OCR MEI AS Paper 1 2019 June — Question 3 4 marks

Exam BoardOCR MEI
ModuleAS Paper 1 (AS Paper 1)
Year2019
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIndices and Surds
TypeRationalize denominator simple
DifficultyModerate -0.8 This is a straightforward rationalizing the denominator question requiring multiplication by the conjugate and simplification of surds. While it involves algebraic manipulation with a parameter k, the technique is standard and mechanical with no problem-solving insight needed. Slightly easier than average due to being a routine application of a well-practiced method.
Spec1.02b Surds: manipulation and rationalising denominators
3 Given that \(k\) is an integer, express \(\frac { 3 \sqrt { 2 } - k } { \sqrt { 8 } + 1 }\) in the form \(a + b \sqrt { 2 }\) where \(a\) and \(b\) are rational expressions in terms of \(k\).
Question 3:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\sqrt{8} = 2\sqrt{2}\)B1 3.1a – soi
\(\frac{(3\sqrt{2}-k)(2\sqrt{2}-1)}{(2\sqrt{2}+1)(2\sqrt{2}-1)}\)M1 1.1a – Multiplying by conjugate. Allow for using \(\sqrt{8}-1\)
Correct denominatorA1 1.1
\(= \frac{12+k}{7} - \frac{3+2k}{7}\sqrt{2}\)A1 [4] 1.1 – Fully correct in the form \(a + b\sqrt{2}\). Condone \(\frac{12+k-(3+2k)\sqrt{2}}{7}\). Do not allow final A1 for \(\frac{12-3\sqrt{2}+k-2k\sqrt{2}}{7}\)
Alternative method:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\sqrt{8} = 2\sqrt{2}\)B1 soi
\(\frac{3\sqrt{2}}{(\sqrt{8}+1)} - \frac{k}{(\sqrt{8}+1)} = \frac{12-3\sqrt{2}}{7} - \left(\frac{-1+2\sqrt{2}}{7}\right)k\)M1 Splitting the fraction into two terms and simplifying each term BC. Allow M1 if both fractions seen
Correct denominator in both terms and \(k\) included correctlyA1
\(= \frac{12+k}{7} - \frac{3+2k}{7}\sqrt{2}\)A1 [4] Rearranging into the form \(a+b\sqrt{2}\) 
## Question 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sqrt{8} = 2\sqrt{2}$ | B1 | 3.1a – soi |
| $\frac{(3\sqrt{2}-k)(2\sqrt{2}-1)}{(2\sqrt{2}+1)(2\sqrt{2}-1)}$ | M1 | 1.1a – Multiplying by conjugate. Allow for using $\sqrt{8}-1$ |
| Correct denominator | A1 | 1.1 |
| $= \frac{12+k}{7} - \frac{3+2k}{7}\sqrt{2}$ | A1 [4] | 1.1 – Fully correct in the form $a + b\sqrt{2}$. Condone $\frac{12+k-(3+2k)\sqrt{2}}{7}$. Do not allow final A1 for $\frac{12-3\sqrt{2}+k-2k\sqrt{2}}{7}$ |

**Alternative method:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sqrt{8} = 2\sqrt{2}$ | B1 | soi |
| $\frac{3\sqrt{2}}{(\sqrt{8}+1)} - \frac{k}{(\sqrt{8}+1)} = \frac{12-3\sqrt{2}}{7} - \left(\frac{-1+2\sqrt{2}}{7}\right)k$ | M1 | Splitting the fraction into two terms and simplifying each term BC. Allow M1 if both fractions seen |
| Correct denominator in both terms and $k$ included correctly | A1 | |
| $= \frac{12+k}{7} - \frac{3+2k}{7}\sqrt{2}$ | A1 [4] | Rearranging into the form $a+b\sqrt{2}$ |

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3 Given that $k$ is an integer, express $\frac { 3 \sqrt { 2 } - k } { \sqrt { 8 } + 1 }$ in the form $a + b \sqrt { 2 }$ where $a$ and $b$ are rational expressions in terms of $k$.

\hfill \mbox{\textit{OCR MEI AS Paper 1 2019 Q3 [4]}}
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Q1 3 Q2 3 Q3 4 Q4 5 Q5 3 Q6 7 Q7 11 Q8 7 Q9 9 Q10 7 Q11 11