ITT: We go through the 3U paper and post up answers. (1 Viewer)

[independantz]

how did u get pi/3 and 5pi/3. i got the quadratic with cos and then when
i factorised it i got (2cos+1)(cos-1), then cos=-1/2 so, the angle should be 2pi/3, 4pi/3 right? i dunno. maybe i stuffed up earlier.

i think you screwed your minus somewhere, it was cos=-1, cos=1/2 and sin=0

 

[lovelle]

thanks for the answerrs
Terry Lee should put them up on the net like today or tomoro

 

[dolbinau]

Yes got all 1/2 right.

 

[scott.syd]

i think you screwed your minus somewhere, it was cos=-1, cos=1/2 and sin=0

ya, i think you're correct, i just re-did it. sigh. stupid minus signs.

 

[bendha]

yeah i put cookiing time instead of the actual time as well

damn that was something i actually knew how to do and im still losing marks

same :'(

 

[bendha]

ya, i think you're correct, i just re-did it. sigh. stupid minus signs.

yea 5 solutions =.= 0, pi/6, 5pi/6, pi,. 2pi .. i forgot 2 solutions fails

 

[ryan-k1990]

Anyone got answers to 3a(ii) and the height of the tower for the 3d trig 6a(ii)

height of tower was 96m

 

[anonomis]

fuck i cant beleive i got that Q6b wrong! i thought they were asking u to prove LHS=RHS so i tried doing that! cost me 3 marks i wudve gotten it otherwise fuckkkk!

 

[SkimDawg]

i'll put the other answers i got, not proofs though, they are too long:
3(a)(ii): -2=<x=<4/3
4(a)(ii): 12.44pm
4(b)(i): 7!= 5040 ways
5(a)(ii): y= 1-(1-2x)^1/2
5(a)(iii): 1/2
5(b): a=2/3 metres T=2pi/3 seconds
6(a)(ii): 96.0m
6(b): 0, 2pi/3, pi, 4pi/3, 2pi
6(c)(i): (p+q)!/(p!q!)
6(c)(ii): (p+q)!/p!q!

Oh shit, i put my solution in so many minutes, not the time..

 

[Poad]

how did u get pi/3 and 5pi/3. i got the quadratic with cos and then when
i factorised it i got (2cos+1)(cos-1), then cos=-1/2 so, the angle should be 2pi/3, 4pi/3 right? i dunno. maybe i stuffed up earlier.

Aha, I did that too.

 

[tacogym27101990]

anyone have a solution to 3ciii

 

[Kearnzo]

anyone have a solution to 3ciii

let theta = y (for simplicity)

dy/dt = (vl)/(l^2 + x^2) = m/4 = v/4l

(vl)/(l^2 + x^2) = v/4l

simplify to 3l^2 = x^2

From the diagram x = ltany

simplifes to tany = + or - sqrt(3)

It is in the range -90 < y < 90

So its plus or minus 60

 

[glockenspiel]

let theta = y (for simplicity)

dy/dt = (vl)/(l^2 + x^2) = m/4 = v/4l

(vl)/(l^2 + x^2) = v/4l

simplify to 3l^2 = x^2

From the diagram x = ltany

simplifes to tany = + or - sqrt(3)

It is in the range -90 < y < 90

So its plus or minus 60

Yep, that's what I got
But in radians
ie -pi/3 or pi/3

But I got 5a wrong. I made a mistake in ii) and got f-1(x)=1-(1-x)^1/2
So I subsequently got some crazy answer for iii)
How many marks do you reckon I'll get?

 

[Kearnzo]

Yep, that's what I got
But in radians
ie -pi/3 or pi/3

But I got 5a wrong. I made a mistake in ii) and got f-1(x)=1-(1-x)^1/2
So I subsequently got some crazy answer for iii)
How many marks do you reckon I'll get?

So did i, its just easier to put everything in degrees. Easier to read.

Hm, depends how much knowledge you showed about the theory related to the question. If you just made a stupid mistake, maybe 1.

 

[timmiitippii]

let theta = y (for simplicity)

dy/dt = (vl)/(l^2 + x^2) = m/4 = v/4l

(vl)/(l^2 + x^2) = v/4l

simplify to 3l^2 = x^2

From the diagram x = ltany

simplifes to tany = + or - sqrt(3)

It is in the range -90 < y < 90

So its plus or minus 60

does that mean for ii) m = v/l ??

i couldnt do ii) so i tried doing iii).
i subbed m/4 into dtheta/dt and subbed tan sumthing into vl/(l^2 + x^2)
and sumthing very complicated = =
then i simplified and got cos theta = 1/2 . (times a bunch of square root v's and l's) and changed my tan^2 + 1 to sec^2 and then i realized that most probably ur gonna need to inverse cos the 1/2 so i guessed that m must equal v/l to cancel out all those square roots.

then i said therefore theta equal plus minus pi/3 and then jsut wrote m = v/l for ii) since its only one mark xD
haha so hopefully i fluked it and got it right

 

[Kearnzo]

does that mean for ii) m = v/l ??

i couldnt do ii) so i tried doing iii).
i subbed m/4 into dtheta/dt and subbed tan sumthing into vl/(l^2 + x^2)
and sumthing very complicated = =
then i simplified and got cos theta = 1/2 . (times a bunch of square root v's and l's) and changed my tan^2 + 1 to sec^2 and then i realized that most probably ur gonna need to inverse cos the 1/2 so i guessed that m must equal v/l to cancel out all those square roots.

then i said therefore theta equal plus minus pi/3 and then jsut wrote m = v/l for ii) since its only one mark xD
haha so hopefully i fluked it and got it right

Yeh, thats what i got.

Haha, that would be awesome if you did. Nice work realising m = v/l to cancel them out.