AQA C1 2011 January — Question 2 5 marks

Exam BoardAQA
ModuleC1 (Core Mathematics 1)
Year2011
SessionJanuary
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIndices and Surds
TypeRationalize denominator simple
DifficultyEasy -1.2 Part (a) is trivial recall of index laws. Part (b) is a standard rationalizing denominator exercise requiring multiplication by conjugate and simplification—routine C1 technique with no problem-solving insight needed, though the algebra requires care.
Spec1.02b Surds: manipulation and rationalising denominators

2
  1. Simplify \(( 3 \sqrt { 3 } ) ^ { 2 }\).
  2. Express \(\frac { 4 \sqrt { 3 } + 3 \sqrt { 7 } } { 3 \sqrt { 3 } + \sqrt { 7 } }\) in the form \(\frac { m + \sqrt { 21 } } { n }\), where \(m\) and \(n\) are integers.

2(a)
AnswerMarks Guidance
\(27\)B1 1
2(b)
AnswerMarks Guidance
\(\frac{4\sqrt{3} + 3\sqrt{7}}{3\sqrt{3} + \sqrt{7}} \times \frac{3\sqrt{3} - \sqrt{7}}{3\sqrt{3} - \sqrt{7}}\)M1
(Numerator =) \(36 + 9\sqrt{21} - 4\sqrt{21} - 21\)m1 expanding numerator condone one slip or omission
(Denominator =) \(20\)B1 must be seen as denominator
\(\frac{15 + 5\sqrt{21}}{20}\)
\(= \frac{3 + \sqrt{21}}{4}\)A1cso \(m = 3, n = 4\) condone \(\frac{3}{4} + \frac{\sqrt{21}}{4}\)
**2(a)**
$27$ | B1 | 1

**2(b)**
$\frac{4\sqrt{3} + 3\sqrt{7}}{3\sqrt{3} + \sqrt{7}} \times \frac{3\sqrt{3} - \sqrt{7}}{3\sqrt{3} - \sqrt{7}}$ | M1 | 

(Numerator =) $36 + 9\sqrt{21} - 4\sqrt{21} - 21$ | m1 | expanding numerator condone one slip or omission

(Denominator =) $20$ | B1 | must be seen as denominator

$\frac{15 + 5\sqrt{21}}{20}$ | | 

$= \frac{3 + \sqrt{21}}{4}$ | A1cso | $m = 3, n = 4$ condone $\frac{3}{4} + \frac{\sqrt{21}}{4}$ | 4

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2
\begin{enumerate}[label=(\alph*)]
\item Simplify $( 3 \sqrt { 3 } ) ^ { 2 }$.
\item Express $\frac { 4 \sqrt { 3 } + 3 \sqrt { 7 } } { 3 \sqrt { 3 } + \sqrt { 7 } }$ in the form $\frac { m + \sqrt { 21 } } { n }$, where $m$ and $n$ are integers.
\end{enumerate}

\hfill \mbox{\textit{AQA C1 2011 Q2 [5]}}