Question 1 - A-Level Maths

AQA FP1 2010 June — Question 1 6 marks

Exam Board	AQA
Module	FP1 (Further Pure Mathematics 1)
Year	2010
Session	June
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Variable acceleration (1D)
Type	Numerical methods for differential equations (step-by-step)
Difficulty	Moderate -0.5 This is a straightforward numerical methods question requiring Euler's method with given step size and initial conditions. While it involves multiple iterations (3 steps), the arithmetic is simple (adding 1 + x³ values), and the method is mechanical with no conceptual challenges. It's easier than average because it's purely procedural application of a standard algorithm, though not trivial since it requires careful arithmetic across multiple steps.
Spec	1.07t Construct differential equations: in context 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

1 A curve passes through the point ( 1,3 ) and satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 1 + x ^ { 3 }$$ Starting at the point ( 1,3 ), use a step-by-step method with a step length of 0.1 to estimate the $y$-coordinate of the point on the curve for which $x = 1.3$. Give your answer to three decimal places.
(No credit will be given for methods involving integration.)

Show mark scheme Show mark scheme source

Question 1:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
At $x=1$: $\frac{dy}{dx} = 1 + 1^3 = 2$	M1	Correct use of step-by-step method
$y(1.1) = 3 + 0.1 \times 2 = 3.2$	A1	Correct first step
At $x=1.1$: $\frac{dy}{dx} = 1 + (1.1)^3 = 2.331$	M1	Correct evaluation of derivative at new point
$y(1.2) = 3.2 + 0.1 \times 2.331 = 3.4331$	A1	Correct second step
At $x=1.2$: $\frac{dy}{dx} = 1 + (1.2)^3 = 2.728$
$y(1.3) = 3.4331 + 0.1 \times 2.728 = 3.706$	A1	Answer to 3 d.p.
Correct method shown throughout	B1	All three steps attempted correctly

# Question 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| At $x=1$: $\frac{dy}{dx} = 1 + 1^3 = 2$ | M1 | Correct use of step-by-step method |
| $y(1.1) = 3 + 0.1 \times 2 = 3.2$ | A1 | Correct first step |
| At $x=1.1$: $\frac{dy}{dx} = 1 + (1.1)^3 = 2.331$ | M1 | Correct evaluation of derivative at new point |
| $y(1.2) = 3.2 + 0.1 \times 2.331 = 3.4331$ | A1 | Correct second step |
| At $x=1.2$: $\frac{dy}{dx} = 1 + (1.2)^3 = 2.728$ | | |
| $y(1.3) = 3.4331 + 0.1 \times 2.728 = 3.706$ | A1 | Answer to 3 d.p. |
| Correct method shown throughout | B1 | All three steps attempted correctly |

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

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

Starting at the point ( 1,3 ), use a step-by-step method with a step length of 0.1 to estimate the $y$-coordinate of the point on the curve for which $x = 1.3$. Give your answer to three decimal places.\\
(No credit will be given for methods involving integration.)

\hfill \mbox{\textit{AQA FP1 2010 Q1 [6]}}

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Q1 6 Q2 6 Q3 5 Q4 8 Q5 6 Q6 11 Q7 10 Q8 10 Q9 13