# Carrotsticks' MX2 HSC 2013 Solutions (1 Viewer)

#### Makematics

##### Well-Known Member
O yea, also with curve sketching, I slightly messed up x intercepts of 2nd graph, but shape is identical, as well as clearly distinguishable y intercept rather than (0,0), what would I get for it? 2/3 or 3/3?
Depends, how exactly did you stuff them up? If they are not between the right x coordinatess then probs one mark off

#### Makematics

##### Well-Known Member
Concensus is that you can, since it's a 1marker (I wrote by symmetry just in case).
Hmm alright, i tried working it out just then for fun, and i think you have to construct another 2 lines and do some other stuff just to find CE. So saying 'similarly' should be fine phew... Was gonna leave it blank because i thought they wouldnt accept 'similarly'

#### hayabusaboston

##### Well-Known Member
>slightly messed up
>will I get 3/3

what do you think
Maybe my mess up wasn't bad, and maybe the shape is rite? Idk I put asymptotes at the same values as carrot, and drew as carrot did ok, I just remember putting the initial part of the graph on the left thru (1,0). Will I lose one for that?

#### RealiseNothing

##### what is that?It is Cowpea
Maybe my mess up wasn't bad, and maybe the shape is rite? Idk I put asymptotes at the same values as carrot, and drew as carrot did ok, I just remember putting the initial part of the graph on the left thru (1,0). Will I lose one for that?
Well your graph isn't right then?

#### hayabusaboston

##### Well-Known Member
Well your graph isn't right then?
k 1 mark off then u'd say? ITS SAME SHAPE PLZ, AND SAME ASYMPTOTES

#### hayabusaboston

##### Well-Known Member
and yea, do I get 3/4 for e^x volumes for a carried on error like replacing 4 with 3?

#### RealiseNothing

##### what is that?It is Cowpea
and yea, do I get 3/4 for e^x volumes for a carried on error like replacing 4 with 3?
Yep 3/4.

#### Makematics

##### Well-Known Member
Fk i would usually be able to scab a mark for the volumes q if it were a school exam. Not this time

#### jyu

##### Member
Can any one explain Q12diii?

#### seanieg89

##### Well-Known Member
yes there is?
Really? Not in the version I have, they must have changed the numbering of things. There is no q12d at all in mine.

Edit: Oh right, this isn't bos trials .

#### RealiseNothing

##### what is that?It is Cowpea
Really? Not in the version I have, they must have changed the numbering of things. There is no q12d at all in mine.
It's the "Prove that BC is parallel to PQ" question.

#### Makematics

##### Well-Known Member
guys i have a question. for the 4 mark volumes, i never like to 'unroll' the shell into a rectangular prism. instead i like to derive the result from fresh using subtraction of the inner and outer radii. when applying this method today, i forgot the height of the shell, and as a result there was no e^x present in my integral AT ALL. how many marks will i lose for it? it's an absolutely terrible mistake, i know...
ended up on 0/4 for this Q. A warning to all students, BE VERY CAUTIOUS WHEN USING ALTERNATIVE METHODS. I also got 1/3 on another Q despite my solution being 100% valid, because I intuitively used a different method to the one they would have liked.

Not too sure if you can get marks even.... but then again I'm not sure.
spot on as usual, RoT

dudeee i will get some marks haha, definitely for at least getting an expression and integrating it.
nope

#### Carrotsticks

##### Retired
ended up on 0/4 for this Q. A warning to all students, BE VERY CAUTIOUS WHEN USING ALTERNATIVE METHODS. I also got 1/3 on another Q despite my solution being 100% valid, because I intuitively used a different method to the one they would have liked.

spot on as usual, RoT

nope
Did you order your actual paper, or just the raw marks?

Usually, when you get 0 for a solution that was 'almost complete' except one expression was missing, it is because that missing expression made the question substantially easier to do.

Also, what was the Q you got 1/3 on, and what method did you use?

#### Makematics

##### Well-Known Member
Did you order your actual paper, or just the raw marks?

Usually, when you get 0 for a solution that was 'almost complete' except one expression was missing, it is because that missing expression made the question substantially easier to do.

Also, what was the Q you got 1/3 on, and what method did you use?
Just the raw marks. Yeah, I know that I oversimplified the question, but at school we have always been told that we would get at least some marks.
I got 1/3 on 15(a)
All I did was let z=x+iy and w=a+ib, and then found the area A of the triangle using 1/2 absinC. Finding the area was somewhat troublesome, but it still worked out for me.

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#### seanieg89

##### Well-Known Member
It would be cool if people could volunteer to upload their papers on a thread here, for the benefit of future students to see how many marks various partial solutions and mistakes are worth.

#### Carrotsticks

##### Retired
Just the raw marks. Yeah, I know that I oversimplified the question, but at school we have always been told that we would get at least some marks.
I got 1/3 on 15(a)
All I did was let z=x+iy and w=a+ib, and it worked out nicely for me.
How did you work out the area of the triangle OZW with Z(x,y) and W(a,b) ? Only fast way I can think of doing is using the determinant of a 2x2, otherwise that would take a pretty long time?

It would be cool if people could volunteer to upload their papers on a thread here, for the benefit of future students to see how many marks various partial solutions and mistakes are worth.
That really would, and it would also educate a lot of the teachers on how marking works in the HSC, because to put it quite bluntly too few of them actually know how it works.

If they all did, then students would no longer need to write those ridiculous conclusions after each induction problem.

#### Makematics

##### Well-Known Member
How did you work out the area of the triangle OZW with Z(x,y) and W(a,b) ? Only fast way I can think of doing is using the determinant of a 2x2, otherwise that would take a pretty long time?
Ok so after redoing the question I realise that it was far from quick and from from elegant, but surely it is mathematically correct?

$\bg_white \\\\A=\frac{1}{2}\left | z \right |\left | w \right |\sin{(\theta-\phi)}\\\\~~~=\frac{1}{2}\left | z \right |\left | w \right |\sin{(\arg{z}-\arg{w})}\\\\Let z=x+iy and w=a+ib\\\\A=\frac{1}{2}(\sqrt{x^2+y^2})(\sqrt{a^2+b^2})\sin{(\tan^{-1}{\frac{y}{x}}-\tan^{-1}{\frac{b}{a}})}\\\\~~~~=\frac{1}{2}(\sqrt{x^2+y^2})(\sqrt{a^2+b^2})[\sin(\tan^{-1}\frac{y}{x})\cos(\tan^{-1}\frac{b}{a})-\sin(\tan^{-1}\frac{b}{a})\cos(\tan^{-1}\frac{y}{x})]\\\\~~~=\frac{1}{2}(\sqrt{x^2+y^2})(\sqrt{a^2+b^2})[(\frac{y}{\sqrt{x^2+y^2}})(\frac{a}{\sqrt{a^2+b^2}})-(\frac{b}{\sqrt{a^2+b^2}})(\frac{x}{\sqrt{x^2+y^2}})]\\\\~~~=\frac{1}{2}(ay-bx)\\\\z\bar{w}-w\bar{z}=(x+iy)(a-ib)-(a+ib)(x-iy)\\\\~~~~~~~~~~~~=ax+by+ayi-bxi-ax-by-bxi+ayi\\\\~~~~~~~~~~~~=2i(ay-bx)\\\\~~~~~~~~~~~~=4iA$

#### Makematics

##### Well-Known Member
Bump. Is my solution valid?