AQA AS Paper 1 2019 June — Question 4 4 marks

Exam BoardAQA
ModuleAS Paper 1 (AS Paper 1)
Year2019
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIndices and Surds
TypeShow surd expression equals value
DifficultyModerate -0.8 This is a routine rationalising the denominator question requiring multiplication by the conjugate and simplification. The algebraic manipulation is straightforward, and the form to match is given explicitly. It's easier than average as it's a standard AS technique with no problem-solving element, though the specific form adds minor complexity beyond basic rationalisation.
Spec1.02b Surds: manipulation and rationalising denominators

Show that \(\frac{\sqrt{6}}{\sqrt{3} - \sqrt{2}}\) can be expressed in the form \(m\sqrt{n} + n\sqrt{m}\), where \(m\) and \(n\) are integers. Fully justify your answer. [4 marks]

Question 4:
AnswerMarks Guidance
4Multiplies by
√3+√2AO1.1a M1
×
√3−√2 √3+√2
=
√18+√12
3−2
√18+ √12
= 1
√9×2+√4×3
= 3
√3+√2
Correctly evaluates denominator to
AnswerMarks Guidance
get 3 – 2 or 1AO1.1b A1
Evaluates numerator, one term
correct
AnswerMarks Guidance
or 3AO1.1b A1
Completes solution CAO
AnswerMarks Guidance
√ 18 𝑜𝑜𝑜𝑜 √12 √2 𝑜𝑜𝑜𝑜 2√3AO2.1 R1
Total4
QMarking Instructions AO
Question 4:
4 | Multiplies by
√3+√2 | AO1.1a | M1 | √6 √3+√2
×
√3−√2 √3+√2
=
√18+√12
3−2
√18+ √12
= 1
√9×2+√4×3
= 3
√3+√2
Correctly evaluates denominator to
get 3 – 2 or 1 | AO1.1b | A1
Evaluates numerator, one term
correct
or 3 | AO1.1b | A1
Completes solution CAO
√ 18 𝑜𝑜𝑜𝑜 √12 √2 𝑜𝑜𝑜𝑜 2√3 | AO2.1 | R1
Total | 4
Q | Marking Instructions | AO | Marks | Typical Solution
Show that $\frac{\sqrt{6}}{\sqrt{3} - \sqrt{2}}$ can be expressed in the form $m\sqrt{n} + n\sqrt{m}$, where $m$ and $n$ are integers.

Fully justify your answer.
[4 marks]

\hfill \mbox{\textit{AQA AS Paper 1 2019 Q4 [4]}}