, but this 1st M mark could be implied by correct expansions of both numerator and denominator.
\(b = -7\)A1
Special case: If numerator is multiplied by \((2+\sqrt{3})\) instead of \((2-\sqrt{3})\), the 2nd M can still be scored for at least 3 of these terms correct: \(10 - 2\sqrt{3} + 5\sqrt{3} - (\sqrt{3})^2\). The maximum score in the special case is 1 mark: M0 M1 A0 A0.
Answer only: Scores no marks.
Alternative method:
AnswerMarks Guidance
\(5 - \sqrt{3} = (a + b\sqrt{3})(2 + \sqrt{3})\)M1 At least 3 terms correct.
\((a + b\sqrt{3})(2 + \sqrt{3}) = 2a + a\sqrt{3} + 2b\sqrt{3} + 3\)M1 Form and attempt to solve simultaneous equations.
\(5 = 2a + 3b\)
\(-1 = a + 2b\)
AnswerMarks
\(a = 13, b = -7\)A1, A1
Total: 4 marks
$\frac{(5-\sqrt{3})}{(2+\sqrt{3})} \times \frac{(2-\sqrt{3})}{(2-\sqrt{3})}$ | M1 | Multiplying top and bottom by $(2-\sqrt{3})$. (As shown above is sufficient.)

$= \frac{10 - 2\sqrt{3} - 5\sqrt{3} + (\sqrt{3})^2}{...}$ (or $= \frac{10 - 7\sqrt{3} + 3}{...}$) | M1 | Attempt to multiply out numerator $(5-\sqrt{3})(2-\sqrt{3})$. Must have at least 3 terms correct.

$= 13 - 7\sqrt{3}$ (Allow $\frac{13-7\sqrt{3}}{1}$) | A1 | Although 'denominator = 1' may be implied, the $13 - 7\sqrt{3}$ must obviously be the final answer (not an intermediate step), to score full marks. (Also M0 M1 A1 A1 is not an option.)

$a = 13$ | A1 | The A marks cannot be scored unless the 1st M mark has been scored, but this 1st M mark could be implied by correct expansions of both numerator and denominator.

$b = -7$ | A1 |

**Special case:** If numerator is multiplied by $(2+\sqrt{3})$ instead of $(2-\sqrt{3})$, the 2nd M can still be scored for at least 3 of these terms correct: $10 - 2\sqrt{3} + 5\sqrt{3} - (\sqrt{3})^2$. The maximum score in the special case is 1 mark: M0 M1 A0 A0.

**Answer only:** Scores no marks.

**Alternative method:**
$5 - \sqrt{3} = (a + b\sqrt{3})(2 + \sqrt{3})$ | M1 | At least 3 terms correct.

$(a + b\sqrt{3})(2 + \sqrt{3}) = 2a + a\sqrt{3} + 2b\sqrt{3} + 3$ | M1 | Form and attempt to solve simultaneous equations.

$5 = 2a + 3b$
$-1 = a + 2b$

$a = 13, b = -7$ | A1, A1 |

**Total: 4 marks**

---
Simplify

$$\frac { 5 - \sqrt { 3 } } { 2 + \sqrt { 3 } } ,$$

giving your answer in the form $a + b \sqrt { } 3$, where $a$ and $b$ are integers.\\

\hfill \mbox{\textit{Edexcel C1 2008 Q3 [4]}}