Maths Extension 2 thoughts (1 Viewer)

[WhoStanLeee]

Not sure if I'm correct but I got (-1)^n.
Its because the product of roots relates to the constant term (independant of x). And since the entire thing went something like x^2n + x^2n-2 + ... + (1-x^2)^n, you only had to consider the last term to get the constant coefficent because all the others were multiplied by some x term.

The x^2n doesn't equal one though... There are more coefficients from the products in the other terms...

 

[lita1000]

Ruse and NSB currently have 11 and 7 100 raws respectively, and I haven't even asked grammar yet.......


Predicted SR cut off will be around 99% raw

 

[Zen2613]

How on earth are people thinking this exam was easy ? This exam was in no way easy, way harder than previous years, except maybe for the early 2000s...

 

[lita1000]

The way you do this using the question:

You want n to go to infinity, this means that there are infinite amount of rows.

So what happens is when you put a black counter in a column, you dont actually change the amount of spaces left in that column since there are an infinite amount of spaces.

So if you want a black counter in every column you just started placing them.

Ways to place first counter? just since there are q columns without a black counter (i.e. all of them)

Second counter? Any column but the one you just chose so

Third counter is and so on.

So the limit is just

I don't know if they will accept Integrand's answer simply because its a fairly easy Q16 and they might want you to use the previous parts. Not sure on this though so don't worry.

Seems like you have a copy of the paper, can u upload?

 

[Drsoccerball]

The x^2n doesn't equal one though... There are more coefficients from the products in the other terms...

No because they were all multiplied by x^2 at least on the outside.

 

[laters]

You had to find the two general solutions and find all the roots and there was only one value with a constant and since the polynomial was even you don't have to put a negative in front.

Not sure if I'm correct but I got (-1)^n.
Its because the product of roots relates to the constant term (independant of x). And since the entire thing went something like x^2n + x^2n-2 + ... + (1-x^2)^n, you only had to consider the last term to get the constant coefficent because all the others were multiplied by some x term.

What I mean was, because all of the terms in the equation in part ii had an x^2n term if you expanded it out???
Like for the x^2n coefficient would be 2nC2 and so on for all the other terms...

and then I added them together and it got real messy real quick

 

[lita1000]

How on earth are people thinking this exam was easy ? This exam was in no way easy, way harder than previous years, except maybe for the early 2000s...

For the top students, there was nothing to distinguish between them (seriously that Q16 was a joke), maybe for high band 5~ low band 6 ers found the middle questions harder than usual?

 

[Drsoccerball]

What I mean was, because all of the terms in the equation in part ii had an x^2n term if you expanded it out???
Like for the x^2n coefficient would be 2nC2 and so on for all the other terms...

Except for the last term which was something like

 

[hawkrider1]

Carrotsticks - when are you posting your solutions? How do you get your hands on the paper now that it's against BOSTES' rules to take the paper out of the exam?

 

[camdaman12]

Not sure if I'm correct but I got (-1)^n.
Its because the product of roots relates to the constant term (independant of x). And since the entire thing went something like x^2n + x^2n-2 + ... + (1-x^2)^n, you only had to consider the last term to get the constant coefficent because all the others were multiplied by some x term.

i got this too, tested with n=4 and it was wrong so i gave up lol

 

[Zen2613]

Yeah but none of them were independent of x, because everything except for the last term was being multiplied by some power of x

 

[Drsoccerball]

Carrotsticks - when are you posting your solutions? How do you get your hands on the paper now that it's against BOSTES' rules to take the paper out of the exam?

Carrotsticks has an exam tomorrow I doubt he can make solutions at the moment.

 

[RealiseNothing]

Ruse and NSB currently have 11 and 7 100 raws respectively, and I haven't even asked grammar yet.......


Predicted SR cut off will be around 99% raw

This happens every year.

In my year (2013) we had like 20 people from those schools claiming 100 raw as well.

Turned out when results came out that only like the top 3 in the state had a 100 HSC mark (which doesnt even mean 100 raw, could be 98 or 99).

They might be fussy with marking and stuff, just wait until results are out is everyone's best bet.

 

[snail489]

Except for the last term which was something like

but you have to divide it by the x^2n coefficient to do product of roots (like, -b/a, c/a, -d/a, etc. - need to divide by the a). And its coefficient was some huge messy shit (it looks like the expression in part iv)

 

[RealiseNothing]

Seems like you have a copy of the paper, can u upload?

Don't have one, I tutor quite a few students so I get the good questions from them (like Q15/16 mostly).

 

[sida1049]

For the top students, there was nothing to distinguish between them (seriously that Q16 was a joke), maybe for high band 5~ low band 6 ers found the middle questions harder than usual?

The likelihood of someone scoring 100/100 in this exam is extremely low, just like with last year's, where the general consensus was that it was relatively easy, and yet no one scored full marks. Top students are distinguished by consistency and prudence, which as exemplified, is extremely rare (more so than being able solve Q16s).

 

[Drsoccerball]

but you have to divide it by the x^2n coefficient to do product of roots (like, -b/a, c/a, -d/a, etc. - need to divide by the a). And its coefficient was some huge messy shit (it looks like the expression in part iv)

lol x^2n isnt a constant?

What you're saying is like "




....

 

[Physicklad]

Did you mention that it was an even polynomial and therefore the last term is the positive of what it is ? I think if they mark strict you may lose a mark.

eyy hoes you gotta remember the leading coefficient wasnt 1 rip

 

[Zen2613]

Did you mention that it was an even polynomial and therefore the last term is the positive of what it is ? I think if they mark strict you may lose a mark.

To be honest, I can't remember if I said the exact word 'even' but I did mentiom something about how there were an odd number of terms, so it was positive. Its over now anyway though so meh.

 

[RealiseNothing]

16)c)iii) is not (-1)^n

Like people have already said, have to divide by co-efficient of x^2n.