OCR Further Pure Core 2 2020 November — Question 2 6 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2020
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeEquation with nonlinearly transformed roots
DifficultyChallenging +1.3 This is a Further Maths question on transformed roots requiring systematic application of Newton's identities and algebraic manipulation. Part (a) involves finding sums of squares and products using Vieta's formulas, then constructing a new equation - a standard FM technique but with multiple careful steps. Part (b) requires recognizing the expression can be simplified using symmetric functions already computed. More demanding than typical A-level due to the algebraic complexity and FM content, but follows well-established procedures without requiring novel insight.
Spec4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots
2 In this question you must show detailed reasoning.
The roots of the equation \(3 x ^ { 3 } - 2 x ^ { 2 } - 5 x - 4 = 0\) are \(\alpha , \beta\) and \(\gamma\).
  1. Find a cubic equation with integer coefficients whose roots are \(\alpha ^ { 2 } , \beta ^ { 2 }\) and \(\gamma ^ { 2 }\).
  2. Find the exact value of \(\frac { \alpha ^ { 2 } \beta ^ { 2 } + \beta ^ { 2 } \gamma ^ { 2 } + \gamma ^ { 2 } \alpha ^ { 2 } } { \alpha \beta \gamma }\).
Question 2:
AnswerMarks Guidance
2(a) DR
u = x2
( )3 ( )2
3 u −2 u −5 u−4 (=0)
3u u−5 u =2u+4⇒u(3u−5)2 =(2u+4)2
u(9u2 – 30u + 25) = 4u2 + 16u + 16 =>
AnswerMarks
9u3 – 34u2 + 9u – 16 = 0B1
M1
M1
AnswerMarks
A13.1a
1.1
1.1
AnswerMarks
3.2aCorrect substitution chosen
Oe
Attempting to make substitution
Rearranging and squaring bs to
remove the square root(s)
AnswerMarks
Rearranging to answeror preparation for substitution by
removing odd powers.
eg x2(3x2 – 5)2 = (2x2 + 4)2...
...and then substituting
u(3u – 5)2 = (2u + 4)2
Equation can be in x
AnswerMarks Guidance
Alternative methodB1 Must include one intermediate
step
DR
 −4 2 16
α2β2γ2 =(αβγ)2 =−  =
 3  9
α2β2+β2γ2+γ2α2
AnswerMarks Guidance
=(αβ+βγ+γα)2−2αβγ(α+β+γ)α2β2+β2γ2+γ2α2
=(αβ+βγ+γα)2−2αβγ(α+β+γ)M1 Writing the expression in terms
of standard symmetrical forms2 5
NB ∑α= , ∑αβ=− ,
3 3
4
αβγ=
3
AnswerMarks Guidance
α2+β2+γ2 =(α+β+γ)2−2(αβ+βγ+γα)M1 Writing the expression in terms
of standard symmetrical forms“2”
2 2 −5 −5 2 4 2 16
u3−   −2×   u2+   −2× ×   u−
3 3   3  3 3  9
34 16
=u3− u2+u− =0⇒9u3−34u2+9u−16=0
AnswerMarks Guidance
9 9A1 Substituting in and rearranging to
answerSubstituting in and rearranging to
answer34
NB ∑α2 = , ∑α2β2 =1
9
[4]
B1
Must include one intermediate
step
Writing the expression in terms
of standard symmetrical forms
Alternative method for last 2 marks:
=
- … -
1 - 1 - 1
4 7 𝑛𝑛−1
1 - 1 1 - 1
5 8 𝑛𝑛−3 𝑛𝑛
1 - 1 1 - 1
6 9 𝑛𝑛−2 𝑛𝑛+1
1 … 1 1
AnswerMarks Guidance
7 𝑛𝑛−1 𝑛𝑛+2M1 At least these terms
M1 can be ft from any A, B
having opposite signs
1
For M1, condone omission of
7
1
or -
n−1
1 1 1 1 1 1
= + + − − −
4 5 6 n n+1 n+2
37 1 1 1
= − − −
AnswerMarks Guidance
60 n n+1 n+2A1 AG. Correct cancellation to AG.
Must have either clear diagonal
cancellations with one explicit
cancellation or less clear
cancellation with numerical and
algebraic cancellation shown or
described
[5]
AnswerMarks
(b)37
= or awrt 0.617
AnswerMarks
60B1
[1]2.2a
Alternative method for last 2 marks:
=
AnswerMarks Guidance
-… -
1
AnswerMarks Guidance
4- 1
7- 1
𝑛𝑛−1
AnswerMarks Guidance
1- 1
81 -
5𝑛𝑛−3 𝑛𝑛
1- 1
91 1
6𝑛𝑛−2 𝑛𝑛+1
1… 1
M1
At least these terms
A1
AG. Correct cancellation to AG.
