OCR MEI Further Pure Core 2021 November — Question 8 9 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2021
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeRoots with special relationships
DifficultyChallenging +1.2 This is a Further Maths question on roots with special relationships, requiring use of Vieta's formulas and algebraic manipulation. The symmetric structure (α, -α, β, 1/β) provides strong constraints that make the problem tractable through systematic application of sum and product of roots, though it requires careful algebra across multiple steps. More challenging than standard A-level but routine for Further Maths students who have practiced this topic.
Spec4.05a Roots and coefficients: symmetric functions
8 The equation \(4 \mathrm { x } ^ { 4 } - 4 \mathrm { x } ^ { 3 } + \mathrm { px } ^ { 2 } + \mathrm { qx } - 9 = 0\), where \(p\) and \(q\) are constants, has roots \(\alpha , - \alpha , \beta\) and \(\frac { 1 } { \beta }\).
  1. Determine the exact roots of the equation.
  2. Determine the values of \(p\) and \(q\).
Question 8:
AnswerMarks Guidance
8(a) 1 9
( )     −   = −
4 
3
𝛼 = (±)
2
1
Sum of roots 1  = + =
2−+1=0
1 i 3 1 1 i 3 
,   = =
2 2 
3 1i 3
so roots are  and
AnswerMarks
2 2M1
A1
M1
A1
AnswerMarks
A12.1
2.2a
2.1
2.1
2.2a
AnswerMarks
Alternative solutionM1
A1
A1
M1
A1
( 1 )
( x ) ( x ) ( x ) x    − + − −
( 1 ) ( 1 )
x 4 x 3 (1 2 ) x 2 2 x 2      = + + + − − + −
 
9 3
2    =  = 
4 2
1
+ =12−+1=0
1 i 3 1 1 i 3 
,   = =
2 2 
3 1  i 3
so roots are  a n d
2 2
[5]
M1
A1
A1
M1
A1
AnswerMarks Guidance
8(b)  
Sum of pairs ( ) 1     = − + + − − +
 
5
=1−2 =−  p=−5
4
( 1) 9 9
Sum of triples =−2 + =− 1=−
 4 4
AnswerMarks
so q=9M1
A1
M1
AnswerMarks
A12.1
2.2a
2.1
AnswerMarks
2.2aUnsimplified
simplified
AnswerMarks
Alternative solutionAward for either 𝑥2 𝑜𝑟 𝑥 coefficient
correct
( 3)( 3)( 1+i 3)( 1−i 3)
x− x+ x− x−
AnswerMarks
2 2 2 2M1
( 9 )
= x 2 − ( x 2 − x + 1 )
4
A1
5 9 9
= x 4 − x 3 − x 2 + x −
4 4 4
A1
AnswerMarks
so p = −5 and q = 9A1
[4]
Award for either 𝑥2 𝑜𝑟 𝑥 coefficient
correct
Question 8:
8 | (a) | 1 9
( )     −   = −
4 
3
𝛼 = (±)
2
1
Sum of roots 1  = + =

2−+1=0
1 i 3 1 1 i 3 
,   = =
2 2 
3 1i 3
so roots are  and
2 2 | M1
A1
M1
A1
A1 | 2.1
2.2a
2.1
2.1
2.2a
Alternative solution | M1
A1
A1
M1
A1
( 1 )
( x ) ( x ) ( x ) x    − + − −

( 1 ) ( 1 )
x 4 x 3 (1 2 ) x 2 2 x 2      = + + + − − + −
 
9 3
2    =  = 
4 2
1
+ =12−+1=0

1 i 3 1 1 i 3 
,   = =
2 2 
3 1  i 3
so roots are  a n d
2 2
[5]
M1
A1
A1
M1
A1
8 | (b) |  
Sum of pairs ( ) 1     = − + + − − +
 
5
=1−2 =−  p=−5
4
( 1) 9 9
Sum of triples =−2 + =− 1=−
 4 4
so q=9 | M1
A1
M1
A1 | 2.1
2.2a
2.1
2.2a | Unsimplified
simplified
Alternative solution | Award for either 𝑥2 𝑜𝑟 𝑥 coefficient
correct
( 3)( 3)( 1+i 3)( 1−i 3)
x− x+ x− x−
2 2 2 2 | M1
( 9 )
= x 2 − ( x 2 − x + 1 )
4
A1
5 9 9
= x 4 − x 3 − x 2 + x −
4 4 4
A1
so p = −5 and q = 9 | A1
[4]
Award for either 𝑥2 𝑜𝑟 𝑥 coefficient
correct
8 The equation $4 \mathrm { x } ^ { 4 } - 4 \mathrm { x } ^ { 3 } + \mathrm { px } ^ { 2 } + \mathrm { qx } - 9 = 0$, where $p$ and $q$ are constants, has roots $\alpha , - \alpha , \beta$ and $\frac { 1 } { \beta }$.
\begin{enumerate}[label=(\alph*)]
\item Determine the exact roots of the equation.
\item Determine the values of $p$ and $q$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2021 Q8 [9]}}
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