HSC 2014 MX2 Marathon (archive) (1 Viewer)

Sy123 · Mar 27, 2014

Re: HSC 2014 4U Marathon

Ikki said:
Facedesk*
I just recapped, I kept trying to think visually but i guess algebraic would be best...
Its just letting z=x+i that threw me off.
So i let z=x+iy.
for i) it is clear that locus is y=1 by equating coeff
ii) Similarly from your expansion y=1 should also be locus for z squared.

How do you let z=x+iy?

Think of it this way

since z=(x^2-1) + 2x i

What is z if x=1? or x=2? or x=3?

You get a series of dots, but they follow a certain curve, this curve is the locus. How do you find this locus given z?

Ikki · Mar 27, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
How do you let z=x+iy?

Think of it this way

since z=(x^2-1) + 2x i

What is z if x=1? or x=2? or x=3?

You get a series of dots, but they follow a certain curve, this curve is the locus. How do you find this locus given z?

If x=1: z^2=2i, If x=2:z^2=3+4i, If x=3: z^2=8+6i

If i let z=x+iy but also x+i
then x^2-y^2+2xyi=x^2-1+2xi
Equating imaginaries, y=1 right? But clearly not the case by substituting...

EDIT: We could just take them as points (0,2) (3,4) and (8,6). SYYYYYYYYYYYYYY Mind is blowing here

Sy123 · Mar 27, 2014

Re: HSC 2014 4U Marathon

Ikki said:
If x=1: z^2=2i, If x=2:z^2=3+4i, If x=3: z^2=8+6i

If i let z=x+iy but also x+i
then x^2-y^2+2xyi=x^2-1+2xi
Equating imaginaries, y=1 right? But clearly not the case by substituting...

EDIT: We could just take them as points (0,2) (3,4) and (8,6). SYYYYYYYYYYYYYY Mind is blowing here

But remember y=/=1 here!

its not z=x+i, its z=(x+i)^2 which is completley different

just imagine its z=t + i instead of x + i just so its not confusing

so z= x+iy = (t+i)^2 = (t^2-1) + 2t i

Hence x=(t^2-1) , y= 2t

t = y/2, x= (y^2/4-1)

Hence y^2 = 4(x+1)

Therefore the locus is the equation y^2 = 4(x+1) in the cartesian plane.

Trebla · Mar 27, 2014

Re: HSC 2014 4U Marathon

Not sure if that is within the HSC syllabus?

Ikki · Mar 27, 2014

Re: HSC 2014 4U Marathon

Ah like parametrics aye, yeh the t thing helped alot
TREBLAAA hi

Sy123 · Mar 27, 2014

Re: HSC 2014 4U Marathon

Trebla said:
Not sure if that is within the HSC syllabus?

Oh I was under the impression it was
(still a good problem though it was not useless)

Sy123 · Mar 27, 2014

Re: HSC 2014 4U Marathon

(Definitely within syllabus)

$\\ $In the complex plane let$ \ z = \cos \theta + i \sin \theta \ $and$ \ 0 < \theta < \frac{\pi}{2} \\ $i) Show that a rhombus is formed with the complex numbers$ \ \ 0,z,z^2, (z+z^2) \\ $ii) By finding$ \ (z^2+z) \ $in modulus-argument form show that$ \\ \cos \theta + \cos 2\theta = 2\cos \frac{\theta}{2} \cos \frac{3\theta}{2}$

Ikki · Mar 28, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
(Definitely within syllabus)

$\\ $In the complex plane let$ \ z = \cos \theta + i \sin \theta \ $and$ \ 0 < \theta < \frac{\pi}{2} \\ $i) Show that a rhombus is formed with the complex numbers$ \ \ 0,z,z^2, (z+z^2) \\ $ii) By finding$ \ (z^2+z) \ $in modulus-argument form show that$ \\ \cos \theta + \cos 2\theta = 2\cos \frac{\theta}{2} \cos \frac{3\theta}{2}$

i) Hmm with this one i simply drew up the unit circle and placed z on the argand diagram between 0 and pi/2.
Similarly, for z^2 the domain extends to pi so I placed z^2 in the second quadrant which also lies in the unit circle.
By the modulus of z is clearly 1 for both numbers and similarly the distance from z and z^2 to (z+z^2)=1 via vector addition.
Therefore all sides of the quadrilateral (0,z,z^2,z+z^2) are equal, therefore it can be classified as a rhombus.

ii) let theta=x for my convienience.
z=cisx
z^2=cis2x (By demoivres)
z+z^2=cisx+cis2x= {cosx + cos2x} + i(sinx+sin2x)

But z^2 also = cos^2x-sin^2x + 2sinxcosxi
Therefore, z+z^2={cos^2x-sin^2x}+cosx+2sinxcosxi

Now observing LHS of what we need to show, we equate the Reals:

ANDDD This is where i'm not sure, am I on the right track?

Sy123 · Mar 28, 2014

Re: HSC 2014 4U Marathon

Ikki said:
i) Hmm with this one i simply drew up the unit circle and placed z on the argand diagram between 0 and pi/2.
Similarly, for z^2 the domain extends to pi so I placed z^2 in the second quadrant which also lies in the unit circle.
By the modulus of z is clearly 1 for both numbers and similarly the distance from z and z^2 to (z+z^2)=1 via vector addition.
Therefore all sides of the quadrilateral (0,z,z^2,z+z^2) are equal, therefore it can be classified as a rhombus.

ii) let theta=x for my convienience.
z=cisx
z^2=cis2x (By demoivres)
z+z^2=cisx+cis2x= {cosx + cos2x} + i(sinx+sin2x)

But z^2 also = cos^2x-sin^2x + 2sinxcosxi
Therefore, z+z^2={cos^2x-sin^2x}+cosx+2sinxcosxi

Now observing LHS of what we need to show, we equate the Reals:

ANDDD This is where i'm not sure, am I on the right track?

