IB Maths HL 2012

Look at the other thread xD, apparently its close to impossible........... I put arctan(x^2) as well initially, but when I tried to confirm using my calculator I god different results. In that case we'd have to sub x^2=u which messes everything up...........

actually what is the question?? i think u guys needa just plug the things in your GDC, and get 1.40...

what i did was expanding the brackets and use partial fractions..... then thru load of crap i got:

1/(4√2) .ln[(x^2+√2x+1)/(x^2-√2x+1)] + 1/2arctan(√2x+1) + 1/2arctan(√2x-1) +c

Reply 101

hkl1994215

Original post by johndoe04

the real ones are out in about 5 weeks, i think.

what do you think about the actual papers? harder or easier than last years and previous years (except 2008: this was very very easy e.g. 3 marks for finding mean from cumulative frequency graph etc)?

Reply 102

hkl1994215

Original post by jc_bach

actually what is the question?? i think u guys needa just plug the things in your GDC, and get 1.40...

what i did was expanding the brackets and use partial fractions..... then thru load of crap i got:

1/(4√2) .ln[(x^2+√2x+1)/(x^2-√2x+1)] + 1/2arctan(√2x+1) + 1/2arctan(√2x-1) +c

I personally think thats what the ib is looking for, using the GDC, in the end paper 2 is a calculator paper

Reply 103

marixmas

Original post by jc_bach

actually what is the question?? i think u guys needa just plug the things in your GDC, and get 1.40...

what i did was expanding the brackets and use partial fractions..... then thru load of crap i got:

1/(4√2) .ln[(x^2+√2x+1)/(x^2-√2x+1)] + 1/2arctan(√2x+1) + 1/2arctan(√2x-1) +c

I think it was a trial and error question, that's what i did and ended up with 1.40.

Reply 104

johndoe04

Original post by hkl1994215

what do you think about the actual papers? harder or easier than last years and previous years (except 2008: this was very very easy e.g. 3 marks for finding mean from cumulative frequency graph etc)?

I'd say it was a lot more difficult overall.

The IB are very......original, I guess with their questions. Kinda annoys the hell out of me in a way. :P

(edited 11 years ago)

Reply 105

MAMDS1993

1

Your coordinator will get them with your breakdowns! There were just 2 qns that bothered me on this paper, one was the integral one, where I did trial and error and didn't get time to do the circle strings question, geometry has never been my strong point! I started out by drawing a straight line to get a rectangle and a right angled triangle..and time was called!

Reply 106

x-tina848

Original post by Hyaline

Look at the other thread xD, apparently its close to impossible........... I put arctan(x^2) as well initially, but when I tried to confirm using my calculator I god different results. In that case we'd have to sub x^2=u which messes everything up...........

Yeah that's exactly what I did. But like you said, it gets really messy with the substitution. :/
I looked up the solution: http://answers.yahoo.com/question/index?qid=20080721154424AAYOrmH
...Seems to me like the cost-benefit for this problem is a bit off. I mean, I really wouldn't have thought it would be this tough for 3 points

Reply 107

dawesbr

RE: The integral of 1/(1+X^4).

Most GDCs should have a solver feature? On my school's GDC, a casio 9750GII, menu 8 has solver (under Equation, after simultaneous and polynomial). Then you just plug in, using the catalogue, integral(1/(1+(X^4)),0,X)=1 and execute and it gives 1.3956... I double checked with graph then finding the definite integral from 0 to 1.3956 and it worked.

Reply 108

hin1994

Guys do you remember what the question / answer was for the complex number question in Section A, Paper 1?

it was something about equating coefficients of a complex number (not the 29 pointer in section B)

(edited 11 years ago)

Reply 109

arra

7

Original post by hin1994

Guys do you remember what the question / answer was for the complex number question in Section A, Paper 1?

it was something about equating coefficients of a complex number (not the 29 pointer in section B)

Yeah. Uhmm. I THINK I got 3 and -5, or something. Not sure actually, but I got it right. For part a) you just equate real and imaginary parts and solve the system. For part b you have to sub in (a + bi) and (a-bi) multiply it out, and solve for a and b by equating real and imaginary parts.

I know I got the right answers, don't remember what exactly I got, though.

Reply 110

hin1994

hmm... I equated coefficients for part A (real and imaginary) and for part B, I made m=a+bi and n=a-bi and solved for that. Should be correct as I remember one of my answers was either 3 or -3. Do you remember the exact question though?

Was it something along the lines of 4(m+n)+3im=16+15i, where part A was find m and n if they were real numbers and part B was find m and n if they were complex conjugates?

Reply 111

arra

7

I think if they are expecting trial and error they should specify it. Even in the calculator section, they want to see appropriate use of a calculator, which would imply some method to the madness rather than stupid trial and error.

Any idiot can do trial and error with a sufficient calculator, and to give question like that without specifying how they expect you to do it or anything really was kinda stupid. I'm all for conceptually challenging questions, even long and tedious questions, but to expect us to realize that we need to trial and error (or integrate something complicated beyond what we normally do) without a few guiding steps is stupid.

Reply 112

Naiomi

OP

1

Who's doing statistics for p3?

And what did you get for the question with two discs in paper 2? 69 something cm?

Reply 113

willamgg1993

Yeah I think I got something like that - 28+13+26.5 ish (can't remember specific values haha)

I came out of the first exam really confident, but everyone else found that harder than the second. But the 2 integration questions on paper 2 (continous dist and last question) were beastly. You shouldn't be expected to use the calculator for trial and error, how is that maths?... :angry:

Reply 114

willamgg1993

And I'm praying they do something sneaky on stats like put algebra in chi squared or type 1 and 2 errors etc. cos anyone can use the calculator to work out distributions and tests, so that means high boundaries

Reply 115

Hyaline

Discrete maths over here..........

Reply 116

lydia.laurent

Original post by Hyaline

Discrete maths over here..........

I'm doing discrete maths too, what proofs do you think they will ask?

Reply 117

Hyaline

No idea....... I can't memorize them all. I just spontaneously try to prove them xD.

Reply 118

lydia.laurent

I've done a few past papers and without answering any of the harder proofs I still manage to get 7s from Graph theory and Euclidean algorithm questions. I think that Discrete is probably the easiest of the all the options (thankfully)