OCR MEI Further Pure Core Specimen — Question 4 5 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
SessionSpecimen
Marks5
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Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeComplex roots with real coefficients
DifficultyStandard +0.3 This is a straightforward Further Maths polynomial question requiring standard techniques: since coefficients are real, the conjugate 1-2i must also be a root, then find the third root using sum of roots, and q using product or substitution. While it's Further Maths content, it's a routine application of well-practiced methods with no novel insight required, making it slightly easier than average overall.
Spec4.02g Conjugate pairs: real coefficient polynomials4.02i Quadratic equations: with complex roots
You are given that \(z = 1 + 2i\) is a root of the equation \(z^3 - 5z^2 + qz - 15 = 0\), where \(q \in \mathbb{R}\). Find • the other roots, • the value of \(q\). [5]
Question 4:
AnswerMarks
4Conjugate root 1 – 2i [because q(cid:143) ]
(cid:11)1 (cid:14) 2i(cid:12) (cid:14) (cid:11)1 – 2i(cid:12) (cid:14)(cid:74)(cid:32) 5
e
AnswerMarks
(cid:74)(cid:32) 3cB1
M1
AnswerMarks
A1i
2.2a
1.1a
AnswerMarks Guidance
1.1or (cid:11)1 (cid:14) 2i(cid:12)(cid:11)1 – 2i(cid:12)(cid:74)(cid:32) 15 Alternative solution
Conjugate root 1 – 2i
[because q(cid:143) ] B1
(z(cid:16)1(cid:16)2i)(z(cid:16)1(cid:14)2i)
(cid:32)z2(cid:16)2z(cid:14)5
z3(cid:16)5z2 (cid:14)qz(cid:16)15
M1A1
(cid:32)(z2 (cid:16)2z(cid:14)5)(z(cid:16)3)
q(cid:32)6(cid:14)5(cid:32)11 M1A1
The other roots are 3 and
(1(cid:16)2i)
p
either S
(1(cid:14)2i)(1(cid:16)2i)(cid:14)3(1(cid:14)2i)(cid:14)3(1(cid:16)2i)(cid:32)q
AnswerMarks
q(cid:32)11M1
A13.1a
1.1FT
or
AnswerMarks
33(cid:16)5(cid:117)32(cid:14)3q(cid:16)15(cid:32)0M1
q(cid:32)11A1
[5]
AnswerMarks Guidance
43 1
5i2 0
5ii1 3
6i1 1
6ii2 0
7i2 0
7iiA1 0
7iiB0 1
1
AnswerMarks Guidance
7iii1 0
7ivA2 1
03
7ivB0 2 
Question 4:
4 | Conjugate root 1 – 2i [because q(cid:143) ]
(cid:11)1 (cid:14) 2i(cid:12) (cid:14) (cid:11)1 – 2i(cid:12) (cid:14)(cid:74)(cid:32) 5
e
(cid:74)(cid:32) 3 | cB1
M1
A1 | i
2.2a
1.1a
1.1 | or (cid:11)1 (cid:14) 2i(cid:12)(cid:11)1 – 2i(cid:12)(cid:74)(cid:32) 15 | Alternative solution
Conjugate root 1 – 2i
[because q(cid:143) ] B1
(z(cid:16)1(cid:16)2i)(z(cid:16)1(cid:14)2i)
(cid:32)z2(cid:16)2z(cid:14)5
z3(cid:16)5z2 (cid:14)qz(cid:16)15
M1A1
(cid:32)(z2 (cid:16)2z(cid:14)5)(z(cid:16)3)
q(cid:32)6(cid:14)5(cid:32)11 M1A1
The other roots are 3 and
(1(cid:16)2i)
p
either S
(1(cid:14)2i)(1(cid:16)2i)(cid:14)3(1(cid:14)2i)(cid:14)3(1(cid:16)2i)(cid:32)q
q(cid:32)11 | M1
A1 | 3.1a
1.1 | FT
or
33(cid:16)5(cid:117)32(cid:14)3q(cid:16)15(cid:32)0 | M1
q(cid:32)11 | A1
[5]
4 | 3 | 1 | 1 | 0 | 5
5i | 2 | 0 | 0 | 0 | 2
5ii | 1 | 3 | 1 | 0 | 5
6i | 1 | 1 | 0 | 0 | 2
6ii | 2 | 0 | 2 | 0 | 4
7i | 2 | 0 | 0 | 0 | 2
7iiA | 1 | 0 | 0 | 0 | 1
7iiB | 0 | 1 | 0 | 0 | n
1
7iii | 1 | 0 | 1 | 0 | 2
7ivA | 2 | 1 | 0 | e
0 | 3
7ivB | 0 | 2 | 0 | 0 | 2
You are given that $z = 1 + 2i$ is a root of the equation $z^3 - 5z^2 + qz - 15 = 0$, where $q \in \mathbb{R}$.

Find
• the other roots,
• the value of $q$. [5]

\hfill \mbox{\textit{OCR MEI Further Pure Core  Q4 [5]}}
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