Math sl paper 1 TZ2

[Nabz]

I think I did better in Paper 2 than paper 1. I did horrible in some of the parts in section B

[patriciaisaias]

hi!

could someone please tell me how did you find the normal to the last question??

[evertonhk94]

I was wondering if anyone remembers how much the question on the quadratic graph was worth in section B? I left it until the end and didn't finish the final part i.e. find the normal. I hope I got most of the marks. What did everyone get for the value of A?

[Andrei D]

Just a reply to the "Statistics" questions from P1.

The Value of X was actually around 17 minutes as you have to take the midpoint of parameters y2<t<y1, so the minimum time to qualify was around 17 minutes. If you looked at the cumulative frequency curve only 3 girls actually met the 17 minute mark in the retry so the actual result is 120 + 3 = 123/200 Alot of people in my class forgot about using the midpoint from the table to find X.

If you think about it its also logical that not that many people would make the retry. I know the questions are theoretical but just saying it would have been weird for the question designer to think that 20/30 people would make it on a retry.

I personally found Paper 1 to be a cakewalk, all the applications were extremely basic. (I literally laughed when I saw that question on the area under the curve using 3pi as the total area etc...)

Paper 2 however was quite the opposite though. I found it annoying that I had forgotten how to use the polysim equation solver so I was unable to find r in the geometric ratio question (I filled the whole page with maths and eventually found a highly simplified version of the equation so I doubt I will lose any more than the A1 mark).

THe final question was also quite confusing because I was used to associating an area maximization with a tangent value of zero so it was a bit confusing.

Edited May 13, 2013 by Andrei D

[Andrei D]

I hadn't thought about the -pi thing actually - I suppose it could be given that they did ask for the intergration of the function and not the area (if I remember correctly), but the minus can't be worth any more than 1 mark, most of it would probably be for working out to use the area of a circle.

Also I got 30 and 70% too! I hope your friend's wrong, what did they think it was?

Glad you all found the last question weird too, did anyone get an answer for the equation of the normal?

His result was 69.5%. In the question it was said that 'girls who run faster than 20 minutes are qualified' or something like that. It would mean that we shouldnt have taken the value for 20 min from the cum freq (that was 20 girls) but the value beneath - 19 girls. Simply, girls whose results was equal to 20 min didnt qualify according to him, therefore we should count 19 girls in our percentage (who for sure run faster than 20 minutes) what would give 69.5%. Ugh, complicated.

The gradient of the normal was -1/27 but I didnt have time to count the equation :<

Oh ok, I see his logic, but I think because the classes were organised as a</=x<b, when plotting a cumulative frequency diagram you would still plot the point for that class at b, here being 20, but that value wouldn't actually include 20. So the point at 20 should represent the number of girls who ran it in <20 minutes. I think anyway, but to be honest statistics has never been my strength.

And at least you should get marks for working out the gradient, I barely wrote anything for that question!

That was by far one of the easiest tests ive ever done.

As for the cum. frequency graph, you were supposed to count from 20 mins and up, getting 20 girls from the second trial + 120 girls form the first trial. This means that 120+20=140, which is 70%..

As for the last questions, the y-coordinate turned out to be -54. The equation of the normal turned out to be y=1/3x-55. Log question was just a gift. The only thing im unsure about is the integration thingy when u had pi and 3pi. I wrote pi on the first one and 2pi on the second one, but i reckon thats wrong. I would have answered -pi and 4pi.. :/

and for quesiton 9, i got the x-coordinate to be pi/2, right?

Man, does nobody get this? On the point BEFORE the cumulative graph question, you found that x (the minimum time to pass) was in the range 14<t<20. Whenever that happens you are supposed to take the MIDPOINT. That automatically tells you that the qualifying time was 17 minutes. (something like that, can't remember exact numbers)

This is all part of the same question, so what you had to do next was plug 17 minutes into the cumulative frequency graph and you find that only THREE of the girls actually passed in the retake. This leads to a total of 120+3/200 passing the qualification time so the answer is actually 61.5%, NOT 69.5%.

[ibsuckzzz]

i mean i though paper 1 was ok apart from the circles graph question. everything else was quite straightforward and i had something like 1/27 for the gradient since i had a really large y coordinate for point P since i just multiplied the values for f(3) and g(3). Is that right? BUT paper 2 was SO hard i think it messed up both my 7 or a chance of getting a 6. I didnt do two whole questions in section A and the last one in 10!!!!! I really hope i get a 6 now!!!!

[Rafie Izwan]

I was expecting paper 1 to be much harder, to be honest. I only had a problem with Q9, because I couldnt find g(4). So I left it out. But then at home I worked it out and got g(4) = 5. Also in Q10 I didnt realise that h' = f'g +g'f (product rule) and simply said h' = f'g'. Such a stupid mistake! Cost me 8 marks

we're in the same boat though! paper 1 was pretty easy and so do the paper 2 except for the last question. But, don't worry my teacher said that they will send complaining letter to the IBO this is because the question is HL since they only give us limited informations to solve the problem..

I just aim for 6 but if I got 7 that is a gift.