Preliminary mathematics marathon (4 Viewers)

[ohexploitable]

New Question:

Find the equations of the four circles which are tangent to the x-axis, the y-axis, and the line x + y = 2.

 

[edmundsung]

another question



find a and b if g(x) is to be continuous for all x.

thanks.

 

[hscishard]

(sin2xcosy-sinycos2x)/sin2x = (sin2ycosx+sinxcos2y)/sin2y

Is it possible to draw out that x = y?

 

[hscishard]

Re: another question

find a and b if g(x) is to be continuous for all x.

thanks.

Is the answer a=4 and b=26?

 

[nikkifc]

Re: another question

find a and b if g(x) is to be continuous for all x.

thanks.

This is not in the scope of the current HSC syllabus. I believe it was taken out in 1981 from the old course.

And yes the answer is (a, b) = (4, 26). Although simple, to do it properly, it requires the use of limits.

 

[ohexploitable]

Re: another question

This is not in the scope of the current HSC syllabus. I believe it was taken out in 1981 from the old course.

And yes the answer is (a, b) = (4, 26). Although simple, to do it properly, it requires the use of limits.

Really? I was taught this stuff...

 

[ohexploitable]

New Question:

Find the equations of the four circles which are tangent to the x-axis, the y-axis, and the line x + y = 2.

.

 

[hscishard]

.

(sin2xcosy-sinycos2x)/sin2x = (sin2ycosx+sinxcos2y)/sin2y

Is it possible to draw out that x = y?

.

Lol. Btw that question is "Is it possible or not" If possible, please post proof

 

[ohexploitable]

.

Lol. Btw that question is "Is it possible or not" If possible, please post proof

Not impossible, involves thinking and simple trig (just socahtoa).

 

[hscishard]

In other words of what you just said:

It's possible but I cbb doing it.

Lol.

 

[ohexploitable]

In other words of what you just said:

It's possible but I cbb doing it.

Lol.

Fine... lol...

For one of the circles...

The two right-angled triangles with sides root 2 and r are congruent (SAS), .: it bisects the 45 deg.

 

[hscishard]

I don't get it. The impossible thing I said was to my question.

I'm pretty sure that was your question..

Are the centres integers or your question?

 

[hscishard]

Fine... lol...

For one of the circles...

The two right-angled triangles with sides root 2 and r are congruent (SAS), .: it bisects the 45 deg.

For that circle, trig wasn't really needed.

Because the circles have tangents x and y axes...
The centres must be on y=x.
The intersection of y=x and x+y=2 is 1,1.

Let centre = (h,h)
Perp distance(radius)=(h-1)root2
Then equation of a circle, (x-h)^2+(y-h)^2=2(h-1)^2
Sub a point (0.h) (Tangent point)

Solve to get h=2+/-root2.
Then you just continue..

 

[ohexploitable]

My attempt

 

[hscishard]

^ well done..

Shit. I was so close to proving my "Extremely difficult" isosceles triangle question..

 

[nikkifc]

Re: another question

Really? I was taught this stuff...

Well then do you have a teacher who has no previous experience in teaching? I could be wrong but it seems like it.

There are two problems with asking this in the HSC course:

1) You don't properly define functions in the HSC. ie. "g(x) is to be continuous" is wrong because it implies that g(x) is a function, when in fact it is the VALUE of the functions of 'g' at a point 'x'.

2) It requires the knowledge of limits and continuity, which was taken out of the new HSC course in 1981. (It used to be in the previous 2F course)

 

[hscishard]

Re: another question

Well then do you have a teacher who has no previous experience in teaching? I could be wrong but it seems like it.

There are two problems with asking this in the HSC course:

1) You don't properly define functions in the HSC. ie. "g(x) is to be continuous" is wrong because it implies that g(x) is a function, when in fact it is the VALUE of the functions of 'g' at a point 'x'.

2) It requires the knowledge of limits and continuity, which was taken out of the new HSC course in 1981. (It used to be in the previous 2F course)

You know your stuff...

Thats a positive definite.

 

[Omnipotence]

How do you place a space in latex eq/n editor?

 

[Trebla]

(sin2xcosy-sinycos2x)/sin2x = (sin2ycosx+sinxcos2y)/sin2y

Is it possible to draw out that x = y?

If x = y then
sin 2x.cos x - sin x.cos 2x = sin 2x.cos x + sin x.cos 2x
=> 2sin x cos 2x = 0
.: sin x = 0 or cos 2x = 0
If sin x = 0 then sin 2x = 0 which would provide an undefined expression (denominator is zero)
If cos 2x = 0 (i.e. x = kπ ± π/4 for integer k) then the equality can hold for x = y
Hence it is possible.

 

[Trebla]

Oh and btw, limits and continuity are in the course, just not treated formally. See section 8.2.