MAY13 Maths SL Paper 2 TZ2

[Kevin Neutrino Lam]

Thanks

Does anyone remember what the geometric sequence question was, like the exact numbers they gave us?

I remember I had 0.9xx (is 0.95 the exact value or correct to two decimal places?)

Your result is correct, it was 0.95. It was really surprised when 'solver' gave an exact value! But I also checked it by drawing the graph so Im 100 % sure.

Hmm.. It was quite strange because I created two equations and plotted them using the graph function on my calculator, and then found the intersection point. It was 0.9 sth but it wasn't an exact value.

Edited May 12, 2013 by Kevin Neutrino Lam

[Alexmwinter]

Poles, Poles everywhere :>>>

Just in case anyone else was looking for answers:

a) ratio from the geometric sequence question: r=0.95

b) common term from the binomial expansion question: a=4

c) m giving the greates area: m=0.707

How did you workout the maximum area for the last part of the last question? I tried to find the derivatives when f(x) and g(x) were equal to each other but I guess that was wrong.

[Pepe]

Poles, Poles everywhere :>>>

Just in case anyone else was looking for answers:

a) ratio from the geometric sequence question: r=0.95

b) common term from the binomial expansion question: a=4

c) m giving the greates area: m=0.707

How did you workout the maximum area for the last part of the last question? I tried to find the derivatives when f(x) and g(x) were equal to each other but I guess that was wrong.

It was a tricky question, because (as far as I know) you didnt have to use the optimalisation or derivatives.

At first, looking at my sketch, I realised that the area is the greatest when functions intersect in x=3. Thus I found the y coordinate of the point of the exponential function when x=3. I dont remember the value but it was something like 2.12.

Then I simply used these coordinates to count m from the second function:

2.12=m*3

m=0.707 (to 3 s.f.)

And thats all! Most probably additional points will be awarded if you recognised that m has to be greater of equal 0.707, because otherwise functions would not intersect in the given domain.

[xxxbiologyxxx]

Poles, Poles everywhere :>>>

Just in case anyone else was looking for answers:

a) ratio from the geometric sequence question: r=0.95

b) common term from the binomial expansion question: a=4

c) m giving the greates area: m=0.707

How did you workout the maximum area for the last part of the last question? I tried to find the derivatives when f(x) and g(x) were equal to each other but I guess that was wrong.

It was a tricky question, because (as far as I know) you didnt have to use the optimalisation or derivatives.

At first, looking at my sketch, I realised that the area is the greatest when functions intersect in x=3. Thus I found the y coordinate of the point of the exponential function when x=3. I dont remember the value but it was something like 2.12.

Then I simply used these coordinates to count m from the second function:

2.12=m*3

m=0.707 (to 3 s.f.)

And thats all! Most probably additional points will be awarded if you recognised that m has to be greater of equal 0.707, because otherwise functions would not intersect in the given domain.

how many points will be awarded for such solution? i also solved it using the graph simply (two sentences of explanation), as the curves must intersect at (3, 2.12) to obtain maximum area. still, my teacher says it's not enough as they want us to use differentiation to get 8 points...

i hope we'll get 5 out of 8

[Pepe]

Poles, Poles everywhere :>>>

Just in case anyone else was looking for answers:

a) ratio from the geometric sequence question: r=0.95

b) common term from the binomial expansion question: a=4

c) m giving the greates area: m=0.707

How did you workout the maximum area for the last part of the last question? I tried to find the derivatives when f(x) and g(x) were equal to each other but I guess that was wrong.

It was a tricky question, because (as far as I know) you didnt have to use the optimalisation or derivatives.

At first, looking at my sketch, I realised that the area is the greatest when functions intersect in x=3. Thus I found the y coordinate of the point of the exponential function when x=3. I dont remember the value but it was something like 2.12.

Then I simply used these coordinates to count m from the second function:

2.12=m*3

m=0.707 (to 3 s.f.)

And thats all! Most probably additional points will be awarded if you recognised that m has to be greater of equal 0.707, because otherwise functions would not intersect in the given domain.

how many points will be awarded for such solution? i also solved it using the graph simply (two sentences of explanation), as the curves must intersect at (3, 2.12) to obtain maximum area. still, my teacher says it's not enough as they want us to use differentiation to get 8 points...

i hope we'll get 5 out of 8

But I cant see how the differentiation can be used here.

And this simple solution must be a good solution so they cant cut any points...

[evertonhk94]

Does anyone recall how much the vectors question in section b was worth, specifically the a (i) (ii) and b, I only finished half of it and wanna get an idea of how many marks it was worth. Also how much did u get for sketching the curve in question 10 a?

[Pedro9604]

I'm happy cause I felt confident with the matrices question, the probability question (sectionA, the question in section B was HORRIBLE), also felt confident about the geometric sequence, and the graph and the area between f(x) and g(x) and a few points in the rest of the questions that i do not answer completly

the probability question in section B was jfdbnskdfjbndklsbnsdlrk I never understand if that s.... was a normal probability or a binomial.... FML

[Ali Aly]

Because I lost about 20 points in the first paper (too tired after history p3) I was extremely motivated during p2 and hope to get over 95% from this one. The only mistake I made was in the last question hopefully (I got m=0.703 while the correct result is m=0.706 I guess).

