| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Apply basic Euler method for differential equations |
| Difficulty | Standard +0.3 This is a straightforward application of two standard numerical methods (Euler's method and a mid-point variant) with given formulas. Students simply substitute values into provided formulas with minimal calculation steps. The function evaluation is simple, and no conceptual understanding of why these methods work is required—just mechanical substitution. |
| Spec | 1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(2, 5) = \frac{2 + 25}{2} = \frac{27}{2} = 13.5\) | M1 | Correct evaluation of \(f(x_0, y_0)\) |
| \(y_1 = 5 + 0.05 \times 13.5 = 5.675\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(2.05, 5.675) = \frac{2.05 + 5.675^2}{2.05} = \frac{2.05 + 32.205625}{2.05} = \frac{34.255625}{2.05} \approx 16.71\) | M1 | Correct evaluation of \(f(x_1, y_1)\) using answer from (a) |
| \(y_2 = y_0 + 2h \cdot f(x_1, y_1) = 5 + 2(0.05)(16.71...)\) | M1 | Correct application of formula with \(y_{r-1} = y_0 = 5\) |
| \(y(2.1) \approx 6.67\) | A1 | Answer to 3 significant figures |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(2, 5) = \frac{2 + 25}{2} = \frac{27}{2} = 13.5$ | M1 | Correct evaluation of $f(x_0, y_0)$ |
| $y_1 = 5 + 0.05 \times 13.5 = 5.675$ | A1 | Correct answer |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(2.05, 5.675) = \frac{2.05 + 5.675^2}{2.05} = \frac{2.05 + 32.205625}{2.05} = \frac{34.255625}{2.05} \approx 16.71$ | M1 | Correct evaluation of $f(x_1, y_1)$ using answer from (a) |
| $y_2 = y_0 + 2h \cdot f(x_1, y_1) = 5 + 2(0.05)(16.71...)$ | M1 | Correct application of formula with $y_{r-1} = y_0 = 5$ |
| $y(2.1) \approx 6.67$ | A1 | Answer to 3 significant figures |
---1 It is given that $y ( x )$ satisfies the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$
where
$$\mathrm { f } ( x , y ) = \frac { x + y ^ { 2 } } { x }$$
and
$$y ( 2 ) = 5$$
\begin{enumerate}[label=(\alph*)]
\item Use the Euler formula
$$y _ { r + 1 } = y _ { r } + h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$
with $h = 0.05$, to obtain an approximation to $y ( 2.05 )$.
\item Use the formula
$$y _ { r + 1 } = y _ { r - 1 } + 2 h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$
with your answer to part (a), to obtain an approximation to $y ( 2.1 )$, giving your answer to three significant figures.\\[0pt]
[3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2015 Q1 [5]}}