OCR Further Pure Core 1 2024 June — Question 4 3 marks

Exam BoardOCR
ModuleFurther Pure Core 1 (Further Pure Core 1)
Year2024
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeEquation with nonlinearly transformed roots
DifficultyChallenging +1.2 This is a Further Maths question on transformed roots requiring knowledge of symmetric functions and Vieta's formulas. Students must find αβγ, α+β+γ, and αβ+βγ+γα from the original equation, then construct symmetric functions of the new roots (which factor as (αβγ)² times the original roots). While systematic, it requires multiple steps and careful algebraic manipulation beyond standard A-level, placing it moderately above average difficulty.
Spec4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots

4 In this question you must show detailed reasoning.
The equation \(2 x ^ { 3 } + 3 x ^ { 2 } + 6 x - 3 = 0\) has roots \(\alpha , \beta\) and \(\gamma\).
Determine a cubic equation with integer coefficients that has roots \(\alpha ^ { 2 } \beta \gamma , \alpha \beta ^ { 2 } \gamma\) and \(\alpha \beta \gamma ^ { 2 }\).

Question 4:
AnswerMarks
4DR
( 32 ) ( 32 )  = − − =
w = 32 x
2 (2w)3+3 (2w)2 +6 (2w)−3(=0)
3 3 3
AnswerMarks
1 6 w 3 + 3 6 w 2 + 1 0 8 w − 8 1 = 0M1
M1
A1
AnswerMarks
[3]1.1
3.1a
AnswerMarks
2.2aFor ( 32 ) .  = − −
Appropriate substitution of the form w ( ) x  = with their value of
(need not be substituted into given cubic for this mark). Condone
reciprocal e.g. x = 32 w .
Allow any integer multiple but must have integer coefficients. Must be an
equation (so must = 0 or terms on both sides of an equation) with all
terms simplified. Condone if in terms of x.
Alternative method
=−3, =3, =−(−3)
AnswerMarks Guidance
2 2=−3, =3, =−(−3)
2 2M1 M1
implied by one correct coefficient in new cubic equation.For at least one of , ,      + + + + correct. Can be
9
( )    + + = −
4
27
()2(++)=
4
8 1
( ) 4  =
AnswerMarks Guidance
1 6M1 For at least two of ( ) 4 ,  ( ) 2 ( ) , ( )        + + + +
correct – not from incorrect values of ,++,++.
9 2 7 8 1
w 3 + w 2 + w − = 0
4 4 1 6
16w3+36w2+108w−81=0
AnswerMarks
A1Allow any integer multiple but must have integer coefficients. Must be an
equation (so must = 0 or terms on both sides of an equation) with all
terms simplified. Condone if in terms of x.
[3]
9
( )    + + = −
4
27
()2(++)=
4
8 1
( ) 4  =
1 6
M1
For at least two of ( ) 4 ,  ( ) 2 ( ) , ( )        + + + +
correct – not from incorrect values of ,++,++.
9 2 7 8 1
w 3 + w 2 + w − = 0
4 4 1 6
16w3+36w2+108w−81=0
AnswerMarks Guidance
QnAnswer Marks
Question 4:
4 | DR
( 32 ) ( 32 )  = − − =
w = 32 x
2 (2w)3+3 (2w)2 +6 (2w)−3(=0)
3 3 3
1 6 w 3 + 3 6 w 2 + 1 0 8 w − 8 1 = 0 | M1
M1
A1
[3] | 1.1
3.1a
2.2a | For ( 32 ) .  = − −
Appropriate substitution of the form w ( ) x  = with their value of
(need not be substituted into given cubic for this mark). Condone
reciprocal e.g. x = 32 w .
Allow any integer multiple but must have integer coefficients. Must be an
equation (so must = 0 or terms on both sides of an equation) with all
terms simplified. Condone if in terms of x.
Alternative method
=−3, =3, =−(−3)
2 2 | =−3, =3, =−(−3)
2 2 | M1 | M1 | For at least one of , ,      + + + + correct. Can be
implied by one correct coefficient in new cubic equation. | For at least one of , ,      + + + + correct. Can be
9
( )    + + = −
4
27
()2(++)=
4
8 1
( ) 4  =
1 6 | M1 | For at least two of ( ) 4 ,  ( ) 2 ( ) , ( )        + + + +
correct – not from incorrect values of ,++,++.
9 2 7 8 1
w 3 + w 2 + w − = 0
4 4 1 6
16w3+36w2+108w−81=0
A1 | Allow any integer multiple but must have integer coefficients. Must be an
equation (so must = 0 or terms on both sides of an equation) with all
terms simplified. Condone if in terms of x.
[3]
9
( )    + + = −
4
27
()2(++)=
4
8 1
( ) 4  =
1 6
M1
For at least two of ( ) 4 ,  ( ) 2 ( ) , ( )        + + + +
correct – not from incorrect values of ,++,++.
9 2 7 8 1
w 3 + w 2 + w − = 0
4 4 1 6
16w3+36w2+108w−81=0
Qn | Answer | Marks | AO | Guidance
4 In this question you must show detailed reasoning.\\
The equation $2 x ^ { 3 } + 3 x ^ { 2 } + 6 x - 3 = 0$ has roots $\alpha , \beta$ and $\gamma$.\\
Determine a cubic equation with integer coefficients that has roots $\alpha ^ { 2 } \beta \gamma , \alpha \beta ^ { 2 } \gamma$ and $\alpha \beta \gamma ^ { 2 }$.

\hfill \mbox{\textit{OCR Further Pure Core 1 2024 Q4 [3]}}