OCR MEI Further Pure Core AS 2019 June — Question 2 3 marks

Exam BoardOCR MEI
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2019
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeQuadratic with transformed roots
DifficultyStandard +0.3 This is a standard Further Maths question on transformed roots requiring application of Vieta's formulas and algebraic manipulation. While it involves multiple steps (finding sum and product of original roots, then transformed roots, then forming new equation), the technique is routine and well-practiced in Further Pure courses, making it slightly easier than average overall but typical for this topic.
Spec4.05b Transform equations: substitution for new roots
2 The roots of the equation \(3 x ^ { 2 } - x + 2 = 0\) are \(\alpha\) and \(\beta\).
Find a quadratic equation with integer coefficients whose roots are \(2 \alpha - 3\) and \(2 \beta - 3\).
Question 2:
AnswerMarks
2y+3  y+3 2 y+3
y =2x−3⇒ x= ⇒3  − +2=0
2  2  2
 y2 +6y+9 y+3
⇒3 − +2=0
 4  2
AnswerMarks
⇒3y2 +16y+29=0M1
A1
AnswerMarks
A11.1a
1.1
AnswerMarks
1.1y+3
subst x=
2
expanding bracket correctly
AnswerMarks
must have ‘=0’or x = (1±√23i)/6
new roots (−8±√23i)/3
B3 if fully correct
Alternative solution
α + β = 1/3, αβ = 2/3
⇒ 2α − 3 + 2β − 3 = 2(α + β) − 6 = −16/3
(2α − 3)(2β − 3) = 4αβ −6(α + β) + 9 = 29/3
New quadratic is y2 + (16/3)y + 29/3 = 0
AnswerMarks
⇒ 3y2 + 16y + 29 = 0attempting sum and product
of 2α − 3 and 2β − 3
sum = −16/3, product = 29/3
must have ‘= 0’
M1
A1
A1
[3]
Alternative solution
α + β = 1/3, αβ = 2/3
⇒ 2α − 3 + 2β − 3 = 2(α + β) − 6 = −16/3
(2α − 3)(2β − 3) = 4αβ −6(α + β) + 9 = 29/3
New quadratic is y2 + (16/3)y + 29/3 = 0
⇒ 3y2 + 16y + 29 = 0
attempting sum and product
of 2α − 3 and 2β − 3
sum = −16/3, product = 29/3
must have ‘= 0’
Question 2:
2 | y+3  y+3 2 y+3
y =2x−3⇒ x= ⇒3  − +2=0
2  2  2
 y2 +6y+9 y+3
⇒3 − +2=0
 4  2
⇒3y2 +16y+29=0 | M1
A1
A1 | 1.1a
1.1
1.1 | y+3
subst x=
2
expanding bracket correctly
must have ‘=0’ | or x = (1±√23i)/6
new roots (−8±√23i)/3
B3 if fully correct
Alternative solution
α + β = 1/3, αβ = 2/3
⇒ 2α − 3 + 2β − 3 = 2(α + β) − 6 = −16/3
(2α − 3)(2β − 3) = 4αβ −6(α + β) + 9 = 29/3
New quadratic is y2 + (16/3)y + 29/3 = 0
⇒ 3y2 + 16y + 29 = 0 | attempting sum and product
of 2α − 3 and 2β − 3
sum = −16/3, product = 29/3
must have ‘= 0’
M1
A1
A1
[3]
Alternative solution
α + β = 1/3, αβ = 2/3
⇒ 2α − 3 + 2β − 3 = 2(α + β) − 6 = −16/3
(2α − 3)(2β − 3) = 4αβ −6(α + β) + 9 = 29/3
New quadratic is y2 + (16/3)y + 29/3 = 0
⇒ 3y2 + 16y + 29 = 0
attempting sum and product
of 2α − 3 and 2β − 3
sum = −16/3, product = 29/3
must have ‘= 0’
2 The roots of the equation $3 x ^ { 2 } - x + 2 = 0$ are $\alpha$ and $\beta$.\\
Find a quadratic equation with integer coefficients whose roots are $2 \alpha - 3$ and $2 \beta - 3$.

\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2019 Q2 [3]}}
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