gunnerfan99
Member
- Joined
- Oct 26, 2019
- Messages
- 48
- Gender
- Male
- HSC
- 2019
fuck, could have done better. Oh well.
fuck, could have done better. Oh well.
Wouldn't it be higher than 90? that is 86%
Oh is it 86 or 81?Wouldn't it be higher than 90? that is 86%
That’s what I didI gave a worded solution for 14biii (the x1<a<x2 question). I am very sure it is valid, but i'm not sure the markers accept that sort of thing. Will I still get full marks if I proved it graphically? Will the marker even bother trying to interpret what I was saying?
Basically, you realise that i) and ii) look quite similar, so you're tempted to take a "cos inverse" of i) to get some relationship of x_0 and x_0-a. However, obviously you can't do this without justification, but you can rigorise it as follows:How did you guys do the 14bi and iii?
hate to be a pain, could you do a written solution of thatBasically, you realise that i) and ii) look quite similar, so you're tempted to take a "cos inverse" of i) to get some relationship of x_0 and x_0-a. However, obviously you can't do this without justification, but you can rigorise it as follows:
u have cos(A)-cos(B)=-2sin(A+B /2)sin(A-B /2), and so if cos(a)=cos(b) then one of A-B or A+B must be an integer multiple of 2pi.
so now you plug in A for x_0 and B for x_0-a, and you get that either a or 2x_0-a is a multiple of 2pi. The first cannot work since a is strictly between 0 and pi (or was it pi/2?), and you can similarly bound 2x_0-a to be strictly between -2pi and 2pi. This means the multiple is 0, so you in fact get 2x_0-a=0, from which it's quite easy to figure out both ii) and iii).
edit: fixed some pronumerals to not reuse stuff
(i) Discriminant: b^2 - 4ac = 0 means one real solution.How did you guys do the 14bi and iii?
That is 14c not 14b but I needed that anyways so thanksBasically, you realise that i) and ii) look quite similar, so you're tempted to take a "cos inverse" of i) to get some relationship of x_0 and x_0-a. However, obviously you can't do this without justification, but you can rigorise it as follows:
u have cos(A)-cos(B)=-2sin(A+B /2)sin(A-B /2), and so if cos(a)=cos(b) then one of A-B or A+B must be an integer multiple of 2pi.
so now you plug in A for x_0 and B for x_0-a, and you get that either a or 2x_0-a is a multiple of 2pi. The first cannot work since a is strictly between 0 and pi (or was it pi/2?), and you can similarly bound 2x_0-a to be strictly between -2pi and 2pi. This means the multiple is 0, so you in fact get 2x_0-a=0, from which it's quite easy to figure out both ii) and iii).
edit: fixed some pronumerals to not reuse stuff
for i is said there is only one intercept of the two given graphs, is this okay?(i) Discriminant
(ii) Then used the fact that x_1 had a different sign to x_2 to prove the inequality.
However, I doubt that this is the correct way to go about it - we'll see I guess...
I'm pretty sure they get counted as an outlier so no NESA won't count their mark iirc.I have a question that someone may have an answer. One of my peers did not turn up for the MX1 paper today. I hope he/she is okay but I need to know, how does this work? Will he/she be counted when they moderate the internal assessment mark? We don’t have a strong team/cohort like yourselves so many of us are ‘riding on this’ to push our ATAR up. Any insight? Thanking you ahead if you have any thoughts posted.
I drew up a graph for the first part, second part I showed f(x1) was negative and f(x2) was positive.(i) Discriminant: b^2 - 4ac = 0 means one real solution.
(ii) Then used the fact that x_1 had a different sign to x_2 to prove the inequality.
However, I doubt that this is the correct way to go about it - we'll see I guess...
Should be. I think I said something about y=x^2 having one real root and then because y=1/x-k has no real roots, if you simultaneously solve, the result should have one real root. Idk if this is true though...for i is said there is only one intercept of the two given graphs, is this okay?
i did same thing for iii
Like for e.g.14c(i) is quite trivial. Then for 14c(ii) you just draw a right angled triangle with (x_0 - alpha) as one of the angles usingAs for the paper. Question 14c seems a bit too easy? I got it out in about 2-3 minutes. Is that the general consensus?
How do you do 14 ci?Like for e.g.14c(i) is quite trivial. Then for 14c(ii) you just draw a right angled triangle with (x_0 - alpha) as one of the angles using
Then just argue about which case to take. Literally everything done in 3/4 lines...
Yeah that, exactly a one liner lolHow do you do 14 ci?
I differentiated and then showed that they are the same point so gradient is equal or something like tht
That's about a borderline E4 however since aligning varies year to year it might just be E4 or might just not be E4, but still very close.What mark do u guys reckon 55/70 will scale up to ?