HSC 2016 MX2 Marathon (archive) (1 Viewer)

Paradoxica · May 29, 2016

Re: HSC 2016 4U Marathon

A simple one.

$$\noindent Solve the following system of equations for $\theta$: $\\ \cos{3\theta} + \cos{2\theta} + \cos{\theta} + 1 = 0 \\ \sin{3\theta} + \sin{2\theta} + \sin{\theta} = 0$

integral95 · May 29, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
A simple one.

$$\noindent Solve the following system of equations for $\theta$: $\\ \cos{3\theta} + \cos{2\theta} + \cos{\theta} + 1 = 0 \\ \sin{3\theta} + \sin{2\theta} + \sin{\theta} = 0$

I'm bored so here's how you do itttt

$Consider$ \ \ 1 + cis(\theta) + cis(2 \theta) + cis(3 \theta) = 0 \\ \\ \frac{1-cis(4 \theta)}{1- cis(\theta)} = 0 \\ \\ $Equating the real parts implies that solving for $ \cos{3\theta} + \cos{2\theta} + \cos{\theta} + 1 = 0 $ is the same as solving for$ \ \ 1 - \cos{4 \theta} = 0 \\ \\ $ While equating the imaginaries means $ \ \ \sin{3 \theta} + \sin{2 \theta} + \sin{\theta} = 0 $ \ \ is the same as $ \sin{4 \theta} = 0 $ We would only take the values of theta that satisfies BOTH equations, and I can't be bothered doing it$

EDIT: I think this isn't right lol.

Paradoxica · May 29, 2016

Re: HSC 2016 4U Marathon

integral95 said:
I'm bored so here's how you do itttt

$Consider$ \ \ 1 + cis(\theta) + cis(2 \theta) + cis(3 \theta) = 0 \\ \\ \frac{1-cis(4 \theta)}{1- cis(\theta)} = 0 \\ \\ $Equating the real parts implies that solving for $ \cos{3\theta} + \cos{2\theta} + \cos{\theta} + 1 = 0 $ is the same as solving for$ \ \ 1 - \cos{4 \theta} = 0 \\ \\ $ While equating the imaginaries means $ \ \ \sin{3 \theta} + \sin{2 \theta} + \sin{\theta} = 0 $ \ \ is the same as $ \sin{4 \theta} = 0 $ We would only take the values of theta that satisfies BOTH equations, and I can't be bothered doing it$

EDIT: I think this isn't right lol.

Consider the excess solutions that do not satisfy the given equation and eliminate them.

integral95 · May 29, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
Consider the excess solutions that do not satisfy the given equation and eliminate them.

Oh wait I was right, is the answer

$4 \theta = 2k \pi $ Where k is an integer but 0 $ \\ \\ \theta = k \frac{\pi}{2}$

EDIT: This is because from the above equation the denominator $1-cis(\theta)$ can't be 0 which is why theta can't be 0, dammit how did I not see that?

Paradoxica · May 29, 2016

Re: HSC 2016 4U Marathon

integral95 said:
I'm bored so here's how you do itttt

Well technically one could also do a triple and double angle expansion, then proceed to solve both polynomials simultaneously...

But that's only if you're feeling masochistic.

integral95 · May 29, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
Well technically one could also do a triple and double angle expansion, then proceed to solve both polynomials simultaneously...

But that's only if you're feeling masochistic.

ewwww no.

Paradoxica · May 29, 2016

Re: HSC 2016 4U Marathon

integral95 said:
ewwww no.

Being the masochistic person I am, here is the alternative bashy solution

$$\noindent$ c= \cos{\theta}, s= \sin{\theta} \\ 4c^3 - 3c + 2c^2 -1 + c +1= 0 \Rightarrow (2c^2 + c -1)c = 0 \\ 3s -4s^3 +2sc + s = 0 \Rightarrow$

The solutions to the cosine equation can be found very easily. These can be substituted into the second equation and then the second equation can be solved for the corresponding solutions.

Paradoxica · Jun 5, 2016

Re: HSC 2016 4U Marathon

$$\noindent Use the AM-GM Inequality to construct an upper bound on $\sqrt[n]{n}$, and hence, using the Squeeze Theorem, show $ \lim_{n \to \infty} \sqrt[n]{n} = 1$

Drsoccerball · Jun 6, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
$$\noindent Use the AM-GM Inequality to construct an upper bound on $\sqrt[n]{n}$, and hence, using the Squeeze Theorem, show $ \lim_{n \to \infty} \sqrt[n]{n} = 1$

lol

seanieg89 · Jun 6, 2016

Re: HSC 2016 4U Marathon

(In this question you need to know how tennis scoring works.)

