| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Substitution to find new equation |
| Difficulty | Challenging +1.2 This is a standard Further Pure 1 question on roots of polynomials requiring systematic application of Vieta's formulas and algebraic manipulation. Part (i) is routine substitution, parts (ii-iii) involve standard techniques (sum of squares from sum and product of roots), and part (iv) requires recognizing that a sum of squares being negative implies complex roots. While it's multi-step and requires careful algebra, it follows predictable patterns typical of FP1 examination questions without requiring novel insight. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(x = 1/y \Rightarrow 2y^4 - 4y^3 - cy^2 - y - 1 = 0\) | M1A1 | [2] |
| (ii) \(\sum \alpha^2 = 1 - 2c\) | M1A1 | |
| \(\sum \alpha^{-2} = 4 + c\) | A1 | |
| (M1 is for use of \(\sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta\) in either part.) | [3] | |
| (iii) \(S = \sum(\alpha - \alpha^{-1})^2 = \sum \alpha^2 + \sum \alpha^{-2} - 8 = -c - 3\) | M1A1√ | |
| A1√ is for adding answers to (ii) correctly and subtracting 8. | [2] | |
| (iv) \(c = -3 \Rightarrow S = 0\) so that if all roots are real then \(\alpha = \pm 1\) | M1A1 CWO | |
| and similarly for \(\beta, \gamma, \delta\) | ||
| This is impossible since e.g., \(\alpha\beta\delta = -2\), or any other contradiction | A1 CWO | [3] |
**(i)** $x = 1/y \Rightarrow 2y^4 - 4y^3 - cy^2 - y - 1 = 0$ | M1A1 | [2] |
**(ii)** $\sum \alpha^2 = 1 - 2c$ | M1A1 |
$\sum \alpha^{-2} = 4 + c$ | A1 |
(M1 is for use of $\sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta$ in either part.) | [3] |
**(iii)** $S = \sum(\alpha - \alpha^{-1})^2 = \sum \alpha^2 + \sum \alpha^{-2} - 8 = -c - 3$ | M1A1√ |
A1√ is for adding answers to (ii) correctly and subtracting 8. | [2] |
**(iv)** $c = -3 \Rightarrow S = 0$ so that if all roots are real then $\alpha = \pm 1$ | M1A1 CWO |
and similarly for $\beta, \gamma, \delta$ | |
This is impossible since e.g., $\alpha\beta\delta = -2$, or any other contradiction | A1 CWO | [3] |
---10 The equation
$$x ^ { 4 } + x ^ { 3 } + c x ^ { 2 } + 4 x - 2 = 0$$
where $c$ is a constant, has roots $\alpha , \beta , \gamma , \delta$.\\
(i) Use the substitution $y = \frac { 1 } { x }$ to find an equation which has roots $\frac { 1 } { \alpha } , \frac { 1 } { \beta } , \frac { 1 } { \gamma } , \frac { 1 } { \delta }$.\\
(ii) Find, in terms of $c$, the values of $\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } + \delta ^ { 2 }$ and $\frac { 1 } { \alpha ^ { 2 } } + \frac { 1 } { \beta ^ { 2 } } + \frac { 1 } { \gamma ^ { 2 } } + \frac { 1 } { \delta ^ { 2 } }$.\\
(iii) Hence find
$$\left( \alpha - \frac { 1 } { \alpha } \right) ^ { 2 } + \left( \beta - \frac { 1 } { \beta } \right) ^ { 2 } + \left( \gamma - \frac { 1 } { \gamma } \right) ^ { 2 } + \left( \delta - \frac { 1 } { \delta } \right) ^ { 2 }$$
in terms of $c$.\\
(iv) Deduce that when $c = - 3$ the roots of the given equation are not all real.
\hfill \mbox{\textit{CAIE FP1 2010 Q10 [10]}}