Higher Level Maths 2009(Paper 1): How was it for you?

QueenOfLeon · 05-06-2009 1:41pm

Sweet wrote: »

For 2(c) I got q>-36/5. Anyone able to confirm this?!

That rings a bell. Except mine involved a square root and an R somewhere..:o

kevogy wrote: »

was question 3 b ii) (-1 0 ) diagonal matrix?
(0 -1 )

-1^17 ?

I got that. That was A^3, so i put it to the power of 5 to get A^15, then multiplied it by A^2 to get A^17. Dont even know if thats allowed!

Davidius wrote: »

I just realised that I never sketched the curve.

Le bollox.

Ah god, sorry, just realised i probably reminded people if they forgot it. Don't worry about it, if it helps i got mad at mine and it looked a bit wrong

xOxSinéadxOx · 05-06-2009 1:41pm

was 1. (c) REALLY REALLY easy for a (c) part or did I just do it wrong?

Amirani · 05-06-2009 1:42pm

Did anyone else get 2 C(ii) in terms of r, and then state that q is greater than or equal to the coeffiecent as r is positive? Sound in any way right?

Nihilist21 · 05-06-2009 1:42pm

kevogy wrote: »

was question 3 b ii) (-1 0 ) diagonal matrix?
(0 -1 )

-1^17 ?

That was the matrix I got, pretty sure it's right. However to raise it to the power of 17, it would have been A^3 5 times which was the diagonal matrix, then multiply by A^2.

kevogy · 05-06-2009 1:43pm

i added up what i thought i got right and put over 600

160
---- multiply by 100
600

around 26 %

is this 26% of the entire maths exam?

MathsManiac · 05-06-2009 1:43pm

r = (5*q^2)/36.

So, for r to be a positive integer, q has to be a multiple of 6. q={6, 12, 18, ...}

Amirani · 05-06-2009 1:44pm

Sweet wrote: »

For 2(c) I got q>-36/5. Anyone able to confirm this?!

Question said that q was positive. I got the square root of 36/5 * r.

.:FuZion:. · 05-06-2009 1:46pm

Not a bad paper.

1(C): C = B + 1???
2(C): I got r = q(p^2)n or something. Dont think its right. And q = (+-) 6?

Toad-Girl · 05-06-2009 1:47pm

xOxSinéadxOx wrote: »

was 1. (c) REALLY REALLY easy for a (c) part or did I just do it wrong?

Ya I thought so too...It was much easier then the first two parts which didn't make much sense...

Am I the only one who thinks it was an unnecessarily difficult paper?? integration in question 3, awkward algebra, grrr matrices (so disappointed with that..) and no proof of negative de moivre or factor!! I'm terrified I've failed it now to be honest. Please tell me someone else found it difficult!!

OxfordComma · 05-06-2009 1:47pm

Just as a matter of interest, what answers did ye get for 1(c)?

I got c=b+1

... and then I realised that the square root of b^2 could be -b as well...

so my answer was c=+/-b+1

Am I way off???

MathsManiac · 05-06-2009 1:48pm

7(c)

(ii) f(2)<0 and f(2.5)>0, so root is between 2 and 2.5. QED.

(iii) One go each at N-R: Ann gets 2.66 and Barry gets 2.58. Since we know the root is less than 2.5, Barry is now closer than Ann.

xOxSinéadxOx · 05-06-2009 1:49pm

Toad-Girl wrote: »

Ya I thought so too...It was much easier then the first two parts which didn't make much sense...

Am I the only one who thinks it was an unnecessarily difficult paper?? integration in question 3, awkward algebra, grrr matrices (so disappointed with that..) and no proof of negative de moivre or factor!! I'm terrified I've failed it now to be honest. Please tell me someone else found it difficult!!

I did find it quite difficult. just thought it was a nice paper if I had studied like, it was definitely fair I was just underprepared.

how do you differentiate sin(3x^2 + x) or whatever it was
it was an (a) part

Piste · 05-06-2009 1:50pm

I thought it was an ok paper. I'm not counting maths for points though, but it's be nice to get a B.

For a lot of the C parts in algebra I just dicked around with the numbers, multiplied them by stuff, backwardsed them, turned them topwise and just chucked down whatever answer I got.

