HSC Mathematics Marathon (2 Viewers)

[UnrealAnchovies]

integrate: 1/ [ x^(1/2) + x^(1/3) ] dx between limits [ 64 .. 1 ]

That was... tedious... and long...
Is the answer: 11 - 6log(3/2)
?

 

[Bored Of Fail 2]

That was... tedious... and long...
Is the answer: 11 - 6log(3/2)
?

yes, very good, that was a harder question in the higher math 1B unsw course notes.

 

[cutemouse]

If lie on a circle through that passes through the origin then prove that the points representing are collinear (ie. lie on a straight line).

I have a solution for this that uses one circle geometry theorem. I'm sure there's a way to do it without using it so I'm interested in other solutions.

 

[cutemouse]

Fully describe the locus of:

2b=10 so b=5

a=4 (from a diagram, at position PS = PS' figure forms a 3,4,5 triangle)

An ellipse with foci at (0,3) and (6,3) and major axis length 5 units and minor axis length 4 units and y axis is the major axis.

 

[deterministic]

If lie on a circle through that passes through the origin then prove that the points representing are collinear (ie. lie on a straight line).

I have a solution for this that uses one circle geometry theorem. I'm sure there's a way to do it without using it so I'm interested in other solutions.

well there are more than 1 circle geometry methods, all along the same lines though. They seem to be the most efficient methods.

 

[jyu]

2b=10 so b=5

a=4 (from a diagram, at position PS = PS' figure forms a 3,4,5 triangle)

An ellipse with foci at (0,3) and (6,3) and major axis length 5 units and minor axis length 4 units and y axis is the major axis.

Did you mean major axis length 10 etc. y axis is the major axis?

ok
describe fully the locus of |z+5| - |z-5| = 8.

 

[UnrealAnchovies]

lol^ thats a tough one, and if you really want you can just not bother about it evaluating it with the limits, because thats just boring calculator work.

If that is too hard heres an easier one, and one which is more likely to come up on the hsc paper

integrate : ( 4x^2-15x +29 ) / [ (x-5)(x^2-4x+13) ] dx

Good question, but I don't think that would appear in a HSC exam. It did take quite a long time.

The answer is: log(x-5)^3(sq root(x^2 - 4x +13)) + 4/ sqroot 13 arctan(x-2)/ sqroot 13

Sorry if it looks messy - I cant find latex!

 

[UnrealAnchovies]

Good question, but I don't think that would appear in a HSC exam. It did take a quite long time.

The answer is: log(x-5)^3(sq root(x^2 - 4x +13)) + 4/ sqroot 13 arctan(x-2)/ sqroot 13

Sorry if it looks messy - I cant find latex!

Crap. I just noticed a mistake with the inverse tan part. The log sum is correct, but the inverse tan one should be: 4/3 arctan(x-2)/ 3

--------

New question:

P(x1, y1) is a point on the hyperbole x^2/a^2 - y^2/b^2 = 1 with foci S, S'

(a) Find the equation of the tangent to the hyperbole at P.
(b) The tangent meets the x-axis in T. Show that PS/PS' = TS/TS'

 

[hscishard]

Crap. I just noticed a mistake with the inverse tan part. The log sum is correct, but the inverse tan one should be: 4/3 arctan(x-2)/ 3

--------

New question:

P(x1, y1) is a point on the hyperbole x^2/a^2 - y^2/b^2 = 1 with foci S, S'

(a) Find the equation of the tangent to the hyperbole at P.
(b) The tangent meets the x-axis in T. Show that PS/PS' = TS/TS'

Oh god. Noones going to type all of the proof.
a) is x(x1)/a^2 - y(y1)/b^2 = 1
b) PS/PS' = (x1 -a/e) / (x1+a/e)
TS/TS' = (ae-a^2/x1) / (ae + a^2/x1)
=(x1 -a/e) / (x1+a/e).

 

[hscishard]

The points O, I, Z and P on the Argand Plane represent the complex numbers
0, 1, z and z + 1 respectively, where z = cosθ + i sinθ is any complex number of
modulus 1, with 0 <θ <π .
(i) Explain why OIPZ is a rhombus.
(ii) Show that (z-1)/(z+1) is purely imaginary.
(iii) Find the modulus of z + 1 in terms of θ .

 

[jyu]

The points O, I, Z and P on the Argand Plane represent the complex numbers
0, 1, z and z + 1 respectively, where z = cosθ + i sinθ is any complex number of
modulus 1, with 0 <θ <π .
(i) Explain why OIPZ is a rhombus.
(ii) Show that (z-1)/(z+1) is purely imaginary.
(iii) Find the modulus of z + 1 in terms of θ .

I think this is done already.

Try this:

Describe fully the locus of |z+5| - |z-5| = 8

 

[blackops23]

hey hey hey how about some conics??

 

[UnrealAnchovies]

hey hey hey how about some conics??

Ugh. I put a conics question up on page 2 (last post).

 

[deterministic]

hey hey hey how about some conics??

The locus question is a conics question in disguise....

 

[ohexploitable]

The locus question is a conics question in disguise....

lol that's a relief, I was wondering why I had no clue how to do them.

 

[UnrealAnchovies]

Sucks not having done complex numbers yet...

 

[hscishard]

I think this is done already.

Try this:

Describe fully the locus of |z+5| - |z-5| = 8

Uhhh it is? Don't think so...send me the link

That uses the theorem |S'P-SP| = 2a for the hyperbola

 

[cutemouse]

Using each of the digits 0,1,2,3,4,5,6,7,8,9 form a number which when increased by one million becomes a perfect square. [NB: There are two possible such numbers]

 

[hscishard]

That's an interesting question...
Expanding that would give, 1-(sum roots)-(sum of two roots at a time) -......+(product of roots)
I think it would eventually give 1+(n-1)= n.
Lol this is because I saw that sum of non real roots = -1 and product of roots = 1

 

[hscishard]

Lol terry lee pg 64...how did I know. Terry lees solution is much better