HSC 2012 MX2 Marathon (archive) (1 Viewer)

[largarithmic]

Re: 2012 HSC MX2 Marathon

Im pretty sure eventually you guys will find that stuff like invariance under reflection/transformation is far more useful and fundamental than stuff like integration by subtitution euclidean geometry itself is useless; analysis when thought slightly geometry is leet though. Theres a really really really awesome way to define continuity actually in terms of sets, regions, open sets etc thats basically thinking about it geometrically (getting on the topology side)

 

[cutemouse]

Re: 2012 HSC MX2 Marathon

You can like, do that integration thing just by considering the geometry of it right; like sending x -> a-x is the same as reflecting the graph about the line x = a/2, so the area under the curve between 0 and a is preserved

Yeah I think I realised during uni maths that it was just integrating starting from the "other end"...

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

In 3rd year, I'm doing Differential Geometry. Looking forward to combining the worlds of geometry and analysis into one.

 

[largarithmic]

Re: 2012 HSC MX2 Marathon

In 3rd year, I'm doing Differential Geometry. Looking forward to combining the worlds of geometry and analysis into one.

Oh awesome, is that an honours course? I assume btw youre taking 2961/2 this semester?

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Oh awesome, is that an honours course? I assume btw youre taking 2961/2 this semester?

Nah it's just a 3rd year topic in the 'Pure' section.

And yeah I am, which tutes you got?

 

[seanieg89]

Re: 2012 HSC MX2 Marathon

Analysis ftw indeed, but symmetry is of universal importance in mathematics.

Side note: I didn't like the diffgeom course so much because the course was nearly completely embedded in R^3. It does make the study of differential geometry on manifolds easier to understand in the long run though i guess.

 

[largarithmic]

Re: 2012 HSC MX2 Marathon

Nah it's just a 3rd year topic in the 'Pure' section.

And yeah I am, which tutes you got?

Ah okay; 1st or 2nd sem? and I'm in mon 12 tute for 2962, and not sure about 2961 - any possible tute clashes with either phys chem or chinese lol, but Ill probably be going to the wed 2pm one - gotta work that out in next week, though.

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Ah okay; 1st or 2nd sem? and I'm in mon 12 tute for 2962, and not sure about 2961 - any possible tute clashes with either phys chem or chinese lol, but Ill probably be going to the wed 2pm one - gotta work that out in next week, though.

This is for sem 1. I don't quite remember which 2961 tute I'm in, I think it was on Wed (or was that my analysis one )

You'll have a fellow BOS'er with you for your monday 2962 tute.

 

[largarithmic]

Re: 2012 HSC MX2 Marathon

This is for sem 1. I don't quite remember which 2961 tute I'm in, I think it was on Wed (or was that my analysis one )

You'll have a fellow BOS'er with you for your monday 2962 tute.

^^

 

[slyhunter]

Re: 2012 HSC MX2 Marathon

Just a friendly reminder:

This is a MX2 Marathon thread for 2012ers. Whilst I have no issue with our uni members correcting errors etc, by answering questions and the like, the purpose of this thread is kinda defeated. Uni students can take it to the uni discipline forums.

Keep this thread for HSC students.

 

[horticulturist]

Re: 2012 HSC MX2 Marathon

find

 

[nightweaver066]

Re: 2012 HSC MX2 Marathon

find

 

[SpiralFlex]

Re: 2012 HSC MX2 Marathon

find

 

[Aesytic]

Re: 2012 HSC MX2 Marathon

b) (i)gradient PQ = (p+q)/2
.'. chord PQ has equation y - ap^2 = [(p+q)/2][x-2ap]
if PQ passes through (2a, 0), then x=2a y = 0 must satisfy the equation of the chord
.'. 0 - ap^2 = [(p+q)/2][2a-2ap]
-2ap^2 = 2ap + 2aq - 2ap^2 - 2apq
divide everything through by 2a
-p^2 = p + q - p^2 - pq
.'. pq = p + q

(ii) midpoint of PQ is (a[p+q], [a/2][p^2 + q^2])
.'. x = a(p+q)
= apq by subbing in pq from part (i)
.'. pq = x/a

.'. y = [a/2][p^2 + q^2]
= [a/2]{[p+q]^2 - 2pq}
= [a/2]{[pq]^2 - 2pq}
= [a/2]{[x^2/a^2] - 2x/a}
= x^2/2a - x
.'. 2ay = x^2 - 2ax
= (x-a)^2 - a^2
2a(y+a/2) = (x-a)^2

