HSC 2017 MX2 Marathon (archive) (1 Viewer)

Kingom · Oct 17, 2016

Re: HSC 2017 4U Marathon

Drsoccerball said:
Did you assume |z| = 1 ?

no i just multiplied by z+1/z+1

Drsoccerball · Oct 17, 2016

Re: HSC 2017 4U Marathon

$\frac{z + 1}{z-1} \cdot \frac{\bar{z} -1}{\bar{z} - 1}$

$\frac{z\bar{z} - z + \bar{z} - 1}{z\bar{z} -z - \bar{z} + 1}$

$$ Take note that : $ z\bar{z} = x^2 + y^2$

$\frac{x^2 + y^2 - 1 - 2iy }{x^2 + y^2 -2x + 1}$

$\frac{x^2 + y^2 -1}{x^2 + y^2 +1 -2x} -2i \cdot \frac{y}{x^2 + y^2 +1 -2x}$

Drsoccerball · Oct 17, 2016

Re: HSC 2017 4U Marathon

leehuan said:
$\text{Let }z,w\in \mathbb{C}$

$\\\text{a) Prove that }|Re(z\overline{w})|<|z||w|\\ \text{b) Simplify }z\overline{z}\\ \text{c) Hence prove the triangle inequality }|z+w| \le |z|+|w|$

Ill get the ball rolling... THE SOCCERBALL ROLLING!!! YOU GET IT?????! pls save me from finals...
i) $$ Let $ z = x + iy $ and $ w = a + ib$

$$ By inspection notice that: $ (ay - bx)^2 > 0$

$a^2y^2 -2abxy +b^2x^2 > 0$

$a^2x^2 +a^2y^2 +b^2x^2 +b^2y^2 > a^2x^2 +2abxy +b^2y^2$

$(a^2 + b^2)(x^2 + y^2) > (ax+by)^2$

$|z|^2|w|^2 > (Re(z\bar{w}))^2$

$|z||w| > |Re(z\bar{w})|$

leehuan · Oct 18, 2016

Re: HSC 2017 4U Marathon

$\text{a) Let }z=r_1(\cos \theta_1+i\sin \theta_1), w=r_2(\cos \theta_2+i\sin \theta_2)$

$\begin{align*}Re(z\overline{w})&=Re(r_1r_2(\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)))\\ &= r_1r_2(\cos (\theta_1+\theta_2))\\ &\le r_1r_2\\ &=|z||w|\end{align*}$

Sorry about the ≤ mistake

InteGrand · Oct 18, 2016

si2136 · Oct 18, 2016

Re: HSC 2017 4U Marathon

Are calculators allowed in the 4u test

Drsoccerball · Oct 18, 2016

Re: HSC 2017 4U Marathon

si2136 said:
Are calculators allowed in the 4u test

Yes.

Green Yoda · Oct 25, 2016

Re: HSC 2017 4U Marathon

Prove De Moivre's Theorem by Mathematical Induction where |z|=1

Paradoxica · Nov 1, 2016

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}}$

Paradoxica · Nov 1, 2016

Re: HSC 2017 4U Marathon

$$\noindent Prove: $ \lim_{t \to 0} \left( 4e^{-t^4} -3e^{-t^3} - 2e^{-t^2} + e^{-t} \right)\log{t} = 0 \\\\ $You may assume without proof that $ \lim_{t \to 0} t^a \log{t} = 0, \\\\$ where $a$ is any positive real number.$ \\\\ $Using Integration by parts and the substitutions: $ x = t^4, y = t^3, z = t^2, \\\\ $Prove: $ \int_0^\infty \frac{4e^{-t^4} - 3e^{-t^3} - 2e^{-t^2} + e^{-t}}{t} \text{d}t = 0$

TaxiCab1729 · Nov 6, 2016

Re: HSC 2017 4U Marathon

For the first part about the limit:

The best I could come up with was that (based on what my teacher said) is that all exponential functions dominate over all polynomial functions which dominate over all log functions. Therefore, based on the second line of the statement, because both g(t) (the function g denotes that left hand expression just before log(t)) and f(t) = t^a both tend to 0, and f(t) dominates over log(t), it must follow that the limit is true because g(t) also tends to 0 and takes precedence over the log function.

Hopefully that made sense...

I'm still trying to practice on latex - I got the second part but I don't think it technically uses parts...

Paradoxica · Nov 6, 2016

Re: HSC 2017 4U Marathon

TaxiCab1729 said:
For the first part about the limit:

The best I could come up with was that (based on what my teacher said) is that all exponential functions dominate over all polynomial functions which dominate over all log functions. Therefore, based on the second line of the statement, because both g(t) (the function g denotes that left hand expression just before log(t)) and f(t) = t^a both tend to 0, and f(t) dominates over log(t), it must follow that the limit is true because g(t) also tends to 0 and takes precedence over the log function.

Hopefully that made sense...

I'm still trying to practice on latex - I got the second part but I don't think it technically uses parts...

You will need to flesh out your explanation more for it to be valid.

The idea is to use the inequalities:

$1+x \le e^x \le \frac{1}{1-x} \quad \forall x<1$

To bound the function behaviour near the origin and squeeze both sides.

