1. ## Looking for solutions

Does anyone have the solutions for this paper: http://4unitmaths.com/2001moriah.pdf ??? It is extremely hard

2. ## Re: Looking for solutions

WTF killer paper. I'm up to the inequality with factorials. No idea...The direction doesn't work if you try to sub the equations into each other

3. ## Re: Looking for solutions

Probably the hardest MX2 paper out there.

I'll ask Sean to help me with it when he starts tutoring me harder 4 unit stuff.
I finished all of 4 unit and I think I understand all the content, all I need to work on from now till the HSC is harder 4 unit stuff like questions 7 and 8 (mechanics, harder 3 unit, etc...), hope Sean can help me develop my skills at tackling those badass questions in the HSC. Because I head what differentiate a top and avarage student is questions 7 and 8!

Do you think I should do HSC past papers or trial past papers, I have like 25 HSC once from 1985, I've done like 7 of them only I need to do at least 2 papers per day but I'm stuck with this paper, it is extremely hard

4. ## Re: Looking for solutions

If you're stuck just keep moving! Come back to it later.

Don't neglect your other subjects though...doing well in worse-scaling subjects + not amazing at 4u is better than amazing at 4u + crap at worse-scaling. If that made any sense.

5. ## Re: Looking for solutions

Ok honestly that's the most titf paper. Up to the transformations...They expect you to expand $\frac{1}{(y-2)^5}$ ok no. And the 5 mark $\phi (x)$ WTF. I'm taking a break.

6. ## Re: Looking for solutions

Originally Posted by asianese
If you're stuck just keep moving! Come back to it later.

Don't neglect your other subjects though...doing well in worse-scaling subjects + not amazing at 4u is better than amazing at 4u + crap at worse-scaling. If that made any sense.
I divided my holidays into 3 parts, 4 days for MX2, 4 days for chemistry and 4 days for physics. I finished MX2 revision 2 days ago, I did 2 chapters per day and about 25 hard questions for each that cover each chapter. I memorised the first 2 modules of chemistry yesterday and today so I'm doing the remaining 2 modules tomorrow and the day after, then spend the remaining 4 days of the holidays doing the 4 modules of physics. I don't have to study for biology because I don't want it to count for my HSC because it sucks. Also English is easy and 3 unit is a joke so I finished it like 3 months ago.
I'm pretty confident with all my subjects so I want to work more on extension 2. My goal from now till the HSC is to develop my skills at solving questions 7 and 8 by doing as many of them as possible so I think I'll get Sean as a tutor to help me achieving that, hope he replies to my PM.
I really wanna be faster at 4 unit, I'll try to do 2 HSC MX2 papers per day so I can hopefully be eventually able to finish a paper in less than 3 hours. I wanna try and do challenge questions from terry lee and Cambridge as well, the challenge questions are similar to questions 7 and 8 so that could help.
I'm not sure whether I should do only HSC past papers or do trials from other schools but trials usually don't have solution

7. ## Re: Looking for solutions

Originally Posted by asianese
Ok honestly that's the most titf paper. Up to the transformations...They expect you to expand $\frac{1}{(y-2)^5}$ ok no. And the 5 mark $\phi (x)$ WTF. I'm taking a break.
where is carrotsticks to help us

8. ## Re: Looking for solutions

The earlier HSC's don't have solutions either. The HSC that was for our syllabus was in 1981 IIRC so you can start there. I have done some of them - they are no where as titf as this moriah one... - less titf, harder lol

9. ## Re: Looking for solutions

Bernoulli Polynomials
WTF ?!?!?

10. ## Re: Looking for solutions

Originally Posted by The Matrix
I divided my holidays into 3 parts, 4 days for MX2, 4 days for chemistry and 4 days for physics.
LOL where is the space for english? :P

11. ## Re: Looking for solutions

Actually Bernoulli Polynomials featured in Sydney Grammar paper - almost the same question but tweaked a little. 1999 SGS - Moriah probably took it from there.

12. ## Re: Looking for solutions

Originally Posted by asianese
Actually Bernoulli Polynomials featured in Sydney Grammar paper - almost the same question but tweaked a little. 1999 SGS - Moriah probably took it from there.
Isn't it like a series of polynomials where they are derivatives/primitives of each other?

13. ## Re: Looking for solutions

Originally Posted by karnbmx
LOL where is the space for english? :P
English and biology are rote learning, no need to study for them, I can memorise everything the day before the test :P

14. ## Re: Looking for solutions

The question isn't particularly difficult, but it is a bit tedious.

For simplicity sake, I will call the polynomial f(x) instead of phi(x).

