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IB HL Specimen Paper for 2014 Syllabus


Dax

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So I just attempted paper 1 for 2014 Syllabus Specimen Paper.
I'm quite conflicted on what to make of it, I thought it was kind of borderline hard, and asked questions which differed from the usual pattern. Plus there were some questions which were quite different from the previous papers like Question 7 and Question 9.
What are your thoughts on this, and if you could please help in solving these questions!
Thanks a lot!

P.S - Just google the specimen paper, and you'll find it

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I won't try it because we're going to have to do it as a second mock 2 weeks from now, and I want to do it honestly to have a good idea of how I am going to do in the final exam. But generally speaking, are specimen papers similar to the actual papers of the year?

If yes, and you thought this specimen paper was harder than the usual, and abnormal, let's hope that the grade boundaries for us will be relatively low. I did read somewhere that because of the new IA format, they can ask you modelling questions on the actual past paper. Did you find that it was the case?

Edit: Actually I just looked at a May 2009 specimen physics paper and the actual paper and there were no identical questions. The questions in paper 1 were similar in that they covered the same topics, but that's always the case on paper 1, while paper 2 and 3 were completely different. So I'm just going to conclude (based on the limited information that I have) that there is no real similarity between the real paper and the specimen paper.

Edited by maroctam
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Yeah Actually, without delving into the question much, there were questions that I believe would've gone for 4-5 marks in past years paper for 1-2 marks. Also, like pre 2008, they asked to comment on certain situations or answers, though it was very little. I think you'd be a better judge of the papers, because I for one am good at maths, but blank our during the exams. But I genuinely felt that this paper was a bit harder than previous 4-5 years paper 1

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Well, that's worrying. I really don't want to look at the papers because I want to genuinely take them as if they were the real exams, so I'll get back to you on them in 2 weeks time. Hopefully other people will reply in the meantime because I'm curious to see if there's a consensus of opinion.

And thank you! Do you generally do well on integration questions but the one in the specimen paper was particularly hard?

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Okay, I asked around my seniors and some tutors, and it seems to be a special substitution that is made in this particular questions case. I generally do well on Calculus (Differentiation, Integration and Differential Equation), I knew this Sum from somewhere because I had seen this in a Non IB Textbook. I'd say this is the first time I'm seeing an Integration like this in a paper, I may be wrong.
I look forward to the other answers, I wanna know if its just me or is it a bit hard for everyone!

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I'll just tell you the question.

Integral of sqrt(4 - x^2), limits 0 and sqrt(2)

This requires a special subsitution of 2sin(a) = x, dx/da = 4cos(a), dx = 4cos(a)da

Then you substitute everything, change the limits and you have the answer!

I'm quite sure I haven't seen this substitution in papers before, but I'm not quite sure. I hope I haven't made a fool of myself XD

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Ah! A trig substitution. I think I've seen a couple of those in past papers before, but I'm unsure. We did the exact same one you described in class though, so I'm familiar with how you would do it, but I'm sure I wouldn't successfully solve it if I hadn't practiced it beforehand: they're pretty hard.

Just by the way, depending on the function you might have to use a different substitution. Generally speaking

√(a2-x2) --> you would substitute x=asin(t)

√(a2+x2) --> you would subsitiute x=atan(t)

This is because of the trig formulae:

For the first scenario, you make use of cos2(x) = 1-sin2(x). So by doing √(a2-a2sin2(t)) you would get acos(t) - which you want because dx=acos(t)dt and you can find the integral of a2cos2(t) relatively easily.

For the second scenario, you make use of 1 + tan2(x) = sec2(x). This follows the same logic as the first scenario. When you do the square root you get asec(t), and then you have dx=asec2(t)dt .... I have no idea how one would integrate a2sec3(t), but whatever.

I'm pretty sure there's a third scenario as well, but I can't remember it right now.

Sorry if what I said makes no sense, please ask if you don't understand something!

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No I got it, I found it in some other textbook. I just checked my Math Notebook, I think we've done something like this extremely long back, but we weren't told of these special substitutions otherwise I would've remembered it.
The integral questions are almost entirely dependent of remembering these substitutions. I think usually, the Questions gave prompts when these questions were given. So I'm quite conflicted if the paper is going to be like this or not?

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Oh sorry, I misunderstood. And yeah, these questions are based solely on memorisation and maybe some skill. Based on some very minimal research:

Edit: Actually I just looked at a May 2009 specimen physics paper and the actual paper and there were no identical questions. The questions in paper 1 were similar in that they covered the same topics, but that's always the case on paper 1, while paper 2 and 3 were completely different. So I'm just going to conclude (based on the limited information that I have) that there is no real similarity between the real paper and the specimen paper.

I don't think that the papers will be very similar. But maybe, since this is the first year with the new syllabus, the style of questions will be comparable? I honestly don't know.

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I just got to know that Specimen Papers are basically a guideline for the teachers to understand what the course will cover. It is not what is expected in the papers. - Kinda like the same thing the quote above said!

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... I have no idea how one would integrate a2sec3(t), but whatever.

I'm pretty sure there's a third scenario as well, but I can't remember it right now...

