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

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1. (Original post by Hyaline)
Oh another question ^^, for the two triangles I got 14.28 and 9.69, is that right?
I think so
2. (Original post by Hyaline)
Oh another question ^^, for the two triangles I got 14.28 and 9.69, is that right?
my friends and i got something like 3.x and 9.68
3. One quick question. If for example, I got something right yet used another method not on the mark scheme would I get full marks, provided i showed my method on paper?

Thanks.
4. (Original post by johndoe04)
One quick question. If for example, I got something right yet used another method not on the mark scheme would I get full marks, provided i showed my method on paper?

Thanks.
I'd hope so xD, for the vectors question I did it without using any row operations because I simply can't........ the answer I got was (1/8, -3/8, 1) which is essentially the same as (1, -3, 8) which is my answer times by 8 ^^.
5. (Original post by hin1994)
my friends and i got something like 3.x and 9.68
Yeah I got 3 something and 9.68
6. (Original post by Hyaline)
Oh another question ^^, for the two triangles I got 14.28 and 9.69, is that right?
Got the same here. A lot of my friends got that too, so it should probably be the right one!
7. Anyone else here doing the sets option?
8. (Original post by dawesbr)
After row reduction (something like R2-R3->R2, R3-2R2->R3) I got the bottom row to be 0,0,0,k+4, meaning there were infinite solutions for k=-4, no solutions for k =/=-4. I then got something like r = 1/4(1,9,0) + (1,-3,8) or something but can't remember exactly. I ended up with the angle between the line and the plane as something around 47.2 degrees I think though?
ok those numbers are looking familiar (1, -3,8) I got that for sure.

47.2 degrees is definitely familiar for the last part

thanks
9. (Original post by kingcoltzan)
Yeah I got 3 something and 9.68
so did I
10. Yesss 47.2 degrees.
11. 47.2 degreeeeeeeeeeeeeees!
12. How could it have been 3.94cm? The two triangles were one with an obtuse angle and one with an acute angle. The angles were 45.7, 66.8 and 67.5 to get 9.68, therefore that's the triangle with the acute. The triangle with the obtuse angle would have a side longer than 9.68 opposite it ^^.

13. (Original post by Hyaline)
How could it have been 3.94cm? The two triangles were one with an obtuse angle and one with an acute angle. The angles were 45.7, 66.8 and 67.5 to get 9.68, therefore that's the triangle with the acute. The triangle with the obtuse angle would have a side longer than 9.68 opposite it ^^.

ar^n only for the first bit. You have to express it as r^n-1 eventually for the summation to infinity to work
14. (Original post by Hyaline)
How could it have been 3.94cm? The two triangles were one with an obtuse angle and one with an acute angle. The angles were 45.7, 66.8 and 67.5 to get 9.68, therefore that's the triangle with the acute. The triangle with the obtuse angle would have a side longer than 9.68 opposite it ^^.

I remember doing the bouncing balls question in class at the start of Y12, and getting it wrong.... It's basically the same as compound interest.

As I saw it (and as I saw in my notes today) u1 becomes 4x0.95... = 3.8 .

Draw it out, it is dropped from 4m, so the max height it reaches after first bounce = u1. Therefore after the 4th bounce, it is 4x0.95x0.95x0.95x0.95 (I think.)

Not sure about the angle one, I rememeber thinking 3.XX was a bit fishy and that I had made I mistake but I just left it.

For the final part of the Geometric sum, did you do sum to infinity to find the distance? ie. u1/ 1-r = answer something like 76m?
15. I put 160, but the answer was 156 ^^ I didn't express it as ar^n-1, everyone I know messed it up (5 in my class of 8 are 2s, 3s and 4s and only 3 of us are predicted 6). We had double it because the ball comes up and down. What was the answer to part b) again, I remember putting 27 or 28 can't remember which.
16. (Original post by Hyaline)
I put 160, but the answer was 156 ^^ I didn't express it as ar^n-1, everyone I know messed it up (5 in my class of 8 are 2s, 3s and 4s and only 3 of us are predicted 6). We had double it because the ball comes up and down. What was the answer to part b) again, I remember putting 27 or 28 can't remember which.
Ahh double it, yeah maybe...idk. it was only 2 marks iirc?

Yeah i think it was 28, you just solve it on GDC then it came to 27.3 or something so n=28. You can always test the value if you are not sure... and iirc 28 worked as it was 0.9X and it asked for the first one below 1m? Whereas 27 was 1.XXm
17. (Original post by kingcoltzan)
I remember doing the bouncing balls question in class at the start of Y12, and getting it wrong.... It's basically the same as compound interest.

As I saw it (and as I saw in my notes today) u1 becomes 4x0.95... = 3.8 .

Draw it out, it is dropped from 4m, so the max height it reaches after first bounce = u1. Therefore after the 4th bounce, it is 4x0.95x0.95x0.95x0.95 (I think.)

Not sure about the angle one, I rememeber thinking 3.XX was a bit fishy and that I had made I mistake but I just left it.

For the final part of the Geometric sum, did you do sum to infinity to find the distance? ie. u1/ 1-r = answer something like 76m?
It is 76m multiplied by 2 (it travels up and down), plus an extra 4m it travels before the first bounce officially "begins" in your formula as the ball is dropped from a height of 4m.
18. Yeah I put 28 I believe. Wait for part b) did you guys use n-1 or not?
19. I got n=28 for the bouncing ball part B

for the total distance...I forgot to double...shoulda read the question properly

but I think I got 70 something without doubling
20. (Original post by Hyaline)
integral of 1/1+x^4 anyone............ I got some long thing with ln and arctan in it, most likely wrong........
http://www.wolframalpha.com/input/?i...281%2Bx%5E4%29

Yep...I think you got it wrong too... :P

For that question you were either supposed to do it by trial and error or use the equation solver on your GDC

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