Looking for solutions (1 Viewer)

deswa1 · Jul 7, 2012

The Matrix said:
"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.

The Matrix · Jul 7, 2012

deswa1 said:
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.

Trebla · Jul 7, 2012

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

deswa1 · Jul 7, 2012

Carrotsticks said:
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.

Carrotsticks · Jul 7, 2012

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)

deswa1 · Jul 8, 2012

Carrotsticks said:
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)

Oh right

. Personally, I hate expanding big things because I always lose terms somewhere but that's just me. BTW, I just did the paper- there are some very nice questions in there (though the mark allocation for some questions is a bit deceiving- some are way easier or harder than their marks suggest).

nightweaver066 · Jul 8, 2012

deswa1 said:
Oh right . Personally, I hate expanding big things because I always lose terms somewhere but that's just me. BTW, I just did the paper- there are some very nice questions in there (though the mark allocation for some questions is a bit deceiving- some are way easier or harder than their marks suggest).

That question 8 circle geometry worth 5 marks was surprisingly easy.

asianese · Jul 8, 2012

Yeah I've tried the paper again with a clearer mind - some good questions. Thanks for the info guys

deswa1 · Jul 8, 2012

nightweaver066 said:
That question 8 circle geometry worth 5 marks was surprisingly easy.

Yep- that was easy marks and the question 7 (I think) mechanics was a straightforward 6 marks.

Nooblet94 · Jul 8, 2012

I had a look through the paper and then did question 8. I found it surprisingly easy in comparison to the rest of the paper. It was as easy, if not easier than some of the earlier questions IMO.

Trebla said:
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

I'm pretty sure your proof for (iii) is circular. You're assuming the result from (ii) is true, but in part (ii) you assume what you're proving in (iii).

Also, for finding the expression for g(x) in part (ii) can't you just replace the n's in $B_{n}(1+x)-B_{n}(x)=nx^{n-1}$ with n+1 rather than having to go through the whole process of differentiating and then integrating? You get the same result.

RealiseNothing · Jul 8, 2012

Did anyone else notice question 3 only has 14 marks?

deswa1 · Jul 8, 2012

RealiseNothing said:
Did anyone else notice question 3 only has 14 marks?

Nope- nice observation. How'd you notice that though?

deswa1 · Jul 8, 2012

medicore said:
I couldn't do that induction one, question 8. part iii D:

Could you do the other parts?

medicore · Jul 8, 2012

I tried them, I don't know where the answer are.

deswa1 · Jul 8, 2012

medicore said:
I tried them, I don't know where the answer are.

Yeah I don't have answers either. If you post your answer for specific questions though, I can check if I got the same thing (not saying I'm always right though).

RealiseNothing · Jul 8, 2012

deswa1 said:
Nope- nice observation. How'd you notice that though?

Not sure actually lol.

seanieg89 · Jul 8, 2012

medicore said:
I tried them, I don't know where the answer are.

happy to do a couple of questions if you need something to compare with. not going to write full solutions to everything though.

deswa1 · Jul 9, 2012

seanieg89 said:
happy to do a couple of questions if you need something to compare with. not going to write full solutions to everything though.

Yeah, could you (or someone else) please do Q3b. I did it but I'm not sure if I interpreted the question correctly or if what I did was valid.

Thanks

Carrotsticks · Jul 9, 2012

Here's something for you to start with because I presume it's part (ii) and (iii) you have trouble with.

jamie300 · Jul 9, 2012

this one is not the hardest MX2 paper, I have a more crzay one

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