# HL Math Paper 2- 24 Hours Later...

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1. Re: HL Math Paper 2- 24 Hours Later...
(Original post by johndoe04)
This was another annoying thing. Substitution wouldn't work. I tried by parts and let dv/dx = 1 and u = 1/1+x^4. It was a long process....
No no it was only 3 marks!!! I'm pretty sure you were supposed to use the equation solver on your GDC or use trial and error. The answer was 1.40
2. Re: HL Math Paper 2- 24 Hours Later...
(Original post by Hyaline)
Yeah I put 28 I believe. Wait for part b) did you guys use n-1 or not?
I put 28 for part b as well.

The formula I had was Hn = 4*(0.95)^n.

The first bounce is 4, the second bounce is 5% less, so on so forth.

Make this a geometric sequence. R = 0.05. U1 = 4

Hence Sinf = 4/(1-0.05) = 4*20 = 80.
3. Re: HL Math Paper 2- 24 Hours Later...
(Original post by Saymyname!)
I got v=arctan(pi/4-t)
Don't you mean v=tan(pi/4-t)... since you tan both sides to get rid of the arctanv
4. Re: HL Math Paper 2- 24 Hours Later...
(Original post by ElationAndPathways)
Don't you mean v=tan(pi/4-t)... since you tan both sides to get rid of the arctanv
Yeah you're absolutely right. I corrected myself at some point in this thread :P
5. Re: HL Math Paper 2- 24 Hours Later...
(Original post by arra)
Hence Sinf = 4/(1-0.05) = 4*20 = 80.
Wait what?!
6. Re: HL Math Paper 2- 24 Hours Later...
(Original post by arra)
I put 28 for part b as well.

The formula I had was Hn = 4*(0.95)^n.

The first bounce is 4, the second bounce is 5% less, so on so forth.

Make this a geometric sequence. R = 0.05. U1 = 4

Hence Sinf = 4/(1-0.05) = 4*20 = 80.
Sorry bro, you might be slightly off on that one. R=0.95, and for that, U1=3.8, at least the was I did it. Sinf then =76m, so Sinf*2+4=156m.
7. Re: HL Math Paper 2- 24 Hours Later...
(Original post by Hyaline)
Wait what?!
We want the sum to infinity. If it is a geometric series described by H = 4(.95)^n, from 0 to infinity, the first term is 4 (obviously, as he drops it from this height) and the common ratio is 0.95.

If abs(R) < 1, we can say that the total sum = u1/(1-r) = 4/(1-0.95) = 4/(0.05) = 4/(1/20) = 80.
8. Re: HL Math Paper 2- 24 Hours Later...
(Original post by Saymyname!)
Sorry bro, you might be slightly off on that one. R=0.95, and for that, U1=3.8, at least the was I did it. Sinf then =76m, so Sinf*2+4=156m.
I see what you're saying. I think I totally forgot to account for the up-distance. ****.
9. Re: HL Math Paper 2- 24 Hours Later...
(Original post by Saymyname!)
Yeah you're absolutely right. I corrected myself at some point in this thread :P
Ooooh ok sorry!! Pretty sure I went through the whole thread but maybe I missed some out :P
10. Re: HL Math Paper 2- 24 Hours Later...
(Original post by arra)
I see what you're saying. I think I totally forgot to account for the up-distance. ****.
relax...that last part was worth only 2 or 3 marks!
11. Re: HL Math Paper 2- 24 Hours Later...
also...how do you guys feel about the normal distribution question?
12. Re: HL Math Paper 2- 24 Hours Later...
RE: The bouncing ball question.

It doesn't matter whether you use n or n-1 so long as you keep that throughout. I used ar^n but specified that n was the number of bounces, which made things clearer, and a was 4. So after 0 bounces, a*r^0 = a = 4, u2 = 3.8 etc. It meant I didn't get confused with my ns, but you could just have easily have used a = 3.8 and n-1. Then when you do the sum to infinity, after you multiply by 2 for the up and down, the only difference is whether you add 4 for the missing drop if you used a = 3.8 or take away four for the additional rise if you used a = 4. Either answer gives Total distance as 156m. (a = 3.8, r = 0.95 => Sinf = 76, Sinf*2 + 4 = 156; a = 4, r = 0.95 => Sinf = 80, Sinf*2 - 4 = 156)

Banban: What was that one?
Last edited by dawesbr; 06-05-2012 at 00:08.
13. Re: HL Math Paper 2- 24 Hours Later...
(Original post by dawesbr)
RE: The bouncing ball question.

It doesn't matter whether you use n or n-1 so long as you keep that throughout. I used ar^n but specified that n was the number of bounces, which made things clearer, and a was 4. So after 0 bounces, a*r^0 = a = 4, u2 = 3.8 etc. It meant I didn't get confused with my ns, but you could just have easily have used a = 3.8 and n-1. Then when you do the sum to infinity, after you multiply by 2 for the up and down, the only difference is whether you add 4 for the missing drop if you used a = 3.8 or take away four for the additional rise if you used a = 4. Either answer gives Total distance as 156m. (a = 3.8, r = 0.95 => Sinf = 76, Sinf*2 + 4 = 156; a = 4, r = 0.95 => Sinf = 80, Sinf*2 - 4 = 156)

Banban: What was that one?
The first question from Part B.
the 14 mark normal distribution qs
14. Re: HL Math Paper 2- 24 Hours Later...
I have a bad feeling part b) for the bouncing ball question was 27............ because I got 28 when I used n-1 -.- dammit. I find all distribution questions easy, its one thing I'm quite good at ^^. For the final part of the Q we had to mix binomial distribution with normal distribution right.
15. Re: HL Math Paper 2- 24 Hours Later...
(Original post by Hyaline)
I have a bad feeling part b) for the bouncing ball question was 27............ because I got 28 when I used n-1 -.- dammit. I find all distribution questions easy, its one thing I'm quite good at ^^. For the final part of the Q we had to mix binomial distribution with normal distribution right.
My answer to part (a) of the bouncing ball question was 27. And for part b I got 156m
16. Re: HL Math Paper 2- 24 Hours Later...
I put n=28 for the bouncing ball one...we should all get error carried forward for that one depending on whether we used n or n-1 so hopefully only penalised once....

for last part of normal distribution I used binomial as well

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Last updated: May 6, 2012
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