Very happy tho ol' cone came up, and I didn't forget my +Cs

I think my differentiation Qs were a bit dodgy, I thought the c part of 7 was a bit weird but I think I worked it out.

Sequences and series was alright, but I messed up the C part (ii) so therefore messed up the (iii) part.

I just know I made a load of little mistakes like getting signs wrong and writing the wrong number and stuff. Might possibly still be able to get a low B.

Amirani · 05-06-2009 1:50pm

xOxSinéadxOx wrote: »

how do you differentiate sin(3x^2 + x) or whatever it was it was an (a) part

Chain rule...

Fringe · 05-06-2009 1:51pm

Anne-Marie!! wrote: »

I got that. That was A^3, so i put it to the power of 5 to get A^15, then multiplied it by A^2 to get A^17. Dont even know if thats allowed!

I just said (A^3)^17/3. -1^(-17/3) is -1. It ends up being the same though.

ayapatrick · 05-06-2009 1:52pm

i spose the damage is done at this stage, no point worryin about it guys!:D

OxfordComma · 05-06-2009 1:52pm

xOxSinéadxOx wrote: »

I did find it quite difficult. just thought it was a nice paper if I had studied like, it was definitely fair I was just underprepared.

how do you differentiate sin(3x^2 + x) or whatever it was
it was an (a) part

You differentiate the sin part and the angle separately...

So you get cos(3x^2+x) and then differentiate what's inside the brackets. So it's cos(3x^2+x)(6x+1)

xOxSinéadxOx · 05-06-2009 1:52pm

niallsparky wrote: »

Chain rule...

yeah but how I think I did it wrong

kevogy · 05-06-2009 1:53pm

xOxSinéadxOx wrote: »

how do you differentiate sin(3x^2 + x) or whatever it was
it was an (a) part

cos(3x^2+x)(6x-1)

not sure how to tidy up

i wrote (6x-1)cos(3x^2+x)

xOxSinéadxOx · 05-06-2009 1:54pm

1fahy4 wrote: »

You differentiate the sin part and the angle separately...

So you get cos(3x^2+x) and then differentiate what's inside the brackets. So it's cos(3x^2+x)(6x+1)

thanks!!

I think that's what I did :cool:

and lads I haven't a clue what any of ye are on about for 2. (c) guess I just wasn't made for the (c) parts

BLARG · 05-06-2009 1:54pm

Nihilist21 wrote: »

That was the matrix I got, pretty sure it's right. However to raise it to the power of 17, it would have been A^3 5 times which was the diagonal matrix, then multiply by A^2.

(A^3)^14 = A^17
The answer I think was
(1 0)
(0 1)

Amirani · 05-06-2009 1:54pm

kevogy wrote: »

cos(3x^2+x)(6x-1)

not sure how to tidy up

i wrote (6x-1)cos(3x^2+x)

That's the way.

Risteard · 05-06-2009 1:54pm

BLARG wrote: »

(A^3)^14 = A^17
The answer I think was
(1 0)
(0 1)

That's what I got.

ayapatrick · 05-06-2009 1:55pm

niallsparky wrote: »

Chain rule...

well **** it anyway!
ah i left it as sin 3x^2 - sin x and diffed it!

OxfordComma · 05-06-2009 1:56pm

****e! I screwed up b(ii) of the matrix question.... Damn! I thought I'd gotten full marks in that question

Piste · 05-06-2009 1:57pm

Oh no, for the chain rule yoke I thought Sinu becomes ucosu and you multiply that by du/dx so I got (6x-1)(3X^2+x)cos(3x^2+x)

kevogy · 05-06-2009 1:57pm

question 7 c part i
how do you u show root is between 2 and 3?

i got dy/dx

solved the quadratic
got x = 1

xOxSinéadxOx · 05-06-2009 1:58pm

kevogy wrote: »

question 7 c part i
how do you u show root is between 2 and 3?

i got dy/dx

solved the quadratic
got x = 1

newton raphson

Piste · 05-06-2009 1:59pm

You find f(2) and f(3) and if the sign changes between them the root lies between them.

Overature · 05-06-2009 1:59pm

kevogy wrote: »

i added up what i thought i got right and put over 600

160
---- multiply by 100
600

around 26 %

is this 26% of the entire maths exam?

its 300 marks per paper. you should have got 160 over 300 or something

Higher Level Maths 2009(Paper 1): How was it for you?

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