TaxiCab1729 · Nov 6, 2016

Re: HSC 2017 4U Marathon

Paradoxica said:
The idea is to use the inequalities:

$1+x \le e^x \le \frac{1}{1-x} \quad \forall x<1$

To bound the function behaviour near the origin and squeeze both sides.

Like, is anyone even gonna get this?

leehuan · Nov 8, 2016

Re: HSC 2017 4U Marathon

TaxiCab1729 said:
Like, is anyone even gonna get this?

This is a marathon. They purposely throw in questions that can get overcomplicated. 100% of the time they are doable using just HSC techniques however if it were the HSC exam you'd be guided step by step

Paradoxica · Nov 12, 2016

Re: HSC 2017 4U Marathon

$$\noindent a) Prove $\binom{n}{0}^2 + \binom{n}{1}^2 + \cdots + \binom{n}{n-1}^2 + \binom{n}{n}^2 = \binom{2n}{n} \\\\ $b) $a,d$ are non-negative real numbers. $a_0, a_1, \cdots , a_{n-1}, a_n,$ are non-negative real numbers in arithmetic progression. $a_0 = a$, and the common difference is $d$. Evaluate: $ \\\\ a_0 \binom{n}{0}^2 + a_1 \binom{n}{1}^2 + \cdots + a_{n-1} \binom{n}{n-1}^2 + a_n \binom{n}{n}^2$

calamebe · Nov 16, 2016

Re: HSC 2017 4U Marathon

Paradoxica said:
$$\noindent a) Prove $\binom{n}{0}^2 + \binom{n}{1}^2 + \cdots + \binom{n}{n-1}^2 + \binom{n}{n}^2 = \binom{2n}{n} \\\\ $b) $a,d$ are non-negative real numbers. $a_0, a_1, \cdots , a_{n-1}, a_n,$ are non-negative real numbers in arithmetic progression. $a_0 = a$, and the common difference is $d$. Evaluate: $ \\\\ a_0 \binom{n}{0}^2 + a_1 \binom{n}{1}^2 + \cdots + a_{n-1} \binom{n}{n-1}^2 + a_n \binom{n}{n}^2$

If anyone wants a hint, consider the expansion (1+x)^2n and what is is equal to for a), and for b), note that nCk=nC(n-k), and group together the like terms. Do consider the even and odd values of n separately for ease.

Paradoxica · Nov 16, 2016

Re: HSC 2017 4U Marathon

calamebe said:
If anyone wants a hint, consider the expansion (1+x)^2n and what is is equal to for a), and for b), note that nCk=nC(n-k), and group together the like terms. Do consider the even and odd values of n separately for ease.

There is no need to consider parity cases if you're at that level.

calamebe · Nov 16, 2016

Re: HSC 2017 4U Marathon

Paradoxica said:
There is no need to consider parity cases if you're at that level.

Eh, it works.

leehuan · Nov 16, 2016

Re: HSC 2017 4U Marathon

$\begin{align*}&\quad \sum_{k=0}^n(a+kd)\binom{n}{k} ^2\\ &= a\binom{2n}{n}+d \sum_{k=0}^nk\binom{n}{k} ^2\end{align*}$

$\text{Meh. Bit lost on how to tackle that without considering parity due to the existence of }\binom{n}{\frac{n}{2}} \\ \text{Odd:}\\ \begin{align*}&\sum_{k=0}^nk\binom{n}{k} ^2\\ &= \binom{n}{1}+2\binom{n}{2}+\dots+\frac{n-1}{2}\binom{n}{\frac{n-1}{2}}+\dots+(n-2)\binom{n}{n-2}+(n-1)\binom{n}{n-1}+n\binom{n}{n}\\ &= n\sum_{k=0}^{\frac{n-1}{2}}\binom{n}{k} \end{align*}\\ \text{but }\sum_{k=0}^{\frac{n-1}2}\binom{n}{k}=\frac{1}{2}\sum_{k=0}^n\binom{n}{k}=2^n$

And I probably did something wrong.

calamebe · Nov 16, 2016

Re: HSC 2017 4U Marathon

leehuan said:
$\begin{align*}&\quad \sum_{k=0}^n(a+kd)\binom{n}{k} ^2\\ &= a\binom{2n}{n}+d \sum_{k=0}^nk\binom{n}{k} ^2\end{align*}$

$\text{Meh. Bit lost on how to tackle that without considering parity due to the existence of }\binom{n}{\frac{n}{2}} \\ \text{Odd:}\\ \begin{align*}&\sum_{k=0}^nk\binom{n}{k} ^2\\ &= \binom{n}{1}+2\binom{n}{2}+\dots+\frac{n-1}{2}\binom{n}{\frac{n-1}{2}}+\dots+(n-2)\binom{n}{n-2}+(n-1)\binom{n}{n-1}+n\binom{n}{n}\\ &= n\sum_{k=0}^{\frac{n-1}{2}}\binom{n}{k} \end{align*}\\ \text{but }\sum_{k=0}^{\frac{n-1}2}\binom{n}{k}=\frac{1}{2}\sum_{k=0}^n\binom{n}{k}=2^n$

And I probably did something wrong.

Yeah you left off the squares. Put the squares on, then use the result from a) and you should be fine.

EDIT: Though I don't see a need to break the terms up as you did in the second line. You can do it easily without breaking it up.

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