OCR MEI FP1 2005 January — Question 5 6 marks

Exam BoardOCR MEI
ModuleFP1 (Further Pure Mathematics 1)
Year2005
SessionJanuary
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeEquation with linearly transformed roots
DifficultyStandard +0.3 This is a standard Further Maths transformation of roots question requiring substitution y = x + 1, so x = y - 1, into the original cubic. The algebraic manipulation is straightforward with no conceptual difficulty beyond the routine technique taught in FP1. Slightly above average difficulty due to being Further Maths content, but this is a textbook exercise.
Spec4.05b Transform equations: substitution for new roots
5 The roots of the cubic equation \(x ^ { 3 } + 2 x ^ { 2 } + x - 3 = 0\) are \(\alpha , \beta\) and \(\gamma\).
Find the cubic equation whose roots are \(\alpha + 1 , \beta + 1\) and \(\gamma + 1\), simplifying your answer as far as you can.
Question 5:
AnswerMarks Guidance
AnswerMark Guidance
\(w = x+1 \Rightarrow x = w-1\)B1 Substitution. For substitution \(w = x-1\) give B0 but then follow through
\(\Rightarrow (w-1)^3 + 2(w-1)^2 + (w-1) - 3 = 0\)M1 Substitute into cubic
\(\Rightarrow w^3 - 3w^2 + 3w - 1 + 2w^2 - 4w + 2 + w - 1 - 3 = 0\)A1, A1 Expansion
\(\Rightarrow w^3 - w^2 - 3 = 0\)A1, [6] Simplifying
Alternative method:
AnswerMarks Guidance
AnswerMark Guidance
\(\alpha + \beta + \gamma = -2\)M1, A1 Attempt to calculate these; All correct
\(\alpha\beta + \beta\gamma + \alpha\gamma = 1\)
\(\alpha\beta\gamma = 3\)
Coefficient of \(w^2 = -1\)B1
\(w = 0\)B1
constant \(= -3\)B1
\(w^3 - w^2 - 3 = 0\)B1, [6] 
# Question 5:

| Answer | Mark | Guidance |
|--------|------|----------|
| $w = x+1 \Rightarrow x = w-1$ | B1 | Substitution. For substitution $w = x-1$ give B0 but then follow through |
| $\Rightarrow (w-1)^3 + 2(w-1)^2 + (w-1) - 3 = 0$ | M1 | Substitute into cubic |
| $\Rightarrow w^3 - 3w^2 + 3w - 1 + 2w^2 - 4w + 2 + w - 1 - 3 = 0$ | A1, A1 | Expansion |
| $\Rightarrow w^3 - w^2 - 3 = 0$ | A1, **[6]** | Simplifying |

**Alternative method:**

| Answer | Mark | Guidance |
|--------|------|----------|
| $\alpha + \beta + \gamma = -2$ | M1, A1 | Attempt to calculate these; All correct |
| $\alpha\beta + \beta\gamma + \alpha\gamma = 1$ | | |
| $\alpha\beta\gamma = 3$ | | |
| Coefficient of $w^2 = -1$ | B1 | |
| $w = 0$ | B1 | |
| constant $= -3$ | B1 | |
| $w^3 - w^2 - 3 = 0$ | B1, **[6]** | |

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5 The roots of the cubic equation $x ^ { 3 } + 2 x ^ { 2 } + x - 3 = 0$ are $\alpha , \beta$ and $\gamma$.\\
Find the cubic equation whose roots are $\alpha + 1 , \beta + 1$ and $\gamma + 1$, simplifying your answer as far as you can.

\hfill \mbox{\textit{OCR MEI FP1 2005 Q5 [6]}}
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