General Thoughts: Mathematics Extension 2 (2 Viewers)

IR

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I finished with around 15-20mims to go so it was a relatively easy paper. Expecting 99 raw.
Same i had like 40 mins to do the circle geo question which was the only one left.. But not sure if got it right :(
 

aDimitri

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Thats it you da man. We are so similar. My best is 93/100 and my worst is 88/100. Hey guys please post up a proof to circle geo ii)
I donno why but some people use angle in semicircle and then used vertically opposite to prove straight line. But that seemed to simple.
you can't use vertically opposite angles, that involves the assumption that they are already collinear. i proved vert opposite of APX and RPC using the sum of the angles about P = 360 and APX = DPQ, YPB = RPC
you end up with APX = RPC therefore collinear

also, i go to penrith :)
 

dan964

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solution to Q16 (c) here it goes

let x=e^u (just a bit easier to work with)
therefore dx = e^u du
#soz no maths symbols

therefore Integral is ue^u/(1+u)^2

hmm. it looks like a quotient rule in reverse, lets stab
e^u/(1+u)
which gives x/(1+ln x)

lost 9 marks minimum probably will lose 11
Q10 - mc, + 5-6 marks for no attempt in Q15 (did attempt part i by drawing diagram), 3 marks Q16, possibly 4.
 
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dan964

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you can't use vertically opposite angles, that involves the assumption that they are already collinear. i proved vert opposite of APX and RPC using the sum of the angles about P = 360 and APX = DPQ, YPB = RPC
you end up with APX = RPC therefore collinear

also, i go to penrith :)
you can provided you use only QPY and XPR. Using sum about angles is good, although I did by taking two straight angles on line XR (same thing though)
 
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IR

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you can't use vertically opposite angles, that involves the assumption that they are already collinear. i proved vert opposite of APX and RPC using the sum of the angles about P = 360 and APX = DPQ, YPB = RPC
you end up with APX = RPC therefore collinear

also, i go to penrith :)
nice bro. u can get this ez
 

IR

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So for the conical pendulum question with toy plane, for the third part did you guys differentiation sin(x)/cos^2(x) or did you guys differentiate the velocity counterpart in respect to d(sin(x)/cos^2(x))/dv?
 

RealiseNothing

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Dont even need substitution for the very last integral. Just add one and subtract one from the numerator, then convert the plus one into x/x and it is reverse quotient rule.
 

rumbleroar

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So for the conical pendulum question with toy plane, for the third part did you guys differentiation sin(x)/cos^2(x) or did you guys differentiate the velocity counterpart in respect to d(sin(x)/cos^2(x))/dv?
.......conical pendulum????
now i'm freaking out about whether or not i actually sat the test LOL

tbh im not sure what I did. I definitely differentiated something (I think it was something from pt ii) and bsed my way through it
 

aDimitri

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you can provided you use only QPY and XPR. Using sum about angles is good, although I did by taking two straight angles on line XR (same thing though)
yes sorry i forgot to mention it's valid to use XPY = QPR, which i did use in my proof
 

IR

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Dont even need substitution for the very last integral. Just add one and subtract one from the numerator, then convert the plus one into x/x and it is reverse quotient rule.
Latex please.
 

noodlesareyum

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It was actually really easy. When I read the paper during reading time, I thought they mixed up the 2u with the 4u paper. I wonder what the cut-off would be... 100/100
 

emilios

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Why is this not valid for the circle geo pt ii)

<APD = 90 (semi circle angle)
<BPC = 90 (semi circle angle)

therefore we have a case where vertically opposite angles are equal. Therefore A,P,C are collinear?

EDIT: Whoops I just assumed B,P and D were collinear. My bad.
 

IR

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Why is this not valid for the circle geo pt ii)

<APD = 90 (semi circle angle)
<BPC = 90 (semi circle angle)

therefore we have a case where vertically opposite angles are equal. Therefore A,P,C are collinear?

EDIT: Whoops I just assumed B,P and D were collinear. My bad.
Dw my friends did the same. Did anyone else use alternate segment theorems and vertically opposite angles and angle in semicircle?
How do you think you went emilios?
 

Ultrafezz

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I felt pretty confident in this exam, Question 11-14 was a breeze, only struggled with the 4 marker.
Stupidly got caught on the circle geometry in q16 and I have no idea why, the solution was simple as anything. Just a bit fried from previous questions I guess. This was my biggest downfall as it wasted my time to get onto the rest of q16.
I'm hoping for 80 raw, more likely 75.
 

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