OCR MEI FP1 2010 June — Question 3 6 marks

Exam BoardOCR MEI
ModuleFP1 (Further Pure Mathematics 1)
Year2010
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeComplex roots with real coefficients
DifficultyModerate -0.3 This is a straightforward application of the complex conjugate root theorem for polynomials with real coefficients, followed by routine algebraic manipulation. Students need to recognize that 1-2j must also be a root, then use sum/product of roots or polynomial division to find the real root and k. While it requires multiple steps, each step follows standard procedures taught in FP1 with no novel problem-solving required.
Spec4.02g Conjugate pairs: real coefficient polynomials4.02j Cubic/quartic equations: conjugate pairs and factor theorem
3 The cubic equation \(2 z ^ { 3 } - z ^ { 2 } + 4 z + k = 0\), where \(k\) is real, has a root \(z = 1 + 2 \mathrm { j }\).
Write down the other complex root. Hence find the real root and the value of \(k\).
Question 3:
AnswerMarks Guidance
AnswerMark Guidance
\(z = 1 - 2j\)B1
\(1 + 2j + 1 - 2j + \alpha = \frac{1}{2}\)M1 Valid attempt to use sum of roots, or other valid method
\(\Rightarrow \alpha = -\frac{3}{2}\)A1 c.a.o.
\(\frac{-k}{2} = -\frac{3}{2}(1-2j)(1+2j) = -\frac{15}{2}\)M1, A1(ft) Valid attempt to use product of roots, or other valid method; correct equation – can be implied
\(k = 15\)A1 [6] c.a.o.
OR:
AnswerMarks Guidance
AnswerMark Guidance
\((z-(1+2j))(z-(1-2j)) = z^2 - 2z + 5\)M1, A1 Multiplying correct factors; correct quadratic, c.a.o.
\(2z^3 - z^2 + 4z + k = (z^2 - 2z + 5)(2z + 3)\)M1 Attempt to find linear factor
\(\alpha = \frac{-3}{2}\)A1(ft)
\(k = 15\)A1 [6] c.a.o. 
# Question 3:

| Answer | Mark | Guidance |
|--------|------|----------|
| $z = 1 - 2j$ | B1 | |
| $1 + 2j + 1 - 2j + \alpha = \frac{1}{2}$ | M1 | Valid attempt to use sum of roots, or other valid method |
| $\Rightarrow \alpha = -\frac{3}{2}$ | A1 | c.a.o. |
| $\frac{-k}{2} = -\frac{3}{2}(1-2j)(1+2j) = -\frac{15}{2}$ | M1, A1(ft) | Valid attempt to use product of roots, or other valid method; correct equation – can be implied |
| $k = 15$ | A1 **[6]** | c.a.o. |

**OR:**

| Answer | Mark | Guidance |
|--------|------|----------|
| $(z-(1+2j))(z-(1-2j)) = z^2 - 2z + 5$ | M1, A1 | Multiplying correct factors; correct quadratic, c.a.o. |
| $2z^3 - z^2 + 4z + k = (z^2 - 2z + 5)(2z + 3)$ | M1 | Attempt to find linear factor |
| $\alpha = \frac{-3}{2}$ | A1(ft) | |
| $k = 15$ | A1 **[6]** | c.a.o. |

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3 The cubic equation $2 z ^ { 3 } - z ^ { 2 } + 4 z + k = 0$, where $k$ is real, has a root $z = 1 + 2 \mathrm { j }$.\\
Write down the other complex root. Hence find the real root and the value of $k$.

\hfill \mbox{\textit{OCR MEI FP1 2010 Q3 [6]}}
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