Question 2:
2 | (a) | DR
u = x2
( )3 ( )2
3 u −2 u −5 u−4 (=0)
3u u−5 u =2u+4⇒u(3u−5)2 =(2u+4)2
u(9u2 – 30u + 25) = 4u2 + 16u + 16 =>
9u3 – 34u2 + 9u – 16 = 0 | B1
M1
M1
A1 | 3.1a
1.1
1.1
3.2a | Correct substitution chosen
Oe
Attempting to make substitution
Rearranging and squaring bs to
remove the square root(s)
Rearranging to answer | or preparation for substitution by
removing odd powers.
eg x2(3x2 – 5)2 = (2x2 + 4)2...
...and then substituting
u(3u – 5)2 = (2u + 4)2
Equation can be in x
Alternative method | B1 | Must include one intermediate
step
DR
 −4 2 16
α2β2γ2 =(αβγ)2 =−  =
 3  9
α2β2+β2γ2+γ2α2
=(αβ+βγ+γα)2−2αβγ(α+β+γ) | α2β2+β2γ2+γ2α2
=(αβ+βγ+γα)2−2αβγ(α+β+γ) | M1 | Writing the expression in terms
of standard symmetrical forms | 2 5
NB ∑α= , ∑αβ=− ,
3 3
4
αβγ=
3
α2+β2+γ2 =(α+β+γ)2−2(αβ+βγ+γα) | M1 | Writing the expression in terms | Condone without factorisation of
of standard symmetrical forms | “2”
2 2 −5 −5 2 4 2 16
u3−   −2×   u2+   −2× ×   u−
3 3   3  3 3  9
34 16
=u3− u2+u− =0⇒9u3−34u2+9u−16=0
9 9 | A1 | Substituting in and rearranging to
answer | Substituting in and rearranging to
answer | 34
NB ∑α2 = , ∑α2β2 =1
9
[4]
B1
Must include one intermediate
step
Writing the expression in terms
of standard symmetrical forms
Alternative method for last 2 marks:
=
- … -
1 - 1 - 1
4 7 𝑛𝑛−1
1 - 1 1 - 1
5 8 𝑛𝑛−3 𝑛𝑛
1 - 1 1 - 1
6 9 𝑛𝑛−2 𝑛𝑛+1
1 … 1 1
7 𝑛𝑛−1 𝑛𝑛+2 | M1 | At least these terms
M1 can be ft from any A, B
having opposite signs
1
For M1, condone omission of
7
1
or -
n−1
1 1 1 1 1 1
= + + − − −
4 5 6 n n+1 n+2
37 1 1 1
= − − −
60 n n+1 n+2 | A1 | AG. Correct cancellation to AG. | Requires joined argument
Must have either clear diagonal
cancellations with one explicit
cancellation or less clear
cancellation with numerical and
algebraic cancellation shown or
described
[5]
(b) | 37
= or awrt 0.617
60 | B1
[1] | 2.2a
Alternative method for last 2 marks:
=
- | … | -
1
4 | - | 1
7 | - | 1
𝑛𝑛−1
1 | - | 1
8 | 1 | - | 1
5 | 𝑛𝑛−3 | 𝑛𝑛
1 | - | 1
9 | 1 | 1 | - | 1
6 | 𝑛𝑛−2 | 𝑛𝑛+1
1 | … | 1 | 1
M1
At least these terms
A1
AG. Correct cancellation to AG.
2 In this question you must show detailed reasoning.\\
The roots of the equation $3 x ^ { 3 } - 2 x ^ { 2 } - 5 x - 4 = 0$ are $\alpha , \beta$ and $\gamma$.
\begin{enumerate}[label=(\alph*)]
\item Find a cubic equation with integer coefficients whose roots are $\alpha ^ { 2 } , \beta ^ { 2 }$ and $\gamma ^ { 2 }$.
\item Find the exact value of $\frac { \alpha ^ { 2 } \beta ^ { 2 } + \beta ^ { 2 } \gamma ^ { 2 } + \gamma ^ { 2 } \alpha ^ { 2 } } { \alpha \beta \gamma }$.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 2 2020 Q2 [6]}}
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