Advanced Maths Predictions/Thoughts (20 Viewers)

[rirish61]

i guessed the answer right but i dont know how to go about it. i found a pattern where like the difference between the numbers increases by the powers of 2 but i dont know how to do it properly

not sure if this is the right way but i'm starting by splitting the thing into two series to hopefully make it easier

 

[rirish61]

did it on paper so I can't upload it but I found the sum of the 400n part by taking the 400 out and using the sum of an AP (a=1, l=10, n=10), then the 2^n part is just the sum of a GP where a=2, r=2 and n=10

so it becomes (400x55) - 2046

 

[bigupsanky]

did it on paper so I can't upload it but I found the sum of the 400n part by taking the 400 out and using the sum of an AP (a=1, l=10, n=10), then the 2^n part is just the sum of a GP where a=2, r=2 and n=10

so it becomes (400x55) - 2046

oh yeah i see what u did thanks man

 

[Clai]

they will never say "don't do more working out" besides "Do not prove" which is just there so you don't waste your time

it was more like "without further calculus". most of the time, ik where to go from there but there are times where an alternative way is kinda obvious so i was just asking for that. but i think i got the hang of it now. thanks!

nvm i was too late dzeeshr addressed it lol

 

[Clai]

apparently this paper is gonna be heavy on calculus and statistical analysis, with financial math and trig being the next 2 popular topics. Always count on an applied trig question, a present/future value question, and a normal loan repayment type financial math question. I think they can ask more questions like, "compare the scatter plots/box and whisker plots" like they did in 2024, which threw a lot of people off so be careful about that. calculus is calm not that much to do there, and same with normal distribution related things asw

ok so ik im late but like stat heavy?? that sucks god i hate stats. but specifically, what stats

 

[Lex Orwell]

does anyone have any idea how to do this one

 

[alphxreturns]

View attachment 49706does anyone have any idea how to do this one

ah i ddi this yesterday

first since the gradient is positive (observed), a is below 2π (if it was at 2π it would be zero, and after 2π it would be negative
so a < 2π
1/2π <1/a

(eliminating A)

cos(a) = ma
as cos(a) must be a bit less than 1
ma < 1
m < 1/a

(eliminating C)

m = rise/run (or cos(a) = ma)
m = cos(a)/a
cos(a) is not yet 1, a is not yet 2π

m = cos(2π-b)/2π-b (where b is some small value)
cos(2π-b) is practically 1
1/2π-b is greater than 1/2π
so m > 1/2π

hence B

this in all honesty prob isnt the way to do it but thats how i worked it out LOL

 

[coolcat6778]

hi guys
i have a random kinda pedantic question
when using the trapezoidal rule or estimates in general, can we use squiggly equal sign instead of the dotted equal sign?
I always use the squiggly one and I'm used to that one, it's on the reference sheet for trapezoidal but all HSC answers use the dotted one.
I remember my math teacher saying somethign about how the squiggly ones aren't reccomended in HSC. Does it matter?
View attachment 49698

Lol. the retarded division equal sign comes from the cambridge textbook

The squiggly line is the STANDARD notation globally.

 

[coolcat6778]

Here, hopefully it makes sense:

View attachment 49702

These questions are pretty much always avoided in the HSC for math advanced. After all, this is just testing algebra manipulation and other useless bullcrap. Just look at the 2024 last question to see what an actual question looks like.

 

[Clai]

View attachment 49706does anyone have any idea how to do this one

is the answer D? my reasoning (not sure if its right):
y = mx passes through (a, cosa) and (0,0).
From this, m (gradient) can be found as cosa/a. In this case, we know cosa <1 since a is a little less than 2 pi (inferred from the graph and using a unit circle). This means cosa/a must be < 1/a (dividing both sides by a). Therefore, m < 1/a (eliminates C).

Now, if we compare 1/a and 1/2pi, a seems to be a little to the left of 2pi (its not quite at the peak), meaning 1/a > 1/2pi (i think since when the denominator is bigger at the bottom, the fraction tends to be smaller). This eliminates A.

I then eliminated last option by substituting 11pi/6, which gave 1/2pi > m.

