Mathematics Extension 1 Exam Predictions/Thoughts (1 Viewer)

worldno17

Active Member
Joined
Oct 24, 2015
Messages
126
Gender
Female
HSC
N/A
Check out the website rawmarks.info to predict your aligned mark.
 

Idkwhattoput

Member
Joined
Mar 14, 2018
Messages
93
Gender
Male
HSC
2019
Uni Grad
2024
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

Member
Joined
Feb 21, 2018
Messages
64
Gender
Male
HSC
2019
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

New Member
Joined
Mar 8, 2016
Messages
19
Gender
Male
HSC
2019
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
 
Last edited:

nasr7601

New Member
Joined
Mar 8, 2019
Messages
10
Gender
Male
HSC
2019
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

Active Member
Joined
Nov 13, 2017
Messages
146
Gender
Male
HSC
2019
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

Active Member
Joined
Sep 27, 2019
Messages
140
Gender
Male
HSC
N/A
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

Active Member
Joined
Sep 27, 2019
Messages
140
Gender
Male
HSC
N/A
(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

Moderator
Moderator
Joined
Feb 24, 2017
Messages
2,025
Location
a VM
Gender
Male
HSC
2018
Uni Grad
2023
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

Member
Joined
Oct 26, 2019
Messages
47
Gender
Male
HSC
2019
(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

Active Member
Joined
Nov 13, 2017
Messages
146
Gender
Male
HSC
2019
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

Moderator
Moderator
Joined
Feb 24, 2017
Messages
2,025
Location
a VM
Gender
Male
HSC
2018
Uni Grad
2023
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...
 
Last edited:

akkjen

Active Member
Joined
Sep 27, 2019
Messages
140
Gender
Male
HSC
N/A
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

Moderator
Moderator
Joined
Feb 24, 2017
Messages
2,025
Location
a VM
Gender
Male
HSC
2018
Uni Grad
2023
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
 

Drdusk

Moderator
Moderator
Joined
Feb 24, 2017
Messages
2,025
Location
a VM
Gender
Male
HSC
2018
Uni Grad
2023
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.
 

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

Top