General Thoughts: Mathematics Extension 2 (1 Viewer)

brachester · Oct 24, 2011

largarithmic said:
This works:

$\frac{p}{1+p} + \frac{q}{1+q} - \frac{r}{1+r} \\ = (1-\frac{1}{1+p}) + (1-\frac{1}{1+q}) - (1-\frac{1}{1+r}) \\ = 1 - \frac{1}{1+p} - \frac{1}{1+q} + \frac{1}{1+r} \\ \ge 1 - \frac{1}{1+p} - \frac{1}{1+q} + \frac{1}{1+p+q+pq} \\ = 1 - \frac{1}{1+p} - \frac{1}{1+q} + \frac{1}{(1+p)(1+q)} \\ \ge (1-\frac{1}{1+p})(1-\frac{1}{1+q}) \ge 0$

is it ok if i just add them up together? it didn't take that long

b3kh1t · Oct 24, 2011

brachester said:
Lol, i was trying to change it into a+iy form, coudn't finish though

Same, how did people get x=-1/2

D94 · Oct 24, 2011

brachester said:
Lol, i was trying to change it into a+iy form, coudn't finish though

Still works though,

$\\\left | 1 + \frac{1}{z} \right | \leq 1 \\ \\\left | \frac{1+z}{z} \right | \leq 1 \\ \\\left | 1+z | \leq | z |\right \ as\ modulus\ is\ positive\ \\ \\\left | 1 + x + iy \right | \leq | x+iy |\right \\ \\\sqrt{(x+1)^2 + y^2} \leq \sqrt{(x)^2 + y^2} \ square\ both\ sides \\ \\(x+1)^2 + y^2 \leq x^2 + y^2 \\(x+1)^2 \leq (x)^2 \\x^2 + 2x + 1 \leq x^2 \\ 2x + 1 \leq 0 \\ x \leq -\frac{1}{2}$

However, since $z$ is in the denominator, $z$ cannot equal zero, so I believe (but not 100% sure) that you need a hole/circle on the x-axis.

zhiying · Oct 24, 2011

brachester said:
is it ok if i just add them up together? it didn't take that long

Yes you can, I think that was the most straight forward way.

So what's everyone satisfied with? Personally I'd be happy with 105, didn't do as well as I would've liked. Damn pressure

largarithmic · Oct 24, 2011

D94 said:
Still works though,

$\\\left | 1 + \frac{1}{z} \right | \leq 1 \\ \\\left | \frac{1+z}{z} \right | \leq 1 \\ \\\left | 1+z | \leq | z |\right \ as\ modulus\ is\ positive\ \\ \\\left | 1 + x + iy \right | \leq | x+iy |\right \\ \\\sqrt{(x+1)^2 + y^2} \leq \sqrt{(x)^2 + y^2} \ square\ both\ sides \\ \\(x+1)^2 + y^2 \leq x^2 + y^2 \\(x+1)^2 \leq (x)^2 \\x^2 + 2x + 1 \leq x^2 \\ 2x + 1 \leq 0 \\ x \leq -\frac{1}{2}$

However, since $z$ is in the denominator, $z$ cannot equal zero, so I believe (but not 100% sure) that you need a whole/circle on the x-axis.

once you get |z+1| <= |z| you can do it straight away by considering it geometrically

also lol riemann hypothesis much?

aichessee · Oct 24, 2011

Oh dear. This is doing nothing for my self esteem. I'm not sure whether this thread is made up of all the geniuses or whether I'm just far stupider than I ever thought, but I found it incredibly difficult. There were quite a few questions that didn't look like anything I'd practised before, and I really struggled with time issues, didn't get to give more than a glance at Q8 just as they said "pens down". I left out quite a few questions too - it's not possible for me to get more than 85. I found it generally much harder than all the past papers.

Still, congratulations to everyone else, it sounds like you all did wonderfully

D94 · Oct 24, 2011

largarithmic said:
once you get |z+1| <= |z| you can do it straight away by considering it geometrically

also lol riemann hypothesis much?

I know, but other users wanted it in x+iy or a+ib form, so yeah. Tbqh, "riemann hypothesis" does not ring a bell; unfortunately, I'm no maths genius

.

Staplerrr · Oct 24, 2011

aichessee said:
Oh dear. This is doing nothing for my self esteem.

