OCR MEI Further Pure Core AS 2018 June — Question 3 5 marks

Exam BoardOCR MEI
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2018
SessionJune
Marks5
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TopicComplex Numbers Arithmetic
TypeSimultaneous equations with complex numbers
DifficultyModerate -0.8 This is a straightforward complex number multiplication requiring expansion of brackets, equating real and imaginary parts, and solving two simple linear equations. It's purely procedural with no problem-solving insight needed, making it easier than average but not trivial since it requires careful algebraic manipulation across multiple steps.
Spec4.02a Complex numbers: real/imaginary parts, modulus, argument4.02e Arithmetic of complex numbers: add, subtract, multiply, divide
Find real numbers \(a\) and \(b\) such that \((a - 3i)(5 - i) = b - 17i\). [5]
Question 3:
AnswerMarks
3(a  3i)(5  i) = 5a  3  15i  ai
 5a  3 = b
15 + a = 17
AnswerMarks
 a = 2, b = 7M1
M1
A1
A1
A1
AnswerMarks
[5]1.1a
1.1
3.1a
1.1
AnswerMarks
1.1expanding and i2 = 1
equating Re and Im parts
soi
o.e. eg 15i + ai = 17i
AnswerMarks
both correctallow 1 sign error (e.g. +3)
(a  3i)(5  i) = 5a  3  15i  ai
 5a  3 = b
15 + a = 17
 a = 2, b = 7
Question 3:
3 | (a  3i)(5  i) = 5a  3  15i  ai
 5a  3 = b
15 + a = 17
 a = 2, b = 7 | M1
M1
A1
A1
A1
[5] | 1.1a
1.1
3.1a
1.1
1.1 | expanding and i2 = 1
equating Re and Im parts
soi
o.e. eg 15i + ai = 17i
both correct | allow 1 sign error (e.g. +3)
(a  3i)(5  i) = 5a  3  15i  ai
 5a  3 = b
15 + a = 17
 a = 2, b = 7
Find real numbers $a$ and $b$ such that $(a - 3i)(5 - i) = b - 17i$. [5]

\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2018 Q3 [5]}}
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Q1 4 Q2 3 Q3 5 Q4 5 Q5 7 Q6 4 Q7 9 Q8 6 Q9 9 Q10 8