** Higher Level Maths Paper 1 2012 Before/After **

ehshup

Dicksboro_man wrote: »

what about q1c (ii)

i used b^2 - 4ac >0 on ( k^2x^2+t)

and since b=0 i got -4k^2t >0

so t<0 ? anyone agree?

that's what i got too, but in my last minute i saw that that only guarantees that the two roots from those two factors are different, but there is a possibility it could be a same as the third root, hence i think you had to place an added restriction, something like t not equal to minus k, which i scribbled at the side of the sheet with no explanation, but i don't know for sure.

DepoProvera

Wanchor wrote: »

In terms of cos(2x)? I mean, in the heat of the exam how does begin to understand that.

I used the identity cos^2x=1/2(1 + cos2x) and rearranged to get cos4x=2cos^2(2x) - 1. Couldn't do the last part though I think I made a mathematical error I assume it should have worked out to a perfect square or something like that to prove is was greater than 0

mcpaddington

BrianC2095 wrote: »

How did you do it? sub answer z^2 +k into the line?

Get the area between (a,a^2+k) and (-a,a^2+k) between the Graf and line.

WestIRL

Wanchor wrote: »

In terms of cos(2x)? I mean, in the heat of the exam how does begin to understand that.

Yeah I agree. It was such a weird question, never seen the likes of it before, and of all the times to see it- THE ACTUAL LEAVING. Trying to get it in terms of cos2x while furiously subbing in every trig formula under the sun to absolutely no avail.

Orange juice

BrianC2095 wrote: »

Did anyone do Q5 seq + series?

Ya! IMO, easiest question on the paper! So happy when I saw teloscoping!
Disappointed with Q6 the cos2A thing, just couldn't figure it out ... did all 8 questions though so hopefully I've scraped an A2/B1.

Hermione Granger

Jan_Itor wrote: »

For 1b(ii) it said you had to find to find cubic equation where the roots were 1 less than the roots of f. I just did f(x-1) worked it out and got a cubic equation is that right?

Am i dont know if this is right but my teacher confirmed it after you take your three roots from the previous answer take one away from each turn them into factors and multiply

BrianC2095

DepoProvera wrote: »

Yeah I did. a) grand b) normal telescoping c) tough but I got b^2=ac then did the standard thing for consec arithmetic terms and it worked out to b^2=ac which had been proved true from earlier

Yea it was a handy enough question i spent 10 mins trying to find A and B until i realised it was just a conjugate :rolleyes:
Yea i did the same and just brought it all to one side and i all cancelled when you put it b^2 =ac:D

Dapics

Jan_Itor wrote: »

For 1b(ii) it said you had to find to find cubic equation where the roots were 1 less than the roots of f. I just did f(x-1) worked it out and got a cubic equation is that right?

It depends.. I just took one away from each of my x-values, brought across the x-vales so i had a factor and then multiplied the three of them out.

RedTexan

Was disappointed coming out, intended on going in and getting every question. But it seems I might have done better than I thought, All right it seems except for two c parts

123 LC

i'm not great at maths but i actually really liked that paper! i think i got all of 7 out (and i never get full answers out)
if i'm lucky i might have gotten a B3, which is more then i'd hoped for! hopfully paper 2 goes well

K_1

Wesc. wrote: »

Yes Yes Yes! I got the out in the last minute haha.


Guys I'm actually in dreamland now I think it's a 100%!

I got root3 over 3 . . .shít.

Wkds

Cosmic.Postman wrote: »

Was there not only one actual value for x? I took it as the common value for x when you worked out the two quadratics, which was 3 I think.

Crap you think I will lose many marks for not stating it was 3?

BrianC2095

mcpaddington wrote: »

Get the area between (a,a^2+k) and (-a,a^2+k) between the Graf and line.

No for Q3cii the points on a colinear line

123 LC

Dapics wrote: »

B almost caught me out, the shnakes asking for an integer! I only just managed to remember what an integer was and figured out it was a trick question

what do u mean by an integer? :O dam i thought i'd got that question right what answers did u get?

Jan_Itor

Dapics wrote: »

It depends.. I just took one away from each of my x-values, brought across the x-vales so i had a factor and then multiplied the three of them out.

The answer i got was x^3 - 5x^2 + 2x +8 ?

mcpaddington

BrianC2095 wrote: »

mcpaddington wrote: »

Get the area between (a,a^2+k) and (-a,a^2+k) between the Graf and line.

No for Q3cii the points on a colinear line

Ah my bad, I thought you were talking about 8cii.

Dicksboro_man

yeah for 6c (iii) i couldnt get a perfect square from my
8cos^2(2x) + 8cos2x +4
:-/

how did we all word 7c (iii) about all further approximations? i just said they were all gonna be the equal to x1 or x2 and wont be getting any closer to the true root.

Dicksboro_man

Jan_Itor wrote: »

The answer i got was x^3 - 5x^2 + 2x +8 ?

same

Zaffy

Project maths paper 1 was fine. Still despise the boxes they give us.

BrianC2095

Dicksboro_man wrote: »

yeah for 6c (iii) i couldnt get a perfect square from my
8cos^2(2x) + 8cos2x +4
:-/

how did we all word 7c (iii) about all further approximations? i just said they were all gonna be the equal to x1 or x2 and wont be getting any closer to the true root.

Just that the values would keep oscillating between 0.5 and 0.75 infinately:D

GarIT

Here are all the scans

Sorry about the writing, I was trying to estimate marks.

emmamurphy233

I was really happy with it until I came on here. For 8.c.ii did anyone get the integral of x^2 + k with limits a and -a and then take away the areas of the triangles?
And did people use De Moivre's theorem for 3.c.i?
I also got a different answer than everyones saying for the last part in 7c. But hopefully that's only 5 marks.. All in all I think I got about 250 in that. If Paper two is nice I could get an A.

Hermione Granger

What did people get for q 1 b

I thought it was just trial and error to get first value and divide this factor to get the other 2

I think I got something like x = 1, -2 and 3 ??/

Wesc.

This is how I went about doing 3 (c) (ii) :

Ok so they're all on the same line right? So I decided to get the equation of the line and subbed in the point (can't remember it but it had k + something in the y or x value) and everything magically cancelled out and I was left with the beautiful number 1! :P

reznov

What did you guys get for the matrix?

RedTexan

Dicksboro_man wrote: »

yeah for 6c (iii) i couldnt get a perfect square from my
8cos^2(2x) + 8cos2x +4
:-/

how did we all word 7c (iii) about all further approximations? i just said they were all gonna be the equal to x1 or x2 and wont be getting any closer to the true root.

That's the answer I got as well, and tried the perfect square. do you think that is correct for that part of the question?

Dapics

123 LC wrote: »

what do u mean by an integer? :O dam i thought i'd got that question right what answers did u get?

An integer is a fraction.

Basically all you had to do was find your three x-values by simply factorsing and removing whats common each time to isolate your x-values..

ei.sderob

Dapics wrote: »

Its a trick question!

All you had to do was divide by two a few times and then let the equation =0 an bring across cos2x...

Beautiful Beautiful paper.... Im so happy.

8) C, was impossible
1) B almost caught me out, the shnakes asking for an integer! I only just managed to remember what an integer was and figured out it was a trick question

It doesn't really make mathematical sense to just "let it equal to zero" though, does it?

leaveiton

Hermione Granger wrote: »

What did people get for q 1 b

I thought it was just trial and error to get first value and divide this factor to get the other 2

I think I got something like x = 1, -2 and 3 ??/

That's exactly what I did/got

Jan_Itor

Dapics wrote: »

An integer is a fraction.
.

An integer is a whole number!