2012 HSC Q15b (1 Viewer)

braintic

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Regarding 2012 HSC Q15b

I know how to do the whole question.
My question is - what does part (iii) have to do with the remainder of the question?
I don't think I've seen an HSC question before where a part does not link in some way to a later part.
But here, the 'real zero' restriction in part (iii) does not apply to the last three parts. (You could make parts (iv) and (v) very trivial using that restriction, but part (vi) is then unprovable.)
What is part (iii) doing there? Surely the HSC examiners weren't just looking for a 2 mark filler.
 

Carrotsticks

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Not at home atm but I'll give a brief answer for the mean time.

The reason why (iii) is required is so we can prove the result for both cases when the polynomial has real and complex roots, as opposed to (iv), which only proves the case when all roots are complex., which will be used for part (iv), where both the real case (proven in iii) and the complex case must be proved and summarised.
 
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braintic

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Prove what result for both cases? I can see how part (iii) COULD be used to prove the result of part (iv) for the real case, but it just stops dead. Unless you are saying that for 2 whole marks we are meant to also consider the real case when doing part (iv).
 

Carrotsticks

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Prove what result for both cases? I can see how part (iii) COULD be used to prove the result of part (iv) for the real case, but it just stops dead. Unless you are saying that for 2 whole marks we are meant to also consider the real case when doing part (iv).
Part (iii) proves that if 2 roots are complex and 2 roots are real, then they still all have modulus 1. Part (iv) proves that regardless of whether the roots are real or complex, then they all have modulus 1. So we were expected to take cases, which I will explain below.

We have a quartic here (more specifically a reciprocal quartic due to the symmetry of the coefficients, which forces a nice condition upon the roots), which would normally have 3 cases.

Case #1: All 4 roots are real.

Case #2: There exists 2 real roots.

Case #3: There exists no real roots.

Note that we cannot have 3 real roots (so 1 complex) or 1 real root (so 3 complex) since all the coefficients are real, so by extension of the Conjugate Root Theorem, there can only exist an even number of complex roots.

What part (iii) is demonstrating is that IF there does exist a real root, it must exist as a double root, which eliminates Case #1 as above. Furthermore, by inspection of the ONLY possible polynomials (if it has a real zero), we see that all roots have modulus 1.

So yes, I AM saying that for 2 marks, students were expected to also consider the real case.

One mark for the complex case (the working out is very short, no more than 2-3 lines), and another mark for the real case (again, the working out is very short).
 

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