1. ## Re: MX2 Exam Thoughts

I didn't do it but I heard from 2 others that it is (3sqrt2-4)/2

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2. ## Re: MX2 Exam Thoughts

Surely the ratio should be less than (1/2)^1/2, right? Since the smaller line is a segment of the line that is (1/2)^1/2 the size of the larger line. I didn't actually do it so I just guessed 1/3sqrt2.
Nah I said 0.5^1/3, not 0.5^1/2

3. ## Re: MX2 Exam Thoughts

Originally Posted by mrbunton
i think i solved part of the poly question, ii and iii. I lost marks in all segments thou. i couldn't solve i) but ii) u use the substitution (B-Y)^2=a^2+bla/a, in which case x=a^2+bla/a; times by a and sub a^3 into what u would get if u sub a into the original polynomial (0=a^3+pa+q or something) then manipulate. after that it is possible to make a the subject, then sub into P(x) then take sum of roots, i made a mistake and although i got an answer it was different. iii) just say if A=B where A and B are arbitrary roots then the LHS of part ii) becomes 0, also the LHS is allays greater or equal to 0, so condition for distinct roots is the RHS being greater then 0.

last question in 15 was the inequality? I took different cases, that was x>=1, x<=1 (yes repeated = twice but doesn't matter), with x less then 1 you don't have to do much

but with x greater then one u use the other expansion provided in previous part and prove using induction that sigma(k=0, n-1)x^k>=n; where x>=1 so the LHS remains positive.

Overall by taking these two cases i proved it to be true

but no marks on circle geo.

Your answer for part iii for the last question may be invalid, cuz it was asking for distinct and real. I did it by letting alpha be real, since one root has to be real, since its a cubic, then let beta and gamma be conjugate pairs, from real co-efficients, and when I ended with a contradiction, that meant that their all real. you can also prove distinct by assuming there's a double root, subbing it back into the equation, and showing that its not equal to 0, showing that there's no double root and hence all distinct.

4. ## Re: MX2 Exam Thoughts

Originally Posted by Cryptical
Your answer for part iii for the last question may be invalid, cuz it was asking for distinct and real. I did it by letting alpha be real, since one root has to be real, since its a cubic, then let beta and gamma be conjugate pairs, from real co-efficients, and when I ended with a contradiction, that meant that their all real. you can also prove distinct by assuming there's a double root, subbing it back into the equation, and showing that its not equal to 0, showing that there's no double root and hence all distinct.
if u let it be imaginary LHS is less then zero. let "a" be real , B=x+iy, Y=x-iy
( ((a-x)+iy) * ((a-x)-iy) )^2 * (2iy)^2 <0
as 2iy creates negative number so the whole thing becomes negative, first two terms are not negative so it still matches the condition for having real and distinct roots.

but i accounted for that in the exam thou but i forgot to put it there

5. ## Re: MX2 Exam Thoughts

What do you all think the cutoff for a 95 aligned will be?

6. ## Re: MX2 Exam Thoughts

Omggg everyone has already typed so muchhh xP

That was arguably harder than last years, muuuuuuch harder!! OMG Q16 LEGIT GETTING 4/15 MAX D: Only got the induction lmao, fuck the geometry and poly is my WORST TOPIC D:

Overall I reckon like MC = ~6-7, Q11 = 14, Q12 = 14, Q13 = 14, Q14 = ~11-12, Q15 = ~12, Q16 = ~4 so I'm hoping >70+, 75 raw would be <3 <3

I'm glad I could get the conics, mechanics and graphing q's because those are my better topics c:

7. ## Re: MX2 Exam Thoughts

Originally Posted by mrbunton
if u let it be imaginary LHS is less then zero. let "a" be real , B=x+iy, Y=x-iy
( ((a-x)+iy) * ((a-x)-iy) )^2 * (2iy)^2 <0
as 2iy creates negative number so the whole thing becomes negative, first two terms are not negative so it still matches the condition for having real and distinct roots.

but i accounted for that in the exam thou but i forgot to put it there
Yeah sweet did pretty much the same

8. ## Re: MX2 Exam Thoughts

Yea i don't know, l do think it was a harder paper then 2017, but it wasn't one of the harder papers where high 60s scale to E4.

