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

rkeano5

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 ??/

Yeah thats correct I think

DepoProvera

RedTexan wrote: »

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?

Glad to see other people were in the same boat.. I couldnt get the perfect square for the life of me so I just wrote that it is >0 thus is always decreasing :P

Hermione Granger

Dapics wrote: »

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..

Take a look at pg 23 of the log tables it explains the word and gives examples. It says that its a whole no. either positive or negative ...

Tankosaur

1// (a) a=3 b=-2 c=5
(b)(i) -2,1,3
(ii) x^3 + x^2 -6x =0
(c) was a proof.
(ii) t is a negative number

2// (a) x=3
(b) (i) -6
(ii) 5x^2+-12x+20
(c) proofs

3// proof
(b)(ii) y=-1 x=3
(c) + and - (root3/2 + i/2)
k=1

4// 2ac/c+4 =b
(b) (i) proof
(ii) root n =1 - 1 (edited for misreading my answer)
(ii) 9
(c) proof

5// (a) x=5
(b) (i) 10C r-1 3^11-r 2^1-r x^23-3r
(ii)76545/8
(ii) 23/3 is not rational therefore 23-3r does not = 0
(c) both (i) and (ii) are proofs

6//24x(4x^2 -1 )^2 Edit: typo
(ii)2/root(9-4x^2)
(b) (i) proof
(ii)6y=x+9
(c)8+4cos4x + 8cos2x
4cos^2 2x + 8cos2x + 8
(2cos2x + 2)^2 >0 thefore all values of fx are increasing

7// proof dy/dx =-1/3
(b) dy/dx = 2y-2x / 4 + 6y -2x
same slopes i.e. m=1/2
(c) different signs for f(1) AND f(0)
3/4 and 1/2
the approximations are repeating

8// x + sin2x / 2 + e^3x /3 + c
(b) 4ln7
(ii) pi/16 - 1/8
(c) a = root k
(ii) area is 5/3 x k^3/2

RedTexan

ei.sderob wrote: »

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

That's not right, you can't just let it equal to 0

Dapics

Hermione Granger wrote: »

Take a look at pg 23 of the log tables it explains the word and gives examples. It says that its a whole no. either positive or negative ...

My bad!
Either way, i got the fecker right, thank god!

Chuchoter

K_1 wrote: »

I got root3 over 3 . . .shít.

I got a half and I thought my method was pretty clever :P So three viable answers

123 LC

Dapics wrote: »

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..

i thought integers couldn't be fractions? they have to be whole numbers i think?

ivanh

Don't worry you guys, I knew we were in for a tricky Paper 1!

Now they'll give us an easy Paper 2 just to try and prove that this Project Maths craic is a good idea...

Dapics

123 LC wrote: »

i thought integers couldn't be fractions? they have to be whole numbers i think?

Yeah, I'm only after confusing my self with it.... soweee!

Anyway, nice damn paper!

Stalin and rugby

Tankosaur wrote: »

1// (a) a=3 b=-2 c=5
(b)(i) -2,1,3
(ii) x^3 + x^2 -6x =0
(c) was a proof.
(ii) t is a negative number

2// (a) x=3
(b) (i) -6
(ii) 5x^2+-12x+20
(c) proofs

3// proof
(b)(ii) y=-1 x=3
(c) + and - (root3/2 + i/2)
k=1

4// 2ac/c+4 =b
(b) (i) proof
(ii) root n =1 - root n
(ii) 9
(c) proof

5// (a) x=5
(b) (i) 10C r-1 3^11-r 2^1-r x^23-3r
(ii)76545/8
(ii) 23/3 is not rational therefore 23-3r does not = 0
(c) both (i) and (ii) are proofs

6//24x(4x^2)^2
(ii)2/root(9-4x^2)
(b) (i) proof
(ii)6y=x+9
(c)8+4cos4x + 8cos2x
4cos^2 2x + 8cos2x + 8
(2cos2x + 2)^2 >0 thefore all values of fx are increasing

7// proof dy/dx =-1/3
(b) dy/dx = 2y-2x / 4 + 6y -2x
same slopes i.e. m=1/2
(c) different signs for f(1) AND f(0)
3/4 and 1/2
the approximations are repeating

8// x + sin2x / 2 + e^3x /3 + c
(b) 4ln7
(ii) pi/16 - 1/8
(c) a = root k
(ii) area is 5/3 x k^3/2

No t was a positive number anything greater than 0

Wesc.

Tankosaur wrote: »

1// (a) a=3 b=-2 c=5
(b)(i) -2,1,3
(ii) x^3 + x^2 -6x =0
(c) was a proof.
(ii) t is a negative number

2// (a) x=3
(b) (i) -6
(ii) 5x^2+-12x+20
(c) proofs

3// proof
(b)(ii) y=-1 x=3
(c) + and - (root3/2 + i/2)
k=1

4// 2ac/c+4 =b
(b) (i) proof
(ii) root n =1 - root n
(ii) 9
(c) proof

5// (a) x=5
(b) (i) 10C r-1 3^11-r 2^1-r x^23-3r
(ii)76545/8
(ii) 23/3 is not rational therefore 23-3r does not = 0
(c) both (i) and (ii) are proofs

6//24x(4x^2)^2
(ii)2/root(9-4x^2)
(b) (i) proof
(ii)6y=x+9
(c)8+4cos4x + 8cos2x
4cos^2 2x + 8cos2x + 8
(2cos2x + 2)^2 >0 thefore all values of fx are increasing

7// proof dy/dx =-1/3
(b) dy/dx = 2y-2x / 4 + 6y -2x
same slopes i.e. m=1/2
(c) different signs for f(1) AND f(0)
3/4 and 1/2
the approximations are repeating

8// x + sin2x / 2 + e^3x /3 + c
(b) 4ln7
(ii) pi/16 - 1/8
(c) a = root k
(ii) area is 5/3 x k^3/2

I got those exact figures.


