# 2012 HSC Extension 2 Mathematics Solutions (1 Viewer)

#### alicekate18

##### New Member
This makes me feel heaps better about the exam. I screwed up that volumes question (which I thought I had anyway) but I'm confident I got most of that multiple choice (I can't remember all of my answers, but I know I got 10 right).

#### lgnorance

##### Member
I don't think that is correct because u1 and u2 are complex numbers (i.e. x and y still contain an imaginary part within u1, u2 and z)

I personally think that locus question is pretty dodgy or there is some sort of typo because it is obvious by inspection that since B1 and B2 are fixed then clearly their midpoint is fixed.
agreed theres a typo al; as p varies the locus doesn't change. in the exam my mind was like 'wtf is this questoin'. checked my working 5 times to ensure that i was right cause its such a double reverse trick :/

i wrote the locus is a "vector" and i had the eqn the same as the correct answer, probs gonna lose one mark cause its actually a point. gayyyyyyyyyyyy

#### Nooblet94

Can't think of any over the top of my head.
$\bg_white (a-b)^2\geq0\\
a^2+b^2\geq 2ab\\
a^2+2ab+b^2 \geq 4ab\\
(a+b)^2 \geq 4ab\\
a+b\geq 2\sqrt{ab}\\
\frac{a+b}{2}\geq \sqrt{ab}$

That's the way I did it back the first time I tried proving that. Your way's far more efficient though.

EDIT: dafuq is up wif mah tex

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#### Trebla

Your TeX code should all be in one line i.e. no physically pressing 'enter' to make a new line

#### someth1ng

##### Retired Nov '14
For inequalities, are you allowed to do it this way

If <line proving is true>
=> Expand/square etc
=> Factorise
=> Forms a perfect square
.'. Must be >0

That's what I did.

#### Carrotsticks

##### Retired
That is usually incorrect because that is going by the converse.

However in this case, that is correct because all steps used are 'iff' steps (both directions work) so by going one way, you have also shown the other direction.

#### someth1ng

##### Retired Nov '14
That is usually incorrect because that is going by the converse.

However in this case, that is correct because all steps used are 'iff' steps (both directions work) so by going one way, you have also shown the other direction.
Okay, that's all cool because when I went to Dux College, they said that was okay as long as you say 'if' and use the 'implies' sign (=>). Just wanted to double check.

Thanks mate.

#### RealiseNothing

##### what is that?It is Cowpea
I don't think that is correct because u1 and u2 are complex numbers (i.e. x and y still contain an imaginary part within u1, u2 and z)

I personally think that locus question is pretty dodgy or there is some sort of typo because it is obvious by inspection that since B1 and B2 are fixed then clearly their midpoint is fixed.
B1 and B2 aren't fixed though? They depend on the position of P as the point B1 is placed such that APB is an isosceles right angled triangle. Hence if P moves, so does B to compensate for this restriction.

#### zeebobDD

##### Member
hey for that locus of B1B2, i got the same expression as you, did you have to write down the locus though? like namewise i wrote straight line lol

#### kingkong123

##### Member
I did my working out and said

$\bg_white \therefore The locus of the midpoint of B_1B_2 is \frac{(u_1+u_2)+i(u_2-u_1)}{2}$

But i didn't specifically say that it is a fixed point. Would i lose any marks?

#### ismeta

##### Member
Oh excellent, it was a fixed point. I was worried I'd messed something up when I found it was a point...

ignore post

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#### NickGero

##### Member
I did my working out and said

$\bg_white \therefore The locus of the midpoint of B_1B_2 is \frac{(u_1+u_2)+i(u_2-u_1)}{2}$

But i didn't specifically say that it is a fixed point. Would i lose any marks?
^This. Except for some reason I said it was the perp bisector of u1 and u2. I really hope they were just looking for the algebra and didn't need the geometric description.

#### Carrotsticks

##### Retired
Okay, that's all cool because when I went to Dux College, they said that was okay as long as you say 'if' and use the 'implies' sign (=>). Just wanted to double check.

Thanks mate.
Actually, it is NOT okay to just say the 'if' sign. Your working out ONLY works if you use the IFF sign, because you are proving the converse with 'iff' statements and thereby proving the forward statement.

Think of it as building a one-way vs two-way street. By arguing the 'if', you have proved one direction and thus constructed a one-way road.

On the other hand, if you had built a two-way road, it doesn't matter whether you started from one end or the other, because it goes both ways anyway!

#### someth1ng

##### Retired Nov '14
Actually, it is NOT okay to just say the 'if' sign. Your working out ONLY works if you use the IFF sign, because you are proving the converse with 'iff' statements and thereby proving the forward statement.

Think of it as building a one-way vs two-way street. By arguing the 'if', you have proved one direction and thus constructed a one-way road.