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HSC 2016 MX2 Marathon (archive) (1 Viewer)

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leehuan

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Re: HSC 2016 4U Marathon

are you sure it can even have a maximum value? Its a min parabola ?
If I try straightforwardly using the triangle inequality directly I get a complex modulus. This is mathematically invalid.

There has to be a corollary involved.
 

Paradoxica

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Re: HSC 2016 4U Marathon

Max occurs when equal



, we can do this because |z|>0, and |z|=0 is an irrelevant solution.

letting a = |z| just coz.

Finding the derivative

When
Since |z|>0, second derivative = a which is positive; hence max T.P. at |z|=1/2
Try again. The answer is not a rational number.
 
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braintic

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Re: HSC 2016 4U Marathon

I've done this without the triangle inequality. Confirming the answer is 1 + sqrt 3.

It also has a minimum value: sqrt 3 - 1
So there is something wrong with the sign in that working.
 

KingOfActing

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Re: HSC 2016 4U Marathon

I've done this without the triangle inequality. Confirming the answer is 1 + sqrt 3.

It also has a minimum value: sqrt 3 - 1
So there is something wrong with the sign in that working.
Oh yeah, damn, modulus can't be negative. So the correct solution would've read
 

braintic

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Re: HSC 2016 4U Marathon

Oh yeah, damn, modulus can't be negative. So the correct solution would've read
Actually, that reasoning will not do.
If that was correct working, you would have to interpret your answer as 0 <= |z| <= .....
I haven't checked the working, but there must be an error there somewhere.
 

KingOfActing

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Re: HSC 2016 4U Marathon

Actually, that reasoning will not do.
If that was correct working, you would have to interpret your answer as 0 <= |z| <= .....
I haven't checked the working, but there must be an error there somewhere.
No, I got that. I know you can't just throw absolute value signs everywhere :p

I meant that if I worked it out properly, that would've been equivalent to the solution.
 

braintic

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Re: HSC 2016 4U Marathon

Alternative solution without directly using the triangle inequality:

Let z = r cis theta and substitute.

Rearrange to write cos^2 theta in terms of r.

Then using the fact that 0 <= cos^2 theta <= 1, solve the resulting inequality to find the range for r^2 and hence for r.
 

InteGrand

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Re: HSC 2016 4U Marathon

A downwards force is not a "lateral" force.

Edit - yes, I know what you mean now.
It's referring to the force up and down the plane, right (like in the direction of friction forces in these Q's that have friction)?
 

hedgehog_7

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Re: HSC 2016 4U Marathon

Given P ( a cos theta , b sin theta ) and Q ( a cos phi , b sin phi ) lie on the ellipse x^2/a^2 + y^2/b^2 = 1

IF PQ subtends a right angle at (a,0) show that ( tan theta/2 ) X ( tan phi/2 ) = -b^2/a^2

How does b (sin theta - sin phi ) / a (cos theta - cos phi) = b/a * ( 2 sin ( (theta - phi) /2 ) ) ) * cos ( theta + phi /2 ) ) / - 2 sin ( (theta - phi) /2 ) ) sin (theta + phi /2 )
 
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