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Reply 20
AnonyMatt
Do you have M1 on Monday then? Or will you not do that until the Summer?
I think it's pretty easy. No differentiation or integration like there was in S1, not that we have any problems with that now that we've done C3 and beyond!
The only problem I have, is that I miss key bits and because I rely on my calculator so much I often make errors when typing numbers in, and end up getting the wrong answer.
I like M2 much more. Conservation of energy and all that, plus oppurtunities to use CALCULUS<3<3<3. :p:

I like FP1, but honestly, it's the hardest module of the lot. FP2 is far easier, with the only hard part being curve sketching. At least, I think so. I haven't looked at the spec in a while. FP3 is easy though.
With FP1, I'm finding the following difficult:
-Differentiation from first principles. As I said, my algebra is rubbish so it takes me ages, and I still can't get it into a form where I can say h = 0, therefore f'(x) = ...
-Reduction to echeleon form. It takes me ages to think of a good way to get three zeros, and I make loads of mistakes getting there.
-Loci. Again, quite alot of algebra. Need to practice it a bit more. When I 'revise' in lessons, I start a paper but because I talk I only get the first few questions done, which are all easy!
-Transformations. I can do them, but I feel like I'm cheating because I have to use the formula book. Once I can remember the matrices we need, I'll feel better. I'm a little iffy on the reflections too, but I can follow mark schemes and see where I went wrong. I'm not finding them that bad really, just haven't done enough.


No M1 on monday :smile: I know what you mean about calculator dependency though, I tend to be decent enough at algebra without it and hate when I have to use it to completely do a question - it took me 10 to 15 minutes to do the Simpson's rule question because of being so extra careful - plus my mind went blank and I forgot which way the formula went around (thank god for formula booklets! I feel no shame in referring to them if I need to and I never realised that transformations were in there! Be having a look for them I will!)

I can help you with first principles if you like - I can scan a page or two or it in to show you how I would do it. I'm not sure what the chances are of it coming up in Feb, but I'm not taking my chances by ignoring it either :o:

You think FP2 and FP3 are easier? I've only done de moivre's theorem and expanding powers of sin and cos, which is fine enough but if anyone's going to make any errors, it would be on that because it's as picky as it gets. I can't really wait to do FP2/3 because I look at the papers and think "how the hell is that even possible :eek:" and it's an awesome feeling once you can do it :biggrin: From what I've looked at, polar coordinates on FP3 look a bit daunting but I like a challenge (as long as it's a nice easy challenge :cool:)

Had a look at induction earlier and have found a new method to do it - it's now so much easier and it's good to have one less thing to worry about!
Natt_105
No M1 on monday :smile: I know what you mean about calculator dependency though, I tend to be decent enough at algebra without it and hate when I have to use it to completely do a question - it took me 10 to 15 minutes to do the Simpson's rule question because of being so extra careful - plus my mind went blank and I forgot which way the formula went around (thank god for formula booklets! I feel no shame in referring to them if I need to and I never realised that transformations were in there! Be having a look for them I will!)

I can help you with first principles if you like - I can scan a page or two or it in to show you how I would do it. I'm not sure what the chances are of it coming up in Feb, but I'm not taking my chances by ignoring it either :o:

You think FP2 and FP3 are easier? I've only done de moivre's theorem and expanding powers of sin and cos, which is fine enough but if anyone's going to make any errors, it would be on that because it's as picky as it gets. I can't really wait to do FP2/3 because I look at the papers and think "how the hell is that even possible :eek:" and it's an awesome feeling once you can do it :biggrin: From what I've looked at, polar coordinates on FP3 look a bit daunting but I like a challenge (as long as it's a nice easy challenge :cool:)

Had a look at induction earlier and have found a new method to do it - it's now so much easier and it's good to have one less thing to worry about!


The only matrix you need to remember is the
1 0 a
0 1 b
0 0 1
Or whatever it is. The one that moves it from (0, 0) to (a, b) or something.

