| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Substitution to find new equation |
| Difficulty | Standard +0.3 This is a standard Further Maths question on transforming polynomial equations using substitution and applying Vieta's formulas. The substitution y=1/x² is explicitly given, making part (i) routine algebraic manipulation. Parts (ii) and (iii) are direct applications of sum and sum-of-products formulas from the new equation. While it requires knowledge of Further Maths content, the execution is methodical with no novel problem-solving required, making it slightly easier than average. |
| Spec | 4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = \frac{1}{x^2} \Rightarrow x = \frac{1}{\sqrt{y}}\) | M1 | Rearranging to make \(x\) the subject |
| \(\frac{1}{y\sqrt{y}} + \frac{2}{y} - 3 = 0 \Rightarrow \frac{1}{y\sqrt{y}} = 3 - \frac{2}{y}\) | M1 | Substituting and squaring |
| \(\Rightarrow 9y^3 - 12y^2 + 4y - 1 = 0\); SR B1 for finding cubic by manipulating roots | A1 | OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{12}{9}\) or \(\frac{4}{3}\) | B1FT |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{\alpha^2\beta^2} + \frac{1}{\beta^2\gamma^2} + \frac{1}{\gamma^2\alpha^2} = \frac{4}{9}\) | B1FT |
## Question 1:
**Part (i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \frac{1}{x^2} \Rightarrow x = \frac{1}{\sqrt{y}}$ | M1 | Rearranging to make $x$ the subject |
| $\frac{1}{y\sqrt{y}} + \frac{2}{y} - 3 = 0 \Rightarrow \frac{1}{y\sqrt{y}} = 3 - \frac{2}{y}$ | M1 | Substituting and squaring |
| $\Rightarrow 9y^3 - 12y^2 + 4y - 1 = 0$; SR **B1** for finding cubic by manipulating roots | A1 | OE |
**Part (ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{12}{9}$ or $\frac{4}{3}$ | B1FT | |
**Part (iii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{\alpha^2\beta^2} + \frac{1}{\beta^2\gamma^2} + \frac{1}{\gamma^2\alpha^2} = \frac{4}{9}$ | B1FT | |
---1 The roots of the cubic equation $x ^ { 3 } + 2 x ^ { 2 } - 3 = 0$ are $\alpha , \beta$ and $\gamma$.\\
(i) By using the substitution $y = \frac { 1 } { x ^ { 2 } }$, find the cubic equation with roots $\frac { 1 } { \alpha ^ { 2 } } , \frac { 1 } { \beta ^ { 2 } }$ and $\frac { 1 } { \gamma ^ { 2 } }$.\\
(ii) Hence find the value of $\frac { 1 } { \alpha ^ { 2 } } + \frac { 1 } { \beta ^ { 2 } } + \frac { 1 } { \gamma ^ { 2 } }$.\\
(iii) Find also the value of $\frac { 1 } { \alpha ^ { 2 } \beta ^ { 2 } } + \frac { 1 } { \beta ^ { 2 } \gamma ^ { 2 } } + \frac { 1 } { \gamma ^ { 2 } \alpha ^ { 2 } }$.\\
\hfill \mbox{\textit{CAIE FP1 2017 Q1 [5]}}