Maths Ext 2 Predictions/Thoughts (1 Viewer)

notme123

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My criticism stands.

But I want to downplay it. It's not a great drama.

He went through quite a bit of hefty algebra and I'm glad he was the one doing it, not me.

Overall I think he did a great job.
how does one even get chosen to do this.
1. go to a prestigious school which will pay the smh to do this article
2. become ranked first in x2
3. profit
 

Siwel

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how does one even get chosen to do this.
1. go to a prestigious school which will pay the smh to do this article
2. become ranked first in x2
3. profit
on the facebook 2 days before they asked ppl, are you 1st in your school for x2?
 

5uckerberg

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ah yes the classic article where boomers question why we still do things like this because it has no value to society. a pattern ive noticed is that top schools always get the opp to do this.
You took the words right out of my mouth.
 

s97127

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For this question, i want to prove 5^n -2^n - 3^n > 0 for n>= 2

(2+3)^n - 2^n - 3^n = 2^(n-1).3 + 2^(n-2).3^2 +....+ 3^(n-1) > 0 (QED)

Is this proof correct?
 

Trebla

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View attachment 33882
For this question, i want to prove 5^n -2^n - 3^n > 0 for n>= 2

(2+3)^n - 2^n - 3^n = 2^(n-1).3 + 2^(n-2).3^2 +....+ 3^(n-1) > 0 (QED)

Is this proof correct?
Almost...except you forgot the binomial coefficients.

Also, generally speaking, proving something is positive to prove it is non-zero doesn’t always work (it works in this case though). You can have expressions that can be positive or negative but is non-zero.
 

Eagle Mum

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Correct I got 100 for x1 but i got nowhere close to a state rank.
I imagine that by scaling extension students up, it’s quite compacted at the top, so amongst the cohort with a scaled mark of 100 would be a range of raw marks, but anyone who achieves a raw mark of 100 deserves to be equal first. Students in different schools who achieve a perfect score in the exam shouldn’t be ranked based on school assessments, since assessments aren’t standardised. Congratulations on your awesome achievement!
 

s97127

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Almost...except you forgot the binomial coefficients.

Also, generally speaking, proving something is positive to prove it is non-zero doesn’t always work (it works in this case though). You can have expressions that can be positive or negative but is non-zero.
Thanks. I forgot about coefficients. Regarding the other comment, it only works if all the terms are either greater/lesser than 0.
 

uart

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same omission of detail as above in the locus question
Also, neither the Itute nor TL solns provided a satisfactory explanation for why the negative solution was invalid in Q14c(ii).


Itute gave no reasoning for selection of the +ive inner surd, while TL reasoned that cos^2() must be positive (but clearly the [] solution is also positive).

My reasoning is that where we subst,
,
in order to get the polynomial, we could equally well have substituted
,
to get the same equation.

So the polynomial is equally valid for solving for, , as well as some other non acute angles (which clearly can be ignored).

So the two acute solutions,

correspond to solutions for cos(pi/10) and cos(3pi/10), and since cos() is a decreasing function on (0,pi), then the larger value must correspond to cos(pi/10).

Did anyone else use this reasoning?
 
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notme123

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Also, neither the Itute nor TL solns provided a satisfactory explanation for why the negative solution was invalid in Q14c(ii).


Itute gave no reasoning for selection of the +ive inner surd, while TL reasoned that cos^2() must be positive (but clearly the [] solution is also positive).

My reasoning is that where we subst,
,
in order to get the polynomial, we could equally well have substituted
,
to get the same equation.

So the polynomial is equally valid for solving for, , as well as some other non acute angles (which clearly can be ignored).

So the two acute solutions,

correspond to solutions for cos(pi/10) and cos(3pi/10), and since cos() is a decreasing function on (0,pi), then the larger value must correspond to cos(pi/10).

Did anyone else use this reasoning?
i didnt justify it at all i think whoops
 

jyu

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Also, neither the Itute nor TL solns provided a satisfactory explanation for why the negative solution was invalid in Q14c(ii).


Itute gave no reasoning for selection of the +ive inner surd, while TL reasoned that cos^2() must be positive (but clearly the [] solution is also positive).

My reasoning is that where we subst,
,
in order to get the polynomial, we could equally well have substituted
,
to get the same equation.

So the polynomial is equally valid for solving for, , as well as some other non acute angles (which clearly can be ignored).

So the two acute solutions,

correspond to solutions for cos(pi/10) and cos(3pi/10), and since cos() is a decreasing function on (0,pi), then the larger value must correspond to cos(pi/10).

Did anyone else use this reasoning?
itute stated cos@=cospi/10=0.951approx so cannot be sqrt((5-sqrt5)/8)
 

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