May 2014 Mathematics HL (Paper 3) Discrete Mathematics

[XeoKnight]

How did it go? I feel terrible. I was pretty good in Papers 1 and 2, expecting high fives, mid sixes, possible low 7, but this just completely destroyed my grade. Did anyone feel happy about it, or perhaps ok? How many of these did you manage to prove? I found a few proofs easy, and tried proof by contradiction for the prime number one lol.

[sushicurry]

I feel the same way. I got most of the proofs I think. Or at least partial points for them. I wasn't able to get the recurrence relation one. That one hung me up for a while.

[Guest]

This paper was horrible. Absolutely disgusting. I don't think I encountered any questions like the ones in the exams in any of my practices except for the extremely obvious ones like question 1. Part (a) of question 4 was ridiculous. And the proofs were crazy weird. I expected only one of these proof questions from the past papers and the specimen paper I did but when I saw that there was 3....... My god. I wrote some BS down so I hope I can get some partial marks. I hope the boundaries are significantly lower.

[Vioh]

This paper went absolutely horribly to me! It took me almost 5 minutes to figure out that I had to substitute n=41 to prove that not all n^2-n+41 is a prime. Don't you guys think that the question relating to the system of equation is a bit weird? They ask us to find the solutions modulo 2, but what does that mean? I even got stuck on the question about proving that all simple connected graphs contain at least a spanning tree. And finally, question 4)a) just killed me. I found a general solution but couldn't make B an integer. Did anybody get that question?

Generally, this exam looks so different from any past exam papers that I have done, including the specimen paper for 2014. Like the IB completely changed the ways they set up the questions, giving me a real shock when I did it.

[Guest]

This paper went absolutely horribly to me! It took me almost 5 minutes to figure out that I had to substitute n=41 to prove that not all n^2-n+41 is a prime. Don't you guys think that the question relating to the system of equation is a bit weird? They ask us to find the solutions modulo 2, but what does that mean? I even got stuck on the question about proving that all simple connected graphs contain at least a spanning tree. And finally, question 4)a) just killed me. I found a general solution but couldn't make B an integer. Did anybody get that question?

Generally, this exam looks so different from any past exam papers that I have done, including the specimen paper for 2014. Like the IB completely changed the ways they set up the questions, giving me a real shock when I did it.

I couldn't figure out that I should have substituted n=41. Goddamnit. Kicked myself so hard after the exam. I could to 4 part b) though. The question with the system of equations seemed completely off topic. I think nearly everyone who took this exam got screwed. FML

[gemmasz]

That paper was horrific. Way too many proofs for my liking, and the recurrence relations question was awful too. Definitely the worst DM paper they've set for a long time!

[Edgod]

@XeoKnight : it wasn't contradiction, it was logic. n2-n+41=not a prime when n=41. Coz yeah... and btw I think the exam was ridiculous as well. Our teacher made us do every exam going back like ten years, we did every example in the book, she taught us so much, but these recurrence relation questions??? SERIOUSLY??? THEY WERE NOWHERE. NOWHERE. OMG. And even though I only missed like 1 or two questions until the recurrence relation bit, that part was worth 17/60 marks, so whoopdiedo I basically cannot get a 6 on this, which was supposed to be the best part of all three. fml. OMG YEAH and that system of equations another 6 points gone...

Edited May 17, 2014 by Edgod

[XeoKnight]

@XeoKnight : it wasn't contradiction, it was logic. n2-n+41=not a prime when n=41. Coz yeah... and btw I think the exam was ridiculous as well. Our teacher made us do every exam going back like ten years, we did every example in the book, she taught us so much, but these recurrence relation questions??? SERIOUSLY??? THEY WERE NOWHERE. NOWHERE. OMG. And even though I only missed like 1 or two questions until the recurrence relation bit, that part was worth 17/60 marks, so whoopdiedo I basically cannot get a 6 on this, which was supposed to be the best part of all three. fml. OMG YEAH and that system of equations another 6 points gone...

Well, you could have proven by contradiction by just giving them any number that wasn't prime but n = 41 makes more sense. You probably did better than me, I did really badly with the proofs so I doubt I got anything good, I had been thinking that the modulo and diophantine equations would come up and had madly revised those as I had forgotten them, which turned out to be useless.

[au785]

Just to add to the atmosphere...

Yep. Horrible Paper.

All classmates including me were freaking out once we were out of that bloody room. That was the worst paper i've done in my whole life and im expecting maximum 50% of the marks. Why? because the question 4 was just so DAMN terrible i think i might get like 4/7 of part b but part a tops 1/10. Did a few of past papers and this just seemed like examiners decided to turn up the volume x4 literally.