Delighted. :P

RedTexan

Stalin and rugby wrote: »

No t was a positive number anything greater than 0

I think the area in 8 was 2k by root k all over 3?

Stalin and rugby

ivanh wrote: »

Don't worry you guys, I knew we were in for a tricky Paper 1!

Now they'll give us an easy Paper 2 just to try and prove that this Project Maths craic is a good idea...

Nope, they'll give a normal paper and if people do badly they'll manipulate the marking scheme. People said that today's paper 1 would be easy too

partyatmygaff

Initial impression is that it's a lot easier than last year's paper. The marking scheme won't be as generous as last year i'd imagine.

Tankosaur

Stalin and rugby wrote: »

No t was a positive number anything greater than 0

My answer and the same for a few I know came out that k= root -t therefore to have real roots t had to be negative. Otherwise k would be complex.

Stalin and rugby

partyatmygaff wrote: »

Initial impression is that it's a lot easier than last year's paper. The marking scheme won't be as generous as last year i'd imagine.

It's always easier when you're not actually doing the exam bro..

RedTexan

For the proof with seqs and series was it correct to get b^2 equal to ac by using r and then subbing your value for b into what you get when you leave the common differences equal?

amymak

Thought that the paper went really really well!
I was able to do every single part and I had enough time to do 7 questions and have 15 minutes to spare.
So the only marks I should lose will be blunders and slips. (My greatest weakness.)
I just hope they don't throw anything horrible at us for paper 2.

Tankosaur

Wesc. wrote: »

I got those exact figures.


Delighted. :P

Bitchin'.

We shall go far together.

Wasn't sure about my integration answer so if two of us have the same then I should get some decent method marks.

aoife..x

well the project maths paper....WTF!! nothing like what we have been doing in school!! also totally different to the sample papers from the SEC! everyone in my year was in complete shock after that! even the ordinary level students think they failed!! FML!

Chuchoter

partyatmygaff wrote: »

Initial impression is that it's a lot easier than last year's paper. The marking scheme won't be as generous as last year i'd imagine.

That was why this was a sneaky paper. I opened it, read through and went oh everything is grand.
THEN I STARTED DOING IT.

Wesc.

RedTexan wrote: »

For the proof with seqs and series was it correct to get b^2 equal to ac by using r and then subbing your value for b into what you get when you leave the common differences equal?

And again, I did the exact same as you! Worked out perfectly, don't know HOW such a question merited a (c) part.

Hermione Granger

Tankosaur wrote: »

1// (a) a=3 b=-2 c=5
(b)(i) -2,1,3
(ii) x^3 + x^2 -6x =0
(c) was a proof.
(ii) t is a negative number

2// (a) x=3
(b) (i) -6
(ii) 5x^2+-12x+20
(c) proofs

3// proof
(b)(ii) y=-1 x=3
(c) + and - (root3/2 + i/2)
k=1

4// 2ac/c+4 =b
(b) (i) proof
(ii) root n =1 - root n
(ii) 9
(c) proof

5// (a) x=5
(b) (i) 10C r-1 3^11-r 2^1-r x^23-3r
(ii)76545/8
(ii) 23/3 is not rational therefore 23-3r does not = 0
(c) both (i) and (ii) are proofs

6//24x(4x^2)^2
(ii)2/root(9-4x^2)
(b) (i) proof
(ii)6y=x+9
(c)8+4cos4x + 8cos2x
4cos^2 2x + 8cos2x + 8
(2cos2x + 2)^2 >0 thefore all values of fx are increasing

7// proof dy/dx =-1/3
(b) dy/dx = 2y-2x / 4 + 6y -2x
same slopes i.e. m=1/2
(c) different signs for f(1) AND f(0)
3/4 and 1/2
the approximations are repeating

8// x + sin2x / 2 + e^3x /3 + c
(b) 4ln7
(ii) pi/16 - 1/8
(c) a = root k
(ii) area is 5/3 x k^3/2

Got almost the exact same answers for all those messed up my proofs and did 8 c in terms of a rather thank k but heres hoping

Wesc.

Tankosaur wrote: »

Bitchin'.

We shall go far together.

Wasn't sure about my integration answer so if two of us have the same then I should get some decent method marks.

I'm a tank at integration.


:P

RedTexan

Wesc. wrote: »

I'm a tank at integration.


:P

what did you get?

Wesc.

RedTexan wrote: »

what did you get?

2.k.root(k) over 3 ... or you could say 2k^3/2 divided by 3.

RedTexan

Wesc. wrote: »

2.k.root(k) over 3 ... or you could say 2k^3/2 divided by 3.

Agreed

Dicksboro_man

Tankosaur wrote: »

1// b
(ii) x^3 + x^2 -6x =0

4// 2ac/c+4 =b
(b)
(ii) root n =1 - root n

got different ans for those:

1bii = x^3-5x^2+2x+8

4a b=4ac/(a+c)

4bi root(n+1) - 1


could be wrong though also i didnt really try 8c and messed up 6 ciii but ah well..

ashh

Is it possible for α(squared) + β(squared) in q2 (b) (i) to be negative?... Or is there some complex root trickery afoot?...