The first is correct, for the second one, try finding (z+z^2) in modulus argument form by considering the vector geometrically (use the first part)

Sy123 · Apr 4, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
(Definitely within syllabus)

$\\ $In the complex plane let$ \ z = \cos \theta + i \sin \theta \ $and$ \ 0 < \theta < \frac{\pi}{2} \\ $i) Show that a rhombus is formed with the complex numbers$ \ \ 0,z,z^2, (z+z^2) \\ $ii) By finding$ \ (z^2+z) \ $in modulus-argument form show that$ \\ \cos \theta + \cos 2\theta = 2\cos \frac{\theta}{2} \cos \frac{3\theta}{2}$

Handwritten Solution

Sy123 · Apr 4, 2014

Re: HSC 2014 4U Marathon

$\\ $What is the domain and range for$ \ \ f(x) = \sin^{-1}x + \sin^{-1} \frac{1}{x}$

braintic · Apr 4, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $What is the domain and range for$ \ \ f(x) = \sin^{-1}x + \sin^{-1} \frac{1}{x}$

Two points for getting this question right.

dunjaaa · Apr 4, 2014

Re: HSC 2014 4U Marathon

Domain: x=±1
Range: y=±pi
I drew each separate graph and they shared the common points of (1,pi/2),(-1,-pi/2) and hence by the summing the graphs, y-ordinate doubles. Everywhere else is undefined, you get 2 dots at (x=1,y=pi) and (x=-1,y=-pi)

Axio · Apr 5, 2014

Re: HSC 2014 4U Marathon

Axio said:
View attachment 30177
The altitudes AP and BQ in an acute-angled triangle ABC meet at H. AP produced cuts the circle through A, B and C at K.
Prove that HP = PK.

View attachment Q.pdf
$\angle BQC = 90^{\circ}\, (given)\\ \therefore \Delta BQC |||\Delta BPH\, (AAA)\\ \angle ACB = \angle AKB\, (angles\: subtended\: by\: same\: segment)\\ \angle BPK = 90^{\circ}\, (given)\\ \therefore \Delta BPK|||\Delta BQC \, (AAA)\\ \therefore \Delta BPK|||\Delta BPH\\ \therefore \frac{PK}{BP}=\frac{HP}{BP}\\ \therefore PK=HP$

Sy123 · Apr 7, 2014

Re: HSC 2014 4U Marathon

$\\ $Find the sum of angles$ \ A, B \ $where$ \ 0 \leq A,B \leq \pi \ \ $and$ \\ \\ \sin A + \sin B = \sqrt{\frac{3}{2}} \\ \cos A + \cos B = \sqrt{\frac{1}{2}}$

VBN2470 · Apr 8, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Find the sum of angles$ \ A, B \ $where$ \ 0 \leq A,B \leq \pi \ \ $and$ \\ \\ \sin A + \sin B = \sqrt{\frac{3}{2}} \\ \cos A + \cos B = \sqrt{\frac{1}{2}}$

$A+B=\frac{\pi}{3}$

hit patel · Apr 9, 2014

Re: HSC 2014 4U Marathon

Volumes Question :

hit patel · Apr 10, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Let$ \ \ I_n = \int_0^{\pi/4} \tan^{2n}x \ dx \\ \\ $i) Show that$ \ \ I_n + I_{n+1} = \frac{1}{2n+1} \ \ \ (n \geq 0) \ \ \ \ \fbox{2} \\ \\ $ii) Show that$ \ \ \ \sum_{n=0}^{N} \frac{(-1)^n}{2n+1} = I_0 + (-1)^N I_{N+1} \ \ \ \fbox{1} \\ \\ $iii) Show that$ \ I_N = \int_0^1 \frac{u^{2N}}{1+u^2} \ du \ \ \ \ \fbox{1} \\ \\ $iv) Hence show that$ \ \ \frac{-1}{2N+1} \leq I_N \leq \frac{1}{2N+1} \ \ \ \ \fbox{2} \\ \\ $v) Prove that$ \ \ \ \ \frac{\pi}{4} = \sum_{n=0}^{\infty} \frac{(-1)^n }{2n+1} \ \ \ \fbox{1}$

Not sure if I'm overdoing it with the guiding thing now

I am getting I_(n-2) + I_n = 1/(2n-1) . Why so different?

Sy123 · Apr 10, 2014

Re: HSC 2014 4U Marathon

hit patel said:
I am getting I_(n-2) + I_n = 1/(2n-1) . Why so different?

If you mean I_(n-1) + I_n = 1/(2n-1) then this is fine since if we just make n -> n+1 we get

I_((n+1) - 1) + I_(n+1) = 1/(2(n+1)-1)

I_n + I_(n+1) = 1/(2n+1)

hit patel · Apr 10, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
If you mean I_(n-1) + I_n = 1/(2n-1) then this is fine since if we just make n -> n+1 we get

I_((n+1) - 1) + I_(n+1) = 1/(2(n+1)-1)

I_n + I_(n+1) = 1/(2n+1)

Sorry its the same thing. Didnt see that. TOo much caffeine.

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