I'll get someone to show me how to set out a first principles question nicely so I can rearrange easily, but thanks. :p:

Yeah I'm gonna enjoy FP2 and FP3 because I like integration and that. But I have to do C4 first, ugh. I already taught myself C4 last year so lessons for the next 4 weeks or something are gonna be boring.
Reply 22
AnonyMatt
The only matrix you need to remember is the
1 0 a
0 1 b
0 0 1
Or whatever it is. The one that moves it from (0, 0) to (a, b) or something.

I'll get someone to show me how to set out a first principles question nicely so I can rearrange easily, but thanks. :p:

Yeah I'm gonna enjoy FP2 and FP3 because I like integration and that. But I have to do C4 first, ugh. I already taught myself C4 last year so lessons for the next 4 weeks or something are gonna be boring.


Ready for FP1 tomorrow?

I don't think it will be that bad at all. Have to think positive :smile: I think given a good cup of tea before it it'll go really well. I know how transformations work now (luckily enough!) but for some reason, whenever they asked to find fixed coordinates I always make mistakes on the simultaneous equations! I don't know why, it's strange but I hope I know what I'm doing when it comes to tomorrow :smile:

Good luck for it. You probably won't need it but I hope you do as well as you want to :biggrin:
Natt_105
Ready for FP1 tomorrow?

I don't think it will be that bad at all. Have to think positive :smile: I think given a good cup of tea before it it'll go really well. I know how transformations work now (luckily enough!) but for some reason, whenever they asked to find fixed coordinates I always make mistakes on the simultaneous equations! I don't know why, it's strange but I hope I know what I'm doing when it comes to tomorrow :smile:

Good luck for it. You probably won't need it but I hope you do as well as you want to :biggrin:


Yeah I think I'm okay with it.
Although I'm making annoying mistakes constantly.
I don't like to write loads of workings because I like to keep my work neat, but I think I'm going to have to write like 5 pages worth for whatever the 'roots of polynomials' question is tomorrow!

I'm still getting pissed off by the reduction to echelon form thing, and the P(x, y) Q(u, v) questions are annoying.

Otherwise, I think I'm fine!

Good luck. :smile:
I shall report back at around 6pm GMT tomorrow. :p:
Reply 24
That's it, I'm definitely getting a low mark in this FP1 exam!!
I could not do so many of the questions like
- the induction one that had factorial on it but part b was great!
- the roots of polynomials was just :facepalm: . I wrote something like alpha = beta, then cube pairs to get q^3..
- the differentiation question i couldn't rearrange properly though for the second part, the recurrence relation was good :yep: (C3 topic)
- the 2x2 matrix question - i forgot how to find the inverse... i did basically all 3x3s revision so i forgot about the 2x2s.
Aside from this the complex numbers were ok, the transformations were also ok, i got lander as = 5 for the consistent question.
But dang i lost quite a few marks, the'll definitely bring down the pass rate
ilock
That's it, I'm definitely getting a low mark in this FP1 exam!!
I could not do so many of the questions like
- the induction one that had factorial on it but part b was great!
- the roots of polynomials was just :facepalm: . I wrote something like alpha = beta, then cube pairs to get q^3..
- the differentiation question i couldn't rearrange properly though for the second part, the recurrence relation was good :yep: (C3 topic)
- the 2x2 matrix question - i forgot how to find the inverse... i did basically all 3x3s revision so i forgot about the 2x2s.
Aside from this the complex numbers were ok, the transformations were also ok, i got lander as = 5 for the consistent question.
But dang i lost quite a few marks, the'll definitely bring down the pass rate


Hmm, I think the induction one came out nicely actually.
If you took out a factor of (k+1)!, then you ended up with (k+1)!(k+2) straight away, so it was quite simple! I think a lot of people will have messed that up though.
I didn't like the roots of polynomials one, but when I did a second attempt on it it came out nicely too.
Lots of people told me they didn't get the a = tanalncoseca.
I think many people didn't simplify f'(x) enough.
I did the 2x2 backwards. I knew that AA^-1 = I, so I tried really hard to get an inverse matrix that would give the identity matrix. After 10mins, I was successful.
I got lamba = 5 too! Those are the questions I usually get wrong too.

