# Thread: HSC 2017 MX2 Marathon

1. ## Re: HSC 2017 4U Marathon

$\noindent Prove: \\\\ \left( x^{n-1} + \frac{1}{x^{n-1}} \right) + 2 \left( x^{n-2} + \frac{1}{x^{n-2}} \right) + \cdots + (n-1) \left( x + \frac{1}{x} \right) + n \equiv \frac{1}{x^{n-1}} \left(\frac{x^n -1}{x-1}\right)^2 \\\\ Hence, or otherwise, prove: \\\\ \cos{(n-1)\theta} + 2\cos{(n-2)\theta} + \cdots + (n-1)\cos{\theta} + \frac{n}{2} \equiv \frac{1}{2} \times \frac{1+\cos{n \theta}}{1+\cos{\theta}}$
Is x a real or complex number?

2. ## Re: HSC 2017 4U Marathon

Originally Posted by frog1944
Is x a real or complex number?
It doesn't matter for the first proof.

3. ## Re: HSC 2017 4U Marathon

Originally Posted by mini8658
For z^5-1=0, the roots are 1, w^2-w^4, and
z^5-1=(z-1)(z-cis2pi/5)(z-cis-2pi/5)(z-cis4pi/5)(z-cis-4pi/5) How do you get back to the LHS from expanding the RHS?
By inspection.

4. ## Re: HSC 2017 4U Marathon

Originally Posted by mini8658
For z^5-1=0, the roots are 1, w^2-w^4, and
z^5-1=(z-1)(z-cis2pi/5)(z-cis-2pi/5)(z-cis4pi/5)(z-cis-4pi/5) How do you get back to the LHS from expanding the RHS?
This is true by definition.
$p(x) = (x-x_0)(x-x_1)...(x-x_n) Where x_i , i = 0,1,2,3...n are roots of p(x)$

5. ## Re: HSC 2017 4U Marathon

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How would you sketch this? It's a HSC question.

6. ## Re: HSC 2017 4U Marathon

Originally Posted by pikachu975
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How would you sketch this? It's a HSC question.
The first locus is a Half- plane locus. You sketch the original locus and you add in the inequality.

The 2nd locus is an Argand locus. It has the ranges of pi/4 and -pi/4.

7. ## Re: HSC 2017 4U Marathon

Originally Posted by pikachu975
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How would you sketch this? It's a HSC question.
The first part is a linear line that is a perpendicular bisector from points that subtend from 0 and 2. (make z=x+iy and you will get a linear eqn)
Second part is just the argument from the origin from -pi/4 to pi on 4

equate the two inequalities

8. ## Re: HSC 2017 4U Marathon

Originally Posted by si2136
The first locus is a Half- plane locus. You sketch the original locus and you add in the inequality.

The 2nd locus is an Argand locus. It has the ranges of pi/4 and -pi/4.
Originally Posted by Rathin
The first part is a linear line that is a perpendicular bisector from points that subtend from 0 and 2. (make z=x+iy and you will get a linear eqn)
Second part is just the argument from the origin from -pi/4 to pi on 4

equate the two inequalities
Oh yeah i get it now thanks!

9. ## Re: HSC 2017 4U Marathon

Originally Posted by Rathin
a linear line

10. ## Re: HSC 2017 4U Marathon

Originally Posted by seanieg89
In sciences a LINE of best fit can be a linear line of best fit or a curve of best fit

11. ## Re: HSC 2017 4U Marathon

Originally Posted by Rathin
In sciences a LINE of best fit can be a linear line of best fit or a curve of best fit
Actually, I don't even know why a line of best fit can be a curve. I would just call a curve of best fit its own seperate thing. (Polynomial of best fit, blah blah)

12. ## Re: HSC 2017 4U Marathon

Originally Posted by Rathin
In sciences a LINE of best fit can be a linear line of best fit or a curve of best fit
That's just hazy high school science definitions. There are various precise definitions of lines/curves of best fit (regression analysis, interpolating polynomials etc), and whenever such an object is named a line it is definitely a line in the usual sense of the word (at least as far as I have seen).

13. ## Re: HSC 2017 4U Marathon

Is there a way to expand it so the cis2kpi/5's cancel out, leaving only the z^5-1?

14. ## Re: HSC 2017 4U Marathon

Quick question: When we are finding the roots of unity (or any complex number) and leaving it in mod-arg form do we have to change the argument to the principle range or just leave it as it as (e.g. for k=5 arg = 15pi/8). The answers in past papers leave it as it is...so not sure what to do.