Question 1 and two are my only hopes...

But this was such a damn surprise. Its a little upheaving too see everybody did bad so the grade boundaries are hopefully coming down this year

Conclusion: Discrete maths is the hardest option now. Escalated quickly from 4th to 1st !@!!!

[Vioh]

I still remember question (4a). They gave us the recurrence relation u_n + 2u_n-1 = 3n - 2 which we then had to show that one of the solutions will have the form u_n = A + nB (A & B are integers). Has anyone found the way to do this yet? Because I'm really really curious about how this question should be approached. Thanks

[Guest]

I hope the grade boundaries are ridiculously low for those who took discrete math this year. 70% for a 7 possibly? Maybe 70% is too low. What do you guys think the boundary will be around this year?

[Vioh]

I just did a quick search on the internet and I believe that the recurrence relation in question (4a) is something called inhomogeneous linear recurrence relation (the link to the website is here http://staff.scem.uws.edu.au/cgi-bin/cgiwrap/zhuhan/dmath/dm_readall.cgi?page=24). But isn't this completely out of the syllabus as the syllabus only mentions "Solution of first- and second-degree linear homogeneous recurrence relations with constant coefficients"?

I hope the grade boundaries are ridiculously low for those who took discrete math this year. 70% for a 7 possibly? Maybe 70% is too low. What do you guys think the boundary will be around this year?

I think 70% may be too low because as far as i know, it has never happened before, not even during the last syllabus change in 2008. But who knows? maybe this year is super super special

[XeoKnight]

I just did a quick search on the internet and I believe that the recurrence relation in question (4a) is something called inhomogeneous linear recurrence relation (the link to the website is here http://staff.scem.uws.edu.au/cgi-bin/cgiwrap/zhuhan/dmath/dm_readall.cgi?page=24). But isn't this completely out of the syllabus as the syllabus only mentions "Solution of first- and second-degree linear homogeneous recurrence relations with constant coefficients"?

I hope the grade boundaries are ridiculously low for those who took discrete math this year. 70% for a 7 possibly? Maybe 70% is too low. What do you guys think the boundary will be around this year?

I think 70% may be too low because as far as i know, it has never happened before, not even during the last syllabus change in 2008. But who knows? maybe this year is super super special

If that's so it would really, really suck. I was thinking we need to know how to do that recurrence relation. Maybe we were supposed to be able to, but if it was an inhomogeneous one (and my tb specifically said there were only homogeneous ones) then it shouldn't count. If it goes extremely terrible for all of us maybe they will make an exception? I don't know, the paper was just really bad.

[au785]

Well there have been years where the physics HL boundaries fall bellow 65% for a 7 ;P and to be fair physics HL is so much easier than math HL - you never find a question worth 5+ marks in physics whereas in maths they always screw us with part A 8 mark-one sentence questions and part B lotsa-marks too questions.

Also, in computer science HL last eyar you needed 67% to get a 7. Assuming you get full marks in the 20% CW, you simply need 47% for the rest 80%; meaning that around 58% of the marks in the papers would do for a 7. LOL

in conclusion i suppose it is possible for the grade boundaries to drop, specially when in previous years when discrete has been much easier the requirements for a 7 in general don't really go above 75%.... Considering this year it went that harsh for everybody, who knows? we'll see =) GOOD LUCK TO YOU ALL

Edited May 18, 2014 by au785

[supericecream]

So there are 5 people in our class who took discrete this year, and basically there's this one guy who is pretty much a solid 7 in almost every math test that we've had. Around 5 minutes before the end of paper 3 I looked up as I was sitting behind him and saw that he was already done and just chilling. Obviously I was bewildered as I looked at my paper and realized how many marks I've pretty much already lost. Later I asked him after the exam how he found the test and he said: "Oh, I only answered half the questions".

Therefore comrades, do not despair. I think the boundaries will be very low this year.

[Guest]

I heard from a math teacher at my school that alot of math teachers from around the world found the paper to be ridiculous and that there was an outrage because the paper left out so much of the option's content in favour of the ridiculously obscure stuff.

I hope the IB would be reasonable enough to listen to the pissed off teachers and really really lower the grade boundary.

