2014 Extension 2 BOS Trial Exam Discussion Thread (1 Viewer)

[Kurosaki]

didn't have time to even reach it

.

 

[Carrotsticks]

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

I'm glad!

The classic proof for the Gaussian Integral is to use double integration (usually by polar coordinates), but I went for the Cartesian approach as it was a lot more HSC friendly.

However, making the concept of a double integral HSC friendly is something quite difficult to do.

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)

Your proof is correct, and what I was looking for.

Actually, I wanted to put the Conics/3D Trig problem in Question 16 (since the last part is quite difficult), but I couldn't because the marks didn't fit. So I decided to go for Circle Geometry, because finding motivation for the steps can be quite tricky. Not many students would have spotted the bolded part (which I assume you know the reason as to why we can say that).

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

I wanted some justification and also we were a bit more generous there to give the students a bit of a break =)

 

[glittergal96]

Hey all, I did the exam today at home. If anyone would like to compare or discuss answers, here are mine:

View attachment bos2014.pdf.

I didn't include the three questions I drew diagrams for (two graphing q's and the circle geometry), I might do them in paint tomorrow lol...don't have a scanner.

I also didn't include every step of my working in this, so please ask if anything is unclear or if I just made sillies/typoes.

 

[aDimitri]

Hey all, I did the exam today at home. If anyone would like to compare or discuss answers, here are mine:

View attachment 31002.

I didn't include the three questions I drew diagrams for (two graphing q's and the circle geometry), I might do them in paint tomorrow lol...don't have a scanner.

I also didn't include every step of my working in this, so please ask if anything is unclear or if I just made sillies/typoes.

did you time yourself?

 

[aDimitri]

Hey all, I did the exam today at home. If anyone would like to compare or discuss answers, here are mine:

View attachment 31002.

I didn't include the three questions I drew diagrams for (two graphing q's and the circle geometry), I might do them in paint tomorrow lol...don't have a scanner.

I also didn't include every step of my working in this, so please ask if anything is unclear or if I just made sillies/typoes.

also 12b says z1z2 =/= -1 so it is true

 

[Tugga]

did you time yourself?

inb4 3 hours

 

[glittergal96]

did you time yourself?

Not strictly, because I was latexing my answers as I did them (Including this time it took a little over 4 hours I think). I estimate I would have finished most but not all of the questions in 3 hours of strict exam conditions.

 

[glittergal96]

also 12b says z1z2 =/= -1 so it is true

isn't though?

 

[aDimitri]

isn't though?

oh woops didn't read your example right!
uh well i proved it in the exam but i guess there must be an exception.
in your example they still lie on the same line that passes through the origin but on opposite sides so the args differ by

 

[glittergal96]

oh woops didn't read your example right!
uh well i proved it in the exam but i guess there must be an exception.
in your example they still lie on the same line that passes through the origin but on opposite sides so the args differ by

Do you remember your proof?

Yeah, I know the args differ by ...I cooked up that counterexample specifically to get the negative when I realised it was possible. I didn't bother examining the situation further to find out how numerous the counterexamples are (and whether or not the extent to which the claimed identity fails is constant.) I am kind of tired now, but I will look at it again tomorrow.

 

[aDimitri]

Do you remember your proof?

Yeah, I know the args differ by ...I cooked up that counterexample specifically to get the negative when I realised it was possible. I didn't bother examining the situation further to find out how numerous the counterexamples are (and whether or not the extent to which the claimed identity fails is constant.) I am kind of tired now, but I will look at it again tomorrow.

i did it with a drawing so not 100% positive. i think it only works if arg(z1)+arg(z2) < pi

edit: nope there are more cases than that

 

[glittergal96]

Yeah, its definitely true and quite straightforward to show if they are both in the first quadrant. That was probably what was intended.

 

[aDimitri]

Yeah, its definitely true and quite straightforward to show if they are both in the first quadrant. That was probably what was intended.

i did it with z2 in second quadrant and it was still fairly straightforward.

if you don't state-rank 4U i will question this world.

 

[glittergal96]

i did it with z2 in second quadrant and it was still fairly straightforward.

if you don't state-rank 4U i will question this world.

Lol appreciate the confidence, I am quite a bit better when doing problems in the comfort of my own room than in exam situations with pressure though so you probably overestimate me.

 

[glittergal96]

i did it with z2 in second quadrant and it was still fairly straightforward.

if you don't state-rank 4U i will question this world.

Ps are you sure it works for z2 in second quadrant? If z1 is in the first quadrant but close to i and z2 is in the second quadrant and close to -1, I think the arguments still differ by pi.

I think that the geometric proof probably breaks when the arguments get big enough that z1z2 gets past -1 on the circle. So arg(z1) + arg(z2) < pi might be the best we could hope for.

 

[aDimitri]

Ps are you sure it works for z2 in second quadrant? If z1 is in the first quadrant but close to i and z2 is in the second quadrant and close to -1, I think the arguments still differ by pi.

I think that the geometric proof probably breaks when the arguments get big enough that z1z2 gets past -1 on the circle. So arg(z1) + arg(z2) < pi might be the best we could hope for.

yeah that's why i mentioned arg(z1) + arg(z2) < pi as the condition before, i noticed that's the only way it worked with z2 in the second quadrant

 

[glittergal96]

Note that my solution to 13biii. is wrong, I made a sign error in the inequalities.

To properly solve this question we can consider the quantity . We have the initial condition of and the equation . This lets us solve for f as a function depending on t with parameters u and T. The particles can only meet if f is eventually negative, so we take the limit as t->inf and enforce that this be negative.

This implies



which simplifies to the desired result upon using the standard equations of motion stuff to evaluate the T-dependent terms.

 

[Hi there444]

Lol appreciate the confidence, I am quite a bit better when doing problems in the comfort of my own room than in exam situations with pressure though so you probably overestimate me.

Na still state rank.

 

[mreditor16]

did anyone have a quick, neat way of doing 11 e) - the locus one with arguments

just curious, mine seems too dodgy...

 

[aDimitri]

did anyone have a quick, neat way of doing 11 e) - the locus one with arguments

just curious, mine seems too dodgy...

i constructed a quadrilateral using the points z, z+1, 1, and 0. and since it's a parallelogram, where the diagonal bisects the base angle, it must be a rhombus -> |z| = 1

obviously with exception z = -1 because then it's not a quadrilateral