HSC 2012 MX2 Marathon (archive) (4 Viewers)

[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

 

[Nooblet94]

Re: 2012 HSC MX2 Marathon

I got

Shit. That's wrong, nightweaver's right. (forgot to finish off one of the inequalities lol)

 

[nightweaver066]

Re: 2012 HSC MX2 Marathon

lol don't sweat it, we all make mistakes.

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Correct, but it's debatable whether there should be equality for the -1/3 side of the inequality.

Reason being d=-1/3 gives the degenerate 'triangle', which is just a straight line.

And the 'axiom of triangle' you mentioned is more famously known as the "Triangle Inequality" haha.

 

[math man]

Re: 2012 HSC MX2 Marathon

but lets all agree, that for a q8 from a previous trial, this question is rather pathetic in that sense, more suited for q4,5,6

 

[nightweaver066]

Re: 2012 HSC MX2 Marathon

 

[Nooblet94]

Re: 2012 HSC MX2 Marathon

but lets all agree, that for a q8 from a previous trial, this question is rather pathetic in that sense, more suited for q4,5,6

Yep, entirely. The only remotely challenging part is knowing to apply the triangle inequality and it's extension 2 maths, so everyone should be able to make basic logical connections like that.

 

[nightweaver066]

Re: 2012 HSC MX2 Marathon

Yep, entirely. The only remotely challenging part is knowing to apply the triangle inequality and it's extension 2 maths, so everyone should be able to make basic logical connections like that.

A question similar to this was in my 2U preliminary half yearly test last year. Only a handful of students got it lol.