Hopefully E4 is around low to mid 70s.

Anybody agree?

9. ## Re: MX2 Exam Thoughts

Don't worry guys it was definitely harder than last year's paper. Hoping for a mark that can be aligned to 85 (don't think I got the E4 haha). I found this better than the trials though surprisingly. I'm sure EVERYONE on here went great!

10. ## Re: MX2 Exam Thoughts

i think i would of really enjoyed this paper if I didn't have my future at stake.

11. ## Re: MX2 Exam Thoughts

Honestly thought it was okay, up until Q16
I am fairly confident I knew how to do basically everything up to Q16, but sorta ran out of time on a few. Honestly feel like I'll be lucky to make like 3/15 on Q16 , but the rest was fairly okay.
Barring silly mistakes, hopefully ~75 raw...?

Anyone wanna guess at E4 cutoff... Really hoping for low 70's...

12. ## Re: MX2 Exam Thoughts

Originally Posted by JaxsenW
Honestly thought it was okay, up until Q16
I am fairly confident I knew how to do basically everything up to Q16, but sorta ran out of time on a few. Honestly feel like I'll be lucky to make like 3/15 on Q16 , but the rest was fairly okay.
Barring silly mistakes, hopefully ~75 raw...?

Anyone wanna guess at E4 cutoff... Really hoping for low 70's...
I would say mid to high 60s, simply because in 2010 64ish aligned to 90, 2013 around 66 aligned to 90.

13. ## Re: MX2 Exam Thoughts

this year’s exam was pretty brutal:
- last 3 multiple choice were a little tricky and weird, but nothing too crazy
- questions 11-14 were pretty tame, nothing too out there
- questions 15 and 16 started throwing some hard ones out, with only a few easy ones - including that sin8θ expansion which was just a time waster really and the induction, however last parts of questions were killers
- minimising angle POQ’ needed a fair bit of algebra, and recognising the sin2θ has to be maximised to minimise the whole expression
- the inequality part ii : needed to take distinct cases for x less than and greater than one, after using first derivative to analyse behaviour of the second bracket (1+x+x^2...)
- part iii needed one to use the substitution x = a/b, so these two later parts seem very hard to even think of how to approach the question
- question 16 triangle: very confusing to approach but similarity, parallelograms and a touch of algebra, then second part should have easily followed
BUT harder 3u = harder harder 2u??
- question 16 polynomials: slight trick to i., part ii was just algebra, an otherwise approach involved expanding everything and simplifying and part iii. needed to be very careful about what sort of argument to express, ie have to prove real and distinct

a whole bunch of tough questions staggered towards the end sooooo pretty hard test guessing low 70s E4 cutoff? and there’s my rant. hoping for that 90+ raw

14. ## Re: MX2 Exam Thoughts

Originally Posted by JaxsenW
Honestly thought it was okay, up until Q16
I am fairly confident I knew how to do basically everything up to Q16, but sorta ran out of time on a few. Honestly feel like I'll be lucky to make like 3/15 on Q16 , but the rest was fairly okay.
Barring silly mistakes, hopefully ~75 raw...?

Anyone wanna guess at E4 cutoff... Really hoping for low 70's...
I'm sure you dude amazing dude - don't worry the paper was hard. My friends said it may be around 70 - 72 this year.

15. ## Re: MX2 Exam Thoughts

Originally Posted by Drdusk
I would say mid to high 60s, simply because in 2010 64ish aligned to 90, 2013 around 66 aligned to 90.
Really hope so. Fingers crossed for 75 ish without silly mistakes so hopefully I make it aha

16. ## Re: MX2 Exam Thoughts

Didn't do aswell as everyone else here feeling 60-70 raw with slight chance of 70+ but not too sure, anyone want to guess how 65 raw would align?