The third situation you're thinking of is x2-a2, for which you would substitute x=asec(theta).

Integrating secant or secant cubed is a pain. I had to work through it for my IA just recently. http://www.math.ubc.ca/~feldman/m121/secx.pdf provides an interesting overview of how to do so.

Edited by Rahul
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  • 3 weeks later...

hahah i just googled ths and...... our teacher used the paper for our mock exams....... it was brutal (for some questions) but generally it was ok...... dunno how the results would turn out

i did a= 2 sin tetha for the substitution....

btw this video should be helpful: http://www.youtube.com/watch?v=vxCVsFoMj3I

question: would the syllabus be harder as years passes by? i looked at way back until 1995, the questions weren't tht hard.. and.. am i only the person who find 2009-2010 harder than 2008? (PS I havent done/am doing in class 2011-2013 papers so idk much about them, will be completing those by these last weeks)

Edited by fufufuufufufufuuu
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We did it as our mock exam in Math HL, and according to my opinion, it is generally harder than the real IB exams because of its nature - it is a specimen paper that introduces the new syllabus. Thus, it includes in itself questions of all difficulty and variety because it will be a model exam for the following 7 years until the silly changes again.

However, I do suggest you all go through it because it had quite a few things that are in the syllabus but that I hadn't generally seen before while doing past papers. And these include things like proving the nth derivative of a function using mathematical induction; trig substitution in integration; application of trigonometry (angles of depression and elevation and etc) as well as in series (compound interest and so on).

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Stronger candidates might have problems with the 8th, 9th, 11th under a) and the 13th question in my opinion. This all depends on the individual and I don't think that a candidate that is aiming for a 7 will fail to get any marks in these questions ( since you failed to solve 2 questions). Even if you fail to gain any marks in them you will end up with 84 marks out of 120. This is 70% of the entire paper and 21% for the final grade. The paper is definitely harder than usual. I would say that it's somewhere around may 2011 TZ1 HLP1 difficulty. I feel that the specimen paper is a little harder . Considering the grade boundaries for the may 2011 paper and the entire examination

https://www.dropbox.com/s/yyx7gy19b521rlm/marks%20range.jpg

https://www.dropbox.com/s/7kxpv6iftangnz4/full%20marks.jpg

Even if you get 42% from paper 1 and 2 and 30% from the IA (15%) and paper 3 (15%) you would still do fine and be able to get a solid 6 or a 7 with 72%.

This is in my opinion the worst case scenario for someone who is aiming for a 7. And I don't think that it's that hard to get at least 8% more for a solid 7 if you can manage to do this much. If you are doing sets, relations and groups it would be crazy to get below 18% or 17% from paper 3.

I personally found 11 a) to be challenging since I had never done this kind of substitution. The 13th a) took me some time however I did manage to do it under the time limit. The 8th and 9th were one of the harder ones on the paper but people who have strong reasoning will solve them quickly. The rest are very similar to past paper questions and someone who has done a lot of past papers should not have problems with them.

Regarding question 9

I am skipping the differentiation

the only way to do this is by inspection

derivative of f(x)= e^x -ex^(e-1)

=> e^x -ex^(e-1)=0 to find roots

=>e^x =ex^(e-1)

=>e^(x-1)=x^(e-1)

=> x-1=(e-1)lnx

the questions states that there are only two roots.

1 is an obvious root because ln1=0 hence LHS 0 and 1-1=0 hence RHS=0

e is also an obvious root because lne=1

If this was on paper 2 you can easily do this by using gdc however there is a hint here and the roots are way to obvious when you simplify the equation

You can sketch the graph by first finding the max and min. You can do this by testing whether the slope is negative or positive on each side of the points. This will tell you where the function increases or decreases. You can find the zero by substituting x=0 in the original equation. The sketch is easy with this info.

The last part follows from the graph

you see that all values for f(x) are larger than 0

=> f(x)>0

=> e^x -x^e>0

=> e^x =x^e

let x= pi

and you prove what is reacquired by substituting

Edited by rinik
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Even if you get 42% from paper 1 and 2 and 30% from the IA (15%) and paper 3 (15%) you would still do fine and be able to get a solid 6 or a 7 with 72%.

This is in my opinion the worst case scenario for someone who is aiming for a 7. And I don't think that it's that hard to get at least 8% more for a solid 7 if you can manage to do this much. If you are doing sets, relations and groups it would be crazy to get below 18% or 17% from paper 3.

You realize that a 42% from paper 1 (20%) and 2 (30%), 30% from the IA (30%) and 15% from paper 3 (20%) means 42%*20%+42%*30%+30%*30%+15%*20%=33% (a 3)

Instead of 42%+15%+15%=72%?

(And a 7 in paper 3 of the sets relations and groups option is actually 42/60 which is 70% or 14% weighted)

You have to get an average of 72% to get a 7 (or a sum of the weighted percentages) instead of a sum of 72% in all papers for the unweighted percentage.

It's possible that I interpreted your post wrong though, if that's the case I'm sorry, but it does seem strange that you said

30% from the IA (15%)

and

If you are doing sets, relations and groups it would be crazy to get below 18% or 17% from paper 3.

which caused my (probably the case) misunderstanding.

