OCR MEI Further Pure Core 2020 November — Question 4 8 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2020
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeEquation with linearly transformed roots
DifficultyStandard +0.8 This is a standard Further Maths question on transformed roots requiring Vieta's formulas and substitution techniques. Part (a) is routine (finding sum of reciprocals = -b/d), but part (b) requires the systematic substitution y = 2x - 1 to derive a new cubic equation, which is more involved than typical A-level work though still a textbook technique for Further Pure.
Spec4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots
4 The roots of the equation \(2 x ^ { 3 } - 5 x + 7 = 0\) are \(\alpha , \beta\) and \(\gamma\).
  1. Find \(\frac { 1 } { \alpha } + \frac { 1 } { \beta } + \frac { 1 } { \gamma }\).
  2. Find an equation with integer coefficients whose roots are \(2 \alpha - 1,2 \beta - 1\) and \(2 \gamma - 1\).
Question 4:
AnswerMarks Guidance
4(a) 2 x 3 ​– 5 x + 7 = 0
​ ​ ​ ​
[α+β+γ = 0],βγ +αγ +αβ =− 5, αβγ =− 7
AnswerMarks
2 2B1
B1
M1
A1
AnswerMarks
[4]1.1a
3.1a
1.1
1.1
PPMMTT
Y420/01 Mark SchemeNovember 2020
4 (b) 2x3 ​– 5x + 7 = 0
​​ ​​
let y = 2x – 1, x = ½ (y + 1) M1 1.1a
​​ ​​ ​​ ​​
M1 1.1 substituting for x (not 2y - 1)
⇒ ​​ ​​
⇒ A2,1,0 1.1,1.1
OR
B1 or by expanding must expand fully for first B1
sum of roots = 2(α + β + γ) – 3 = −3
​ ​ ​ ​ ​​ (x−2α +1))(x −2β+1)(x−2γ+1)
(2α–1)(2β−1)+(2β−1)(2γ−1)+(2γ−1)(2α-1) ​​ ​ ​ ​​ ​ ​ ​​
​ ​ ​ ​ ​ ​ ​​ ​​ ​ ​ B1
=4(αβ+βγ+γα) – 4(α+β+γ) + 3 = −10 + 3 = −7
​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​​
(2α–1)(2β−1)(2γ−1)
​ ​ ​ ​ ​​ B1
= 8αβγ−4(αβ+βγ+γα)+2(α+β+γ)−1 = −19
​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​​
⇒ B1ft must be an equation
[4]
5 (a) A is [5a, 0], B1 1.1 SC Coordinates reversed
​ ​
B is [3a, ½π] B1 1.1 (θ, r ) award B1B0
​ ​ ​ ​
[2]
5 (b) cos (−θ) = cos θ M1 2.4 accept even function
​ ​ ​
so the value of r for −θ is the same as for θ A1 2.2a
​​ ​ ​ ​
[2]
8
AnswerMarks Guidance
4(b) 2x3 ​– 5x + 7 = 0
​​ ​​
let y = 2x – 1, x = ½ (y + 1)
​​ ​​ ​​ ​​
⇒
⇒
OR
sum of roots = 2(α + β + γ) – 3 = −3
​ ​ ​ ​ ​​
(2α–1)(2β−1)+(2β−1)(2γ−1)+(2γ−1)(2α-1)
​ ​ ​ ​ ​ ​ ​​ ​​ ​ ​
=4(αβ+βγ+γα) – 4(α+β+γ) + 3 = −10 + 3 = −7
​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​​
(2α–1)(2β−1)(2γ−1)
​ ​ ​ ​ ​​
= 8αβγ−4(αβ+βγ+γα)+2(α+β+γ)−1 = −19
​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​​
