Discussion and Suggested Solutions - Mathematics 2017

[TheSpicyMeatBall]

just checked and i cant see page 17 of your exam 

[Joeystrippa]

I'm not sure if it has been posted here yet but I think there is a different way to do 16a) ii about Sin(alpha)=Sin(Beta).

I just let dL/Dx=0 and then moved the negative part of the equation over so it was x/square root (x^2+25) = 9-x/square root (49+(9-x)^2) and then from the triangles if you just found Sin(alpha) and Sin(beta) they were equal to the left and right hand respectively using Sin= opposite/hypotenuse (using pythagorus from i for the hypotenuse) so that LHS=RHS and therefore sin(alpha)=sin(beta). Idk that what worked for me and was much less work.

[OliverN]

The question about the three isosceles triangles (15A) the common answer is 36deg but did anyone else also get 108deg as an answer for obtuse-angled triangles? I am fairly sure there are two answers but curious what the general consensus on this is and if NESA is likely to recognise both answers.

I don't think that could be possible since you need to fit three of the angles, plus an extra, into 180 degrees for them to be parallel right? Could you perhaps elaborate further?

If you visualise the triangles you have to line up the base side (lets call that AB) to be 180deg from the first side right, you probably think of it where you rotate the base side down making x smaller than what they present in the diagram but if you rotate it up and around making x bigger, the two sides will eventually become parallel as well the side will me upside down (so it will read as BA) but it's still parallel I would have thought. I will try to add some images to help visualise but I am not sure how this works.

[jamonwindeyer]

I'm looking at a raw mark of between 85-89 depending on how hard I'm marked and I'm 2/14 in the cohort. But 1st and 2nd are quite close and then there's a huge drop off. Do you guys think I could still slide/align into a band 6?

I reckon you are a good chance

in question 14 ciii you reckon you should have used the rounded of value for k as that gives you a value of 880 years while using the complete value of k gives you 870 years
do you think both answers may end up being valid

I think both will be marked correct (poor question design to have that actually make a difference tbh) -> It is common practice to use the rounded value when it is given to you, but obviously using exact is also correct, so I think they'll pay both

[jamonwindeyer]

I'm not sure if it has been posted here yet but I think there is a different way to do 16a) ii about Sin(alpha)=Sin(Beta).

I just let dL/Dx=0 and then moved the negative part of the equation over so it was x/square root (x^2+25) = 9-x/square root (49+(9-x)^2) and then from the triangles if you just found Sin(alpha) and Sin(beta) they were equal to the left and right hand respectively using Sin= opposite/hypotenuse (using pythagorus from i for the hypotenuse) so that LHS=RHS and therefore sin(alpha)=sin(beta). Idk that what worked for me and was much less work.

Yep, definitely a way better approach than my rushed brute force method and definitely correct (what they would have been anticipating people to do)

I suppose on the plus side, I had almost zero work to do for (iii) and (iv), but I definitely still shot myself in the foot

If you visualise the triangles you have to line up the base side (lets call that AB) to be 180deg from the first side right, you probably think of it where you rotate the base side down making x smaller than what they present in the diagram but if you rotate it up and around making x bigger, the two sides will eventually become parallel as well the side will me upside down (so it will read as BA) but it's still parallel I would have thought. I will try to add some images to help visualise but I am not sure how this works.

I see your images! Interesting perspective - I mean I personally think it warps the way the problem was presented a tad too much (although not to scale, the diagrams are still indicators of the arrangement of the problem), but it is definitely technically correct. I'd be interested to see if they gave the marks to someone who just did this! If you did both, well then you've got full marks anyway

PS - What did you use for those images? Look slick af 

[Shoesta]

What do you think the band 6 cut-off will be for this paper?

[jamonwindeyer]

What do you think the band 6 cut-off will be for this paper?

I think it will be mid to high 70's

[Stefan Kolar]

can somebody please send me a link to the paper.

[jamonwindeyer]

can somebody please send me a link to the paper.

NESA will upload it, based on how they've been going, tomorrow

[Stefan Kolar]

ok, i want to see the questions,  i actually think i did pretty well. Anyway do you have any tips for the upcoming physics exam? What things should i focus on?

[~BK~]

NESA will upload it, based on how they've been going, tomorrow

well, nesa just put on the modern history one so i'd say jamon, your prediction would be right
just soo glad to have that paper behind me, i've been trying not to worry too much about what my mark will be 
annnd, happy 21st jamon?

[~BK~]

ok, i want to see the questions,  i actually think i did pretty well. Anyway do you have any tips for the upcoming physics exam? What things should i focus on?

good on ya stefan?!?
that sorta q prolly best put on this thread: https://atarnotes.com/forum/index.php?topic=164552.0

[jamonwindeyer]

ok, i want to see the questions,  i actually think i did pretty well. Anyway do you have any tips for the upcoming physics exam? What things should i focus on?

This guide might help!

annnd, happy 21st jamon?

Thanks heaps!

[avalansh7]

I'm not sure if it has been posted here yet but I think there is a different way to do 16a) ii about Sin(alpha)=Sin(Beta).

I just let dL/Dx=0 and then moved the negative part of the equation over so it was x/square root (x^2+25) = 9-x/square root (49+(9-x)^2) and then from the triangles if you just found Sin(alpha) and Sin(beta) they were equal to the left and right hand respectively using Sin= opposite/hypotenuse (using pythagorus from i for the hypotenuse) so that LHS=RHS and therefore sin(alpha)=sin(beta). Idk that what worked for me and was much less work.

Yeah i did that too.

[kevvdev]

Hey jamon, for question 16a) ii, instead of proving sin(alpha)=sin(beta), i proved tan(a)=tan(b), does that work as well?
Thanks!