HSC 2014 MX2 Marathon (archive) (4 Viewers)

[Immortality]

Re: HSC 2014 4U Marathon

Sketch the locus of z on an Argand Diagram of

 

[dunjaaa]

Re: HSC 2014 4U Marathon

I'm not sure if this is right (haven't touched complex in a while)
arg(z-2)+arg(z+2)=pi
arg((z-2)(z+2))=pi
arg(z^2-4)=pi
Thus, the complex number z^2-4 must be purely real
Let z=x+iy
z^2-4 = (x^2-y^2-4)+2ixy
Demand: Im(z)=0
2xy=0
xy=0
Pair of intersecting lines (x=0 ; y=0)

 

[dunjaaa]

Re: HSC 2014 4U Marathon

Next Question:

 

[Immortality]

Re: HSC 2014 4U Marathon

I'm not sure if this is right (haven't touched complex in a while)
arg(z-2)+arg(z+2)=pi
arg((z-2)(z+2))=pi
arg(z^2-4)=pi
Thus, the complex number z^2-4 must be purely real
Let z=x+iy
z^2-4 = (x^2-y^2-4)+2ixy
Demand: Im(z)=0
2xy=0
xy=0
Pair of intersecting lines (x=0 ; y=0)

This is not entirely correct. Think about domain/range restrictions!

 

[dunjaaa]

Re: HSC 2014 4U Marathon

?, chat is bugged

 

[dunjaaa]

Re: HSC 2014 4U Marathon

A+B = 2pi/3 ?

 

[Immortality]

Re: HSC 2014 4U Marathon

View attachment 30306?, chat is bugged

very good (Y)

Tho technically it should be

 

[dunjaaa]

Re: HSC 2014 4U Marathon

Sketch the curve: 6cos(x^2-y^2)=pi

 

[RealiseNothing]

Re: HSC 2014 4U Marathon

View attachment 30306?, chat is bugged

Inequality signs like < need a space before/after so that everything shows up.

 

[dunjaaa]

Re: HSC 2014 4U Marathon

Here's a good work through complex numbers question (part (e) meant to be arrows --> on top)

 

[jyu]

Re: HSC 2014 4U Marathon

Curve sketching, don't hate <a href="http://www.codecogs.com/eqnedit.php?latex=Sketch\,\,&space;the\,\,&space;curve\,\,&space;x^2&space;-9xy+y^2&space;=1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?Sketch\,\,&space;the\,\,&space;curve\,\,&space;x^2&space;-9xy+y^2&space;=1" title="Sketch\,\, the\,\, curve\,\, x^2 -9xy+y^2 =1" /></a>

unnecessary use of latex

The equation shows symmetry in x and y. Interchanging x and y yields the same equation, the inverse is the same relation.
Rotate the relation by 45 deg anticlockwise, the new relation will be symmetrical about the y-axis.
Sketch the new relation 11x^2/2 -7y^2/2=1, a hyperbola.
Rotate the hyperbola by 45 deg clockwise to obtain the graph of x^2-9xy+y^2=1

 

[braintic]

Re: HSC 2014 4U Marathon

The equation shows symmetry in x and y. Interchanging x and y yields the same equation, the inverse is the same relation.
Rotate the relation by 45 deg anticlockwise, the new relation will be symmetrical about the y-axis.
Sketch the new relation 11x^2/2 -7y^2/2=1, a hyperbola.
Rotate the hyperbola by 45 deg clockwise to obtain the graph of x^2-9xy+y^2=1

Other than converting x^2/a^2 - y^2/b^2 =1 to xy=c^2, rotating entire curves is not in the course.
Just checking - how did you rotate the curve? Not using matrices I hope?
Just differentiate implicitly and go from there.

 

[seanieg89]

Re: HSC 2014 4U Marathon

Other than converting x^2/a^2 - y^2/b^2 =1 to xy=c^2, rotating entire curves is not in the course.
Just checking - how did you rotate the curve? Not using matrices I hope?
Just differentiate implicitly and go from there.

You can use complex numbers to obtain a formula for rotations at an MX2 level.

 

[dunjaaa]

Re: HSC 2014 4U Marathon

View attachment RotationofAxes.pdf Here's what I followed to sketch that. It shows the process of deriving the angle of rotation plus the new coordinates on the shifted x-y axis, pretty doable in 4u tbh

 

[dunjaaa]

Re: HSC 2014 4U Marathon

Yeah can also be done by complex numbers, that's how my teacher derived the formulae for the rectangular hyperbola

 

[braintic]

Re: HSC 2014 4U Marathon

You can use complex numbers to obtain a formula for rotations at an MX2 level.

There is no mention of that in the syllabus - it is just the way that all textbooks have chosen to "prove that the hyperbola with equation xy=(1/2)a^2 is the hyperbola x^2 - y^2 = a^2 referred to different axes."

Rotating points using complex number is one thing - rotating whole curves is something else.
Not that I think it is a difficult concept to get across, but I am quite confident that if this was ever to appear in an HSC exam, there would be a LOT of hand-holding. For example, the algebra needed to eliminate the parameters in this particular question (although it appears in textbooks) would not be asked in an exam without a lead-in.

 

[Sy123]

Re: HSC 2014 4U Marathon

 

[VBN2470]

Re: HSC 2014 4U Marathon

The function is not defined for . Is there a value that could be given for that would make the function continuous at 2? Give reasons for your answer.

 

[Carrotsticks]

Re: HSC 2014 4U Marathon

There is no mention of that in the syllabus - it is just the way that all textbooks have chosen to "prove that the hyperbola with equation xy=(1/2)a^2 is the hyperbola x^2 - y^2 = a^2 referred to different axes."

Other than converting x^2/a^2 - y^2/b^2 =1 to xy=c^2, rotating entire curves is not in the course.
Just checking - how did you rotate the curve? Not using matrices I hope?
Just differentiate implicitly and go from there.

Rotating points using complex number is one thing - rotating whole curves is something else.

Not that I think it is a difficult concept to get across, but I am quite confident that if this was ever to appear in an HSC exam, there would be a LOT of hand-holding. For example, the algebra needed to eliminate the parameters in this particular question (although it appears in textbooks) would not be asked in an exam without a lead-in.

You can think of rotating a curve as rotating an infinitely large set of points that happen to lie along some curve.

Also, rotating the curve via matrix rotation is the same (in terms of the mechanics) as the methods using complex numbers. It's just a more concise way of doing it.

 

[Sy123]

Re: HSC 2014 4U Marathon