Maths SL P2 TZ2

[candypop]

Oh, and the ratio of the sequence. I got 0.4 was that right?

yes that's correct!

[shinnmoso]

I really did hate the paper 2 this year, It was just difficult. After the exam, we joked that someone has accidentally wrote the SL thinking it was the HL paper.

Though I was kinda glad that I get to use the N-spire CX for the exam this year. It is so much easier than the original N-spire.

Now I'm thinking about retaking the exam if I don't make a 6...

[TheregretfulIBstudent]

I thought paper 2 TZ2 was absolutely ridiculous, i'm terrified that i'm gonna fail math and therefore my diploma because of it... didn't help that i had a high fever, i completely blanked - don't think i got over 20 points, if even that. And i was very well prepared, knew 12 papers by heart when i went in. I just think it was plain unfair of the IB to give us a paper like that. I cried yesterday for two hours and again today.

I'm in the same boat as you! I prepared soooo much for math, doing as many past papers I could do so that I had an idea of what I should do in the exam but the minute I sat for it and opened paper 1 and (day after) paper 2, I just cried my eyes out. It was wayyyyy too hard! But we'll be fine, don't worry Even if you do fail math, you still get the diploma! You are allowed to fail a subject as long as you don't get a 1 and you still achieve the minimum of 24 points AND your HL's are 12. So breathe and do well on your other exams!

Edited May 6, 2012 by TheregretfulIBstudent

[Guest CasaNova]

damnn... i thought about getting a 5 or 6 in Math SL, but now, i will be glad if i am going to have 3

[HoolaBaloola]

I thought the last question was an absolute nightmare! Other than that everything seemed great with a few errors in sequences and conditional probability and a tiny bit in vectors (got 96.5 degrees instead of 95-95.5) lolz.

I'm surprised that people couldn't solve the derivatives and the three simultaenous equations! It was straightforward!

[ro_1293x]

I think the only one that I stupidly might have done wrong is the acceleration, displacement question? Well I thought I did it correctly but a few of my classmates disagreed.

The acceleration question was a negative answer, can't remember at all what I got. All I know was it was the derivative of the velocity function and I plugged in the value for t tht was given. BUT for displacement you always find integral of the velocity function that much I know but by the way the question was phrased I thought that maybe they meant just from the integral and plug in the value of t, which is what my classmates did. BUT they give you no values to use to solve for the c value you get when integrating, right? So I thought okay well I usually confuse these anyway, so I found the area under the curve from 0 to the value given and got some decent sounding answer. My friend apparently plugged in the value of t and just left it as Something + c, but they didn't ask for it in terms of c... right?

[ro_1293x]

Yea it was more difficult than paper 1

The second part of the probability question was difficult

I was getting a value more than 1

I just realized after the exam that we have to find the intersection of both probabilities

Since the first part was asking was asking about probability of at least one

And the second part asked for at most three, given that at least one already happened

Then the intersection is 1, and 2

I think the answer is

(binompdf(30,0.05,1)+binompdf(30,0.05,2))/(1-binompdf(30,0.05,0) = 0.7608

I found the answer as P(x=1) + P(X=2), thats it. Because they ask for what is the probability of getting at MOST 2 but at least 1. Hence 1 or 2... so add them.

[HoolaBaloola]

I think the only one that I stupidly might have done wrong is the acceleration, displacement question? Well I thought I did it correctly but a few of my classmates disagreed.

The acceleration question was a negative answer, can't remember at all what I got. All I know was it was the derivative of the velocity function and I plugged in the value for t tht was given. BUT for displacement you always find integral of the velocity function that much I know but by the way the question was phrased I thought that maybe they meant just from the integral and plug in the value of t, which is what my classmates did. BUT they give you no values to use to solve for the c value you get when integrating, right? So I thought okay well I usually confuse these anyway, so I found the area under the curve from 0 to the value given and got some decent sounding answer. My friend apparently plugged in the value of t and just left it as Something + c, but they didn't ask for it in terms of c... right?

It was definite integration, there is no c. The limits was from 0 to 1.6

EDIT:I think it was like 7.5 or something, 7.x. I got this answer, consistent with the answers on IBsurvival

Edited May 6, 2012 by HoolaBaloola

[Beastern]

I failed the question with a, b, c in Section B, also, did not have enough time for the tenth question... How many points was the section B?

[Vincent Casey F]

Oh, and the ratio of the sequence. I got 0.4 was that right?

I got the same!

used PolySmlt2 and got 1, 0.4, and a bunch of really big numbers and I thought r=1 can't be right and those big numbers don't makes sense, so I wrote 0.4 too

[IB>mindf**k]

by the way what was the limit for the last inequality on q10 p2 think i got between 0 and 1.22 rads but excluding them, does any one know the question and how it was to be solved

Edited May 6, 2012 by IB>mindf**k

[DeBrogliez]

by the way what was the limit for the last inequality on q10 p2 think i got between 0 and 1.22 rads but excluding them, does any one know the question and how it was to be solved

I did the opposite

Between 1.22 and pi/2

[IB>mindf**k]

by the way what was the limit for the last inequality on q10 p2 think i got between 0 and 1.22 rads but excluding them, does any one know the question and how it was to be solved

I did the opposite

Between 1.22 and pi/2

Do you remember the question, as i am not sure, was it sth ike 0<2.6sin(theta)-2(theta), correct me if i am wrong

[ro_1293x]

I think the only one that I stupidly might have done wrong is the acceleration, displacement question? Well I thought I did it correctly but a few of my classmates disagreed.