I am sorry to burst your bubble but the last question:

m= (1/4)*e approximately 0.68 (to two decimals)

[Pepe]

Because I lost about 20 points in the first paper (too tired after history p3) I was extremely motivated during p2 and hope to get over 95% from this one. The only mistake I made was in the last question hopefully (I got m=0.703 while the correct result is m=0.706 I guess).

I am sorry to burst your bubble but the last question:

m= (1/4)*e approximately 0.68 (to two decimals)

Any calculations please?

[Ali Aly]

Because I lost about 20 points in the first paper (too tired after history p3) I was extremely motivated during p2 and hope to get over 95% from this one. The only mistake I made was in the last question hopefully (I got m=0.703 while the correct result is m=0.706 I guess).

I am sorry to burst your bubble but the last question:

m= (1/4)*e approximately 0.68 (to two decimals)

Any calculations please?

g'(x)=f'(x) (when g(x) is tangent to f(x) that counts as a point of intersection)

hence m=(1/4)e^(x/4)

hence ((1/4)e^(x/4))*x=e^(x/4) (substitute m into g(x) and equate to f(x))

hence x=4 (using intersect or which ever method you prefer)

hence m=(1/4)e^(4/4)=(1/4)e^(1)= 0.68 (to two decimal places)

TaDa

[murgutschui]

g'(x)=f'(x) (when g(x) is tangent to f(x) that counts as a point of intersection)

hence m=(1/4)e^(x/4)

hence ((1/4)e^(x/4))*x=e^(x/4) (substitute m into g(x) and equate to f(x))

hence x=4 (using intersect or which ever method you prefer)

hence m=(1/4)e^(4/4)=(1/4)e^(1)= 0.68 (to two decimal places)

TaDa

Hey Ali, I'm not sure who's got the bubble here..

I did the same thing as you, by saying that for the area to be maximised both f(x)=g(x) and f'(x)=g'(x) have to be true, and then using these simultaneous equations to find x and m, and it gave me the e/4 as well -TaDa.

I felt really smart, until about five minutes after the exam, because then I realized that the the limits for f(x) and g(x) of -3<x<3 apply for all the sub-questions, so f(x) is not even defined for x=4, and as 3=x is the highest possible value of x, R ist largest for where f(3)=g(3), which is the case when m=0.707.

I spoke to my maths teacher about it, he said they will most likely only take away 2 or 3 marks. I'm not completely sure, but another teacher told me that the IB send a letter to the teachers in which they acknowledged some badly worded question in this years exam, which can only be p2-10.c).

Anyway, if you got that far you probably have nothing to worry about.

[Ali Aly]

g'(x)=f'(x) (when g(x) is tangent to f(x) that counts as a point of intersection)

hence m=(1/4)e^(x/4)

hence ((1/4)e^(x/4))*x=e^(x/4) (substitute m into g(x) and equate to f(x))

hence x=4 (using intersect or which ever method you prefer)

hence m=(1/4)e^(4/4)=(1/4)e^(1)= 0.68 (to two decimal places)

TaDa

Hey Ali, I'm not sure who's got the bubble here..

I did the same thing as you, by saying that for the area to be maximised both f(x)=g(x) and f'(x)=g'(x) have to be true, and then using these simultaneous equations to find x and m, and it gave me the e/4 as well -TaDa.

I felt really smart, until about five minutes after the exam, because then I realized that the the limits for f(x) and g(x) of -3<x<3 apply for all the sub-questions, so f(x) is not even defined for x=4, and as 3=x is the highest possible value of x, R ist largest for where f(3)=g(3), which is the case when m=0.707.

I spoke to my maths teacher about it, he said they will most likely only take away 2 or 3 marks. I'm not completely sure, but another teacher told me that the IB send a letter to the teachers in which they acknowledged some badly worded question in this years exam, which can only be p2-10.c).

Anyway, if you got that far you probably have nothing to worry about.

This solution maximizes the area, I believe that they wanted us to disregard the domain in 10 b)

It is worth noting that I did not solve the question like that, I came up with this solution 5 min before and was unable to transfer it to the booklet.

I assumed that the last possible intersection point had to be x=3 (as that was the stated domain), but i fear that I am mistaken, because it does not make sense to put 8 points on a question with a solution in which you had to guess the x coordinate and then give an answer.

[Pepe]

g'(x)=f'(x) (when g(x) is tangent to f(x) that counts as a point of intersection)

hence m=(1/4)e^(x/4)

hence ((1/4)e^(x/4))*x=e^(x/4) (substitute m into g(x) and equate to f(x))

hence x=4 (using intersect or which ever method you prefer)

hence m=(1/4)e^(4/4)=(1/4)e^(1)= 0.68 (to two decimal places)

TaDa

Hey Ali, I'm not sure who's got the bubble here..

I did the same thing as you, by saying that for the area to be maximised both f(x)=g(x) and f'(x)=g'(x) have to be true, and then using these simultaneous equations to find x and m, and it gave me the e/4 as well -TaDa.

