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March 29, 2024, 09:41:49 pm

Author Topic: Compilation of solutions to HARD past HSC papers (4U)  (Read 22267 times)  Share 

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RuiAce

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Re: Compilation of solutions to HARD past HSC papers (4U)
« Reply #15 on: October 22, 2017, 02:36:55 pm »
+3


Looking back, I remember having an annoying time thinking about this question as well. The thing is, if we had only the one equation \(\sin x = \sin (-\alpha)\) then we would actually have another case.

The whole point of the other equation was to eliminate the wrong case. When I think about it now, jumping straight to \( \theta = -\alpha\) whilst not rejecting the other case is a mistake. But it is certainly true that once you've rejected it, the only viable case is \(\theta = -\alpha\)

frog1944

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Re: Compilation of solutions to HARD past HSC papers (4U)
« Reply #16 on: October 22, 2017, 05:41:04 pm »
0
Oh, ok. That makes sense. Great! Thanks RuiAce :)

RuiAce

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Re: Compilation of solutions to HARD past HSC papers (4U)
« Reply #17 on: October 22, 2017, 05:44:32 pm »
+1
Pretty glad you asked that one though, because I finally resolved a confusion I had last year now because of it aha

RuiAce

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Re: Compilation of solutions to HARD past HSC papers (4U)
« Reply #18 on: December 13, 2017, 03:31:17 pm »
+2


_________________________________________________________________




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This is where things get hard. It is not obvious at all that exhaustion is what we require to complete our proof. We will need to do some working backwards to figure out why.


Working backwards


And that's that case sorted.

So the question now reduces to how do we prove that, GIVEN we've already eliminated the possibility of \( |aq+bp| = 0\)?

As it turns out, we rely on what \(a,b,p,q\) all are.



_________________________________________________________________


Note: The statement \(e > 0\) is certainly safe to assume.





Remember that \(r=e\) for our purposes.
In part iii), we have claimed that for any choice of integers \(a\) and \(b\), we must have \( |a+be| = 0 \) or \( |a+be| \ge \frac1q \). What we've just done, is shown that the integers \(a_{3q}\) and \(b_{3q}\) don't satisfy either of the above. This is because the above inequality can be dissected into \(|a+be| \neq 0\) AND \( |a+be| < \frac{1}{q} \).

(Remark: This is my preferred way of doing this question. There is another alternate method very similar to this, but I find it hard to explain without using intuition beyond the MX2 student's capability. I will remark that the proof's conclusion is a bit nicer though - instead of contradiction part iii, we contradict the elementary statement \(\text{a number cannot be smaller and greater than another number at the same time}\))
« Last Edit: December 13, 2017, 08:34:40 pm by RuiAce »

RuiAce

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Re: Compilation of solutions to HARD past HSC papers (4U)
« Reply #19 on: September 28, 2018, 05:32:39 pm »
+2


Can of course, still be done with the usual distance formula if you wish.
_______________________________________________________________

This part more or less requires a lot of angle-chasing. It's not immediately obvious that \( \angle ACP = \alpha - \beta\), but firstly the presence of \(AC\) and \(AP\) should've hinted at narrowing our focus to \( \triangle ACP\), and also the weird expressions should've hinted at trigonometry.





(I didn't feel like adding labels because it would look complicated, but apologies in that I wasn't bothered to copy out the diagram either.)

_______________________________________________________________


It may be worth noting that \(CP\) does depend on the position of \(P\) due to the problem of \(AP\). It's just that upon multiplying, we can take advantage of the fact that \( AP\times PB\) is equal to the constant value of \(b^2\). Also, the reason behind why \( \beta\) is constant is because it's uniquely determined by the asymptotes, which are also determined by the hyperbola itself.
_______________________________________________________________



_______________________________________________________________

One may be tempted to jump straight into bashing it (finding the equation of the tangent, finding any intercepts and explicitly using the midpoint formula). But given that it was only a 1 mark question, it had to relate to part iv) somehow.


Note that this is equivalent to saying that the points \(P\) and \(Q\) must coincide...

« Last Edit: September 28, 2018, 05:37:41 pm by RuiAce »

RuiAce

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Re: Compilation of solutions to HARD past HSC papers (4U)
« Reply #20 on: November 21, 2018, 01:35:21 pm »
+3
I have completed the solutions to the 2020 sample paper uploaded by NESA recently to the new syllabus resources. This was obviously unnecessary as they already had solutions at the end of the paper. It is only here for the sake of completeness.

(The images zoom in heaps if you tap on them.)

Multiple Choice


Short Answers


All NESA resources are accessible here for the new MX2. Links to the other maths syllabus materials are also available. Note that the formula sheet has also been updated and whilst less well formatted, it covers a larger range of material.

Note that the sample paper is not a full paper. Only 30 marks of short response questions have been provided, however you will be given 90 marks worth of those in the actual exam.

Note that HSC-2019 students do not need to worry about this post. However there are some questions that are still relevant to your current syllabus, if you want that extra practice.
« Last Edit: November 21, 2018, 01:39:41 pm by RuiAce »

RuiAce

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Re: Compilation of solutions to HARD past HSC papers (4U)
« Reply #21 on: December 18, 2019, 11:30:02 pm »
0
I had a look at my solution to 2001 Q7b and I'm kinda disappointed by it. It looks like i went down a needlessly convoluted path somewhere. I'm swapping out for the following set of handwritten solutions.
Click me