OCR MEI FP1 2013 June — Question 6 7 marks

Exam BoardOCR MEI
ModuleFP1 (Further Pure Mathematics 1)
Year2013
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeEquation with linearly transformed roots
DifficultyStandard +0.8 This is a standard Further Maths transformation of roots question requiring systematic application of substitution and algebraic manipulation. While it involves multiple steps (substituting y = x/3 + 1, rearranging to x = 3(y-1), substituting into the original equation, and simplifying), the technique is well-practiced in FP1. It's harder than typical A-level questions due to the algebraic complexity and being Further Maths content, but remains a textbook exercise without requiring novel insight.
Spec4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots

6 The cubic equation \(x ^ { 3 } - 5 x ^ { 2 } + 3 x - 6 = 0\) has roots \(\alpha , \beta\) and \(\gamma\). Find a cubic equation with roots \(\frac { \alpha } { 3 } + 1 , \frac { \beta } { 3 } + 1\) and \(\frac { \gamma } { 3 } + 1\), simplifying your answer as far as possible.

Question 6:
AnswerMarks
\(w = \frac{x}{3} + 1 \Rightarrow 3(w-1) = x\)M1
\(x^3 - 5x^2 + 3x - 6 = 0\)
AnswerMarks Guidance
\(\Rightarrow (3(w-1))^3 - 5(3(w-1))^2 + 3(3(w-1)) - 6 = 0\)M1 Substituting
A1Correct
\(\Rightarrow 27(w^3 - 3w^2 + 3w - 1) - 45(w^2 - 2w + 1) + 9w - 15 = 0\)
AnswerMarks Guidance
\(\Rightarrow 27w^3 - 126w^2 + 180w - 87 = 0\)A3 FT \(x = 3w+3\), \(3w \pm 1\), \(-1\) each error
\(\Rightarrow 9w^3 - 42w^2 + 60w - 29 = 0\)A1 cao
OR
AnswerMarks Guidance
In original equation \(\sum\alpha = 5\), \(\sum\alpha\beta = 3\), \(\alpha\beta\gamma = 6\)M1A1 all correct for A1
New roots \(A, B, \Gamma\):
AnswerMarks Guidance
\(\sum A = \frac{\sum\alpha}{3} + 3\), \(\sum AB = \frac{\sum\alpha\beta}{9} + \frac{2}{3}\sum\alpha + 3\)M1 At least two relations attempted
\(AB\Gamma = \frac{\alpha\beta\gamma}{27} + \frac{\sum\alpha\beta}{9} + \frac{\sum\alpha}{3} + 1\)A3 Correct \(-1\) each error FT their 5,3,6
Fully correct equationA1 cao, accept rational coefficients here
[7]
## Question 6:
$w = \frac{x}{3} + 1 \Rightarrow 3(w-1) = x$ | M1 |
$x^3 - 5x^2 + 3x - 6 = 0$
$\Rightarrow (3(w-1))^3 - 5(3(w-1))^2 + 3(3(w-1)) - 6 = 0$ | M1 | Substituting
| A1 | Correct
$\Rightarrow 27(w^3 - 3w^2 + 3w - 1) - 45(w^2 - 2w + 1) + 9w - 15 = 0$
$\Rightarrow 27w^3 - 126w^2 + 180w - 87 = 0$ | A3 | FT $x = 3w+3$, $3w \pm 1$, $-1$ each error
$\Rightarrow 9w^3 - 42w^2 + 60w - 29 = 0$ | A1 | cao

**OR**

In original equation $\sum\alpha = 5$, $\sum\alpha\beta = 3$, $\alpha\beta\gamma = 6$ | M1A1 | all correct for A1
New roots $A, B, \Gamma$:
$\sum A = \frac{\sum\alpha}{3} + 3$, $\sum AB = \frac{\sum\alpha\beta}{9} + \frac{2}{3}\sum\alpha + 3$ | M1 | At least two relations attempted
$AB\Gamma = \frac{\alpha\beta\gamma}{27} + \frac{\sum\alpha\beta}{9} + \frac{\sum\alpha}{3} + 1$ | A3 | Correct $-1$ each error FT their 5,3,6
Fully correct equation | A1 | cao, accept rational coefficients here
**[7]**

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6 The cubic equation $x ^ { 3 } - 5 x ^ { 2 } + 3 x - 6 = 0$ has roots $\alpha , \beta$ and $\gamma$. Find a cubic equation with roots $\frac { \alpha } { 3 } + 1 , \frac { \beta } { 3 } + 1$ and $\frac { \gamma } { 3 } + 1$, simplifying your answer as far as possible.

\hfill \mbox{\textit{OCR MEI FP1 2013 Q6 [7]}}