| Exam Board | OCR MEI |
|---|---|
| Module | Further Pure Core AS (Further Pure Core AS) |
| Year | 2019 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Complex roots with real coefficients |
| Difficulty | Standard +0.3 This is a standard Further Maths question on complex roots with real coefficients. Part (a) requires substituting z=i and using the conjugate root theorem to find two constants—straightforward algebra. Part (b) involves factoring the quartic using the known roots, which is routine for this topic. While it requires multiple steps and careful algebraic manipulation, it follows a well-established method with no novel insight needed. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem4.02g Conjugate pairs: real coefficient polynomials |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (a) | DR |
| Answer | Marks |
|---|---|
| ⇒a=−2, b=2 | M1 |
| Answer | Marks |
|---|---|
| [4] | 2.1 |
| Answer | Marks |
|---|---|
| 2.2a | substituting z = i into eqn |
| Answer | Marks |
|---|---|
| AG | condone verification |
| Answer | Marks | Guidance |
|---|---|---|
| roots i, −i, α, β | or c + id, c − id | |
| Σα = α + β = 2 | B1 | 2c = 2 |
| Σαβ = 1 + αβ = 3 ⇒ αβ = 2 | B1 | ⇒ 1 + c2 + d2 = 3 ⇒ c = d |
| Σαβ γ = α + β = −a ⇒ a = −2 | B1 | or using 1+i, 1−i from |
| α2 − 2a + 2 = 0 AG | not established allow | |
| αβγδ = αβ = b ⇒ b = 2 | B1 | or using 1+i, 1−i AG |
| Answer | Marks | Guidance |
|---|---|---|
| (z + i)(z − i) = z2 + 1 | B1 | used (see below) |
| z4 − 2z3 + 3z2 + az + b = (z2 + cz + d)(z2 + 1) | M1 | comparing coeffs |
| z3: c = −2, z2: d + 1 = 3, so d = 2 | A1 | finding c, d |
| z: a = c = −2, constants: b = d = 2 | A1 | verifying a, b |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (b) | DR |
| Answer | Marks |
|---|---|
| roots are i, −i,1+i, 1−i | B1 |
| Answer | Marks |
|---|---|
| B1 | 1.1a |
| Answer | Marks |
|---|---|
| 1.1 | equating coeffs or long divn |
| Answer | Marks |
|---|---|
| or (z − 1 − i)(z − 1 + i) | working backwards: |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (b) | Alternative solution |
| Another root is −i | B1 | |
| Other roots α, β | M1 | or c + id, c − id |
| i + (−i) + α + β = 2 ⇒ α + β = 2 | M1 | or c + id + c − id = 2 |
| i(−i)αβ = 2 ⇒ αβ = 2 | M1 | or (c + id)(c − id) = 2 |
| ⇒ α2 − 2α + 2 = 0 | A1 | or c2 + d2 = 2 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | M1 | c = 1, d = 1 |
| roots are 1+i, 1−i, i, −i | B1 |
Question 8:
8 | (a) | DR
i4 −2i3+3i2 +ai+b=0
⇒1+2i−3+ai+b=0
⇒2+a=0, −2+b=0
⇒a=−2, b=2 | M1
A1
M1
E1
[4] | 2.1
1.1
2.1
2.2a | substituting z = i into eqn
equating real and im parts
AG | condone verification
Alternative solution
roots i, −i, α, β | or c + id, c − id
Σα = α + β = 2 | B1 | 2c = 2
Σαβ = 1 + αβ = 3 ⇒ αβ = 2 | B1 | ⇒ 1 + c2 + d2 = 3 ⇒ c = d
Σαβ γ = α + β = −a ⇒ a = −2 | B1 | or using 1+i, 1−i from | if 1+i, 1−i roots used but
α2 − 2a + 2 = 0 AG | not established allow
αβγδ = αβ = b ⇒ b = 2 | B1 | or using 1+i, 1−i AG | B0 B0 B1 B1
[4]
Alternative solution
(z + i)(z − i) = z2 + 1 | B1 | used (see below) | may be inferred from
z4 − 2z3 + 3z2 + az + b = (z2 + cz + d)(z2 + 1) | M1 | comparing coeffs | multiplication or
z3: c = −2, z2: d + 1 = 3, so d = 2 | A1 | finding c, d | long division
z: a = c = −2, constants: b = d = 2 | A1 | verifying a, b
[4]
8 | (b) | DR
Another root is −i
(z+i)(z−i)= z2 +1
z4 −2z3+3z2 −2z+2=(z2 +1)(z2 +cz+d)
z3 terms: −2=c
constants: 2=1×d ⇒d =2
2± −4
z2 −2z+2=0⇒ z =
2
roots are i, −i,1+i, 1−i | B1
B1
M1
A1
A1
M1
B1 | 1.1a
3.1a
1.1
1.1
1.1
1.1
1.1 | equating coeffs or long divn
solving their quad factor = 0
or (z − 1 − i)(z − 1 + i) | working backwards:
(z−1−i)(z−1+i)=z2−2z+2
B1
(z2+1)(z2−2z+2)=…
= z4−2z3+3z2−2z+2 B1
max 4 marks
[7]
8 | (b) | Alternative solution
Another root is −i | B1
Other roots α, β | M1 | or c + id, c − id
i + (−i) + α + β = 2 ⇒ α + β = 2 | M1 | or c + id + c − id = 2
i(−i)αβ = 2 ⇒ αβ = 2 | M1 | or (c + id)(c − id) = 2
⇒ α2 − 2α + 2 = 0 | A1 | or c2 + d2 = 2
2± −4
⇒α= = 1 ± i
2 | M1 | c = 1, d = 1
roots are 1+i, 1−i, i, −i | B1
PPMMTT
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© OCR 20198 In this question you must show detailed reasoning.
You are given that i is a root of the equation $z ^ { 4 } - 2 z ^ { 3 } + 3 z ^ { 2 } + a z + b = 0$, where $a$ and $b$ are real constants.
\begin{enumerate}[label=(\alph*)]
\item Show that $a = - 2$ and $b = 2$.
\item Find the other roots of this equation.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2019 Q8 [11]}}