AQA FP1 2008 January — Question 2 5 marks

Exam BoardAQA
ModuleFP1 (Further Pure Mathematics 1)
Year2008
SessionJanuary
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNewton-Raphson method
TypeAlgorithm or flowchart implementation
DifficultyModerate -0.3 This is a straightforward application of Euler's method with given starting point, step length, and function. It requires only mechanical iteration (two steps) with no conceptual difficulty beyond knowing the basic formula y_{n+1} = y_n + h·f(x_n). The calculation is routine for FP1 level, though the 6 significant figures requirement demands careful arithmetic.
Spec1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

2 A curve satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 ^ { x }$$ Starting at the point \(( 1,4 )\) on the curve, use a step-by-step method with a step length of 0.01 to estimate the value of \(y\) at \(x = 1.02\). Give your answer to six significant figures.

Question 2:
AnswerMarks Guidance
Working/AnswerMarks Guidance
\(0.01(2^1)\) added to value of \(y\)M1 Variations possible here
So \(y(1.01) \approx 4.02\)A1 PI
Second increment is \(0.01(2^{1.01})\)m1
\(\approx 0.020\,139\)A1
So \(y(1.02) \approx 4.040\,14\)A1
Total5
## Question 2:

| Working/Answer | Marks | Guidance |
|---|---|---|
| $0.01(2^1)$ added to value of $y$ | M1 | Variations possible here |
| So $y(1.01) \approx 4.02$ | A1 | PI |
| Second increment is $0.01(2^{1.01})$ | m1 | |
| $\approx 0.020\,139$ | A1 | |
| So $y(1.02) \approx 4.040\,14$ | A1 | |
| **Total** | **5** | |

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2 A curve satisfies the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 ^ { x }$$

Starting at the point $( 1,4 )$ on the curve, use a step-by-step method with a step length of 0.01 to estimate the value of $y$ at $x = 1.02$. Give your answer to six significant figures.

\hfill \mbox{\textit{AQA FP1 2008 Q2 [5]}}