We are given that f(x) is divisible by x^3, meaning it can be expressed in the form f(x) = x^3 ( ax^2 + bx + c )

We are also given that g(x) = f(x) - 1 is divisible by (x-1)^3, so we can say that g(x) = (x-1)^3 (dx^2 + ex + f)

So we have two equations essentially:

$\\ f(x) = x^3 (ax^2 + bx + c) \qquad (1) \\\\ f(x) - 1 = (x-1)^3 (dx^2 + ex + f) \\\\ f(x) = (x-1)^3 (dx^2 + ex + f) +1 \qquad (2)$

Express both (1) and (2) as polynomials of degree 5 by expanding and factorising etc, then equate their coefficients of x^5, x^4, ... , x, then constants.

Then a whole bunch of very basic simultaneous equations will lead you to finding a, b and c, thus finding the polynomial f(x).

15. ## Re: Looking for solutions

For part Q3c) since x and y are positive integers and given
x - 1 > y
then
(x - 1)! > y!
Hence
(x - 1)!(x - 1) > y(y!)
x(x - 1)! - (x - 1)! > ((y + 1) - 1)y!
x! - (x - 1)! > (y + 1)! - y!
x! + y! > (x - 1)! + (y + 1)!

16. ## Re: Looking for solutions

Originally Posted by asianese
WTF killer paper. I'm up to the inequality with factorials. No idea...The direction doesn't work if you try to sub the equations into each other
Not sure if you ended up solving this but I think this is how you do it:

$x! + y! > (x-1)! + (y+1)!$

Divide through by (x-1)!, since we know it must be positive and thus the sign doesn't change:

$x + \frac{y!}{(x-1)!} > 1 + \frac{(y+1)!}{(x-1)!}$

Now here is where we use the condition they have given:

$x-y>1$

Re-arranging, we can say:

$x - 1 > y$

Now since they are both positive integers, we can say that the minimum difference of 'x' and 'y' is 2. And hence, if we take this minimum difference:

$x - 1 = y + 1$

Now on the RHS of our inequality from before, we can say that for a minimum difference between x and y, the following is true:

$\frac{(y+1)!}{(x-1)!} = 1$

Substituting this into the RHS makes the RHS = 2.

So we have:

$x + \frac{y!}{(x-1)!} > 1 + 1$

But using the condition:

$x - y > 1$

And the fact that x and y are both positive integers, we can once again use the minimum value of y to be 1, so that:

$x - 1 > 1$

$x > 2$

Substituting this minimum into our inequality gives:

$2 + \frac{y!}{(x-1)!} > 1 + 1$

Which holds true if the remaining fraction is greater than 0, which trivially it is given both the numerator and denominator are positive.

So we have proven it true for the minimum difference of x and y, and trivially it must be true for all x and y given that they are positive integers. Actually I might justify why:

We know that the minimum value of x is 2 (shown above), so we can say that the inequality is (without assuming a minimum difference):

$2 + \frac{y!}{(x-1)} > 1 + \frac{(y+1)!}{(x-1)!}$

Now the fraction in the RHS has a maximum value of 1, which is when there is a minimum difference between x and y. So the RHS has a maximum value of 2. But the LHS has a value of 2 already due to the minimum value of x being 2. And if we add on the fraction in the LHS, then the LHS will exceed the RHS and hence the inequality holds.

17. ## Re: Looking for solutions

Originally Posted by Trebla
For part Q3c) since x and y are positive integers and given
x - 1 > y
then
(x - 1)! > y!
Hence
(x - 1)!(x - 1) > y(y!)
x(x - 1)! - (x - 1)! > ((y + 1) - 1)y!
x! - (x - 1)! > (y + 1)! - y!
x! + y! > (x - 1)! + (y + 1)!
Or you could do this lol.

18. ## Re: Looking for solutions

Haha poor Realise. I noticed people said 'titf'. What does 'titf' mean?

19. ## Re: Looking for solutions

Originally Posted by Carrotsticks
Haha poor Realise. I noticed people said 'titf'. What does 'titf' mean?
"Taking It Too Far", makes sense, hehe...
I wanna have a go at this paper when I'm done with chemistry!
Or I think I should do HSC past papers for now since this one doesn't seem to be as relevant to the HSC...

