Maths Extension 2 Predictions/Thoughts (2 Viewers)

thomas mcnamee

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I’d say that’s probably a high band e3 in this paper which is the equivalent of a ≈99.5 ATAR (depends a lot on how high though because the marks tend to fall off hard in the band 5 range)
what you reckon about 84-88
 

carrotsss

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what do u predict min e4 raw is?
it’s hard to judge impartially myself (personally I found it a bit harder than 2022 but likely due to the exam pressure) but based on what everyone is saying I’d estimate mid-high 60s
 

kms1234

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last year scaling was quite insane. I believe a 65 went up too a 91. The general consensus from what I've heard is that this exam was as hard or at least similar to last year. Thus, I would say that 65 most likely scales up to an e4. Also, you might have forgotten carrott, but what aligned do you think a 73 would be considering that last year 76 went to a 94
(all stats are from the rawmarks database)
 

carrotsss

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last year scaling was quite insane. I believe a 65 went up too a 91. The general consensus from what I've heard is that this exam was as hard or at least similar to last year. Thus, I would say that 65 most likely scales up to an e4. Also, you might have forgotten carrott, but what aligned do you think a 73 would be considering that last year 76 went to a 94
(all stats are from the rawmarks database)
As I said it’s quite hard for me impartially judge the difficulty given that I say the past hsc exams in more laid back (though still timed) conditions, but I think that as a paper it was slightly easier than 2022, hence the mid-high 60s cutoff estimation. Hence, a 73 should be 91-93 (though I’d lean more towards the lower side of that)
 

kms1234

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Both fizzysoda and mok's solutions used different methods to get to the answer. Would the hsc cater to the variety of different methods used. I am a but worried that my answer may deviate from the worked solutions
 

carrotsss

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Both fizzysoda and mok's solutions used different methods to get to the answer. Would the hsc cater to the variety of different methods used. I am a but worried that my answer may deviate from the worked solutions
That’s fine as long as you used a valid method
 

tywebb

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Both fizzysoda and mok's solutions used different methods to get to the answer. Would the hsc cater to the variety of different methods used. I am a but worried that my answer may deviate from the worked solutions
This is extension 2 right? Multiple valid methods are plentiful in extension 2, maybe less so in the Standard courses.

So there are no surprises that there be variations in methods for extension 2 solutions.
 

notme123

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another solution to question 16c:

xz + yw has to be an obtuse rotation away from z which is a region spanned by the complex numbers iz and -z. in x and y, this is somewhere in the third quadrant which would be easier to show with a diagram but i cbb just take it as fact.

let a and b be positive real numbers.

the first boundary of this region is that zx+yw = -az where a is positive real --> x + y*w/z = -a
Taking imaginary parts, y*Im(w/z) = 0 --> y = 0
Taking real parts x + y*Re(w/z) = -a --> x = -a, meaning that x < 0 since a is positive real.
Hence one of the boundaries of the region is the line y = 0, x < 0 or the x axis on the left of the cartesian plane'
Another way to do this step is looking at zx+yw = -az, z and w are linearly independent vectors so you can instantly go to x = -a and w = 0.

the second boundary of the region is zx+yw = -biz where b is positive real --> x + y w/z = -bi
Taking real parts --> x + y*Re(w/z) = 0 --> y = -x/Re(w/z) --> |y| = |-x/Re(w/z)| --> -y = -x/|Re(w/z)| --> y = x/|Re(w/z)| where x < 0, y < 0

The complete solution is the region bounded by S = { (x,y) \in R^2 : y < 0, x < 0, y > x/|Re(w/z)|}
 

Hudz777

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me being rank 1 of 1 💀 (failed the exam tho anyways...)
aye same, absolutely bummed it. Even stuffed up the partial fractions… idk how, I wrote 2ln2 - 3ln3 instead of 2ln3 - 3ln3. I’m actually done.
 

sneak11

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does anyone know how the 4marks in the partial fraction integration in q11 might have been allocated - like what would u need to do to get each mark?
 

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