BOS Trials 2013 Extension 2 Discussion Thread (2 Viewers)

RealiseNothing

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c) iii) This is just an integration by parts, I won't go through most of the working out as it is just straight forward IBP. However:



Let and thus (given)

Let and thus

Put these into the IBP formula and evaluate the , then factor out to obtain the result of:



c)iv) Now we know that , hence we can deduce that:



Evaluating the upper and lower bounds gives:



Now we know that so we evaluate using the limits and we get the required result:



c)v) As we get .

Now as we established above, is a finite as it is bounded between two finite values.

Thus

c)vi) We know now that as we get and so from part c)ii) we get:



Also we have:



Substituting in our known value for the sum of the reciprocal odd squares, and factorising out a quarter from the last summation gives:





 
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Sy123

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b)

Let so and and



Now the first part is 0 and so we get:

from



c)i) Let

From part a)ii) we get:





Substituting in the limits for the first part gives the required result of:



c)ii) Simply let



But note that



Divide everything by and take out the minus sign from the summation yields the result:

I think there is a mistake here, when you do:



The 'summand' value isn't correct, it should be:



Then, when you sum the series out, all the even k cancel out, which leaves us with sum from k=1 to n 1/(2k-1)^2

So Makematics there isn't a typo there.
 

RealiseNothing

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I think there is a mistake here, when you do:



The 'summand' value isn't correct, it should be:



Then, when you sum the series out, all the even k cancel out, which leaves us with sum from k=1 to n 1/(2k-1)^2

So Makematics there isn't a typo there.
Yep it isn't a typo (the top of the summation should be 'n', my bad I typed it up wrong).

I did what you are saying in the actual exam yesterday, but when I was talking to carrot about it afterwards he said just substituting it in like that will work.

Edit: Now that I think about it, maybe carrot was saying the exact same thing we both did, I just misinterpreted what he said. I'll fix it now anyway.
 

Carrotsticks

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Realise, just to point out a small thing. You said that since it is bounded by two finite values, it is finite. The sine function is a basic counter example.
 

RealiseNothing

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Realise, just to point out a small thing. You said that since it is bounded by two finite values, it is finite. The sine function is a basic counter example.
What would be a better way to explain why it is finite then?
 

RealiseNothing

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Realise, just to point out a small thing. You said that since it is bounded by two finite values, it is finite. The sine function is a basic counter example.
Oh right, it is still finite, just not a finite value. It's just a problem with the wording right.
 

seanieg89

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If anyone needs solutions to a couple of particular questions before Carrot and Trebla release their official solutions, I am willing to write mine up.
 
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I am really interested in seeing the actual HSC paper this year after this bomb of a paper.
 

cutemouse

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I don't think Q11(c) is really in the syllabus because it involves mappings (although it can be done by arguments/circle geom as well). I think, in fact, the syllabus explicitly excludes these types of locus questions.
 

Carrotsticks

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I don't think Q11(c) is really in the syllabus because it involves mappings (although it can be done by arguments/circle geom as well). I think, in fact, the syllabus explicitly excludes these types of locus questions.
Not quite sure what you mean by involving mappings. Mappings (rotational, scaling etc) are most certainly in the syllabus. The problem does indeed require the substitution w=1/z, but students need not understand the concept of a mapping. They just need to know that |w|=|w-k| yields a straight line, which is most certainly in the syllabus as a standard locus problem. Once w=1/z is made (which is a fairly intuitive substitution to make), the rest is just complex number algebra and manipulations.

The following is taken from the syllabus:

Capture.JPG

That is very clearly a mapping of the unit circle (which is a curve, notice in my exam that we are merely mapping points, which is def. in the syllabus) via a Mobius transform.

tl;dr the syllabus 'forbids' mappings of curves -> curves, but it allows mappings points -> points, which is what Q11(c) was.
 

cutemouse

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Not quite sure what you mean by involving mappings.
I meant complex maps.

I still think there's serious question as to whether it is in the syllabus or not.

Anyway, isn't mapping curves -> curves the same thing as mapping points -> points? I.e. In the former case aren't we mapping the points of a curve to another point of a curve etc? Oo
 

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