2026 HSC CHAT (10 Viewers)

[Study to success]

These were the types of questions we did in class bc ‘they’re too easy’

My class never did these sort of ones. Or maybe we did and I just forgot but their not too bad

 

[jane1820]

My class never did these sort of ones. Or maybe we did and I just forgot but their not too bad

Its just logic: ‘applying previous knowledge to new concepts’ (quoting my teacher)

 

[cristal]

Ok. I’ll prob practice smth like this towards prelim or next year

definitely don't have a fat prelim and HSC folder
definitely don't have 2 more

 

[melanie_o]

HSC '25er infiltrating your GOATED chat as 2025 chat is lacking, and is lowkey making y12 so anticlimactic .

 

[Study to success]

View attachment 47382
definitely don't have a fat prelim and HSC folder
definitely don't have 2 more

What do u have on there

 

[cristal]

What do u have on there

I have a lot of math prelims, stuff off THSC and ace ( so wouldn't count that) and a lot of things I've dug up from BoS.
For just math I don't have as much advanced stuff but I have some revision things I believe idk

 

[cristal]

Are there are topic stuff you want I can check
Can also try generating a paper from past HSC questions

 

[NotBamboo]

@Shavi Masee
got it with the help of AI cus i forget about turning point silly me

AND CHECKING WITH DESMOS DRUMROLL PLEASE


@cheesynooby
can u teach me how to use the element of symbol for domain and range thanks

 

[Study to success]

Are there are topic stuff you want I can check
Can also try generating a paper from past HSC questions

Thanks. Maybe some stuff on linear, non linear and polynomials cause I have a topic test on that first week back and maybe some stuff for term 2 like calculus. Also we should all put some more stuff on the drive cause it is kinda empty

 

[NotBamboo]

domain:
1. solve the quadratic and identify its concavity to find the x values where it is negative e.g. if it has x intercepts at x = -2 and 2, and is concave up the quadratic is negative between x = -2 and 2. since u cant square root a negative and cant divide by 0, that gives u the domain (x < -2, x > 2)

range:
2. find the range of the quadratic by finding the y coord of the vertex. e.g. vertex at (1, 3) and concave down = range: y ≤ 3
3. square root that (keeping in mind u can't square root a negative)-> 0 ≤ y ≤ sqrt3
4. find the reciprocal of that (a graph of y = 1/x could be useful - imagine the range of 1/x if you could only plug in values between 0 and sqrt3 (from step 3) -> y ≥ 1/sqrt3

silly she needed help with the ii)

also can u show me a faster way of doing it, refer to previous post
if this in my exam its gonna use a buncha time

 

[cristal]

@Shavi Masee

View attachment 47383

Wtf is this

 

[NotBamboo]

Wtf is this

my working?
i even had to clean it up and add lil comments for u guys
ur welcome

i can explain any part of it if u dont understand

 

[cristal]

my working?
i even had to clean it up and add lil comments for u guys
ur welcome

i can explain any part of it if u dont understand

I don't understand anything
I'll work through it first myself just to understand
But why differentiation btw?

 

[Shavi Masee]

Can't thank you guys rn I'm boiling in the heattt. Will get to you soon
Thank you bamboo (ive got ur rock!!!)and cheesynooby, and everyone elsee

 

[Study to success]

Can't thank you guys rn I'm boiling in the heattt. Will get to you soon
Thank you bamboo (ive got ur rock!!!)and cheesynooby, and everyone elsee

Are u sitting in the sun

 

[cheesynooby]

@cheesynooby
can u teach me how to use the element of symbol for domain and range thanks

wut
like this?

I don't understand anything
I'll work through it first myself just to understand
But why differentiation btw?

yeah why differentiation

 

[NotBamboo]

I don't understand anything
I'll work through it first myself just to understand
But why differentiation btw?

At one point the gradient is 0
When u differentiate it, the y value of it will be 0 because original function's gradient is 0, making the x easy to find.
This is called a critical point
Usually, u would use the second derivative to find the nature of the turning point (being a max)
But here since I already know a vague shape of the graph, I know that this will be the local maximum, so I only need to use first derivative
now that i have the x value by subbing in y=0, i can plug it into the original function to get the y value of the point as well yayyyyy!!!!
THE END.

 

[Shavi Masee]

Are u sitting in the sun

Ya

 

[NotBamboo]

View attachment 47388
Ya

enjoyyyyyy
we can discuss this later
that beach is relatively empty ngl

 

[cristal]

View attachment 47388
Ya

Omg the water looks rlly nice


@NotBamboo I still don't understand
Idk everything much abt derivatives yet and all the first and second stuff and I don't know what's a local maximum is (do you mean like the turning point)

Also it's only like chapter 3 right you wouldn't really be using that yet right