HSC 2015 MX2 Marathon (archive) (8 Viewers)

glittergal96 · Jun 13, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Let the polynomial$ \ P(x) \ $be a monic cubic polynomial with non-negative real roots$ \ a,b,c$

$\\ $i) Find$ \ P(a) + P(b) + P(c) \ $in terms of$ \ a,b,c$

$\\ $ii) Prove for non-negative reals$ \ x,y,z \ $that$ \ x^2 + y^2 + z^2 \geq xy + xz + zx$

$\\ $iii) Hence prove$ \ \frac{a^3 + b^3 + c^3}{3} \geq abc \ $for$ \ a,b,c \geq 0$

i) 0, as these are roots of the polynomial.

ii) Expand $\frac{1}{2}((x-y)^2+(x-z)^2+(y-z)^2)\geq 0.$

iii) $3LHS=(a^2+b^2+c^2)(a+b+c)-a^2(b+c)-b^2(a+c)-c^2(a+b)\\ \geq (ab+ac+bc)(a+b+c)-a^2(b+c)-b^2(a+c)-c^2(a+b)=3abc.$

FrankXie · Jun 13, 2015

Re: HSC 2015 4U Marathon

lol part (i) is too simple to be true and is irrelevant, maybe it means find P(a+b+c) i guess ?

Edit: oh he actually meant P(a)+P(b)+P(c), and he wanted this result used for part (iii): P(x)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc, and sub. x=a, b, c and use P(a)+P(b)+P(c)=0, so
$3abc=(a^3+b^3+c^3)-(a+b+c)(a^2+b^2+c^2)+(a+b+c)(ab+bc+ca)$
and using the result of part (ii), the desired result follows.

Trebla · Jun 13, 2015

Re: HSC 2015 4U Marathon

Guys, would appreciate it if you let the current HSC students have some breathing room to participate in this marathon.

Drsoccerball · Jun 13, 2015

Re: HSC 2015 4U Marathon

$$ Prove that $ \lim _{x\rightarrow 0}}xlnx =0$

Trebla · Jun 13, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$$ Prove that $ _{lim_{h->0}}xlnx =0$
EDIT: can someone show me how to code that better?

$$ Prove that $ \lim_{h\rightarrow 0}x\ln x =0$

porcupinetree · Jun 13, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$$ Prove that $ \lim _{x\rightarrow 0}}xlnx =0$

https://imgur.com/4t4ZTxf

The equation x^3 + 3x + 2 has roots A, B and Y. Find the equation with roots A + 1/A, B + 1/B and Y + 1/Y.

Drsoccerball · Jun 13, 2015

Re: HSC 2015 4U Marathon

porcupinetree said:
https://imgur.com/4t4ZTxf

The equation x^3 + 3x + 2 has roots A, B and Y. Find the equation with roots A + 1/A, B + 1/B and Y + 1/Y.

That solution nice

For this question you can do the long method (which i would do for this question)
$x^3 -x^2( $ sum of roots $ ) +x( $ Multiplication of 2's $ )-($ Multiplication of roots $)$
But there has to be a more elegant way

Trebla · Jun 13, 2015

Re: HSC 2015 4U Marathon

porcupinetree said:
https://imgur.com/4t4ZTxf

Can you explain why

$\displaystyle\lim_{x\rightarrow 0} x^x = 1$

Drsoccerball · Jun 13, 2015

Re: HSC 2015 4U Marathon

Trebla said:
Can you explain why

$\displaystyle\lim_{x\rightarrow 0} x^x = 1$

0^0 =1
EDIT: my bad D:

porcupinetree · Jun 13, 2015

Re: HSC 2015 4U Marathon

Trebla said:
Can you explain why

$\displaystyle\lim_{x\rightarrow 0} x^x = 1$

I wasn't familiar with a formal proof of this until a few minutes ago when I looked it up - it can apparently be proven using L'Hopital's rule. However, when I first encountered the original question (lim x -> 0 of xlnx) a while ago, I was happy to accept that it was 1, because I had seen a graph of the function before.

