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

awhir

for Q7(c) (ii) if u said that the tangents were asemptotes and there for if they were perpendicular one would cut the an axis which is impossible for an assemptoe.

would that get me anything ? :P

Piste

Actually for the same question, to prove that no tangents wer perpendicular I just subbed in two points and was like "eh yeah, these aren't perpendicular...>_>" which was a bit sketchy but I was hoping I'd get at least SOMETHING.

Also anyone remember the part in Q4 about an expression for the middle term of the series? How did you get it?

~Candy~

awhir wrote: »

for Q7(c) (ii) if u said that the tangents were asemptotes and there for if they were perpendicular one would cut the an axis which is impossible for an assemptoe.

would that get me anything ? :P

lol..that's excually what i wrote...woohooo...not sure if its right but i am with you

hmm but i got the limits wrong...silly mistake -___-

AndyWhite

WolfForager wrote: »

After a small bit of tidying up, divide x+(1-c) into x^2+(5c-5)x-6b^2

You'll end up getting something like -6b^2+(6c-6)=0 (that's not what i came up with i'm just too lazy to do it again :P) or something like that. It didn't really work out for me at the very end, but apparently it didn't for the majority of people.

Thats what i got too. you just had to move it around a bit then and get c in terms of b.

meathawk

For question 4(c) part ii the one where you have to find the middle term
what I did was (2m+1)/2 and let that equal to n

so it'd be U(2m+1)/2=ar^n-1
then I just subbed in for n and got ar^m-1/2

ya-ba-da-ba-doo

U could use factor thrm for Q1c, although i said c+1 was a factor when it should have been c-1.. anyone know will i just get a blunder or will loads be taken off seein that its at the start of the question?

WolfForager

ya-ba-da-ba-doo wrote: »

U could use factor thrm for Q1c, although i said c+1 was a factor when it should have been c-1.. anyone know will i just get a blunder or will loads be taken off seein that its at the start of the question?

By factor you mean root right? c-1 is a root whilse x-c+1 is a factor.

ya-ba-da-ba-doo

WolfForager wrote: »

By factor you mean root right? c-1 is a root whilse x-c+1 is a factor.

Sorry, meant root, as in i put f(c+1)=0 and solved when it should have been f(c-1)=0... Hopefully its just a blunder im kicking myself over it!

andyman

SligoBrewer wrote: »

There was no misprint ffs.

Ok, ok calm down! Sorry I didn't read 12 pages of a boards.ie thread when I've an English exam in the morning :rolleyes:

Antamojo

F*ck that, I'll have to pull off a miracle on Paper II to try and get C or at least a pass.

nDakota

It's been a while since i thought about the exam, and for good reasons
Q1 should be a 50/50
Q2 was nice up until the C part where i was completely thrown
Q3 not the complex numbers i revised
Q6 c(ii) is my only real concern there
Q7 bit sketchy in areas especially the C part involving Anne and Barry. I'm glad im the not the only one who felt like swapping the names around to make my answer look more attractive
Q8 ... cant even remember, dont think i did great, didn know the proof but i did do a nice sketch of a cone:) and i nearly got b(ii) out just kept getting stuck.

All in all.. not great but not bad.. hope paper 2 is handy to bring up the marks

bstoran

thing about the asymptote question was that dy/dx (tangent to curve) was -2 over a squared term.....ie any value wood b minus....minus by minus cant equal -1 for perpendicular!!!

straight_As

meathawk wrote: »

For question 4(c) part ii the one where you have to find the middle term
what I did was (2m+1)/2 and let that equal to n

so it'd be U(2m+1)/2=ar^n-1
then I just subbed in for n and got ar^m-1/2

Would the middle term not just be (m+1)th term?

m terms before and m terms after?

xOxSinéadxOx

bstoran wrote: »

thing about the asymptote question was that dy/dx (tangent to curve) was -2 over a squared term.....ie any value wood b minus....minus by minus cant equal -1 for perpendicular!!!

sounds right to me!

galway.gaa

bstoran wrote: »

thing about the asymptote question was that dy/dx (tangent to curve) was -2 over a squared term.....ie any value wood b minus....minus by minus cant equal -1 for perpendicular!!!

i said the exact same thing for that:D

Nimur90

timmywex wrote: »

Just back in now, did all the questions,

thought in general it was grand. Few tricky parts.


Question 1 and 2, the very end of them was tricky to get as required.


Made a slip on c part 1 in question 3,


Question 4 part c was tricky enough. Confusingly phrased


Question 5 part c had a complex change of base, i got it but i imagine alot didnt even notice it!.

Question 6 was grand in general

Question 7 was slightly confusing in part c i would think

Question 8 was grand.


Hopefully got an A1 in the paper overall!:cool:

You did all the questions??? The whole thing was grand except for question 3. How the hell do you do that integration question on complex numbers? ah well its over now...

Crocodilius

Sweet wrote: »

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

Yeah, i think it worked out that q^2 = 36r/5 so q = at least 6/(5)^1/2 as r is a positive integer. that's what i got but it probably is wrong. I tried to work the question out using undetermined coefficients, is that what you did?