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# AQA FP3 June 2015 Unofficial Mark Scheme watch

1. anyone have the paper for this?
2. (Original post by Lau14)
There's some bits missing/not sure on and the wording isn't right but these are the questions as far as I know
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1. f(x,y) = (y2 + x)/x
f(2) = 5
(a) Use Euler with h = 0.05 to find an approximation for y(2.05) [2]
(b) Use the formula yr+1 = yr-1 + 2hf(xr, yr) to obtain an approximation for y(2.1), giving your answer to three significant figures [3]

2. Use an integrating factor to solve dy/dx + (tanx)y = tan3xsecx, given that y = 2 when x = π/4. [9]

3. (a)(i) Write down the expansion of ln(1 + 2x) up to and including terms in x4 [1]
(a)(ii) Find the first two non-zero terms of ln[(1 + 2x)(1 – 2x)] and write down the range of validity for this expansion [3]
(b) Find the value of Limx>0 [(3x – x√(9 + x))/ln[(1 + 2x)(1 – 2x)]] [4]

4. (a) Why is the integral (range 0 – infinity) of (x-2)e-2x improper? [1]
(b) Find the value of the integral, showing the limiting process used [6]

5. (a) Find the general solution of d2y/dx2 + 6dy/dx + 9 = (-36?)sin2x [7]
(b)(i) Given that y = f(x) is a solution of the equation, y(0) = 0 and y’(0) = 0, show that y’’(0) = 0 [1]
(b)(ii) Find the first two non-zero terms of the expansion of f(x) [3]

6. x = et, y is a function of x.
(a) Show that the substitution x = et turns the differential equation (?) into d2y/dt2 -2dy/dt + 2y = 4t2 + 5e-t [7]
(b) Hence find the general solution of the equation (?) [10]

7. (a) Find the area of r = 1 + cos2Θ, range –π/2 ≤ Θ ≤ π/2 [5]
(b)(i) The curve 1 + sinΘ intersects the curve 1 + cos2Θ at the origin and point A. Find the polar coordinates of A [4]
(b)(ii) A horizontal line through A intersects the curve 1 + cos2Θ again at B. Show that the length of OB is (√13 + 1)/4 [6]
(b)(iii) Find the length of AB [3]
Hey there, Do you have the exam paper or at least know what (?) is on question 6
3. Hi, I'm currently revision for this as I am sitting it in the summer.
Looking at this paper, can some one help me with question 2 please ?
I got my general solution to be -
y= (tan^4x + c ) / 4secx

But when I come to find c I get -1 + 8 root 2
Can some one explain where the 7 comes from please ?
4. (Original post by Roxanne18)
Hi, I'm currently revision for this as I am sitting it in the summer.
Looking at this paper, can some one help me with question 2 please ?
I got my general solution to be -
y= (tan^4x + c ) / 4secx

But when I come to find c I get -1 + 8 root 2
Can some one explain where the 7 comes from please ?
I think you've re-arranged the equation incorrectly, after integrating you should get y*sec x=(tan^4 x)/4 + c, then substitute in y = 2 and x = pi/3 and you should get 7/4 for your + c. It's always best to substitute in your initial conditions straight after you've integrated. That's true for both C4 and FP3.
5. (Original post by TheLifelessRobot)
I think you've re-arranged the equation incorrectly, after integrating you should get y*sec x=(tan^4 x)/4 + c, then substitute in y = 2 and x = pi/3 and you should get 7/4 for your + c. It's always best to substitute in your initial conditions straight after you've integrated. That's true for both C4 and FP3.
Okay thank you, I shall try that now (:

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