HSC 2015 MX2 Marathon ADVANCED (archive) (1 Viewer)

[FrankXie]

Re: HSC 2015 4U Marathon - Advanced Level

no the angle between a fixed line and a fixed plane is a constant, which is defined as the angle between this line and its projection onto the plane.

 

[braintic]

Re: HSC 2015 4U Marathon - Advanced Level

so R is only 1 point on the plane (rather than a circle on the plane p)

Is Q allowed to lie on the normal to the plane at P?
In that particular case, don't you get a circle (with radius equal to the height of Q above the plane)?

 

[Sy123]

Re: HSC 2015 4U Marathon - Advanced Level

no the angle between a fixed line and a fixed plane is a constant, which is defined as the angle between this line and its projection onto the plane.

Oh well I meant any line that lies on the plane

 

[Sy123]

Re: HSC 2015 4U Marathon - Advanced Level

Is Q allowed to lie on the normal to the plane at P?
In that particular case, don't you get a circle (with radius equal to the height of Q above the plane)?

Oh ok yea that's another solution then

 

[FrankXie]

Re: HSC 2015 4U Marathon - Advanced Level

Is Q allowed to lie on the normal to the plane at P?
In that particular case, don't you get a circle (with radius equal to the height of Q above the plane)?

This solution is included in my solution, which is the case where \alpha=90degree, and the maximum is square root of 2, when R lies on the circle.

 

[dan964]

Re: HSC 2015 4U Marathon - Advanced Level

just for something interesting
which I don't know what it'll look like
sketch the locus of
|z| = arg (z)

 

[seanieg89]

Re: HSC 2015 4U Marathon - Advanced Level

just for something interesting
which I don't know what it'll look like
sketch the locus of
|z| = arg (z)

It's just an anticlockwise spiral. You could use calculus to tell you what the gradient of the start of the curve at (0,0) is.

 

[braintic]

Re: HSC 2015 4U Marathon - Advanced Level

If H is the foot of the perpendicular to the plane, I believe the point R we are looking for lies on PH produced such that PQ=PR.
Is that correct?

In any case, it is definitely not H.

 

[braintic]

Re: HSC 2015 4U Marathon - Advanced Level

My solution: View attachment 3D Maximisation.pdf

 

[Sy123]

Re: HSC 2015 4U Marathon - Advanced Level

 

[Kaido]

Re: HSC 2015 4U Marathon - Advanced Level

(c=cos) (k=desired integer)
c^2x+(2c^2x-1)^2+(4c^3x-3cx)^2=1
->c^2x(16c^4x-20c^2x+6)=0 (expand and factorise)

c^2x=0; x=pi/2 (+kpi)

->16c^4x-20c^2x+6=0
Let c^2x=m
->16m²-20m+6=0 (solve for m, resub c and yeah...)

x=pi/6,-pi/6,pi/4,-pi/4,5pi/6,-5pi/6,3pi/4,-3pi/4 (+k2pi)

(Am tired atm, check my solutions and correct me lol)

 

[dan964]

Re: HSC 2015 4U Marathon - Advanced Level

tried doing it with cosine, but stuffed up somewhere in my working. So this is the question redone using sine.

can someone else post the next question, all mine are either predictable or have errors in them.

 

[HeroicPandas]

Re: HSC 2015 4U Marathon - Advanced Level

(not mine)

 

[Sy123]

Re: HSC 2015 4U Marathon - Advanced Level

(c=cos) (k=desired integer)
c^2x+(2^2x-1)^2+(4c^3x-3cx)^2=1
->c^2x(16c^4x-20c^2x+6)=0 (expand and factorise)

c^2x=0; x=pi/2 (+kpi)

->16c^4x-20c^2x+6=0
Let c^2x=m
->16m²-20m+6=0 (solve for m, resub c and yeah...)

x=pi/6,-pi/6,pi/4,-pi/4,5pi/6,-5pi/6,3pi/4,-3pi/4 (+k2pi)

(Am tired atm, check my solutions and correct me lol)

tried doing it with cosine, but stuffed up somewhere in my working. So this is the question redone using sine.

can someone else post the next question, all mine are either predictable or have errors in them.

My solution was:



Which can be solved easily by re-arranging and taking cos(4x) common

 

[FrankXie]

Re: HSC 2015 4U Marathon - Advanced Level

My solution was:



Which can be solved easily by re-arranging and taking cos(4x) common

How? Note that you have constant term other than zero.

 

[seanieg89]

Re: HSC 2015 4U Marathon - Advanced Level

How? Note that you have constant term other than zero.

If u is cos(2x), this equation is cubic in u with zero constant coefficient.

 

[FrankXie]

Re: HSC 2015 4U Marathon - Advanced Level

If u is cos(2x), this equation is cubic in u with zero constant coefficient.

This way I knew. but I thought Sy meant something else?

 

[Chlee1998]

Re: HSC 2015 4U Marathon - Advanced Level

Since no one has psoted a new question, here's one:

given A+B+C+D = 4, prove that 4/(ABCD) >= A/B +B/C + C/D + D/A

 

[dan964]

Re: HSC 2015 4U Marathon - Advanced Level

I would first group RHS on common denominator.
then it would be heaps easier

 

[InteGrand]

Re: HSC 2015 4U Marathon - Advanced Level

Since no one has psoted a new question, here's one:

given A+B+C+D = 4, prove that 4/(ABCD) >= A/B +B/C + C/D + D/A

I think we need to assume that , because the inequality does not always hold otherwise (e.g if ).