Mathematics Extension 2 Predictions (1 Viewer)

[tgone]

As in draw a triangle to prove Am-Gm for 2? Got scammed. Normally all the geometrical proofs I see are all functions and areas under the curve tbh

yeah same with all the ones I see- considering that 'hard' questions typically like to combine syllabus points ... but no, it's literally just 'hey you remember pythagoreas... yeah, now draw it'

 

[okkk]

yeah same with all the ones I see- considering that 'hard' questions typically like to combine syllabus points ... but no, it's literally just 'hey you remember pythagoreas... yeah, now draw it'

View attachment 36614

If I open up Q16 on Thursday and they give us a munted inequality and write prove geometrically I'm gonna ask for a refund

 

[tgone]

another inequality q from our trial

 

[tgone]

If I open up Q16 on Thursday and they give us a munted inequality and write prove geometrically I'm gonna ask for a refund

lmao literally... i'll just like draw them a nice triangle and leave the question

 

[okkk]

View attachment 36615

another inequality q from our trial

This one is not too bad actually, I can do this one. The one that I've attached is the inequality that threw off most people in our yr in our trials and it's what I thought geometric inequalities meant lmao

 

[tgone]

This one is not too bad actually, I can do this one. The one that I've attached is the inequality that threw off most people in our yr in our trials and it's what I thought geometric inequalities meant lmao

yeah - it's not too bad but under exam conditions the notation looks a little scary I guess, especially in like q14-15 (I forget)

the other two that threw people off in ours was part (ii) of this (part i is free)


and also the later parts of this

 

[okkk]

yeah - it's not too bad but under exam conditions the notation looks a little scary I guess, especially in like q14-15 (I forget)

the other two that threw people off in ours was part (ii) of this (part i is free)
View attachment 36617

and also the later parts of this

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Ok complex is chill but that vectors one is weird cos in my mind, aren't there an infinite amount of values for R that trace an annulus about the sphere?

 

[Trebla]

View attachment 36615

another inequality q from our trial

Check out Q6a) from 2005 HSC…

 

[tgone]

Check out Q6a) from 2005 HSC…

whoa, very nice- very similar sum as well -time do to all the 200X HSC papers hahah

 

[d1zzyohs]

Real talk.
if you are cis(theta) gang.
if you are exponential gang.
if you are Cartesian form gang.

cis(theta) is the only right answer, and the only one that is correct.
why?
because cis(theta) is easier to read, graphically easier to understand, and has very nice properties of cos and sins.
exponential is too small and you can misread the angle. cringe.

This one is not too bad actually, I can do this one. The one that I've attached is the inequality that threw off most people in our yr in our trials and it's what I thought geometric inequalities meant lmao

cssa. i didn't get that one . was actually pretty straight forward after you graphed everything with hypothetical constants.

 

[tgone]

Real talk.
if you are cis(theta) gang.
if you are exponential gang.
if you are Cartesian form gang.

cis(theta) is the only right answer, and the only one that is correct.
why?
because cis(theta) is easier to read, graphically easier to understand, and has very nice properties of cos and sins.
exponential is too small and you can misread the angle. cringe.

cssa. i didn't get that one . was actually pretty straight forward after you graphed everything with hypothetical constants.

exponential form gang

and yeah- I had a go at the q just then and after graphing it becomes clear that |cbrta-cbrt(a-b)| will have a greater magnitude than |cbrt(a)-cbrt(a+b)| due to the concavity? not sure if that’s the right path, but it seems to work;

|cbrta-cbrt(a-b)|>|cbrta-cbrt(a+b)|
cbrta-cbrt(a-b) > cbrt(a+b)-cbrta
2cbrta>cbrt(a+b)+cbrt(a-b)

 

[okkk]

exponential form gang

and yeah- I had a go at the q just then and after graphing it becomes clear that |cbrta-cbrt(a-b)| will have a greater magnitude than |cbrt(a)-cbrt(a+b)| due to the concavity? not sure if that’s the right path, but it seems to work;

|cbrta-cbrt(a-b)|>|cbrta-cbrt(a+b)|
cbrta-cbrt(a-b) > cbrt(a+b)-cbrta
2cbrta>cbrt(a+b)+cbrt(a-b)

Yeah, I think this does work, I mean in the exam I did it funkily by finding trapezium areas. The suggested probably makes the must intuitive sense with concavity honestly

 

[tgone]

Yeah, I think this does work, I mean in the exam I did it funkily by finding trapezium areas. The suggested probably makes the must intuitive sense with concavity honestly

yeah that suggested does look like it makes the most sense; I don’t know if I would have thought of it in an exam, my way just seemed the most obvious to me since the concavity implies it increases at a decreasing rate… but suggested sltn is probably more rigorous than my method— although, it uses the same assumption essentially so I think mine would still work

 

[tgone]

Yeah, I think this does work, I mean in the exam I did it funkily by finding trapezium areas. The suggested probably makes the must intuitive sense with concavity honestly

how did you do it with trapeziums though, I’m curious?

 

[ExtremelyBoredUser]

View attachment 36615

another inequality q from our trial

Oh ye i remember doing that question in the caringbah paper, it was a cool method to use for the n = k+1 part as opposed to the regular integration way

 

[okkk]

how did you do it with trapeziums though, I’m curious?

It's the same assumption but I said that the area of two trapeziums would be greater than just the one trapezium cos it's concave down so that mid point on the graph will be above the line drawn for one trapezium. It was the only thing that came into my mind during the exam, kinda funny how I basically did the suggested solution but made it unnecessarily complicated with areas but I guess in an exam you take what you can get

 

[tgone]

It's the same assumption but I said that the area of two trapeziums would be greater than just the one trapezium cos it's concave down so that mid point on the graph will be above the line drawn for one trapezium. It was the only thing that came into my mind during the exam, kinda funny how I basically did the suggested solution but made it unnecessarily complicated with areas but I guess in an exam you take what you can get

lmao i have no idea how you got to that but not the suggested thing… then again i’ve done some stupidly complicated stuff in exams that isn’t required so I can’t really talk

 

[okkk]

lmao i have no idea how you got to that but not the suggested thing… then again i’ve done some stupidly complicated stuff in exams that isn’t required so I can’t really talk

Well, I'm sure we'll activate our big brains for the nesa overlords . Not much prep I can do today so I'm just gonna lay back for a bit, good luck for tomorrow : )

 

[josh312]

I would much prefer a mechanics q16 than an inequalities one

Mathimatical induction questions with inequalities will be my downfall

 

[notme123]

Real talk.
cis(theta) is the only right answer, and the only one that is correct.
why?
because cis(theta) is easier to read, graphically easier to understand, and has very nice properties of cos and sins.
exponential is too small and you can misread the angle. cringe.

if you ever do maths at uni they will burn you alive if you use cis form. cis is very much a high school thing. i think the ib uses cis more religiously than hsc. i agree its easier to read and putting pi in exponent can lead to clutter tho but exponential form is more universally used.

also clarifying: I mean if you use cis(pi/3), or something which is sacrilig. the eulars formula is fine, e^ix = cosx + isinx, because its like the most important formula in complex