HSC 2014 MX2 Marathon (archive) (3 Viewers)

Sy123 · Mar 25, 2014

Re: HSC 2014 4U Marathon

$\\ $By considering complex numbers show that$ \\ \\ \tan^{-1} \frac{7}{17} + 2\tan^{-1} \frac{1}{5} = \frac{\pi}{4}$

Carrotsticks · Mar 25, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $By considering complex numbers show that$ \\ \\ \tan^{-1} \frac{7}{17} + 2\tan^{-1} \frac{1}{5} = \frac{\pi}{4}$

Might need more of a lead-up for students.

dunjaaa · Mar 25, 2014

Re: HSC 2014 4U Marathon

Screen shot 2014-03-25 at 3.27.30 PM.png

Sy123 · Mar 25, 2014

Re: HSC 2014 4U Marathon

Carrotsticks said:
Might need more of a lead-up for students.

My bad

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$\\ $a) Consider the polynomial for real$ \ \ x, A, B \ \ P(x) = Ax^3+Bx^2+Bx+A \\ \\ $i) Show that$ \ \ \ P(x) = (x+1)(Ax^2-(A-B)x + A) \\ \\ $ii) For what values of$ \ A,B \ $does the polynomial have 3 real roots$ \\ \\ $iii) (extension) If$ \ -1 \leq A \leq 1 \ \ $and$ \ \ -1 \leq B \leq 1 \ \ \ $Find the region in the$ \\ A$-$B \ $plane such that$ \ P(x) \ $has 3 real roots$$

dunjaaa · Mar 25, 2014

Re: HSC 2014 4U Marathon

(i)By inspection x=-1 is a factor of P(x), by polynomial division we get the required result
(ii) For 3 real roots, we set the discriminant >=0 and we get B<=-A
(iii) The region is bounded by lines B>=-1, A>=-1, B<=-A

Sy123 · Mar 25, 2014

Re: HSC 2014 4U Marathon

dunjaaa said:
(i)By inspection x=-1 is a factor of P(x), by polynomial division we get the required result
(ii) For 3 real roots, we set the discriminant >=0 and we get B<=-A
(iii) The region is bounded by lines B>=-1, A>=-1, B<=-A

In (ii) and (iii) you are missing another solution

dunjaaa · Mar 25, 2014

Re: HSC 2014 4U Marathon

B>=3A?

Sy123 · Mar 25, 2014

Re: HSC 2014 4U Marathon

dunjaaa said:
B>=3A?

Yep, however B=> 3A and -A => B are only the first half of hte solutions

I'm guessing you got to the point in that

$\triangle = (-A-B)(3A-B) \geq 0$

So both brackets either have to be negative or both brackets have to be positive. You have only covered one of those 2 cases.

Sy123 · Mar 25, 2014

Re: HSC 2014 4U Marathon

$\\ $Let$ \ \ 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

dunjaaa · Mar 25, 2014

Re: HSC 2014 4U Marathon

so careless tonight sigh..
(ii) First case: B>=3A, B<=-A (both positive)
Second Case: B>=-A, B<=3A (both negative)
(iii)First Case: A>=-1, B>=-1, B>=3A, B<=-A
Second Case: A<=1, B<=1, B>=-A, B<=3A

Carrotsticks · Mar 25, 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

Even for the average 4U student, that would be too difficult.

Sy123 · Mar 25, 2014

Re: HSC 2014 4U Marathon

dunjaaa said:
so careless tonight sigh..
(ii) First case: B>=3A, B<=-A (both positive)
Second Case: B>=-A, B<=3A (both negative)
(iii)First Case: A>=-1, B>=-1, B>=3A, B<=-A
Second Case: A<=1, B<=1, B>=-A, B<=3A

Yea something like that

Carrotsticks said:
Even for the average 4U student, that would be too difficult.

Ah well, ok

Sy123 · Mar 25, 2014

Re: HSC 2014 4U Marathon

$\\ $For some complex$ \ z \ $and real$ \ x \ $let$ \ z=x+i \ \ $describe the locus in the complex plane of$ \\ $i)$ \ z \\ $ii)$ \ z^2$

Ikki · Mar 26, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $For some complex$ \ z \ $and real$ \ x \ $let$ \ z=x+i \ \ $describe the locus in the complex plane of$ \\ $i)$ \ z \\ $ii)$ \ z^2$

I'm probably going to sound stupid, but ok z=x+i.
Locus of z=x+i, is that just all of the plane? 0.0 Liek all of x values right, so adding i doesn't really do much...
Man i need to revise my complex.
K, sub any x
Then will lie in y=i. So just line y=i? Horizontal line at i?

Sy123 · Mar 26, 2014

Re: HSC 2014 4U Marathon

Ikki said:
I'm probably going to sound stupid, but ok z=x+i.
Locus of z=x+i, is that just all of the plane? 0.0 Liek all of x values right, so adding i doesn't really do much...
Man i need to revise my complex.
K, sub any x
Then will lie in y=i. So just line y=i? Horizontal line at i?

Yep the line y=1 (When using x,y we are talking about reala numbers so it doesn't make sense to say y=i)

Think of it this way

z is some number on the complex plane, where the real part can be anything (x is real), but its imaginary part has to always be 1, this is of course y=1

Try the second one

Sy123 · Mar 27, 2014

Re: HSC 2014 4U Marathon

$\\ $i) Find the area underneath the curve$ \ f(x) = \frac{1}{x} \ $for$ \ 1 \leq x \leq n \\ $ii) Find the volume when the area in (i) is rotated about the$ \ x$-axis by 360^{\circ}$ \\ $iii) What happens when$ \ n \to \infty \ $what does this mean?$$

Carrotsticks · Mar 27, 2014

Re: HSC 2014 4U Marathon

http://en.wikipedia.org/wiki/Gabriel's_Horn

Ikki · Mar 27, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
Yep the line y=1 (When using x,y we are talking about reala numbers so it doesn't make sense to say y=i)

Think of it this way

z is some number on the complex plane, where the real part can be anything (x is real), but its imaginary part has to always be 1, this is of course y=1

Try the second one

Ok, z^2=(x+i)^2
Using the same logic, it is the line squared so a parabola with vertex at y=1?

Sy123 · Mar 27, 2014

Re: HSC 2014 4U Marathon

Ikki said:
Ok, z^2=(x+i)^2
Using the same logic, it is the line squared so a parabola with vertex at y=1?

So $z^2 = (x+i)^2 = (x^2-1) + i \cdot 2x$

What kind of parabola would it be? How did you know it was a parabola?

Ikki · Mar 27, 2014

Re: HSC 2014 4U Marathon

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.

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