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Maths Ext 2 Predictions/Thoughts (2 Viewers)

ExtremelyBoredUser

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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.
 

ohnose

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

Paradoxica

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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.
 

Paradoxica

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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

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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

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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?
 

Paradoxica

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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.
 

Alistruggles

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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)
 

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