OCR Further Pure Core AS 2021 November — Question 2 4 marks

Exam BoardOCR
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2021
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeSubstitution to find new equation
DifficultyStandard +0.3 This is a standard Further Maths technique requiring the substitution y = x + 1 (so x = y - 1) into the given cubic, then expanding and simplifying. While it involves algebraic manipulation across multiple steps, it's a routine textbook exercise with a well-established method that students practice regularly. Slightly above average difficulty due to the algebraic work required, but no novel insight needed.
Spec4.05b Transform equations: substitution for new roots
2 The equation \(2 x ^ { 3 } + 3 x ^ { 2 } - 2 x + 5 = 0\) has roots \(\alpha , \beta\) and \(\gamma\).
Use a substitution to find a cubic equation with integer coefficients whose roots are \(\alpha + 1 , \beta + 1\) and \(\gamma + 1\).
Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\(u = x + 1\)B1
\((u-1)^3 = u^3 - 3u^2 + 3u - 1\) used in solutionM1 Attempt to expand using binomial, 4 terms. Follow through on their \(u = x+1\)
\(2x^3 + 3x^2 - 2x + 5 = 0 \Rightarrow 2(u^3 - 3u^2 + 3u - 1) + 3(u^2 - 2u + 1) - 2(u-1) + 5 = 0\)M1 Substituting into equation. Allow if no "\(= 0\)". Must attempt expanding \((u-1)^3\) and \((u-1)^2\). Follow through on their \(u=x+1\)
\(2u^3 - 3u^2 - 2u + 8 = 0\)A1 Must be an equation. For correct equation found using sums and products of roots, allow SC2 (method required was dictated in question) 
## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $u = x + 1$ | B1 | |
| $(u-1)^3 = u^3 - 3u^2 + 3u - 1$ used in solution | M1 | Attempt to expand using binomial, 4 terms. Follow through on their $u = x+1$ |
| $2x^3 + 3x^2 - 2x + 5 = 0 \Rightarrow 2(u^3 - 3u^2 + 3u - 1) + 3(u^2 - 2u + 1) - 2(u-1) + 5 = 0$ | M1 | Substituting into equation. Allow if no "$= 0$". Must attempt expanding $(u-1)^3$ and $(u-1)^2$. Follow through on their $u=x+1$ |
| $2u^3 - 3u^2 - 2u + 8 = 0$ | A1 | Must be an equation. For correct equation found using sums and products of roots, allow SC2 (method required was dictated in question) |

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2 The equation $2 x ^ { 3 } + 3 x ^ { 2 } - 2 x + 5 = 0$ has roots $\alpha , \beta$ and $\gamma$.\\
Use a substitution to find a cubic equation with integer coefficients whose roots are $\alpha + 1 , \beta + 1$ and $\gamma + 1$.

\hfill \mbox{\textit{OCR Further Pure Core AS 2021 Q2 [4]}}
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Q1 5 Q2 4 Q3 6 Q4 7 Q5 8 Q6 6 Q7 5 Q8 6 Q9 13