Maths Ext 2 Predictions/Thoughts (8 Viewers)

[notme123]

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]

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?

 

[notme123]

on the facebook 2 days before they asked ppl, are you 1st in your school for x2?

i did not see this and i refresh facebook every 2 seconds

wait which fb page

 

[Siwel]

i did not see this and i refresh facebook every 2 seconds

wait which fb page

the 3u and 4u one

 

[notme123]

the 3u and 4u one

whatttt i did not know those existed

 

[Siwel]

whatttt i did not know those existed

they do, but honestly no one uses them

 

[5uckerberg]

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]

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?

 

[AlphaBetaThetaGamma]

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?

I'm really confused at it. Let me see if latex works.
By first glanace no. wat about < .

 

[Trebla]

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]

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]

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.

 

[harrowed2]

Here are the solutions by itute : https://www.itute.com/wp-content/uploads/2021-NSW-ESA-Mathematics-Extension-2-Solutions.pdf

 

[Paradoxica]

Here are the solutions by itute : https://www.itute.com/wp-content/uploads/2021-NSW-ESA-Mathematics-Extension-2-Solutions.pdf

they didn't put an open circle around z=-π, since it's not clear whether the inclusion of the curve from below takes precedence or not (it doesn't)

 

[tywebb]

Terry Lee's solutions: http://advancedmathematics.com.au/Resources/2021 Ext2.pdf

 

[Paradoxica]

Terry Lee's solutions: http://advancedmathematics.com.au/Resources/2021 Ext2.pdf

same omission of detail as above in the locus question

 

[uart]

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?

 

[notme123]

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]

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)

 

[Paradoxica]

itute stated cos@=cospi/10=0.951approx so cannot be sqrt((5-sqrt5)/8)

ah yes the engineer's method