AnswerMarks
⇒M1
M1
A2,1,0
B1
B1
B1
B1ft
AnswerMarks
[4]1.1a
1.1
AnswerMarks
1.1,1.1substituting for x (not 2y - 1)
​​ ​​
or by expanding
(x−2α +1))(x −2β+1)(x−2γ+1)
​​ ​ ​ ​​ ​ ​ ​​
AnswerMarks
must be an equationmust expand fully for first B1 
Question 4:
4 | (a) | 2 x 3 ​– 5 x + 7 = 0
​ ​ ​ ​
[α+β+γ = 0],βγ +αγ +αβ =− 5, αβγ =− 7
2 2 | B1
B1
M1
A1
[4] | 1.1a
3.1a
1.1
1.1
PPMMTT
Y420/01 Mark SchemeNovember 2020
4 (b) 2x3 ​– 5x + 7 = 0
​​ ​​
let y = 2x – 1, x = ½ (y + 1) M1 1.1a
​​ ​​ ​​ ​​
M1 1.1 substituting for x (not 2y - 1)
⇒ ​​ ​​
⇒ A2,1,0 1.1,1.1
OR
B1 or by expanding must expand fully for first B1
sum of roots = 2(α + β + γ) – 3 = −3
​ ​ ​ ​ ​​ (x−2α +1))(x −2β+1)(x−2γ+1)
(2α–1)(2β−1)+(2β−1)(2γ−1)+(2γ−1)(2α-1) ​​ ​ ​ ​​ ​ ​ ​​
​ ​ ​ ​ ​ ​ ​​ ​​ ​ ​ B1
=4(αβ+βγ+γα) – 4(α+β+γ) + 3 = −10 + 3 = −7
​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​​
(2α–1)(2β−1)(2γ−1)
​ ​ ​ ​ ​​ B1
= 8αβγ−4(αβ+βγ+γα)+2(α+β+γ)−1 = −19
​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​​
⇒ B1ft must be an equation
[4]
5 (a) A is [5a, 0], B1 1.1 SC Coordinates reversed
​ ​
B is [3a, ½π] B1 1.1 (θ, r ) award B1B0
​ ​ ​ ​
[2]
5 (b) cos (−θ) = cos θ M1 2.4 accept even function
​ ​ ​
so the value of r for −θ is the same as for θ A1 2.2a
​​ ​ ​ ​
[2]
8
4 | (b) | 2x3 ​– 5x + 7 = 0
​​ ​​
let y = 2x – 1, x = ½ (y + 1)
​​ ​​ ​​ ​​
⇒
⇒
OR
sum of roots = 2(α + β + γ) – 3 = −3
​ ​ ​ ​ ​​
(2α–1)(2β−1)+(2β−1)(2γ−1)+(2γ−1)(2α-1)
​ ​ ​ ​ ​ ​ ​​ ​​ ​ ​
=4(αβ+βγ+γα) – 4(α+β+γ) + 3 = −10 + 3 = −7
​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​​
(2α–1)(2β−1)(2γ−1)
​ ​ ​ ​ ​​
= 8αβγ−4(αβ+βγ+γα)+2(α+β+γ)−1 = −19
​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​​
⇒ | M1
M1
A2,1,0
B1
B1
B1
B1ft
[4] | 1.1a
1.1
1.1,1.1 | substituting for x (not 2y - 1)
​​ ​​
or by expanding
(x−2α +1))(x −2β+1)(x−2γ+1)
​​ ​ ​ ​​ ​ ​ ​​
must be an equation | must expand fully for first B1
4 The roots of the equation $2 x ^ { 3 } - 5 x + 7 = 0$ are $\alpha , \beta$ and $\gamma$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { 1 } { \alpha } + \frac { 1 } { \beta } + \frac { 1 } { \gamma }$.
\item Find an equation with integer coefficients whose roots are $2 \alpha - 1,2 \beta - 1$ and $2 \gamma - 1$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2020 Q4 [8]}}
This paper (16 questions)
View full paper
Q1 6 Q2 6 Q3 4 Q4 8 Q5 8 Q6 4 Q7 6 Q8 9 Q9 8 Q10 7 Q11 8 Q12 8 Q13 9 Q14 11 Q15 17 Q16 25