.'. locus of M is a parabola with vertex (a, -a/2) and focal length 2a

i might've done something wrong, i don't feel confident with this answer :/

 

[nightweaver066]

Re: 2012 HSC MX2 Marathon

b) (i)gradient PQ = (p+q)/2
.'. chord PQ has equation y - ap^2 = [(p+q)/2][x-2ap]
if PQ passes through (2a, 0), then x=2a y = 0 must satisfy the equation of the chord
.'. 0 - ap^2 = [(p+q)/2][2a-2ap]
-2ap^2 = 2ap + 2aq - 2ap^2 - 2apq
divide everything through by 2a
-p^2 = p + q - p^2 - pq
.'. pq = p + q

(ii) midpoint of PQ is (a[p+q], [a/2][p^2 + q^2])
.'. x = a(p+q)
= apq by subbing in pq from part (i)
.'. pq = x/a

.'. y = [a/2][p^2 + q^2]
= [a/2]{[p+q]^2 - 2pq}
= [a/2]{[pq]^2 - 2pq}
= [a/2]{[x^2/a^2] - 2x/a}
= x^2/2a - x
.'. 2ay = x^2 - 2ax
= (x-a)^2 - a^2
2a(y+a/2) = (x-a)^2

.'. locus of M is a parabola with vertex (a, -a/2) and focal length 2a

i might've done something wrong, i don't feel confident with this answer :/

Looks right.

Accidently posted 3U questions here..

Do the next question!

 

[kingkong123]

Re: 2012 HSC MX2 Marathon

The other question was posted so for (c)
The degree of the remainder is always one less than the degree of the divisor. Therefore it is of degree 1.

<a href="http://www.codecogs.com/eqnedit.php?latex=\\P(3)=3 \textup{ and }P(4)=4\\\\\textup{Let }P(x)=Q(x)(x-4)(x-3)@plus;(ax@plus;b)\\P(3)=3a@plus;b=3 \to (1)\\P(4)=4a@plus;b=4\rightarrow (2)\\\\(1)-(2)\Rightarrow \therefore a=1\\\textup{Sub into }(1)\Rightarrow b=0\\\\\therefore \textup{Remainder} =x" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\P(3)=3 \textup{ and }P(4)=4\\\\\textup{Let }P(x)=Q(x)(x-4)(x-3)+(ax+b)\\P(3)=3a+b=3 \to (1)\\P(4)=4a+b=4\rightarrow (2)\\\\(1)-(2)\Rightarrow \therefore a=1\\\textup{Sub into }(1)\Rightarrow b=0\\\\\therefore \textup{Remainder} =x" title="\\P(3)=3 \textup{ and }P(4)=4\\\\\textup{Let }P(x)=Q(x)(x-4)(x-3)+(ax+b)\\P(3)=3a+b=3 \to (1)\\P(4)=4a+b=4\rightarrow (2)\\\\(1)-(2)\Rightarrow \therefore a=1\\\textup{Sub into }(1)\Rightarrow b=0\\\\\therefore \textup{Remainder} =x" /></a>

 

[deswa1]

Re: 2012 HSC MX2 Marathon

New question:

The lengths of the sides of a triangle are the first three terms of an arithmetic progression with common difference d. If the first term is 1, find the set of possible values of d.

 

[math man]

Re: 2012 HSC MX2 Marathon

New question:

The lengths of the sides of a triangle are the first three terms of an arithmetic progression with common difference d. If the first term is 1, find the set of possible values of d.

this question i think was from the 2010 sydney tech q8 trial paper, which was shockingly easy for q8.

 

[nightweaver066]

Re: 2012 HSC MX2 Marathon

New question:

The lengths of the sides of a triangle are the first three terms of an arithmetic progression with common difference d. If the first term is 1, find the set of possible values of d.

Taking a swing at it.

Let the sides of the triangle be 1, 1 + d and 1 + 2d.

2 + d >= 1 + 2d (axiom of a triangle)
1 - d >= 0

d <= 1

Similarly, 2 + 3d >= 1
1 + 3d >= 0

d >= -1/3

2 + 2d >= 1 + d

1 + d >= 0

d >= -1

therefore -1/3 <= d <= 1

 

[Nooblet94]

Re: 2012 HSC MX2 Marathon

New question:

The lengths of the sides of a triangle are the first three terms of an arithmetic progression with common difference d. If the first term is 1, find the set of possible values of d.

I got