[SYL]

I still remember question (4a). They gave us the recurrence relation u_n + 2u_n-1 = 3n - 2 which we then had to show that one of the solutions will have the form u_n = A + nB (A & B are integers). Has anyone found the way to do this yet? Because I'm really really curious about how this question should be approached. Thanks

 

If my memory serves correct, question 4(b) had  the recurrence relation as u_n = 2u_n-1 + 3n - 2. So after days of sporadic attempts at searching for a website/video on how to approach this type of recurrence relation, I finally found it on http://www.macs.hw.ac.uk/~jack/F12LE2/c4a.pdf, on Page 8~9, under the header 4.3 Solving Inhomogenous Recurrence Relations.

 

Based on what I saw, you have to let u_n = pn + q, where p and q are constants. We have to find p and q.

 

pn + q = 2 [ p( n -1) + q] + 3n - 2

pn + q = 2pn -2p +2q + 3n -2

 0 = n (p+3) -2p + q - 2

 

Hence p + 3 = 0 , p = -3

 

Then 4 + q = 0 , q = -4,      giving u_n = - 3n - 4.

 

Also Vloh, inhomogenous recurrence relations are not entirely out of the syllabus - only the special case  u_n = au_n-1 + b is included ( source: 2014 syllabus). So therefore, the term b should clearly have been a constant and not a function of n, otherwise they would've written u_n = au_n-1 + g(n). So either IB was deliberately ambiguous, which I doubt, or the person who wrote the exam misinterpreted? So yeah, part b) and c) of question 4 should definitely not be counted   

Edited June 25, 2014 by SYL

[SYL]

For question 4c) where they give u₁ = 7, you plug it into u_n = c(2ⁿ) - 3n - 4 and you'll obtain c = 7, giving the final answer to be u_n = 7(2ⁿ) - 3n - 4.

 

If you want to check, you can sub in n =  2 and 3 into u_n = 7(2ⁿ) - 3n - 4, giving u₂ = 18, u₃ = 43.

 

From the original equation  u₃ = 2u_3-1 + 3(3) - 2

LHS = 43

RHS = 2 x 18 + 9 -2 = 36 +7 = 43

 

LHS = RHS. So, its correct

Edited June 25, 2014 by SYL

[Vioh]

Also Vloh, inhomogenous recurrence relations are not entirely out of the syllabus - only the special case  u_n = au_n-1 + b is included ( source: 2014 syllabus). So therefore, the term b should clearly have been a constant and not a function of n, otherwise they would've written u_n = au_n-1 + g(n). So either IB was deliberately ambiguous, which I doubt, or the person who wrote the exam misinterpreted? So yeah, part b) and c) of question 4 should definitely not be counted   

 

Don't you mean that the question (4a) shouldn't be counted? because question 4a (as far as i remember) was u_n = 2u_n-1 + 3n - 2, which can be basically written as u_n = au_n-1 + g(n), where g(n) = 3n-2. Thus it wasn't part of the syllabus.

 

SYL, it seems that you are right about the method of solving this recurrence relation. All we have to do is to substitute u_n = A + nB (as the question suggested) into the relation u_n = 2u_n-1 + 3n - 2, and then solve for A, & B. I was pretty stupid to not think of this during the exam, cuz this question was worth 10 points. However, I think the IB is a bit ambiguous in the way they set up the question. I remembered that the question was quoted as: "one of the solutions will have the form u_n = A + nB where A & B are integers", which made me misunderstand into thinking that only one solution has that form while the rest of the solutions will have a different form.

 

By the way SYL, could you tell me the way that question (4b) and (4c) were quoted in the exam? I think these two last questions were part of the syllabus, but they were written in a quite weird ways. It has been a month since the exams, so my memory is a little bit blurred right now

[SYL]

 

Don't you mean that the question (4a) shouldn't be counted? because question 4a (as far as i remember) was u_n = 2u_n-1 + 3n - 2, which can be basically written as u_n = au_n-1 + g(n), where g(n) = 3n-2. Thus it wasn't part of the syllabus.

 

By the way SYL, could you tell me the way that question (4b) and (4c) were quoted in the exam? I think these two last questions were part of the syllabus, but they were written in a quite weird ways. It has been a month since the exams, so my memory is a little bit blurred right now

 

 

 

nah you mixed up 4a) and 4b). Question 4 a) was where you had to find the general solution of u_n-2u_n-1 = 0, which is a 1st order homogenous r.r., and is therefore in line with the syllabus. But It's probably only 1~2 marks, also the answer was u_n = c(2ⁿ)

 

It was question 4b) that asked for a solution with the integers A & B.  Finally question 4c) said: hence find the particular(<-not sure if that was the exact word) solution given u₁ = 7. And since part b and c were connected, they're both out of the syllabus.

 

then 5 was the 2nd order homogenous r.r. - which we all anticipated