# Maths Ext 2 Predictions/Thoughts (1 Viewer)

#### vishnay

##### God
far out other than q16 that exam was such a silly mistakes exam

#### ExtremelyBoredUser

##### Bored Uni Student
thx to the dude who provided the copy of the paper. kinda thrown off by this m/c a bit. Ended up picking C in the end just felt like it looks like a regular projectile motion graph.....probs got it wrong tho, anyone got any ideas?View attachment 33774

I just thought abt centripetal force, which would have to be to the centre, and acceleration would have to be in the same direction since its moving in a circular motion.

#### Trebla

How did you guys do 16a?
Cheated a bit with using the 'otherwise' option, but still acceptable lol

#### notme123

##### Well-Known Member
Cheated a bit with using the 'otherwise' option, but still acceptable lol

View attachment 33775
part i you can just do |xi +yj + zk|=1<=|xi|+|yj|+|zk| = |x|+|y|+|z| since you can multiply lengths of the unit vectors i j and k which are 1

#### ohnose

##### New Member
What would a raw mark of 73 or 74 ish scale to roughly?

#### AndyDandy

##### New Member
how much raw for 80 scaled u reckon

##### -insert title here-
How then?
Suppose T(2n) = H(m) for some integer m.

2n(2n+1)/2 = m(2m-1)

2n²+n = 2m²-m

m+n = 2m²-2n² = 2(m-n)(m+n)

Since m and n are positive integers, we can divide.

1 = 2(m-n)

This is impossible, as integers cannot have a difference of ½.

Therefore, the even triangular numbers are never equal to any hexagonal numbers.

#### AndyDandy

##### New Member
What would a raw mark of 73 or 74 ish scale to roughly?
85+
edit: probably close to 90 or low

##### -insert title here-
Here is the region for which Re(z) ≥ Arg(z)

#### DrugLord69

##### Member
Nah mate i reckon i'm gonna ace tf out the test

#### AndyDandy

##### New Member
Nah mate i reckon i'm gonna ace tf out the test
u contributed to the god-tier scaling, only thing u aced

##### -insert title here-
Meme Solution for 15. (d).

If the result were true, this would be a non-trivial solution to Fermat's Last Theorem for every integer n>2.

Hence, the result must be false for all n>2.

The result is clearly false for n=2 by manually checking.

QED

#### vishnay

##### God
Meme Solution for 15. (d).

If the result were true, this would be a non-trivial solution to Fermat's Last Theorem for every integer n>2.

Hence, the result must be false for all n>2.

The result is clearly false for n=2 by manually checking.

QED
i did that in the exam

#### CM_Tutor

##### Moderator
Moderator
Meme Solution for 15. (d).

If the result were true, this would be a non-trivial solution to Fermat's Last Theorem for every integer n>2.

Hence, the result must be false for all n>2.

The result is clearly false for n=2 by manually checking.

QED
I posted a version of that in the discussion in the MX2 forum... I wonder how markers will evaluate anything like it?

#### zakoht

##### New Member
also how did you do the triangle and hexagon numbers
did you figure this one out? i also got lost with this lol and ran outta time to look at it properly

#### notme123

##### -insert title here-
I WAS ABOUT TO DO THIS IS THIS THE CORRECT ANSWER??? IF IT IS IM GOING TO BE HANGING FROM THE CHANDELIER
I can't say. I know that for proving the irrationality of the nth root of 2, the proof of fermat's last theorem does in fact require some elementary irrationality results (including this very thing) so it would be circular reasoning. This would need the consultation of an expert in algebraic number theory to resolve.

##### -insert title here-
i think a bit too much is sketched in there
specifying the branches of the function is a pain so i'm leaving that as an exercise to the reader

#### JesusChriiii

##### Member
what you guys reckon a raw of 70-75 will align to?

#### Alistruggles

##### Member
did you figure this one out? i also got lost with this lol and ran outta time to look at it properly
i'm not the person u quoted and idk if this would be accepted but i noted that they want u to prove for every odd number and an odd number can be represented by 2n - 1 so....subbing 2n - 1 into the Triangle equation actually gets u the hexagonal equation, therefore hexagonal number??? seems way too simple lol (lmk if you think this would get any marks)