2014 Extension 2 BOS Trial Exam Discussion Thread (2 Viewers)

[Carrotsticks]

Hi all,

Feel free to discuss any questions/answers to the 2014 Extension 2 BOS Trial in this thread! Were there any questions you couldn't get out? Any questions in particular that you liked/disliked?

I'll answer any questions you have about the paper in general, but I'll let the users discuss solutions in here. Myself and Trebla will give hints if students have difficulty with any problems =)

Official solutions will be posted in this thread some time next week.

The link to the exam is here: https://www.dropbox.com/s/27ibdwia0y40o6h/2014 BOS Trial Mathematics Extension 2.pdf?dl=0

 

[Kurosaki]

The derangements and inequalities in the last question was fantabulous =).

Could you give me a pointer regarding 14 bi? I spent maybe half an hour trying to figure out/visualise how it worked before moving on.
I'm just not sure how you get the term in O_O.

 

[Carrotsticks]

The derangements and inequalities in the last question was fantabulous =).

Could you give me a pointer regarding 14 bi? I spent maybe half an hour trying to figure out/visualise how it worked before moving on.
I'm just not sure how you get the term in O_O.

Try to find the value of k for that particular x value, because it changes depending on where x is.

 

[Kurosaki]

Oh wow I'm pretty blind - did not see that AT ALL. OK, that makes a lot more sense to me now. Thanks for the assist. I should be able to get out the rest now that I've laid that demon to rest.

 

[aDimitri]

Oh wow I'm pretty blind - did not see that AT ALL. OK, that makes a lot more sense to me now. Thanks for the assist. I should be able to get out the rest now that I've laid that demon to rest.

this was by far my favourite question in any exam i have ever sat

 

[mreditor16]

I really have to ask why that circle geometry question is there at Question 16?
solution:
i) AMD is similar to CMB (AA)
Thus, CM/AM = CB/AD = CG/AF
And angle MCB = MAD (lol)
Thus FMA is similar to GMC (SAS)
ii) We know that OMX = OMY = 90 since M is midpoint of XY
So we just prove that POQ is isosceles.
construct lines OQ and MG, FM and PO
LEt angle MGC = MFA = alpha
Then angle MFO = MGO =90-alpha
which means that OPM = OQM = 90 - alpha
Thus the triangle POQ is isosceles and this means M bisects PQ since it's altitude.


Was this question there for free marks lol? Or was it put in because it's rather long (with all the reasoning and stuff)

 

[aDimitri]

I really have to ask why that circle geometry question is there at Question 16?
solution:
i) AMD is similar to CMB (AA)
Thus, CM/AM = CB/AD = CG/AF
And angle MCB = MAD (lol)
Thus FMA is similar to GMC (SAS)
ii) We know that OMX = OMY = 90 since M is midpoint of XY
So we just prove that POQ is isosceles.
construct lines OQ and MG, FM and PO
LEt angle MGC = MFA = alpha
Then angle MFO = MGO =90-alpha
which means that OPM = OQM = 90 - alpha
Thus the triangle POQ is isosceles and this means M bisects PQ since it's altitude.


Was this question there for free marks lol? Or was it put in because it's rather long (with all the reasoning and stuff)

you can't just make the assumption that CB/AD = CG/AF can you?

 

[Kurosaki]

this was by far my favourite question in any exam i have ever sat

But, but, combinatorics!

 

[aDimitri]

But, but, combinatorics!

didn't have time to even reach it

 

[aDimitri]

also 14ci is still wrong T_T
should be nx^(n+1)

edit: fixed now

 

[Chlee1998]

you can't just make the assumption that CB/AD = CG/AF can you?

you can because CB = 2 x GC (since OG is perpendicular to CB)
and similarly AD = 2 x AF same reason

Therefore CB / AD = (2x GC) / (2x AF) = GC/AF

this is becasue the 2s cancel out.
Do you understand?

Or do you mean that I should've posted this working?

 

[Tugga]

you can't just make the assumption that CB/AD = CG/AF can you?

I think you can, because of perpendicular bisector thingo from centre. I didn't get it anyway LOL, left like 15 mins for q16 :S...

also @chlee, it's not that easy of a proof to see, you could be underestimating your inability or just happened to have had the clarity to see it on the day it's sometimes like that with circle geo...

also, just looked at q13a again and it came to me immediately damnn

 

[aDimitri]

you can because CB = 2 x GC (since OG is perpendicular to CB)
and similarly AD = 2 x AF same reason

Therefore CB / AD = (2x GC) / (2x AF) = GC/AF

this is becasue the 2s cancel out.
Do you understand?

Or do you mean that I should've posted this working?

oh ffs i tried to do that in the exam but couldn't find a reason why CB/AD = CG/AF
man this could have pulled me into the potential high 60s

 

[aDimitri]

I think you can, because of perpendicular bisector thingo from centre. I didn't get it anyway LOL, left like 15 mins for q16 :S...

also @chlee, it's not that easy of a proof to see, you could be underestimating your inability or just happened to have had the clarity to see it on the day it's sometimes like that with circle geo...

also, just looked at q13a again and it came to me immediately damnn

i couldn't do ii

i understand that is has something to do with |P(i)| but i don't really understand it

 

[Tugga]

i couldn't do ii

i understand that is has something to do with |P(i)| but i don't really understand it

Are we allowed to spoil answers in here or...?

 

[aDimitri]

also the other thing i didn't understand is why 14.b.i & ii were worth 2 marks each for like 30 seconds of work

 

[Tugga]

Are we allowed to spoil answers in here or...?

okay nvm just read the first post :L

You change the polynomial so that the roots are c_q-i etc.

i.e. replace z with z+i

then consider product of roots, which would simply be simply P(i). So now we take modulus of everything and we get:

|prod.ofroots|<1

but from part i), the real roots (or the transformed real roots), are all >=1,

hence, the product of all the modulus of transformed complex roots are <1

then part iii) just uses the conjugate root theorem I think.
At least one of |c_q-i| <1, meaning distance from i is less than 1.
the conjugate of c_q is simply c_q mirrored by the x-axis, so the distance from -i is less than 1
so for some c_q' (the conjugate of c_q which satisfies |c_q-i|<1) |c_q'+1|<1

and by conjugate root theorem, c_q' is also a root, so the statement is true.


In the test I only got up to the bit where i transformed the roots then skipped ahead.

 

[faisalabdul16]

Does anyone happen to remember the polynomial they got for 12c?

The one with roots being 1/alpha + 1/beta, etc etc

 

[Carrotsticks]

Please share answers in here!

I'm currently pre-occupied atm so I won't have time to respond to questions until tomorrow.

But for the mean time, I would really appreciate it if students could guide other students/provide their solutions for now until I can further explain why we did XYZ in the exam.

 

[aDimitri]

Does anyone happen to remember the polynomial they got for 12c?

The one with roots being 1/alpha + 1/beta, etc etc

or something vaguely similar to that not sure about the signs.
i actually didn't get to finish it in the exam