Maths Extension 2 Predictions/Thoughts (1 Viewer)

thomas mcnamee

New Member
Joined
May 21, 2023
Messages
26
Gender
Male
HSC
2023
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

New Member
Joined
May 7, 2022
Messages
4,752
Gender
Male
HSC
2023
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

New Member
Joined
Oct 18, 2023
Messages
13
Gender
Male
HSC
2023
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

New Member
Joined
May 7, 2022
Messages
4,752
Gender
Male
HSC
2023
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

New Member
Joined
Oct 18, 2023
Messages
13
Gender
Male
HSC
2023
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

New Member
Joined
May 7, 2022
Messages
4,752
Gender
Male
HSC
2023
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

dangerman
Joined
Dec 7, 2003
Messages
1,735
Gender
Undisclosed
HSC
N/A
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

Well-Known Member
Joined
Apr 14, 2020
Messages
1,002
Gender
Male
HSC
2021
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

Member
Joined
Jul 29, 2023
Messages
84
Gender
Male
HSC
2023
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

New Member
Joined
Nov 19, 2021
Messages
10
Gender
Male
HSC
2023
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?
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top