MX2 Marathon (1 Viewer)

altSwift

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Re: HSC 2018 MX2 Marathon

Ah I remember seeing telescopic stuff in an old MX2 marathon as a hard Q7 or Q8

Are there any resources in particular to practice Q7/Q8 level questions apart from past hsc papers?
 

altSwift

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Re: HSC 2018 MX2 Marathon

I'm not sure what you did on this line, I'm guessing its in regard to the fact that Im(1) = 0



Also, did you mean:

 

InteGrand

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Re: HSC 2018 MX2 Marathon

Ah I remember seeing telescopic stuff in an old MX2 marathon as a hard Q7 or Q8

Are there any resources in particular to practice Q7/Q8 level questions apart from past hsc papers?
One thing (probably not the only thing) you can try is searching up past years' HSC 4U marathons here on BOS (including the Advanced Level ones, though these are typically harder than HSC style Q's or would be broken into multiple steps if asked in the HSC exam).
 

altSwift

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Re: HSC 2018 MX2 Marathon

That does not follow. You can verify it for yourself by converting the conjugates to reciprocals and clearing them.
Sorry if I'm missing something obvious, but just to clarify





When you take the imaginary part of the sum of GP formula, does that instantly disregard the z^0 (=1) in the original series becuase Im(1) = 0? It seems weird because you use a = 1 for the formula, maybe I'm thinking about this too much haha
 

InteGrand

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Re: HSC 2018 MX2 Marathon

Sorry if I'm missing something obvious, but just to clarify





When you take the imaginary part of the sum of GP formula, does that instantly disregard the z^0 (=1) in the original series becuase Im(1) = 0? It seems weird because you use a = 1 for the formula, maybe I'm thinking about this too much haha








 
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fan96

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Re: HSC 2018 MX2 Marathon

Is the answer ?

(I intend to post a writeup later)
 

fan96

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Re: HSC 2018 MX2 Marathon



Converting to polar form,



Since ,



The cosine function is even, so the first term on the LHS and RHS are equal. Cancelling,



Since the sine function is odd, for all .



And clearly this is only true when .

So therefore,
 
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altSwift

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Re: HSC 2018 MX2 Marathon



Converting to polar form,



Since ,



The cosine function is even, so the first term on the LHS and RHS are equal. Cancelling,



Since the sine function is odd, for all .



And clearly this is only true when .

So therefore,
Nice!











 

mrbunton

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Re: HSC 2018 MX2 Marathon

2u integration+circle geo
i made this question myself in paint so dont mind the bad production quality.
edit: look for the question which has the circle with one edge on -5 instead of -3. ignore the question where the circle touches -3.
integral composite graph.png
 

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fan96

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Re: HSC 2018 MX2 Marathon

i) define the graph as



Let the lower bound be such that .

Now,



So for the integral to be zero,



Rearranging,



Using the quadratic formula,



Since ,

So



(One other solution is also )
_______________________________________________

ii) The area of the semicircle is , and

The graph doesn't seem to be defined for .

Clearly , so if one of the bounds is , the only bound that will make the integral zero is .
 

mrbunton

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Re: HSC 2018 MX2 Marathon

2^2*pi/2 = 2pi >6 therefore ii) is possible. remember it has radius 2 not 1
also u got the gradient of the second line for i) incorrect, getting the gradient wrong also affected the y-intercept.

noticing k=5 was impressive although.
 
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fan96

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Re: HSC 2018 MX2 Marathon

2^2*pi/2 = 2pi >6 therefore ii) is possible. remember it has radius 2 not 1
It's a semicircle with diameter from -3 to -1 isn't it? So the diameter has length 2 and radius 1.

also u got the gradient of the second line for i) incorrect, getting the gradient wrong also affected the y-intercept.
Oops. The line is .

So the new equation is



And the quadratic is now which gives .
 

mrbunton

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Re: HSC 2018 MX2 Marathon

rip the circle was meant to have radius 2; i changed it now so the circle touches -5 to -1.
edit: i keep writing questions incorrectly; i check next time btw.
 
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