# Edexcel's P4

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17 years ago
#1
How did people find it? I thought some of it was alright. Couldn't do the first question (we weren't taught about 'method of differences' :/), the x^3 - 27 thing, integration on the polar co-ordinate question or showing there was only one value for which arg z = pi/4. Question 6 was very nice, as was the last one.
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17 years ago
#2
the arg(z) one became obvious to me after i'd left the exam room. if the argument is pi/4, then the imaginary part has to equal the real part. this annoyed me somewhat because its worth 6 marks :\

i didn't understand the first part of that x^3 - 27 =0 thing until someone explained it to me later. you basically just say that x^3 - 27 = (x-3)(x^2+Ax+9) and then calculate A and it comes out to be 3 or something.

the polar co-ordinates one i did do, i just took the area from 0->D and doubled it for the large shape and from D->pi for the smaller shape and doubled it.
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17 years ago
#3
Aye, I never knew there were 3 cube roots of 27!

The polar co-ordinates one I was dead chuffed with, cos I managed to get it out. (first time the examiners seen a big smiley face in a maths exam I wonder ...).

U had to add the area of the bigger one from 0 - pi/3 to the area of the smaller one from pi/3 - pi. something like that nehu.

Twas my last exam today, can't say it's come soon enough, Smiling Rob
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17 years ago
#4
Originally posted by davem
the arg(z) one became obvious to me after i'd left the exam room. if the argument is pi/4, then the imaginary part has to equal the real part. this annoyed me somewhat because its worth 6 marks :\
Oh **** .
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17 years ago
#5
Oh and for the differential equation on question 8, did you get K = 2?
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17 years ago
#6
Originally posted by Unregistered
Oh and for the differential equation on question 8, did you get K = 2?
Wooooo .. yay!

Rob
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17 years ago
#7
Originally posted by Rob
Wooooo .. yay!

Rob
Hehe, I was worried for a minute while I was doing it, cos I thought nothing would cancel and I'd be left with a string of terms. Lol.
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17 years ago
#8
Originally posted by Unregistered
Oh and for the differential equation on question 8, did you get K = 2?
Wow i did get at least one mark!
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17 years ago
#9
It was quite a good paper. I liked the Differential Equations question, and I thought the polar coordinates question (finding the area) was the hardest.

What did you get for that last question (sketching the graph).
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17 years ago
#10
What I got ... was probably very wrong, but here goes...

infrom the left (like an asymptote) up to a maximum at t=0, corssing at y=1, then down, to a minimum at t = 1.86, y = -1.35, then back up and off to the right....

forgotten the points where it cross the x axes tho, somethin like 1 and 2, or am I thinking of the other question involving intervals .. mm

Rob
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17 years ago
#11
For the points where the graph crossed the x-axis, I got x = 1/2 and x = 1. The equation was like y = (-3t + 1 + 2t^2)e^3t.

e^3t can't be 0.

2t^2 - 3t + 1 = 0
(2t - 1)(t - 1).

Something like that anyway, I can't remember if that was the exact equation or not.
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17 years ago
#12
Originally posted by Unregistered
For the points where the graph crossed the x-axis, I got x = 1/2 and x = 1. The equation was like y = (-3t + 1 + 2t^2)e^3t.

e^3t can't be 0.

2t^2 - 3t + 1 = 0
(2t - 1)(t - 1).

Something like that anyway, I can't remember if that was the exact equation or not.
It was [f(t)] e^-3t

I got a normal e^x shape for t<0 and after that a min at t=0 (y=1) and an intersection at t=1/2. Another min at t=7/8 (???) and y= negative, then another intersection at t=1 and y increasing for t increasing.
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17 years ago
#13
ARGHHHHHHHHHHHHHHHHHHHH .
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17 years ago
#14
Question 1 was all proof (answer to a was in part b)

I forgot how to do 2 which wasn't good-since inequalities are usually easy marks, wrote down x>-0.5 though, so should get some marks. No graphics calculator also made it more difficult.

3-fine I got -3/2 +- (3/2)*root(3)*i for b

4 fine b)1.875-->2
c)Something in between those 2 above

5a root(5)/2
b-here was the problem got 1 or -6, can't see why it can't be both-someone please explain!

6 a -(1/2)(X^2 + 1)*e^(-x^2)

b x^3y= -(1/2)(X^2 + 1)*e^(-x^2) + C

7a B(3a,0) and A(5a,0)

b (4a, pi/3)

c Stuffed it up-too long got the 49pi but not the other one. Used same method as Rob, but to no avail

8a 2 (gives you it in part d)
b (A + Bt)*e^3t + 2t^2*e^3t
c A=3 and B=8
d crosses t-axis at t=0.5 and t=1, y-axis at (0,1)
min point at (5/6, -e^(5/2)/9)
Sketch is rubbsih, but includes above info-hopefully it'll be alright!

*All of above are my answers, and not necessarily correct
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17 years ago
#15
Pretty much the same as what I got.
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17 years ago
#16
Originally posted by Unregistered
Pretty much the same as what I got.
Cool,
I actually wrote the answers down on the question paper-so this is exactly what I got. I'd quite like to know how I did.
Anyone anyone else get anything different from me?
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17 years ago
#17
What was question 5.b?
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