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Math SL P2 TZ2


helenpark

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Hi everyone. how did you do on your math paper 2? I live in Asia so I am guessing I am doing TZ2? erm,.. I personally think that paper 2 is tougher than paper 1.. however both papers are still more simple that past year papers. I left our 2 questions though, worth 24 points T_T anyone did the trapezium thing? I really had no idea how to work it out.

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hey guys... how was the paper2 TZ2...???

i think it was very calculative (obviously... it was a calculator paper haha :D )

but i thought the last question especially the last part c or d may be was very confusing.... they asked that john wants windows of same area (A) but different angles. so find all the possible vales of A.

confusion no.1- which angle were they taking about?? whether he wants different angles in 1 window itself and such two windows OR two windows with different angles from each other but each of them having the same base angle... (i hope i am clear with my question :blink: )

confusion no. 2- if john wants windows with same A how can we find all the possible different areas??? they all are the same!!

actually please someone help me to understand this the question!! :read:

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Did ANYONE get the very last question, 10.C ? It was worth 7 points and no one at my school seemed to get it.

I had 30 minutes left for the whole of last question, 10, which was 16 points and I'm pretty sure I only got about 2 points on it in total.

The rest of the paper was very easy.

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How did everyone find Paper 2 for TZ2? It's much harder in comparison to Paper 1, especially in Section B! I lost the whole 7 marks on the very last question regarding finding different values of A! I couldn't even figure out how to start that question! But anyway, the question itself was weirdly phrased. I remember reading it like 10 times trying to figure out what it means but still have no clue what it was asking! It was something like the windows had to have the same areas with different values of theta and find all values of A.

It's so contradictory! How can the area be the same when there are different values!?!?! IB needs to dump this question!

I don't think anyone did. At least I have never heard of anyone in my community that could anyway. Haha including myself.

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That trapezium question was absolutely impossible, i couldn't do any of it! I think it was a really really difficult paper, one of the hardest in the past five years.

The last question was extremely hard and i only had 10minutes left to do the whole thing, i ended up missing it out because i just didnt have enough time and it was so damn hard. In general the paper was a lot harder than paper 1

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Well, precisely in 10/c the question was "find different values FOR A" and this way A=5 remains, and the angles had to be found again. Personally I did it in a way that the 2 solutions of b, were subtracted from 180 degrees. (As sin (x) = sin (180 - x))

This is my theory :P

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I agree that the last question was most challenging,

for last part, I simply graphed the equation for the area on the calc, and observed where two different values of x (the angles in the given interval) share the same y coordinates (the corresponding area!) :)

Edited by pigsonthewing
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I found paper 2 pretty normal, definitely harder than P1 (TZ2) but I think it went pretty well. Regarding 10, it was definitely the trickiest question in the paper, but still not too hard. Copypaste from FB by someone else on 10c (I did the same thing):

"is that the 7 mark question? i think what u had to do (thats what i did) was first show the graph and show that at x= π/2 (90º) ,which is the boudary they gave in the question, y=4 (show by substituting into the equation), thats when the graph starts having 2 x-values for the same value for y..and then calculate where the max point of the graph is (i got 5.20) and then these are the values for which A has two possible values for theta, so 4≤A<5.20"

The wording was definitely pretty confusing overall.

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You had to use binomial theorem and basically add the probability of obtaining 3 5s, probability of obtaining 4 5s, and probability of obtaining 5 5s together. IIRC the probability of obtaining 5 in one throw was 4/16, so probability of obtaining 3 5s would be (5 nCr 3)*(4/16)^3*(12/16)^2 and so on. It could be done using binompdf or binomcdf on a calculator :)

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I actually thought of doing exactly that but of course I read the domain to be 2pi somehow which is very annoying giving me a wrong graph -.-... That wording is ridiculou though! How can that be in the exam and be phrased so damn badly took forever to even understand! Just really annoying. Doesnt the IB check their questions?

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does anyone know how to do the last question of question 9. it was a probability question and its something about rolling a fair 4 sided dice 4 times max and getting a sum of 5 within 3 or more throws.

Yeah, you need to use binomialpdf on your calculator. There were 4 trials, the desired outcome had the probability of 4/16 and X=3 and 4.

Basically, you have to plug in your calculator Binompdf(4, (4/16), 3) and Binompdf (4, (4/16), 4) and the sum of the two numbers is the answer.

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does anyone know how to do the last question of question 9. it was a probability question and its something about rolling a fair 4 sided dice 4 times max and getting a sum of 5 within 3 or more throws.

Yeah, you need to use binomialpdf on your calculator. There were 4 trials, the desired outcome had the probability of 4/16 and X=3 and 4.

Basically, you have to plug in your calculator Binompdf(4, (4/16), 3) and Binompdf (4, (4/16), 4) and the sum of the two numbers is the answer.

Were there 4 trials? Lol, what a stupid mistake for me to make :D oh well...

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does anyone know how to do the last question of question 9. it was a probability question and its something about rolling a fair 4 sided dice 4 times max and getting a sum of 5 within 3 or more throws.

Yeah, you need to use binomialpdf on your calculator. There were 4 trials, the desired outcome had the probability of 4/16 and X=3 and 4.

Basically, you have to plug in your calculator Binompdf(4, (4/16), 3) and Binompdf (4, (4/16), 4) and the sum of the two numbers is the answer.

Were there 4 trials? Lol, what a stupid mistake for me to make :D oh well...

I made the exact same dumb mistake on the exam. Taking 5 trials.. that sum of 5 thing must have gotten me confused.

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What was the right answer for the standar deviation????

I got something like 2.38 don't remember really

and also for the vector question in section B the coordinates of the points of intersection of lines L1 and L2

I think i got a similar answer for the standard deviation

for the vector equation you had to put the points equal to each other so:

r = a+tb

i cant remember what the co-ordinates were exactly but for example

r = (i-j+k) +t (2i +2j +5k) = L1

r = (5i -2j+3k) +s (4i-j+3k) = L2

So to figure out the co-ordinate of the point of intersection you would have to find t and s by doing the following:

1+2t = 5+4s

1+2t = -2-s

1+5t = 3+3s

and use algebra to solve. then once you have found s and t you can substitute it into the original vector equation and find the point of intersection.

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does anyone know how to do the last question of question 9. it was a probability question and its something about rolling a fair 4 sided dice 4 times max and getting a sum of 5 within 3 or more throws.

Yeah, you need to use binomialpdf on your calculator. There were 4 trials, the desired outcome had the probability of 4/16 and X=3 and 4.

Basically, you have to plug in your calculator Binompdf(4, (4/16), 3) and Binompdf (4, (4/16), 4) and the sum of the two numbers is the answer.

Were there 4 trials? Lol, what a stupid mistake for me to make :D oh well...

I made the exact same dumb mistake on the exam. Taking 5 trials.. that sum of 5 thing must have gotten me confused.

Same... oh well, I assume that will only lose me one misread mark, still pretty annoying.

@iber2011; 1-var stats L1,L2 (where L1 is the values, L2 is the frequency).

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