Yeah, the pass rate will be way low. Probably 55/75 for an A.

I'm gonna post my answers I think...
Reply 26
AnonyMatt
Hmm, I think the induction one came out nicely actually.
If you took out a factor of (k+1)!, then you ended up with (k+1)!(k+2) straight away, so it was quite simple! I think a lot of people will have messed that up though.
I didn't like the roots of polynomials one, but when I did a second attempt on it it came out nicely too.
Lots of people told me they didn't get the a = tanalncoseca.
I think many people didn't simplify f'(x) enough.
I did the 2x2 backwards. I knew that AA^-1 = I, so I tried really hard to get an inverse matrix that would give the identity matrix. After 10mins, I was successful.
I got lamba = 5 too! Those are the questions I usually get wrong too.

Yeah, the pass rate will be way low. Probably 55/75 for an A.

I'm gonna post my answers I think...


Yeah I didn't like the roots queation at all but I am fairly sure I got everything else perfectly due to the amount of show thats :biggrin:
lamba = 5, other than that I had a little problem with the a= tanalncoseca due to the fact I forgot what the differemtial of cosec was :P but in the end it was in the formula booklet lol
Reply 27
That was a very very hard exam! When I first looked at the paper I thought it was some kind of joke! With a bit of time though, it wasn't too bad but some of the questions were beyond what I knew and this went side by side with a lot of mistakes.

My answers are here:

1a] simple enough to prove it's a root.
b] the other roots were 1-2i and -2

2a/b] I tried answering but I was genuinely not sure what to do in this.

3] complex numbers. According to me, 8 multiplied by 2 is 10! this meant the answer was wrong, and so was the modulus and argument. I had -9/5 + 2i, I thought something was wrong. I did it twice as well. Dear me :L

4] λ\lambda = 5 for me as well. The general solution was

x = 55α5\frac{5-5\alpha}{5} and y = 1α3\frac{-1-\alpha}{3} I think but I'm sure my answer in the actual exam for x was wrong.

5] I don't know at all. I did something like (x-a)(x-a)(x-b) = x3x+r=0x^3 - x + r = 0 and ended up with a ton of powered integers. I've never been that stumped on an exam before.

6a] Induction - was fine until I got to n=k+1 and then for some reason which I myself don't know I just went straight to what I was trying to prove and said it was right. I thought "this is a bit too simple" at the time; now I know why!

b] Got n(n+1)2n(n+1)^2

7a] (cosecx)x(lncosecxxtanx)(cosecx)^x (lncosecx - \frac{x}{tanx}) because the two cosecs I had cancelled to that but I certainly got there, I worked backwards and then rewrote it out again :smile:

b] simple enough

c] Put the calculator into radians mode, cosec = 1/sin and got 0.3986 in the end. Might be okay with that.

8a] Could not really do much wrong with that
EDIT: Clockwise. Yep I think that caught be out (I'm blind because I read it as anticlockwise) however I just followed the given matrix so it might not matter. Got at least 3 of the 5 for that though.

b] a bit weird - had something like x = x - 1 and I said that doesn't work, so they aren't fixed points.

c] I just did it and came out as 2x' + y' = 0, but in the exam I had a 1 in there. I think I was wrong when I did it first.