20. ## Re: Looking for solutions

Originally Posted by Trebla
For part Q3c) since x and y are positive integers and given
x - 1 > y
then
(x - 1)! > y!
Hence
(x - 1)!(x - 1) > y(y!)
x(x - 1)! - (x - 1)! > ((y + 1) - 1)y!
x! - (x - 1)! > (y + 1)! - y!
x! + y! > (x - 1)! + (y + 1)!
http://home.fuse.net/wolfonenet/FTP/...athematics.pdf is this the technique you used? Looks like they use it in English too, hehe

21. ## Re: Looking for solutions

Originally Posted by The Matrix
"Taking It Too Far", makes sense, hehe...
I wanna have a go at this paper when I'm done with chemistry!
Or I think I should do HSC past papers for now since this one doesn't seem to be as relevant to the HSC...
Do these papers- the harder papers are the ones that really test you and as you seem to make 1500 threads about needing helping with Q7's and 8's- I'm telling you do HARD papers.

22. ## Re: Looking for solutions

Originally Posted by deswa1
Do these papers- the harder papers are the ones that really test you and as you seem to make 1500 threads about needing helping with Q7's and 8's- I'm telling you do HARD papers.
1500 ?!?!? I have no thread that is explicitly asking for help regarding questions 7 - 8.
Ok, I'll do this one once I'm done with the HSC papers that I'm doing at the moment.

23. ## Re: Looking for solutions

For the Bernoulli Polynomials part don't be intimidated by the jargon and notation. It might be a little unusual and unfamiliar but it can be solved with the tools you already have.

(i) From condition 2

$B'_1(x) = 1\times B_0(x)\\\\\Rightarrow B_1(x) = x + c$

But from condition 3

$\displaystyle\int_0^1 B_1(x)\,dx = 0\\\\\Rightarrow \left[\dfrac{x^2}{2}+cx\right]_0^1 = 0\\\\\Rightarrow c = -\dfrac{1}{2}$

(ii) From condition 2

$g'(x) = B'_{n+1}(x+1) - B'_{n+1}(x)\\\\\Rightarrow g'(x) = (n + 1)(B_n(x+1) - B_n(x))\\\\\Rightarrow g'(x) = (n + 1)nx^{n-1}$

Hence

$g(x) = (n+1)x^n + c$

To find c is a little tricky, first note that from condition 2 (after replacing n with n + 1 and integrating both sides from 0 to 1) and then applying condition 3

$\displaystyle\int_0^1 B'_{n+1}(x)\,dx = (n+1)\displaystyle\int_0^1 B_n(x)\,dx = 0\\\\\Rightarrow B_{n+1}(1) - B_{n+1}(0) = 0\\\\\Rightarrow g(0) = 0$

from the definition of g(x)

We can now evaluate c which is zero, hence

$g(x) = (n+1)x^n$

(iii) The first case for n = 1 is obvious using part (i)

Assume that for an arbitary n

$B_n(x+1)-B_n(x) = nx^{n-1}$

Now consider

$B_{n+1}(x+1) - B_{n+1}(x)\\\\ = g(x) \\\\= (n+1)x^n$

Note that the use of the assumption in the inductive step was already proved in part (ii) so I've sort of shortcutted it, hence the result holds by induction since it holds for n = 1

24. ## Re: Looking for solutions

Originally Posted by Carrotsticks
The question isn't particularly difficult, but it is a bit tedious.

For simplicity sake, I will call the polynomial f(x) instead of phi(x).

We are given that f(x) is divisible by x^3, meaning it can be expressed in the form f(x) = x^3 ( ax^2 + bx + c )

We are also given that g(x) = f(x) - 1 is divisible by (x-1)^3, so we can say that g(x) = (x-1)^3 (dx^2 + ex + f)

So we have two equations essentially:

$\\ f(x) = x^3 (ax^2 + bx + c) \qquad (1) \\\\ f(x) - 1 = (x-1)^3 (dx^2 + ex + f) \\\\ f(x) = (x-1)^3 (dx^2 + ex + f) +1 \qquad (2)$

Express both (1) and (2) as polynomials of degree 5 by expanding and factorising etc, then equate their coefficients of x^5, x^4, ... , x, then constants.

Then a whole bunch of very basic simultaneous equations will lead you to finding a, b and c, thus finding the polynomial f(x).
The way I did it is nicer I think:

We know that f(x) is of the form ax^5+bx^4+cx^3=0 (as x^3 is a root).
Define g(x) to be the other function -> g(x)=ax^5+bx^4+cx^3-1
Now we know that (x-1) is a triple root -> Differentiate twice and you end up with three equations. Sub in x=1 to each of them (because x-1 is a factor) and you get three simulataneous with three variables (a,b,c). Solve these.

25. ## Re: Looking for solutions

That works too! Just tried it out of curiosity to see how long it was, ended up getting about a page, same as the other. Most of the working out was from solving the 3 simultaneous equations. My one had a lot more expansions but the simultaneous equations when equating were really nice (like 1-f = 0 etc)

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