15. ## Re: HSC 2017 4U Marathon

Originally Posted by Rathin
Quick question: When we are finding the roots of unity (or any complex number) and leaving it in mod-arg form do we have to change the argument to the principle range or just leave it as it as (e.g. for k=5 arg = 15pi/8). The answers in past papers leave it as it is...so not sure what to do.
I always converted it to the principle range, technically you'd be right, but it's better to be safe.

16. ## Re: HSC 2017 4U Marathon

Originally Posted by Rathin
Quick question: When we are finding the roots of unity (or any complex number) and leaving it in mod-arg form do we have to change the argument to the principle range or just leave it as it as (e.g. for k=5 arg = 15pi/8). The answers in past papers leave it as it is...so not sure what to do.
I like to use -2, -1, 0, 1, 2 instead of 0, 1, 2, 3, 4 because it'll always give you arguments in the principal argument range.

Just a side note it's principal argument not principle

17. ## Re: HSC 2017 4U Marathon

$\noindent Let n = 2m be a positive even integer and let x_{r} = \sin \left(\frac{\pi r^{2}}{2n}\right) for r = 0,1,\ldots, n-1. For integers j\in \left\{0,1,\ldots, n-1\right\}, define \xi_{j} = \sum_{r = 0}^{n-1}x_{r} \mathrm{cis}\left(-\frac{2\pi j r}{n}\right). Find and prove a relationship between:$

$a) \xi_{j} and \xi_{n-j} \quad (0\leq j \leq n)$

$b) \xi_{j} and \xi_{m - j} \quad (0\leq j \leq m).$

18. ## Re: HSC 2017 4U Marathon

$\noindent Show that the roots of the quadratic equation az^{2} + bz + c = 0 (a,b,c real, a\neq 0) both have negative real parts if and only if a,b,c are either all positive or all negative.$

19. ## Re: HSC 2017 4U Marathon

Originally Posted by InteGrand
$\noindent Show that the roots of the quadratic equation az^{2} + bz + c = 0 (a,b,c real, a\neq 0) both have negative real parts if and only if a,b,c are either all positive or all negative.$
This is probably a vague explanation but...

In the case where both roots are real.
The sum of roots = -b/a, which is negative if a,b are either both positive or both negative. This implies that at least one of the roots is negative.
Product of roots = c/a, if the roots are both negative, then the product must be positive therefore a,c must be both positive or both negative.

If the roots are complex, they must be in conjugate pairs due to real coefficients, this means that the real parts must be the same.
So if the real part is negative, then the sum of roots must be negative, and the product of roots must be positive.

20. ## Re: HSC 2017 4U Marathon

here's a nice binomial followed by a harder binomial
133.PNG
134.PNG

21. ## Re: HSC 2017 MX2 Marathon

This geometry question doesn't involve any advanced knowledge but requires a bit of creativity.

ABCD is a quadrilateral with three equal sides AB,BC and CD. Show that the mid-point of AD lies on a circle with diameter BC if and only if the area of ABCD is a quarter of the product of its diagonals.

22. ## Re: HSC 2017 MX2 Marathon

A question posted in the 2016 MX2 marathon by Paradoxica that wasn't answered:
$\lim_{x \to 0} \frac{\sqrt{1- \cos x^2}}{1 - \cos x}$

23. ## Re: HSC 2017 MX2 Marathon

Originally Posted by frog1944
A question posted in the 2016 MX2 marathon by Paradoxica that wasn't answered:
$\lim_{x \to 0} \frac{\sqrt{1- \cos x^2}}{1 - \cos x}$
It approaches infinity.

lim x--> 0, sin x / (1-cos x)

= lim x--> 0 (1+cos x) / sin x

Apply l'hopital, sin^2 x, sin^4 x, all approach infinity.

Therefore the limit approaches infinity.

24. ## Re: HSC 2017 MX2 Marathon

Originally Posted by si2136
It approaches infinity.

lim x--> 0, sin x / (1-cos x)

= lim x--> 0 (1+cos x) / sin x

Apply l'hopital, sin^2 x, sin^4 x, all approach infinity.

Therefore the limit approaches infinity.

$\noindent With the question written the way it is, it looks more like \cos\left(x^{2}\right) inside the radical than \left(\cos x\right)^{2} (if the the latter were intended, it would normally be written like \cos^{2} x).$

25. ## Re: HSC 2017 MX2 Marathon

Suppose you have n points on a circle such that no three distinct chords coincide at any single point.

How many regions do the nC2 chords divide the interior of the circle into?

(You don't have to provide rigorous proof for this question if you can guess the answer correctly).

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