Mathematics Extension 1 Exam Predictions/Thoughts (2 Viewers)

[gunnerfan99]

92

fuck, could have done better. Oh well.

 

[akkjen]

How did you guys do the 14bi and iii?

 

[akkjen]

90*

Wouldn't it be higher than 90? that is 86%

 

[worldno17]

Check out the website rawmarks.info to predict your aligned mark.

 

[Idkwhattoput]

I 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?

 

[AHafza]

Wouldn't it be higher than 90? that is 86%

Oh is it 86 or 81?

 

[AHafza]

I 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?

That’s what I did

 

[englishessayssuck]

How did you guys do the 14bi and iii?

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:
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
another edit: whoops I answered the wrong question this is solution to 14c not 14b soz

 

[nasr7601]

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:
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

hate to be a pain, could you do a written solution of that

 

[Heresy]

How did you guys do the 14bi and iii?

(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...

 

[akkjen]

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:
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

That is 14c not 14b but I needed that anyways so thanks

 

[akkjen]

(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...

for i is said there is only one intercept of the two given graphs, is this okay?
i did same thing for iii

 

[Drdusk]

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'm pretty sure they get counted as an outlier so no NESA won't count their mark iirc.

As for the paper. Question 14c seems a bit too easy? I got it out in about 2-3 minutes. Is that the general consensus?

 

[gunnerfan99]

(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...

I drew up a graph for the first part, second part I showed f(x1) was negative and f(x2) was positive.

 

[Heresy]

for i is said there is only one intercept of the two given graphs, is this okay?
i did same thing for iii

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...

 

[Drdusk]

As for the paper. Question 14c seems a bit too easy? I got it out in about 2-3 minutes. Is that the general consensus?

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...

 

[akkjen]

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...

How do you do 14 ci?
I differentiated and then showed that they are the same point so gradient is equal or something like tht

 

[Drdusk]

How do you do 14 ci?
I differentiated and then showed that they are the same point so gradient is equal or something like tht

Yeah that, exactly a one liner lol

 

[jeception177]

Wait so k=5 is wrong?
if so would I lose 1 mark on that or more?

 

[Drdusk]

What mark do u guys reckon 55/70 will scale up to ?

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.