HSC 2015 MX2 Marathon (archive) (1 Viewer)

[Drsoccerball]

Re: HSC 2015 4U Marathon

Correction: bx, not bc

its bc

 

[Drsoccerball]

Re: HSC 2015 4U Marathon

x^2+bx+x? or x^2+bx+cx?

Solve simultaneously by subbing 2,-2,3 and setting =0 ? pre-guess

its really easy just look at the question for a bit and it will pop up

 

[Ekman]

Re: HSC 2015 4U Marathon

Sounds good. Anyways next question:
The lines x=2 x= -2 y=3 are asymptotes to the curve with the equation
y=(ax^2)/((x)^2 +bc+c)
By using the above information show that:
y= (3x^2)/((x)^2 -4)

For those who still haven't understood what was going on, just find the limits of the equation y = (3x^2)/((x)^2 -4)
Since the denominator cant be equal to 0, thus x cant be equal to +2, -2
Lim x to infinity y = 3
Thus x = 2, x = -2 and y =3 are the three asymptotes.

EDIT:
Plus the question doesn't need to mention y=(ax^2)/((x)^2 +bc+c) since it directly asks for y= (3x^2)/((x)^2 -4).

 

[Drsoccerball]

Re: HSC 2015 4U Marathon

For those who still haven't understood what was going on, just find the limits of the equation y = (3x^2)/((x)^2 -4)
Since the denominator cant be equal to 0, thus x cant be equal to +2, -2
Lim x to infinity y = 3
Thus x = 2, x = -2 and y =3 are the three asymptotes.

EDIT:
Plus the question doesn't need to mention y=(ax^2)/((x)^2 +bc+c) since it directly asks for y= (3x^2)/((x)^2 -4).

Correct but the way i did it was as you said found limits for y value a=3 and wrote the x assymptotes as (x-2)(x+2) and then realized that b=0 and c is equal to -4 which gives us the equation

 

[braintic]

Re: HSC 2015 4U Marathon

its bc

It actually doesn't make any difference whether it's bx, bc, pi, or 3+7i

 

[Drsoccerball]

Re: HSC 2015 4U Marathon

It actually doesn't make any difference whether it's bx, bc, pi, or 3+7i

Because b is zero ahahah

 

[Drsoccerball]

Re: HSC 2015 4U Marathon

New question :
z is a point in the first quadrant of the Argand diagram which lies on the circle |z-3|. Given arg(z)= theta, find arg(z^2 -9z +18) in terms of theta

 

[braintic]

Re: HSC 2015 4U Marathon

New question :
z is a point in the first quadrant of the Argand diagram which lies on the circle |z-3|. Given arg(z)= theta, find arg(z^2 -9z +18) in terms of theta

The circle |z-3| = what?

 

[FrankXie]

Re: HSC 2015 4U Marathon

Because b is zero ahahah

what about b=1,c=-2; or b=-2,c=4; or b=pi, c=-4/(pi+1), etc.

 

[Drsoccerball]

Re: HSC 2015 4U Marathon

=3 typo

 

[integral95]

Re: HSC 2015 4U Marathon

New question :
z is a point in the first quadrant of the Argand diagram which lies on the circle |z-3|. Given arg(z)= theta, find arg(z^2 -9z +18) in terms of theta

EDIT:yeah I'm not too sure really haha

 

[RealiseNothing]

Re: HSC 2015 4U Marathon

EDIT:yeah I'm not too sure really haha

This is right if

But he didn't actually say what it equals.

 

[integral95]

Re: HSC 2015 4U Marathon

The circle |z-3| = what?

=3 typo

This is right if

But he didn't actually say what it equals.

lel the guy had to correct it in the following post, but sweet I still got it

 

[Drsoccerball]

Re: HSC 2015 4U Marathon

EDIT:yeah I'm not too sure really haha

correct

 

[Drsoccerball]

Re: HSC 2015 4U Marathon

I don't know why im the only one posting questions but here:

 

[Drsoccerball]

Re: HSC 2015 4U Marathon

ignore the < br/ > notsure why its happening

 

[FrankXie]

Re: HSC 2015 4U Marathon

ignore the < br/ > notsure why its happening

i think it is because you input "enter" in the "tex" environment.

Here I post my question:

 

[Drsoccerball]

Re: HSC 2015 4U Marathon

i think it is because you input "enter" in the "tex" environment.

Here I post my question:

still left mine unanswered

 

[Drsoccerball]

Re: HSC 2015 4U Marathon

i think it is because you input "enter" in the "tex" environment.

Here I post my question:

 

[Drsoccerball]

Re: HSC 2015 4U Marathon

ARGGGGGG stupid mistake 13 not 31