OCR MEI FP1 2014 June — Question 5 7 marks

Exam BoardOCR MEI
ModuleFP1 (Further Pure Mathematics 1)
Year2014
SessionJune
Marks7
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 knowledge of relationships between roots and coefficients, plus systematic substitution. While methodical, it involves multiple steps (finding the transformation y = 3x - 1, substituting x = (y+1)/3, and simplifying) and is more demanding than typical A-level questions, placing it moderately above average difficulty.
Spec4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots
5 The roots of the cubic equation \(3 x ^ { 3 } - 9 x ^ { 2 } + x - 1 = 0\) are \(\alpha , \beta\) and \(\gamma\). Find the cubic equation whose roots are \(3 \alpha - 1,3 \beta - 1\) and \(3 \gamma - 1\), expressing your answer in a form with integer coefficients.
Question 5:
Either method:
AnswerMarks Guidance
\(y = 3x-1 \Rightarrow x = \dfrac{y+1}{3}\)M1* Change of variable, condone \(\dfrac{y-1}{3}\), \(\dfrac{y}{3}\pm 1\)
\(\Rightarrow 3\left(\dfrac{y+1}{3}\right)^3 - 9\left(\dfrac{y+1}{3}\right)^2 + \left(\dfrac{y+1}{3}\right) - 1 = 0\)M1dep* A1 Substitute into cubic expression. Correct
Correct coefficients in cubic expression (may be fractions)A3ft ft their substitution (−1 each error)
\(\Rightarrow y^3 - 6y^2 - 12y - 14 = 0\)A1 [7] cao. Must be equation with integer coefficients
Or method:
\(\alpha+\beta+\gamma = \dfrac{9}{3} = 3\)
\(\alpha\beta+\alpha\gamma+\beta\gamma = \dfrac{1}{3}\)
AnswerMarks Guidance
\(\alpha\beta\gamma = \dfrac{1}{3}\)M1 A1 All three root relations, condone incorrect signs. All correct
Let new roots be \(k, l, m\) then:
\(k+l+m = 3(\alpha+\beta+\gamma)-3 = 6\)
\(kl+km+lm = 9(\alpha\beta+\alpha\gamma+\beta\gamma)-6(\alpha+\beta+\gamma)+3 = -12\)
AnswerMarks Guidance
\(klm = 27\alpha\beta\gamma - 9(\alpha\beta+\beta\gamma+\beta\gamma)+3(\alpha+\beta+\gamma)-1 = 14\)M1 A3ft Using \((3\alpha-1)\) etc in \(\sum k, \sum kl, klm\), at least two attempted. One each for \(6, -12, 14\), ft their \(3, \frac{1}{3}, \frac{1}{3}\)
\(\Rightarrow y^3 - 6y^2 - 12y - 14 = 0\)A1 [7] cao. Must be equation with integer coefficients 
## Question 5:

**Either method:**

$y = 3x-1 \Rightarrow x = \dfrac{y+1}{3}$ | M1* | Change of variable, condone $\dfrac{y-1}{3}$, $\dfrac{y}{3}\pm 1$

$\Rightarrow 3\left(\dfrac{y+1}{3}\right)^3 - 9\left(\dfrac{y+1}{3}\right)^2 + \left(\dfrac{y+1}{3}\right) - 1 = 0$ | M1dep* A1 | Substitute into cubic expression. Correct

Correct coefficients in cubic expression (may be fractions) | A3ft | ft their substitution (−1 each error)

$\Rightarrow y^3 - 6y^2 - 12y - 14 = 0$ | A1 **[7]** | cao. Must be equation with integer coefficients

**Or method:**

$\alpha+\beta+\gamma = \dfrac{9}{3} = 3$

$\alpha\beta+\alpha\gamma+\beta\gamma = \dfrac{1}{3}$

$\alpha\beta\gamma = \dfrac{1}{3}$ | M1 A1 | All three root relations, condone incorrect signs. All correct

Let new roots be $k, l, m$ then:

$k+l+m = 3(\alpha+\beta+\gamma)-3 = 6$

$kl+km+lm = 9(\alpha\beta+\alpha\gamma+\beta\gamma)-6(\alpha+\beta+\gamma)+3 = -12$

$klm = 27\alpha\beta\gamma - 9(\alpha\beta+\beta\gamma+\beta\gamma)+3(\alpha+\beta+\gamma)-1 = 14$ | M1 A3ft | Using $(3\alpha-1)$ etc in $\sum k, \sum kl, klm$, at least two attempted. One each for $6, -12, 14$, ft their $3, \frac{1}{3}, \frac{1}{3}$

$\Rightarrow y^3 - 6y^2 - 12y - 14 = 0$ | A1 **[7]** | cao. Must be equation with integer coefficients

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5 The roots of the cubic equation $3 x ^ { 3 } - 9 x ^ { 2 } + x - 1 = 0$ are $\alpha , \beta$ and $\gamma$. Find the cubic equation whose roots are $3 \alpha - 1,3 \beta - 1$ and $3 \gamma - 1$, expressing your answer in a form with integer coefficients.

\hfill \mbox{\textit{OCR MEI FP1 2014 Q5 [7]}}
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