AQA FP1 2013 January — Question 1 5 marks

Exam BoardAQA
ModuleFP1 (Further Pure Mathematics 1)
Year2013
SessionJanuary
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypeNumerical methods for differential equations (step-by-step)
DifficultyModerate -0.3 This is a straightforward numerical methods question requiring Euler's method with given starting point and step size. It involves routine substitution into dy/dx = f(x,y) over two steps with basic arithmetic. While it requires careful calculation, it's a standard textbook exercise with no conceptual difficulty or problem-solving insight needed, making it slightly easier than average.
Spec1.09f Trapezium rule: numerical integration

1 A curve passes through the point (1,3) and satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x } { 1 + x ^ { 3 } }$$ Starting at the point ( 1,3 ), use a step-by-step method with a step length of 0.1 to estimate the value of \(y\) at \(x = 1.2\). Give your answer to four decimal places.

Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(h y'(1) = 0.1 \times y'(1) = 0.05\)M1 Attempt to find \(h y'(1)\). PI by e.g. 3.05 for \(y(1.1)\)
\(y(1.1) \approx 3 + 0.05 = 3.05\)A1
\(y(1.2) \approx y(1.1) + 0.1 \times y'(1.1) = 3.05 + 0.1 \times y'(1.1)\)m1 Attempt to find \(y(1+0.1)+0.1\times y'(1+0.1)\); must see evidence of calculation if correct ft value not obtained
\(\approx 3.05 + 0.1 \times \frac{1.1}{1+1.1^3}\)A1F OE; ft on value; PI
\(\approx 3.0972\) (to 4 d.p.)A1 Must be 4 dp
## Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $h y'(1) = 0.1 \times y'(1) = 0.05$ | M1 | Attempt to find $h y'(1)$. PI by e.g. 3.05 for $y(1.1)$ |
| $y(1.1) \approx 3 + 0.05 = 3.05$ | A1 | |
| $y(1.2) \approx y(1.1) + 0.1 \times y'(1.1) = 3.05 + 0.1 \times y'(1.1)$ | m1 | Attempt to find $y(1+0.1)+0.1\times y'(1+0.1)$; must see evidence of calculation if correct ft value not obtained |
| $\approx 3.05 + 0.1 \times \frac{1.1}{1+1.1^3}$ | A1F | OE; ft on value; PI |
| $\approx 3.0972$ (to 4 d.p.) | A1 | Must be 4 dp |

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1 A curve passes through the point (1,3) and satisfies the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x } { 1 + x ^ { 3 } }$$

Starting at the point ( 1,3 ), use a step-by-step method with a step length of 0.1 to estimate the value of $y$ at $x = 1.2$. Give your answer to four decimal places.

\hfill \mbox{\textit{AQA FP1 2013 Q1 [5]}}