Edited by ssy
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Even if you get 42% from paper 1 and 2 and 30% from the IA (15%) and paper 3 (15%) you would still do fine and be able to get a solid 6 or a 7 with 72%.

This is in my opinion the worst case scenario for someone who is aiming for a 7. And I don't think that it's that hard to get at least 8% more for a solid 7 if you can manage to do this much. If you are doing sets, relations and groups it would be crazy to get below 18% or 17% from paper 3.

You realize that a 42% from paper 1 (20%) and 2 (30%), 30% from the IA (30%) and 15% from paper 3 (20%) means 42%*20%+42%*30%+30%*30%+15%*20%=33% (a 3)

Instead of 42%+15%+15%=72%?

(And a 7 in paper 3 of the sets relations and groups option is actually 42/60 which is 70% or 14% weighted)

You have to get an average of 72% to get a 7 (or a sum of the weighted percentages) instead of a sum of 72% in all papers for the unweighted percentage.

It's possible that I interpreted your post wrong though, if that's the case I'm sorry, but it does seem strange that you said

30% from the IA (15%)

and

If you are doing sets, relations and groups it would be crazy to get below 18% or 17% from paper 3.

which caused my (probably the case) misunderstanding.

I meant to say 21% from paper 1 and 21% from paper 2 out of a maximum of 30% for each which is 42% for the final grade

84 marks out of 120 is 70% of the paper and 70% out of 30% is 0.7*0.3=0.21 or 21%

When I said 30% form the IA and paper 3 I meant that you can get 15% out of a maximum of 20% from each which gives 30% for the final grade.

in this scenario you would end up with 72% for the final grade. 0.7*0.3 +0.7*0.3+ 0.15 +0.15=0.72 or 72%

Paper 1(30%)

Paper 2 (30%)

Paper 3(20%)

IA (20%)

Sorry if if previous post was confusing. I just read it and you can definitely interpret it in different ways

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Even if you get 42% from paper 1 and 2 and 30% from the IA (15%) and paper 3 (15%) you would still do fine and be able to get a solid 6 or a 7 with 72%.

This is in my opinion the worst case scenario for someone who is aiming for a 7. And I don't think that it's that hard to get at least 8% more for a solid 7 if you can manage to do this much. If you are doing sets, relations and groups it would be crazy to get below 18% or 17% from paper 3.

You realize that a 42% from paper 1 (20%) and 2 (30%), 30% from the IA (30%) and 15% from paper 3 (20%) means 42%*20%+42%*30%+30%*30%+15%*20%=33% (a 3)

Instead of 42%+15%+15%=72%?

(And a 7 in paper 3 of the sets relations and groups option is actually 42/60 which is 70% or 14% weighted)

You have to get an average of 72% to get a 7 (or a sum of the weighted percentages) instead of a sum of 72% in all papers for the unweighted percentage.

It's possible that I interpreted your post wrong though, if that's the case I'm sorry, but it does seem strange that you said

30% from the IA (15%)

and

If you are doing sets, relations and groups it would be crazy to get below 18% or 17% from paper 3.

which caused my (probably the case) misunderstanding.

I meant to say 21% from paper 1 and 21% from paper 2 out of a maximum of 30% for each which is 42% for the final grade

84 marks out of 120 is 70% of the paper and 70% out of 30% is 0.7*0.3=0.21 or 21%

When I said 30% form the IA and paper 3 I meant that you can get 15% out of a maximum of 20% from each which gives 30% for the final grade.

in this scenario you would end up with 72% for the final grade. 0.7*0.3 +0.7*0.3+ 0.15 +0.15=0.72 or 72%

Paper 1(30%)

Paper 2 (30%)

Paper 3(20%)

IA (20%)

Sorry if if previous post was confusing. I just read it and you can definitely interpret it in different ways

Woops I'm sorry, I did interpret your post wrong. I thought you meant 30% from the IA and 30% from paper 3 which made me confused. :doh:

(And I mixed up the weightings for IA and paper 1...)

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I've only tried paper 1 thus far (will be doing paper 2 and 3 in the upcoming days) and it was hard, definitely harder than other past papers I've tried. The most difficult questions in my opinion are 7, 8, 10b and 13b and c.

Question 7 was difficult because we hadn't studied row operations in class properly, and despite having practiced a lot in my own free time, it doesn't come intuitively to me at all. I also found 10b) difficult because i didn't notice that tan(75) = 1/tan(15). So I just used one of the identities in the data booklet, and I obviously didn't get the points. And for 13b) i seem to have gotten the gist of things (said that in the interval from -infinity to 3 there are no maxima or minima so it must be one-to-one) but I don't think i phrased it properly to get the points necessary.

I actually have a question: if we solve the systems of 3 equations and 3 unknowns using matrices, will we still get the points? because the formula for the distance of a point to a plane and point to a line are not on the data booklet or the syllabus, but they still accept them on the exams. would it be the same for matrices? (for example using the determinant to see how many solutions there are, or using the inverse to find a solution)

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