9a] u=1+x2y2u = 1 + x^2 - y^2 and v=2xyv = 2xy

b] ended up with u=13v4u = 1 - \frac{3v}{4}

I hope the grade boundaries are lower than 55! Maybe 50 would be fairer :biggrin: I've had nothing but A grades in maths so far but it's probably going to change now. Very glad it's over though, I think I put the wrong candidate number on one of the answer booklets as well! Just one of those days :smile:
1.
a) f(1+2i) = 0
b) (x-z)(z-z*)(x-k) = f(x)
blah blah blah
k = -2
2.
a) detA = -1
A^-1 = something. I worked it out like, (7 x ?) - (5 x -?) = 1
Kind of turned upside down.
b) X = BA^-1
Dunno what I got. Something like 4, 6, 12, 14... I have no idea really.
3.
a)z = -3 + 2i
I think?
Can't remember the question.
b)
|z| = root13
Arg(z) = 146.3 degrees (or -33.6)
4.
Reduction to Echelon form.
I did:
r2 - 2r1 (I think)
Then r3 - 4r1
That gave r2 = r3, so lamba - 4 = 1 or something
Giving lamba = 5
b) Not sure what I wrote for this. Something rubbish I bet!
5.
Hated this. Got there eventually though.
(x-a)^2(x-b) = f(x)
-(2a + b) = 0
b = -2a
Sub -2a into all b, giving r = f(a), p = g(a)
Cube and square to get the given result.
6.
a) n = 1, true.
Assume n = k is true.
n = k + 1: (k+1)! - 1 + (k+1)(k+1)!
= (k+1)!(1 + (k + 1)) - 1
= (k+2)! - 1
Blah Blah Blah, it's true.
b) I got something like
1/2*n(n+1)(8n + 5)
Seems wrong, ugh.
7.
a) lnf(x) = xlncosecx
f'(x)f(x)^-1 = lncosecx - xcosecxcotx(cosecx)^-1
= lncosecx - xcotx
f'(x) = [(cosecx)^x][lncosecx - xcotx]
b) When f'(x) = 0, x = a
lncoseca - acota = 0
lncoseca = acota
a = lncoseca(cota)^-1
a = tanalncoseca
c)
a0 = 0.5
a1 = ...
a2 = ...
a3 = ...
a ~ 0.3896 or something I think?
8.
a) T = C(BA)
A =
0 1 0
1 0 0
0 0 1
B =
1 0 1
0 1 -1
0 0 1
C = (I think this one caught a lot of people out because of the clockwise rotation)
0 1 0
-1 0 0
0 0 1
T = required result.
b)
T[x, y, 1] = [x, y, 1]
Multiplying out, we get x = x - 1
which implies 0 = 1
The equations have no solution, therefore there are no fixed points.
c)
T[2x + 1, y, 1] = [x', y', 1]
Multiplying out gives something.
Answer: y' = -x'
9.
This one came out nicely I think. I usually crash and burn with these...
a)
w = 1 + z^2
u + iv = 1 + x^2 + 2iy - y^2
u = 1 + x^2 - y^2
v = 2y
b)
y = 2x (I think?)
u = 1 + x^2 - 4x^2
= 1 - 3x^2
v = 4x
x = 4/v
u = 1 - 48/v^2
v^2u = v^2 - 48
v^2(u - 1) = -48
v^2 = -48/(u-1)
I don't think I needed to simplify that far, I just like my equations to be y = f(x). It may even be wrong!


Thoughts:
By far the most difficult FP1 exam of this specification. However, this is good. It will sort the ones who learn methods from the ones that know how to apply the maths. I kind of liked the synoptic aspect of the C3 question too!
The exam hall was giggling as the papers were being collected, because so many people had been staring at each other for so long after having given up.
I'm hoping that I did well enough that it will be okay if I've made slight calculation errors, because obviously the boundaries will be low!
Natt_105
That was a very very hard exam! When I first looked at the paper I thought it was some kind of joke! With a bit of time though, it wasn't too bad but some of the questions were beyond what I knew and this went side by side with a lot of mistakes.

x = 55α5\frac{5-5\alpha}{5} and y = 1α3\frac{-1-\alpha}{3} I think but I'm sure my answer in the actual exam for x was wrong.

b] Got n(n+1)2n(n+1)^2

c] Put the calculator into radians mode, cosec = 1/sin and got 0.3986 in the end. Might be okay with that.

c] I just did it and came out as 2x' + y' = 0, but in the exam I had a 1 in there. I think I was wrong when I did it first.