17. ## Re: MX2 Exam Thoughts

Originally Posted by jryl
Didn't do aswell as everyone else here feeling 60-70 raw with slight chance of 70+ but not too sure, anyone want to guess how 65 raw would align?
dont worry man, i lost 40 marks

18. ## Re: MX2 Exam Thoughts

Originally Posted by mrbunton
i think i solved part of the poly question, ii and iii. I lost marks in all segments thou. i couldn't solve i) but ii) u use the substitution (B-Y)^2=a^2+bla/a, in which case x=a^2+bla/a; times by a and sub a^3 into what u would get if u sub a into the original polynomial (0=a^3+pa+q or something) then manipulate. after that it is possible to make a the subject, then sub into P(x) then take sum of roots, i made a mistake and although i got an answer it was different. iii) just say if A=B where A and B are arbitrary roots then the LHS of part ii) becomes 0, also the LHS is allays greater or equal to 0, so condition for distinct roots is the RHS being greater then 0.

last question in 15 was the inequality? I took different cases, that was x>=1, x<=1 (yes repeated = twice but doesn't matter), with x less then 1 you don't have to do much

but with x greater then one u use the other expansion provided in previous part and prove using induction that sigma(k=0, n-1)x^k>=n; where x>=1 so the LHS remains positive.

Overall by taking these two cases i proved it to be true

but no marks on circle geo.
Could you elaborate on your solution to part (ii) of the polynomials? Btw here is my solution for the part (i):

$x^3 + px + q = 0$
$(\beta - \gamma)^2 = \beta^2 - 2\beta\gamma + \gamma^2$
$= \alpha^2 + \beta^2 + \gamma^2 - \alpha^2 - 2\dfrac{\alpha\beta\gamma}{\alpha}$
$= (\alpha+\beta+\gamma)^2 - 2(\alpha\beta + \alpha\gamma + \beta\gamma) + \dfrac{2q}{\alpha}$
$= -2p + \dfrac{2q}{\alpha} - \alpha^2$
Now,
$\alpha^3 + p\alpha + q = 0$
$p = \dfrac{-q-\alpha^3}{\alpha} = -\alpha^2 - \dfrac{q}{\alpha}$
Sub in...
$= 2\alpha^2 + \dfrac{2q}{\alpha} +\dfrac{2q}{\alpha} - \alpha^2$
$= \alpha^2 + \dfrac{4q}{\alpha}$

19. ## Re: MX2 Exam Thoughts

It felt easier than previous years.. in the earlier parts. It would seem that most questions were either standard or very difficult, with little middle ground.
Several questions posed a considerable conceptual challenge though! Reverse engineering the result typically did the trick.
I'm fairly sure the ratio in the triangle question was (3sqrt(2)/2)-2 btw

Hoping for 95 raw

20. ## Re: MX2 Exam Thoughts

did anyone get (1/3)^n for first part of prob. feels good.

21. ## Re: MX2 Exam Thoughts

Originally Posted by charmingman
did anyone get (1/3)^n for first part of prob. feels good.
Yeah I did

22. ## Re: MX2 Exam Thoughts

Originally Posted by charmingman
did anyone get (1/3)^n for first part of prob. feels good.
Yep sure did

23. ## Re: MX2 Exam Thoughts

Q1-14 were normal but Q16 and end of Q15 were hard

hoping for 88-93 raw

24. ## Re: MX2 Exam Thoughts

How does one find out one's raw mark? Does it come with the scaled mark/contribution thing in mid December ?

25. ## Re: MX2 Exam Thoughts

Originally Posted by Carius
How does one find out one's raw mark? Does it come with the scaled mark/contribution thing in mid December ?
have to buy the raw marks report AND hsc results check (cant buy them separately) for like \$52

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