Answers (1 Viewer)

[Golbez]

Re: 回复: Re: Answers

well my solution is on the other thread, but logically p/q can't equal zero because p doesn't ever equal zero.





lim n--> infinity of 2 (sin pi/2n)^2

as n approaches infinity, pi/2n approaches zero. But lim x--> 0 of sin x = x

therefore it is equal to 2 (pi/2n)^2
= pi^2/2

Sorry, I still don't quite agree with you.

I think you might be mixed up with

x-> 0 of sin(x)/x = x

Because as x->0, sin(x) -> 0.

Then again maybe I'm just stupid?

 

[roadrage75]

i don't believe that that is possible.

because part (i) + 2part (ii) should = 1

 

[roadrage75]

i don't believe that that is possible.

because part (i) + 2part (ii) should = 1

 

[roadrage75]

im pretty sure its .38 for a, and .31 ish for B

 

[fishy89sg]

for i) i got .26.... LOL

 

[AMorris]

For the probability question:

(i) = 0.36 Exact fraction is (12C3*12C3)/(24C6) (consider them all as different i.e. R_1, R_2... then we have 12C3 ways of choosing the red ones, 12C3 ways of choosing the yellow ways. In the whole sample space there are 24C6 ways of choosing 6 balls)

(ii) = 0.32

P(>3 red) + P(>3 yellow) + P(3 red, 3 yellow) = 1 (these are all options)

now because red and yellow are indistinguishable P(>3 red) = P(>3 yellow)

so 2P(>3 red) = 1 - P(3R, 3Y) = 1 - .36 = 0.64
so P(>3 red) = 0.32

 

[Argen1447]

actually, gao, i didnt type it right, i got 2e - e^2/2 - 1/2, which is almost what you got

that is def right, well either that or im retarded

x-> 0 of sin(x)/x = x

Because as x->0, sin(x) -> 0.

Then again maybe I'm just stupid?

well when x is very small sinx is approx equal to x, thats a rule

and as n approaches infinity 2pi/n approaches 0 i.e. 'very small'

( the answer is def (pi^2)/2 )


in Q1. e) i think that integral is way too easy for 4 marks (the simplification is given) and i found that the answer could be simplified SIGNIFICANTLY so im just guessing but i think thats where the 4th mark went


also, Q2. c) it says "Give a geometrical description of the locus P as z varies"

i got y=(+-)3^(1/2)x and said "as z varies P moves along y".. but im not sure if thats what they were asking exactly or if im completely right, what did you guys get?


and, OMG i cant believe i didnt notice the 4 went missing in for iii) >.<


AND (lol long post), anyone wna post their explanied solution to the probability Q?


FINALLY, for the raindrop question ( Q6. b) v) ),

anyone else think it was a bit unfair that we had to assume the cloud was high enough for the raindrop to be able to reach the terminal velocity before it hits the ground? Ofcourse logically it makes sense, but mathamatically its something we shouldnt ASSUME.. like it was a bit left field, questions usually arent like that


ANYWAY, cheers

(LOL one more query! in Q6. a), did anyone use part ii) for part iii) or iv) ? because otherwise i) and ii) werent related to iii) and iv).. another weird thing in the exam..)

 

[AMorris]

For the geometrical description of P. The locus is a circle centred at (1,0) with radius 1 but with a hole at the origin (0,0). Equation is y^2 + (x-1)^2 = 1. The hole exists because 1/0 is undefined (not sure if they will penalise this though).

For the raindrop question, I think thats why they phrased the question as estimate. However I must admit that its a very general-ish sort of question in a 4u exam.

and for 6(a) i used part ii in (iv) to prove that (n+2)/2^(n-1) was 0 as n -> infinity however its possible to not have to use that fact (but i dunno, perhaps to be rigorous otherwise you might've needed calculus to show that that limit was 0)

 

[roadrage75]

thumbs up for the probability, i totally agree!

but, im not sure if there's a hole at the origin for the locus of P....

 

[jah_lu]

http://www.hsccoaching.com/Resources/2007 Ext2.pdf if anyone wants them

 

[AMorris]

I'm pretty sure there is because 1/0 + 1/0 =/= 1 so (0,0) can't be in the locus of P.

other tricky questions were the 1(e) which I'm not sure if you had to simplify the whole way down to arctan(3/4) or not but considering it's worth 4 marks then there is that possibility.