The acceleration question was a negative answer, can't remember at all what I got. All I know was it was the derivative of the velocity function and I plugged in the value for t tht was given. BUT for displacement you always find integral of the velocity function that much I know but by the way the question was phrased I thought that maybe they meant just from the integral and plug in the value of t, which is what my classmates did. BUT they give you no values to use to solve for the c value you get when integrating, right? So I thought okay well I usually confuse these anyway, so I found the area under the curve from 0 to the value given and got some decent sounding answer. My friend apparently plugged in the value of t and just left it as Something + c, but they didn't ask for it in terms of c... right?

It was definite integration, there is no c. The limits was from 0 to 1.6

EDIT:I think it was like 7.5 or something, 7.x. I got this answer, consistent with the answers on IBsurvival

Yes I got 7 something thanks, it was like 3 marks but it was the only one I unsure about.

[ro_1293x]

by the way what was the limit for the last inequality on q10 p2 think i got between 0 and 1.22 rads but excluding them, does any one know the question and how it was to be solved

I did the opposite

Between 1.22 and pi/2

Do you remember the question, as i am not sure, was it sth ike 0<2.6sin(theta)-2(theta), correct me if i am wrong

I remember it being between 0 and 1.22 rad because yes something like that was the inequality but it's basically what is on your graph that you would have just used and its all the values of x where y is still positive, because if you solve the inequality that is what you will find and after 1.22 the graph goes below the x-axis.

[IB>mindf**k]

Yea it was more difficult than paper 1

The second part of the probability question was difficult

I was getting a value more than 1

I just realized after the exam that we have to find the intersection of both probabilities

Since the first part was asking was asking about probability of at least one

And the second part asked for at most three, given that at least one already happened

Then the intersection is 1, and 2

I think the answer is

(binompdf(30,0.05,1)+binompdf(30,0.05,2))/(1-binompdf(30,0.05,0) = 0.7608

I found the answer as P(x=1) + P(X=2), thats it. Because they ask for what is the probability of getting at MOST 2 but at least 1. Hence 1 or 2... so add them.

i think i used same method as debrogeliez as it was a conditional probability type of question with given that...

[Ageha]

by the way what was the limit for the last inequality on q10 p2 think i got between 0 and 1.22 rads but excluding them, does any one know the question and how it was to be solved

I did the opposite

Between 1.22 and pi/2

Do you remember the question, as i am not sure, was it sth ike 0<2.6sin(theta)-2(theta), correct me if i am wrong

I remember it being between 0 and 1.22 rad because yes something like that was the inequality but it's basically what is on your graph that you would have just used and its all the values of x where y is still positive, because if you solve the inequality that is what you will find and after 1.22 the graph goes below the x-axis.

I got confused by this question.

The question asks to use the graph of f(x) to find the values, so I thought it might had to do with the maximum point on the graph...

I placed the boundaries between the maximum and 1.22...

Although I think it should have been 0 and 1.22.... ...Not sure though..

[Ageha]

Yea it was more difficult than paper 1

The second part of the probability question was difficult

I was getting a value more than 1

I just realized after the exam that we have to find the intersection of both probabilities

Since the first part was asking was asking about probability of at least one

And the second part asked for at most three, given that at least one already happened

Then the intersection is 1, and 2

I think the answer is

(binompdf(30,0.05,1)+binompdf(30,0.05,2))/(1-binompdf(30,0.05,0) = 0.7608

Don't you multiply for intersection???

Yea i thought of that too!

But those are not independent events, are they?

Uggh...I only thought of that after the exam....

Although the only way to find the intersection between them was to multiply them, no?

[IB>mindf**k]

Yea it was more difficult than paper 1

The second part of the probability question was difficult

I was getting a value more than 1

I just realized after the exam that we have to find the intersection of both probabilities

Since the first part was asking was asking about probability of at least one

And the second part asked for at most three, given that at least one already happened

Then the intersection is 1, and 2

I think the answer is

(binompdf(30,0.05,1)+binompdf(30,0.05,2))/(1-binompdf(30,0.05,0) = 0.7608

Don't you multiply for intersection???

Yea i thought of that too!

But those are not independent events, are they?

Uggh...I only thought of that after the exam....

Although the only way to find the intersection between them was to multiply them, no?

But do yo really need to multiply for intersection, we know that p(x<=2) (or sth nt sure)=p(x=0)+p(x=1)+p(x=2)

and p(x=>1)=p(x=1)+p(x=2)+p(x=3)+....

So "intersection" between these two sets are p(x=1) and p(x=2), hence by adding p(x=1)+p(x=2) intersection is found

Edited May 6, 2012 by IB>mindf**k

[Ageha]

Yea it was more difficult than paper 1

The second part of the probability question was difficult

I was getting a value more than 1

I just realized after the exam that we have to find the intersection of both probabilities

Since the first part was asking was asking about probability of at least one

And the second part asked for at most three, given that at least one already happened

Then the intersection is 1, and 2

I think the answer is

(binompdf(30,0.05,1)+binompdf(30,0.05,2))/(1-binompdf(30,0.05,0) = 0.7608

Don't you multiply for intersection???

Yea i thought of that too!

But those are not independent events, are they?

Uggh...I only thought of that after the exam....

Although the only way to find the intersection between them was to multiply them, no?

But do yo really need to multiply for intersection, we know that p(x<=2) (or sth nt sure)=p(x=0)+p(x=1)+p(x=2)

and p(x=>1)=p(x=1)+p(x=2)+p(x=3)+....

So "intersection" between these two sets are p(x=1) and p(x=2)

Hmm...Good point...Didn't really pay attention to that...

I found the probability of P(x>1) using GDC so I didn;t really pay attention, but it surely does make sense