# Edexcel's P4

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#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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#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.

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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#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

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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#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 :\**

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#6

*Originally posted by Unregistered*

**Oh and for the differential equation on question 8, did you get K = 2?**

Rob

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#7

*Originally posted by Rob*

**Wooooo .. yay!**

Rob

Rob

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#8

*Originally posted by Unregistered*

**Oh and for the differential equation on question 8, did you get K = 2?**

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#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).

What did you get for that last question (sketching the graph).

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#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

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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#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.

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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#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.

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.

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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#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

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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#16

*Originally posted by Unregistered*

**Pretty much the same as what I got.**

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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