I couldn't agree more :/. I'm the only one doing ext.2 at my school so i've never really had anyone to work with or compare myself too. Still, i'm guessing most of these people go to Sydney schools? That makes me feel better

.

michaeljennings · Oct 24, 2011

aichessee said:
Oh dear. This is doing nothing for my self esteem. I'm not sure whether this thread is made up of all the geniuses or whether I'm just far stupider than I ever thought, but I found it incredibly difficult. There were quite a few questions that didn't look like anything I'd practised before, and I really struggled with time issues, didn't get to give more than a glance at Q8 just as they said "pens down". I left out quite a few questions too - it's not possible for me to get more than 85. I found it generally much harder than all the past papers.

Still, congratulations to everyone else, it sounds like you all did wonderfully

A lot of people here are extremely good at maths ext 2 and will probably get marks of like 98+. Dont feel bad if youve realised youve gotten a question wrong cos it would be extremely hard to get everything right. Just be happy maths extension 2 is over!!

4025808 · Oct 24, 2011

michaeljennings said:
A lot of people here are extremely good at maths ext 2 and will probably get marks of like 98+. Dont feel bad if youve realised youve gotten a question wrong cos it would be extremely hard to get everything right. Just be happy maths extension 2 is over!!

Also given that tonnes of the state actually suck at maths (seriously).

largarithmic · Oct 24, 2011

D94 said:
I know, but other users wanted it in x+iy or a+ib form, so yeah. Tbqh, "riemann hypothesis" does not ring a bell; unfortunately, I'm no maths genius .

wikipedia it, the line Re(z) = -1/2 is very significant lol

D94 · Oct 24, 2011

largarithmic said:
wikipedia it, the line Re(z) = -1/2 is very significant lol

Interesting, good read, but mind boggling.

twoey · Oct 24, 2011

aichessee said:
Oh dear. This is doing nothing for my self esteem. I'm not sure whether this thread is made up of all the geniuses or whether I'm just far stupider than I ever thought, but I found it incredibly difficult. There were quite a few questions that didn't look like anything I'd practised before, and I really struggled with time issues, didn't get to give more than a glance at Q8 just as they said "pens down". I left out quite a few questions too - it's not possible for me to get more than 85. I found it generally much harder than all the past papers.

Still, congratulations to everyone else, it sounds like you all did wonderfully

I concur. I recently found put my school is ranked in the 500s, so being 1/4 in MX2 dooesn't count for much... anyway, I'm hoping for high E3.

brachester · Oct 24, 2011

zhiying said:
Yes you can, I think that was the most straight forward way.

So what's everyone satisfied with? Personally I'd be happy with 105, didn't do as well as I would've liked. Damn pressure

105 is freaking gooddddddddd!!! i'm hoping for a low E4, i;m confident that i'll get higher than 70/120, maybe around 75, but it still depends on this year E4 cutoff though

khfreakau · Oct 24, 2011

so who remembered to put open circles on their graph in 3ai? i know i certainly did

michaeljennings · Oct 24, 2011

khfreakau said:
so who remembered to put open circles on their graph in 3ai? i know i certainly did

I did =)

MATHSCAPE · Oct 24, 2011

khfreakau said:
so who remembered to put open circles on their graph in 3ai? i know i certainly did

heh, glad I did.

4025808 · Oct 24, 2011

khfreakau said:
so who remembered to put open circles on their graph in 3ai? i know i certainly did

I did, but I put it at 0 instead of 2/pi T_T and then I drew the graph approaching 0 instead of 2/pi. looks like one mark will be deducted T_T

lpodnano · Oct 24, 2011

4025808 said:
I did, but I put it at 0 instead of 2/pi T_T and then I drew the graph approaching 0 instead of 2/pi. looks like one mark will be deducted T_T

Haha! Omg I did that at first and then I started doing trial and error with my calculator to make sure my graph was approaching the right direction and then i realised

4025808 · Oct 24, 2011

lpodnano said:
Haha! Omg I did that at first and then I started doing trial and error with my calculator to make sure my graph was approaching the right direction and then i realised

far out I should have done that T_T
next time, use calculator to find limits and find where the curve is going

General Thoughts: Mathematics Extension 2 (1 Viewer)

Member

Member

New Member

Active Member

Member

New Member

New Member

New Member

Active Member

Well-Known Member

Member

New Member

New Member

Member

Member

Active Member

Member

Well-Known Member

5eva alone

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)