Thus, this gives m < 1/2pi < 1/a (D)

tbh i got a little confused towards the end so i wont be surprised if im wrong

oh nvm alphxreturns got to it first. itd be great if u could explain where my reasoning went wrong lol

 

[splatcat]

View attachment 49706does anyone have any idea how to do this one

You can find m to be -sin(a) from differentiating cos and then use the point gradient formula and the fact that the line goes through the origin to find y-cosa = m(x-a)
y=mx-ma+cosa
which means that ma=cosa
-asina=cosa
1/a =-tana
It can be seen that m>0 which implies -sina > 0 which means a is in the 3rd or 4th quadrant
further, since a is positive, -tana > 0 which means that a is in the 4th quadrant in combination with the above. It is also known that a != 2pi as the gradient at that point doesn't make such a tangent possible.
so we get
3pi/2 < a < 2pi
which implies
2/3pi > 1/a > 1/2pi which rules out option A

From before we have ma =cosa
m = cosa/a
since a != 2pi, cosa < 1 but positive, so cosa/a = m < 1/a

This narrows it down to B and D, so we need to determine if m < 1/2pi
It can be seen that for m = cosa/a to be greater than 1/2pi, a needs to be very close to 2pi, which doesn't seem likely given the graph, so I would put D

EDIT:
Using online calculators to solve for a where 1/a = -tana in the domain yields a value slightly greater than 6.1, meaning that B is actually the correct answer, this is a little tricky and I'll give it a bit more thought about proving that cosa/a > 1/2pi

 

[99.95dreams]

explaining math solutions through typing must be one of the most commendable things i've seen in a while

 

[Socialism]

it might be difficult to learn, but this site does have laTex compatibilty!!


i think you can convert your regular text to latex with conversion websites online?

 

[coolcat6778]

Symbolab, type stuff in https://www.symbolab.com/ then just copy and paste the textbox, you get the latex code

 

[splatcat]

So if we consider x, where x = 2pi - a, this gives 0<x<pi/2
From before tana=-1/a
tan(2pi -x) = -1/a
-tanx = -1/a
tanx = 1/a
Since x>0 and x is in the first quadrant, tanx > x
subbing in from prior
1/a > x
since a > 3pi/2
1/a < 2/3pi
1/pi>2/3pi > 1/a > x
x < 1/pi

Using the useful inequality, 1-cos(x) <= (x^2)/2 (This can technically be proven with calculus learnt in advanced)
from before we know that x^2/2 < x/2pi by changing an x to 1/pi
1-cosx < x/2pi
1-x/2pi < cosx
Subbing in x = 2pi - a
1-(2pi-a)/2pi < cos(2pi -a)
1-1 + a/2pi < cosa
a/2pi < cosa
1/2pi < cosa/a =m

 

[vigstar]

View attachment 49706does anyone have any idea how to do this one

View attachment 49706does anyone have any idea how to do this one

 

[bigupsanky]

ok so ik im late but like stat heavy?? that sucks god i hate stats. but specifically, what stats

i think mainly normal distribution and other probability stuff, they cant ask too much on the scatter plots

 

[bigupsanky]

These questions are pretty much always avoided in the HSC for math advanced. After all, this is just testing algebra manipulation and other useless bullcrap. Just look at the 2024 last question to see what an actual question looks like.

cuz that was one of the goofiest questions ever

 

[Lex Orwell]

ah i ddi this yesterday

first since the gradient is positive (observed), a is below 2π (if it was at 2π it would be zero, and after 2π it would be negative
so a < 2π
1/2π <1/a

(eliminating A)

cos(a) = ma
as cos(a) must be a bit less than 1
ma < 1
m < 1/a

(eliminating C)

m = rise/run (or cos(a) = ma)
m = cos(a)/a
cos(a) is not yet 1, a is not yet 2π

m = cos(2π-b)/2π-b (where b is some small value)
cos(2π-b) is practically 1
1/2π-b is greater than 1/2π
so m > 1/2π

hence B

this in all honesty prob isnt the way to do it but thats how i worked it out LOL

Thank u

 

[Lex Orwell]

ah i ddi this yesterday

first since the gradient is positive (observed), a is below 2π (if it was at 2π it would be zero, and after 2π it would be negative
so a < 2π
1/2π <1/a

(eliminating A)

cos(a) = ma
as cos(a) must be a bit less than 1
ma < 1
m < 1/a

(eliminating C)

m = rise/run (or cos(a) = ma)
m = cos(a)/a
cos(a) is not yet 1, a is not yet 2π

m = cos(2π-b)/2π-b (where b is some small value)
cos(2π-b) is practically 1
1/2π-b is greater than 1/2π
so m > 1/2π

hence B

this in all honesty prob isnt the way to do it but thats how i worked it out LOL

I get the reasoning, just couldn't do it in an exam, the rest of 2021 multies where fine, credit to you mate, a solution is a solution