I felt really smart, until about five minutes after the exam, because then I realized that the the limits for f(x) and g(x) of -3<x<3 apply for all the sub-questions, so f(x) is not even defined for x=4, and as 3=x is the highest possible value of x, R ist largest for where f(3)=g(3), which is the case when m=0.707.

I spoke to my maths teacher about it, he said they will most likely only take away 2 or 3 marks. I'm not completely sure, but another teacher told me that the IB send a letter to the teachers in which they acknowledged some badly worded question in this years exam, which can only be p2-10.c).

Anyway, if you got that far you probably have nothing to worry about.

This solution maximizes the area, I believe that they wanted us to disregard the domain in 10 b)

It is worth noting that I did not solve the question like that, I came up with this solution 5 min before and was unable to transfer it to the booklet.

I assumed that the last possible intersection point had to be x=3 (as that was the stated domain), but i fear that I am mistaken, because it does not make sense to put 8 points on a question with a solution in which you had to guess the x coordinate and then give an answer.

I agree with murgutshui, x cannot be equal 4 because it is not in the domain therefore this answer seems to be wrong.

If you look at the past papers there were some similar questions that gave you 7-8 points just for the substitution of the maximum value (coordinates at x=3 in our case) because at first you need to notice that you actually need to do it and then know to which equation to substitute etc.

[jinye]

I talked to my math teacher yesterday and she informed me that she had received an email from IB regarding the SL math p1 and p2 exams. Apparently, there were mistakes on BOTH exams..........i.e. things we weren't able to do or were not answerable.

Really? I didn't see any question which was not answerable... Not saying that I aced the exam, because I did not (in fact I'm praying for a 6 lol) but I think there wasn't any major unanswerable question...

But, who cares? If that means more points to us, then YUPIIII!

[Paddy94]

I can confirm these rumours, my centre has also been alerted. Apparently in Paper 1 the latter stage of one of the questions near the end of Section A (which I didn't get near to!) needed a calculator - something to do with Cos. In paper 2, there were issues with the first and last questions in Section B, forgetting the numbers, not the middle one with probability and stuff. The first was possible but beyond our syllabus by some distance, the second impossible. So pretty major stuff it seems. Normally they'd re-distribute the marks etc, but given the severity there's talk they could instead have to award predicted grades balanced with coursework marks and forget the papers altogether.

[jinye]

I can confirm these rumours, my centre has also been alerted. Apparently in Paper 1 the latter stage of one of the questions near the end of Section A (which I didn't get near to!) needed a calculator - something to do with Cos. In paper 2, there were issues with the first and last questions in Section B, forgetting the numbers, not the middle one with probability and stuff. The first was possible but beyond our syllabus by some distance, the second impossible. So pretty major stuff it seems. Normally they'd re-distribute the marks etc, but given the severity there's talk they could instead have to award predicted grades balanced with coursework marks and forget the papers altogether.

That would be SO FREAKING AWESOME!

But not going to happen ahaha

[Faleleevka]

Oh please let it be that!!!!

[Pepe]

I can confirm these rumours, my centre has also been alerted. Apparently in Paper 1 the latter stage of one of the questions near the end of Section A (which I didn't get near to!) needed a calculator - something to do with Cos. In paper 2, there were issues with the first and last questions in Section B, forgetting the numbers, not the middle one with probability and stuff. The first was possible but beyond our syllabus by some distance, the second impossible. So pretty major stuff it seems. Normally they'd re-distribute the marks etc, but given the severity there's talk they could instead have to award predicted grades balanced with coursework marks and forget the papers altogether.

Thats strange because most of my friends did the cosine question and got cosx=0 while I had something strange like 0=cosx-2x-1. Maybe I was right? What did you have guys?

Probability was strange I do agree. I also made a mistake in the last question.

Since I dont believe in this awarding predicted grades (did it happen once in the past?) could you please explain to me how the redistribution of the marks works?

Personally I do not like the situation because it would mean boundries going up (if people who did the whole question in the wrong way are given the same amount of points like people who at least tried and knew the method...)

Edited May 18, 2013 by Pepe

[maereth]

I think predicted grades are used when the papers are lost... That would be unfair tbh... I saw on a biology markscheme that when 1 question is canceled then the marks for it are given depending on the rest of the questions. There was a table which said if you have more than 12 points you get 3 for this question if you have 8-11 you get 2 etc... which is also unfair but whatever Anyway I hope that they won't disregard the papers but only adjust the marking scheme and that I'll get a 7

[Pepe]

I think predicted grades are used when the papers are lost... That would be unfair tbh... I saw on a biology markscheme that when 1 question is canceled then the marks for it are given depending on the rest of the questions. There was a table which said if you have more than 12 points you get 3 for this question if you have 8-11 you get 2 etc... which is also unfair but whatever Anyway I hope that they won't disregard the papers but only adjust the marking scheme and that I'll get a 7

But if they do it this way the boundaries would be higher not lower so that less people will get 7, am I right?

Oh God, I want my results now!