9a] u=1+x2y2u = 1 + x^2 - y^2 and v=2xyv = 2xy

b] ended up with u=13v4u = 1 - \frac{3v}{4}

I hope the grade boundaries are lower than 55! Maybe 50 would be fairer :biggrin: I've had nothing but A grades in maths so far but it's probably going to change now. Very glad it's over though, I think I put the wrong candidate number on one of the answer booklets as well! Just one of those days :smile:



I agree with your y = general solution, but I don't think I got the x = one. I'm terrible at that though. I've never attempted a question on it, I just know you make z = alpha and then get y and x in terms of alpha.

Hmm, do you have the paper? I didn't get what you did for the summation question.

I agree with your u = , but how did you get the v = ?
Your solution to part b looks much nicer than mine. I guess you got part a right then, and I didn't!

Paul1242
Yeah I didn't like the roots queation at all but I am fairly sure I got everything else perfectly due to the amount of show thats :biggrin:
lamba = 5, other than that I had a little problem with the a= tanalncoseca due to the fact I forgot what the differemtial of cosec was :P but in the end it was in the formula booklet lol


AahahAHhaHAHahHAhahA!

With the cosec thing, I looked it up in the formula booklet.

It wasn't until I tried part b that I found out something had gone terribly wrong.

I wrote down the integral. -.-
Fixed it though.
Reply 30
AnonyMatt
I agree with your y = general solution, but I don't think I got the x = one. I'm terrible at that though. I've never attempted a question on it, I just know you make z = alpha and then get y and x in terms of alpha.

Hmm, do you have the paper? I didn't get what you did for the summation question.

I agree with your u = , but how did you get the v = ?
Your solution to part b looks much nicer than mine. I guess you got part a right then, and I didn't!



AahahAHhaHAHahHAhahA!

With the cosec thing, I looked it up in the formula booklet.

It wasn't until I tried part b that I found out something had gone terribly wrong.

I wrote down the integral. -.-
Fixed it though.


I've got the paper! I'll do the locus question first, and then the summation in a short while to see if you would agree with it.

complex numbers blah blah pqxyv and w = 1+z21+z^2

So we let w = u + iv and z = x + iy

u+iv=1+(x+iy)2u + iv = 1 + (x+iy)^2
u+iv=1+x2+2xyiy2u + iv = 1 + x^2 + 2xyi - y^2 - here I've expanded the square and got a minus y squared because i squared is minus one.

Now take the real and imaginary solutions, u and v.

u = 1+x2y21 + x^2 - y^2
v = 2xy2xy and they're the expressions I obtained.

The second part I found hard, I won't lie - and if I had this right, I'd be shocked, but anyway:

the line is y = 2x, and we want the cartesian equation.

from the above I thought well how about v=4x2v = 4x^2?

back into u:

u = 1+x2(2x)21 + x^2 - (2x)^2
u = 1+x2v1 + x^2 - v because v = 4x24x^2

I put x squared = v/4

and then the equation I got was u=13v4u = 1 - \frac{3v}{4} which could well be correct but is just as likely to be wrong!
Natt_105
I've got the paper! I'll do the locus question first, and then the summation in a short while to see if you would agree with it.

complex numbers blah blah pqxyv and w = 1+z21+z^2

So we let w = u + iv and z = x + iy

u+iv=1+(x+iy)2u + iv = 1 + (x+iy)^2
u+iv=1+x2+2xyiy2u + iv = 1 + x^2 + 2xyi - y^2 - here I've expanded the square and got a minus y squared because i squared is minus one.

Now take the real and imaginary solutions, u and v.

u = 1+x2y21 + x^2 - y^2
v = 2xy2xy and they're the expressions I obtained.


The second part I found hard, I won't lie - and if I had this right, I'd be shocked, but anyway:

the line is y = 2x, and we want the cartesian equation.

from the above I thought well how about v=4x2v = 4x^2?

back into u:

u = 1+x2(2x)21 + x^2 - (2x)^2
u = 1+x2v1 + x^2 - v because v = 4x24x^2

I put x squared = v/4

and then the equation I got was u=13v4u = 1 - \frac{3v}{4} which could well be correct but is just as likely to be wrong!


GOD DAMNIT
I AM SO STUPID.