In a basic model of a tennis match, every point is won by the server with probability p and the receiver with probability 1-p.

a) What is the probability of winning a service game in terms of p?

b) Suppose it is known that both sides win equally often from the game score of 0-30. Approximate the value of p that this condition forces.

c) Approximate the probability of winning a service game in terms of the p found in part b.

(Use Wolfram Alpha if you like for the approximations, unless you are feeling particularly algebraically macho and would like to do it using something like Newton's.)

seanieg89 · Jun 7, 2016

Re: HSC 2016 4U Marathon

Suppose you have never heard of exponentials or logarithms before.

Define for positive reals x, the function $L(x):=\int_1^x \frac{dt}{t}.$

Prove:

a) L(1)=0
L(xy)=L(x)+L(y)
L(x^r)=rL(x) for rational r
L is differentiable in its domain and L'(1)=1.

b) Prove that L is the unique function on the positive reals with the properties in a.

c) Prove that L is invertible and denote its inverse function by E(x), what is the domain of E?

d) Prove that E(r)=E(1)^r for rational r. Prove that E'(0)=1. (You may assume without proof that E is differentiable).

e) Suppose another differentiable function A(x) satisfies A(r)=A(1)^r for all rationals r (with A(1) not equal to E(1)). Show that A'(0) is not equal to 1.

hedgehog_7 · Jun 12, 2016

Re: HSC 2016 4U Marathon

The area between y = 2x +1 and y = (x+1)^2 is rotated about y = -1 with x=0 as the boundary axis. Find the volume

Paradoxica · Jun 19, 2016

Re: HSC 2016 4U Marathon

a) For any natural n, find and classify the solutions to the following equation:

$\frac{x^n-1}{x-1}=\frac{x^n+1}{x+1}$

b) Repeat the above for all non-natural integers n.

KingOfActing · Jun 19, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
a) For any natural n, find and classify the solutions to the following equation:

$\frac{x^n-1}{x-1}=\frac{x^n+1}{x+1}$

b) Repeat the above for all non-natural integers n.

For natural $n$ , by "cross multiplying":

$n = 0 \implies \text{ No solution or }x = 0\text{ depending on your definition of } 0^0\\n = 1 \implies x \neq \pm 1\\n \geq 2 \implies x \in \{0\}\cup\{z : z^{n-1} = 1 \wedge z \neq \pm 1\}$

In the case $n < 0:$

$n = -1 \implies \text{No solution}\\n < -1 \implies x \in \{z : z^{1-n} = 1 \wedge z \neq \pm 1\}$

Notice also that the solution for some negative n is the same as the solution for -n + 2.
E.g. n = -2 has the solutions to z^3 = 1 which is the same as n = 4.

Paradoxica · Jun 19, 2016

Re: HSC 2016 4U Marathon

KingOfActing said:
For natural $n$ , by "cross multiplying":

$n = 0 \implies \text{ No solution or }x = 0\text{ depending on your definition of } 0^0\\n = 1 \implies x \neq \pm 1\\n \geq 2 \implies x \in \{0\}\cup\{z : z^{n-1} = 1 \wedge z \neq \pm 1\}$

In the case $n < 0:$

$n = -1 \implies \text{No solution}\\n < -1 \implies x \in \{z : z^{1-n} = 1 \wedge z \neq \pm 1\}$

Notice also that the solution for some negative n is the same as the solution for -n + 2.
E.g. n = -2 has the solutions to z^3 = 1 which is the same as n = 4.

-1 is not necessarily a solution for some values of n anyway, but I think I can accept that non-intersection.

KingOfActing · Jun 19, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
-1 is not necessarily a solution for some values of n anyway, but I think I can accept that non-intersection.

It just becomes vacuously true if it isn't even in the original solution set

Paradoxica · Jun 19, 2016

Re: HSC 2016 4U Marathon

KingOfActing said:
It just becomes vacuously true if it isn't even in the original solution set

also your second set should read n-1 but whatever.

Katebate · Jun 25, 2016

Re: HSC 2016 4U Marathon

In mechanics, banked circular tracks, if there is not lateral force stated, do we just assume it isn't there?

leehuan · Jun 25, 2016

Re: HSC 2016 4U Marathon

Katebate said:
In mechanics, banked circular tracks, if there is not lateral force stated, do we just assume it isn't there?

What do you mean by lateral force?

You are meant to consider friction, normal reaction and gravity when analysing banked tracks, to which the overall effect is uniform circular motion

InteGrand · Jun 25, 2016

Re: HSC 2016 4U Marathon

leehuan said:
What do you mean by lateral force?

You are meant to consider friction, normal reaction and gravity when analysing banked tracks, to which the overall effect is uniform circular motion

By lateral force, Katebate most likely meant the friction force (which is a lateral force, i.e. up or down the bend).

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