Also Q5d leaves a lot of scope for leniency/strictness with how much they wanted you to prove the lengths of everything (like did u have to prove ^CAD was pi/5 or did you have to prove AD = AC) but considering its only worth 2 marks I'd expect them to be fairly lenient on that sort of thing.

in Q7c(ii) you couldnt use similar triangles because it wasn't the included angle and so the easiest way to prove that was by saying that their cosines were equal.

Overall I'd expect them to probably have to be fairly strict in their marking because the exam had a lot of show questions and because it wasn't too hard, unfortunately that will probably stuff a lot of good people up because they will slip up by not including every single detail. (e.g. for Q3d(ii) they might want u to say 0 < w < rt(...) instead of just w < rt(...) though I hope they don't).

 

[bikapika]

For the probability question:

(i) = 0.36 Exact fraction is (12C3*12C3)/(24C6) (consider them all as different i.e. R_1, R_2... then we have 12C3 ways of choosing the red ones, 12C3 ways of choosing the yellow ways. In the whole sample space there are 24C6 ways of choosing 6 balls)

(ii) = 0.32

P(>3 red) + P(>3 yellow) + P(3 red, 3 yellow) = 1 (these are all options)

now because red and yellow are indistinguishable P(>3 red) = P(>3 yellow)

so 2P(>3 red) = 1 - P(3R, 3Y) = 1 - .36 = 0.64
so P(>3 red) = 0.32

yeah but dont u have to take account into the order of withdrawing balls?

 

[AMorris]

Nup. Technically you can so the probability is (12C3*12C3*6!)/(24P6) but it is exactly the same so it doesn't really matter.

Alternatively you can do it the really bashy way:

(12/24)*(11/23)*(10/22)*(12/21)*(11/20)*(10/19)*6C3

(The last term is there for the ordering of which there aer 6C3 ways)

They all come out to the same answer however so it doesn't really matter.

 

[Starcraftmazter]

http://www.hsccoaching.com/Resources/2007 Ext2.pdf if anyone wants them

That made me quite depressed. The stupid mistakes I made

 

[lifesucks]

AMorris you genius... i remembered your estimated mark in the other post (117), and im not so suprised now. the probability answer you gave is same as mine, yeh!

my answers to non-proof questions:
1d. 5/12
1e. pi/2 (simplified to tan(-1)2 + tan(-1)1/2)
3c. 2pi
4c. 2e - 1/2 - (e^2)/2
5ai. 0.36, wasnt sure at all but then Amorris came along!!! you sexy thing!!
5aii. 0.32, see comment above
5cii. pi/2 (thats my wrong answer. after exchanging some *kind* words with friends, im sure it's 2pi now)
6aiv. 4
6bv. terminal velocity is 7
8ciii.pi^2/2

hopefully you can see my name along side morris in the paper...

EDIT: I just checked all my answers against the awesome file posted by jah_lu, and it turns out i will get 116. im fairly happy with that....

 

[kony]

in Q7c(ii) you couldnt use similar triangles because it wasn't the included angle and so the easiest way to prove that was by saying that their cosines were equal.

however since the angle given is a right angle, i believe it doesn't need to be the included angle for a similarity test?

much like the RHS congruency proof.

 

[joshuajspence]

however since the angle given is a right angle, i believe it doesn't need to be the included angle for a similarity test?

much like the RHS congruency proof.

omg... i hope you are right...

can someone confirm?

if 2 triangles have 2 sides in proportion and are both right angled triangles, are they similar?

 

[AMorris]

They are similar but I have not heard of that as a possible similarity proof.

As far as I know, there are three main proofs of similarity: AA, SAS, SSS whereas there are 4 for conguency: SSS, AAS, SAS, RHS. However the lack of RHS in the similarity system is basically just an arbitrary choice so you would have to refer to the syllabus for whether they would accept it.

 

[kony]

well you just said so yourself, given that the cos of the angles are equal, the angles themselves must also be equal. (and hence similar since you have a right angle already)

 

[joshuajspence]

They are similar but I have not heard of that as a possible similarity proof.

As far as I know, there are three main proofs of similarity: AA, SAS, SSS whereas there are 4 for conguency: SSS, AAS, SAS, RHS. However the lack of RHS in the similarity system is basically just an arbitrary choice so you would have to refer to the syllabus for whether they would accept it.

surely for 2 marks tho.... you could get one mark for recognising they are similar... and 1 mark for the conclusion that the angles are equal...

i don't know... i just think it 4 unit they aren't as fussy on the tedious proofs