| Exam Board | OCR MEI |
|---|---|
| Module | Further Pure Core AS (Further Pure Core AS) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Equation with linearly transformed roots |
| Difficulty | Standard +0.8 This is a standard Further Maths transformation of roots question requiring systematic application of substitution techniques and relationship between roots and coefficients. Part (a) involves a linear transformation y = (x-1)/2, requiring substitution and algebraic manipulation to find the new equation. Part (b) requires recognizing that the transformed equation likely has simpler roots (possibly integers) that can be found by inspection or factor theorem, then back-transforming. While methodical, it requires multiple steps, careful algebra, and insight to connect the transformation to solving the original equation—placing it moderately above average difficulty. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (a) | DR |
| Answer | Marks |
|---|---|
| 8y3+32y=0 or y3+4y=0 | M1 |
| Answer | Marks |
|---|---|
| A1 | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | condone rearrangement slips |
| Answer | Marks | Guidance |
|---|---|---|
| Sum of new roots = 12 ( 3 ) 12 ( 3 3 ) 0 − = − = | B1 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 4 4 | M1 | attempt to find sum of pairs or product of new roots |
| Product = 18 ( 1 ) 18 (1 7 1 9 3 1 ) 0 − + − = − + − = | A1 | either correct (from correct working) |
| Equation is y 3 + 4 y = 0 | A1 | must be an equation |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (b) | DR |
| Answer | Marks |
|---|---|
| roots of original cubic are 1. 1 + 4i, 1 − 4i | B1ft |
| Answer | Marks |
|---|---|
| A1 | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | ft their cubic in y (but not solved BC) |
| Answer | Marks | Guidance |
|---|---|---|
| f(1) = 0 x −1 is a factor | B1 | |
| x3 − 3x2 + 19 x − 17= (x − 1)(x2 −2x + 17) | M1 | factorising by inspection or long division |
| x = 1, 1 + 4i or 1 − 4i | A1 |
Question 4:
4 | (a) | DR
let y = 12 ( x − 1 ) x = 2y + 1
( 2 y + 1 ) 3 − 3 ( 2 y + 1 ) 2 + 1 9 ( 2 y + 1 ) − 1 7 [ = 0 ]
8y3+12y2+6y+1−12y2−12y−3+38y+19−17=0
8y3+32y=0 or y3+4y=0 | M1
M1
A1
A1 | 1.1
1.1
1.1
1.1 | condone rearrangement slips
substituting for x in the cubic
expanding (2y + 1)3 correctly
must be an equation
Alternative solution
Sum of new roots = 12 ( 3 ) 12 ( 3 3 ) 0 − = − = | B1 | B1 | sum of new roots = 0 | sum of new roots = 0
Sum of pairs = 1(−2+3)= 1(19−6+3)=4
4 4 | M1 | attempt to find sum of pairs or product of new roots
Product = 18 ( 1 ) 18 (1 7 1 9 3 1 ) 0 − + − = − + − = | A1 | either correct (from correct working)
Equation is y 3 + 4 y = 0 | A1 | must be an equation
[4]
4 | (b) | DR
y = 0, 2i, −2i
x = 2y + 1
roots of original cubic are 1. 1 + 4i, 1 − 4i | B1ft
M1
A1 | 1.1
1.1
1.1 | ft their cubic in y (but not solved BC)
Alternative solution
f(1) = 0 x −1 is a factor | B1
x3 − 3x2 + 19 x − 17= (x − 1)(x2 −2x + 17) | M1 | factorising by inspection or long division
x = 1, 1 + 4i or 1 − 4i | A1
[3]4 In this question you must show detailed reasoning.
The roots of the cubic equation $x ^ { 3 } - 3 x ^ { 2 } + 19 x - 17 = 0$ are $\alpha , \beta$ and $\gamma$.
\begin{enumerate}[label=(\alph*)]
\item Find a cubic equation with integer coefficients whose roots are $\frac { 1 } { 2 } ( \alpha - 1 ) , \frac { 1 } { 2 } ( \beta - 1 )$ and $\frac { 1 } { 2 } ( \gamma - 1 )$.
\item Hence or otherwise solve the equation $x ^ { 3 } - 3 x ^ { 2 } + 19 x - 17 = 0$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2024 Q4 [7]}}