AQA FP1 2011 June — Question 1 5 marks

Exam BoardAQA
ModuleFP1 (Further Pure Mathematics 1)
Year2011
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypeNumerical methods for differential equations (step-by-step)
DifficultyModerate -0.5 This is a straightforward numerical methods question requiring Euler's method with given initial conditions and step size. It involves routine substitution and arithmetic over 2 steps, which is simpler than typical A-level calculus problems that would require analytical integration or more complex reasoning.
Spec1.09f Trapezium rule: numerical integration

1 A curve passes through the point \(( 2,3 )\) and satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { 2 + x } }$$ Starting at the point \(( 2,3 )\), use a step-by-step method with a step length of 0.5 to estimate the value of \(y\) at \(x = 3\). Give your answer to four decimal places.

Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Attempt at \(0.5 \times y'(2) (= 0.25)\)M1 Other variations are allowed
\(y(2.5) \approx 3.25\)A1
\(y(3) \approx 3.25 + 0.5\, y'(2.5)\)m1
\(\approx 3.25 + 0.2357(0)\)A1F PI; OE; ft c's value for \(y(2.5)\)
\(\approx 3.4857\)A1 4 dp needed
Total5
## Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempt at $0.5 \times y'(2) (= 0.25)$ | M1 | Other variations are allowed |
| $y(2.5) \approx 3.25$ | A1 | |
| $y(3) \approx 3.25 + 0.5\, y'(2.5)$ | m1 | |
| $\approx 3.25 + 0.2357(0)$ | A1F | PI; OE; ft c's value for $y(2.5)$ |
| $\approx 3.4857$ | A1 | 4 dp needed |
| **Total** | **5** | |

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1 A curve passes through the point $( 2,3 )$ and satisfies the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { 2 + x } }$$

Starting at the point $( 2,3 )$, use a step-by-step method with a step length of 0.5 to estimate the value of $y$ at $x = 3$. Give your answer to four decimal places.

\hfill \mbox{\textit{AQA FP1 2011 Q1 [5]}}