I expanded (x+iy)(x+iy) wrongly.

I did + 2iy, rather than + 2ixy. :frown::frown::frown::frown::frown::frown::frown::frown::frown::frown::frown::frown::frown:

So that's -1 from part a, and probably -1/-2 from part b.

Nooooooooooooooo!


Your answer is correct.
Reply 32
The summation question was to get the sum of r(3r + 1).

I got 3[1/6n(n+1)(2n+1)] + [1/2n(n+1)]

n/2 out as a fractor:

n2\frac{n}{2}(2n2+4n+2)(2n^2 + 4n + 2) but only after simplifying it first if you can follow what's been done here :smile:

it can all be written after more simplifying down as n/2 (2[n+1])^2 which simplifies to n(n+1)^2.

It didn't ask to express it as a product of linear factors, otherwise we take out a factor of (n+1) which we didn't have to do on this one. It just said to "simplify" your answer so I have to believe I'm pretty confident on this one - you might have had the right answer though, but it could have been written a different way depending on if it's linearly expressed or not. :p:

Don't worry though - on question 2, I was in a rush to finish (had left this one till last) and I decided that my X matrix would be a multiple of the other two and got
-41 -34
-21 -6

as you can guess, I got a wrong inverse before. I think I might have done the multiplication wrong as well. As a matter of fact, I used the inverse for this question when I didn't even need that, apparently. Double fail :frown:

So actually the worst part of this exam for me... was the multiplication :s:
Natt_105
The summation question was to get the sum of r(3r + 1).

I got 3[1/6n(n+1)(2n+1)] + [1/2n(n+1)]

n/2 out as a fractor:

n2\frac{n}{2}(2n2+4n+2)(2n^2 + 4n + 2) but only after simplifying it first if you can follow what's been done here :smile:

it can all be written after more simplifying down as n/2 (2[n+1])^2 which simplifies to n(n+1)^2.

It didn't ask to express it as a product of linear factors, otherwise we take out a factor of (n+1) which we didn't have to do on this one. It just said to "simplify" your answer so I have to believe I'm pretty confident on this one - you might have had the right answer though, but it could have been written a different way depending on if it's linearly expressed or not. :p:

Don't worry though - on question 2, I was in a rush to finish (had left this one till last) and I decided that my X matrix would be a multiple of the other two and got
-41 -34
-21 -6

as you can guess, I got a wrong inverse before. I think I might have done the multiplication wrong as well. As a matter of fact, I used the inverse for this question when I didn't even need that, apparently. Double fail :frown:

So actually the worst part of this exam for me... was the multiplication :s:


I see where I went wrong of the summation. As usual, I said 3/6 = 2.
FML.
That's -2 marks on that.
70/75 max for me. :frown:

No wonder mommy and daddy don't love me. D:
Reply 34
AnonyMatt
I see where I went wrong of the summation. As usual, I said 3/6 = 2.
FML.
That's -2 marks on that.
70/75 max for me. :frown:

No wonder mommy and daddy don't love me. D:


hahaha :p:

70/75 raw marks may well get you 100% :biggrin: I've done a count myself and the maximum I think I'm getting is... well, anywhere between 50 and 56 out of 75. It could go either way between a B or an A so fingers crossed :biggrin:
Natt_105
hahaha :p:

70/75 raw marks may well get you 100% :biggrin: I've done a count myself and the maximum I think I'm getting is... well, anywhere between 50 and 56 out of 75. It could go either way between a B or an A so fingers crossed :biggrin:


I'm sure you'll have gotten an A, even without reading your answers.
It's clear you're an A grade student!

And even if you don't, you'll probably do that much better in FP2.
I'm pretty sure just about everyone has a 'dip' in their results at FP1 with WJEC.
Natt_105
hahaha :p:

70/75 raw marks may well get you 100% :biggrin: I've done a count myself and the maximum I think I'm getting is... well, anywhere between 50 and 56 out of 75. It could go either way between a B or an A so fingers crossed :biggrin:


So did you get the results you wanted? :smile:

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