| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive stationary point equation |
| Difficulty | Standard +0.3 This is a straightforward calculus question requiring quotient rule differentiation, setting dy/dx = 0 and algebraic manipulation to reach a given form, then applying fixed point iteration. All steps are standard A-level techniques with clear guidance (the equation form is given, iteration method is specified). Slightly easier than average due to the scaffolding provided. |
| Spec | 1.07q Product and quotient rules: differentiation1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Use quotient rule: \(\frac{dy}{dx} = \frac{3x^2(2-5x) - (8+x^3)(-5)}{(2-5x)^2}\) | M1 | Correct quotient rule structure |
| \(\frac{dy}{dx} = \frac{3x^2(2-5x)+5(8+x^3)}{(2-5x)^2}\) | A1 | Correct unsimplified form |
| Curve crosses x-axis when \(8 + x^3 = 0\), so \(x = -2\) | M1 | Set numerator of y = 0 |
| Gradient \(= \frac{3(4)(12)+5(0)}{144} = \frac{144}{144} = 1\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Set \(\frac{dy}{dx} = 0\): \(6x^2 - 15x^3 + 40 + 5x^3 = 0\) | M1 | |
| \(-10x^3 + 6x^2 + 40 = 0\) rearranged | M1 | |
| \(x^2 = 0.6x + 4x^{-1}\), hence \(x = \sqrt{0.6x + 4x^{-1}}\) | A1 | Completion of proof shown |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(x_1 = 1\) (or suitable starting value) | M1 | Use iterative formula |
| Successive iterations shown to 5 s.f. | M1 | At least 3 iterations shown |
| \(x \approx 1.69\) (3 s.f.) | A1 | cao |
## Question 6(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Use quotient rule: $\frac{dy}{dx} = \frac{3x^2(2-5x) - (8+x^3)(-5)}{(2-5x)^2}$ | M1 | Correct quotient rule structure |
| $\frac{dy}{dx} = \frac{3x^2(2-5x)+5(8+x^3)}{(2-5x)^2}$ | A1 | Correct unsimplified form |
| Curve crosses x-axis when $8 + x^3 = 0$, so $x = -2$ | M1 | Set numerator of y = 0 |
| Gradient $= \frac{3(4)(12)+5(0)}{144} = \frac{144}{144} = 1$ | A1 | cao |
## Question 6(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Set $\frac{dy}{dx} = 0$: $6x^2 - 15x^3 + 40 + 5x^3 = 0$ | M1 | |
| $-10x^3 + 6x^2 + 40 = 0$ rearranged | M1 | |
| $x^2 = 0.6x + 4x^{-1}$, hence $x = \sqrt{0.6x + 4x^{-1}}$ | A1 | Completion of proof shown |
## Question 6(iii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $x_1 = 1$ (or suitable starting value) | M1 | Use iterative formula |
| Successive iterations shown to 5 s.f. | M1 | At least 3 iterations shown |
| $x \approx 1.69$ (3 s.f.) | A1 | cao |6\\
\includegraphics[max width=\textwidth, alt={}, center]{6694ccc1-c8b1-42a7-8b21-829a89af74c9-08_732_807_258_667}
The diagram shows the curve with equation $y = \frac { 8 + x ^ { 3 } } { 2 - 5 x }$. The maximum point is denoted by $M$.\\
(i) Find an expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and determine the gradient of the curve at the point where the curve crosses the $x$-axis.\\
(ii) Show that the $x$-coordinate of the point $M$ satisfies the equation $x = \sqrt { } \left( 0.6 x + 4 x ^ { - 1 } \right)$.\\
(iii) Use an iterative formula, based on the equation in part (ii), to find the $x$-coordinate of $M$ correct to 3 significant figures. Give the result of each iteration to 5 significant figures.\\
\hfill \mbox{\textit{CAIE P2 2019 Q6 [9]}}