Squar3root · Jun 13, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$$ Prove that $ \lim _{x\rightarrow 0}}xlnx =0$

I remmeber this one from math semester 1 last year because there was an unmotivated trick behind it

kawaiipotato · Jun 13, 2015

Re: HSC 2015 4U Marathon

Trebla said:
$$ Prove that $ \lim_{h\rightarrow 0}x\ln x =0$

$Sketching the curve, y = 1/x$

$$ Upper rectangle $ > \int_{1}^{\sqrt{x}} \frac{dx}{x} > 0$

$0< \int_1^{\sqrt{x}} \frac{dx}{x} < \sqrt{x} - 1$

$0 < ln(\sqrt{x}) < \sqrt{x} - 1$

$0 < ln(x) < 2(\sqrt{x} - 1)$

$Multiplying by x$

$0 < xln(x) < 2(x\sqrt{x} - x)$

$\lim_{x\rightarrow 0} 0 < xln(x) < 2(x\sqrt{x} - x)$

$= 0 < xln(x) < 0$

$By Squeeze Theorem, xlnx approaches 0 as x approaches 0$

lol was in my integration hw

InteGrand · Jun 13, 2015

Re: HSC 2015 4U Marathon

Without L'Hôpital,

We can start off with the limit $\lim _{t\to \infty}t\mathrm{e}^{-t} = 0^+$ (this is allowed to be assumed for HSC I'm pretty sure, but then again, it has L'Hôpital built into it of course).

Then apply the substitution $x=\mathrm{e}^{-t}\Longleftrightarrow t=-\ln x$ , and $x\to 0^+$ as $t\to \infty$ , so $\lim _{t\to \infty}t\mathrm{e}^{-t} = 0^+ \Rightarrow \lim _{x\to 0^+} (-\ln x \cdot x)=0^+ \Rightarrow \lim_{x\to 0^+}(x\ln x)=0^-$ .

Drsoccerball · Jun 14, 2015

Re: HSC 2015 4U Marathon

porcupinetree said:
I wasn't familiar with a formal proof of this until a few minutes ago when I looked it up - it can apparently be proven using L'Hopital's rule. However, when I first encountered the original question (lim x -> 0 of xlnx) a while ago, I was happy to accept that it was 1, because I had seen a graph of the function before.

Well i guess we don't need to prove the limit since ive seen this graph before

Sy123 · Jun 14, 2015

Re: HSC 2015 4U Marathon

$\\ $Show that for real$ \ x \ $and positive integer$ \ n \ $that$ \ e^x \geq \left(1 + \frac{x}{n} \right)^n$

(edit: this inequality only works for x > - n if n is even)

Ekman · Jun 14, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Show that for real$ \ x \ $and positive integer$ \ n \ $that$ \ e^x \geq \left(1 + \frac{x}{n} \right)^n$

$$ Prove: $ e^x \geq \left(1 + \frac{x}{n} \right)^n \Rightarrow x \geq \ln{\left(1 + \frac{x}{n} \right)^n}\Rightarrow x - n\ln{\left(n + x \right)} + n\ln{n} \geq 0$

$$ Let $ f(x) = x - n\ln{\left(n + x \right)} + n\ln{n} \therefore f'(x)=1-\frac{n}{n+x}$

$x=0 $ is the minimum value $ \therefore f(0) = 0-n\ln{n} + n\ln{n} = 0$

$\therefore f(x) \geq 0 $ for all real values of x $$

$$ Hence: $ x - n\ln{\left(n + x \right)} + n\ln{n} \geq 0$

$\therefore e^x \geq \left(1 + \frac{x}{n} \right)^n$

Sy123 · Jun 15, 2015

Re: HSC 2015 4U Marathon

$\\ $Let$ \ x_1,x_2, \dots, x_n \ $be non-negative real numbers$ \\ \\ $i) Explain why$ \ \left( \sum_{i = 1}^n x_i \right)^2 = \sum_{k=1}^{n} \sum_{j=1}^n x_j x_k \\ \\ $ii) Hence or otherwise prove$ \ n \sum_{i =1}^n x_i^2 \geq \left(\sum_{i=1}^n x_i \right)^2$

Sy123 · Jun 18, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Let$ \ x_1,x_2, \dots, x_n \ $be non-negative real numbers$ \\ \\ $i) Explain why$ \ \left( \sum_{i = 1}^n x_i \right)^2 = \sum_{k=1}^{n} \sum_{j=1}^n x_j x_k \\ \\ $ii) Hence or otherwise prove$ \ n \sum_{i =1}^n x_i^2 \geq \left(\sum_{i=1}^n x_i \right)^2$

$\\ $i)$ \ \left( \sum_{i=1}^n x_i \right)^2 = \left(\sum_{i=1}^n x_i \right) \left(\sum_{i=1}^n x_i \right) \\ = x_1 \left(\sum_{i=1}^n x_i \right) + x_2 \left(\sum_{i=1}^n x_i \right) + \dots + x_n \left(\sum_{i=1}^n x_i \right) \\ = \sum_{k=1}^n x_k \left(\sum_{j=1}^n x_j \right) \\ = \sum_{k=1}^{n} \sum_{j=1}^n x_j x_k$

$\\ $ii)$ \ x_k^2 + x_j^2 \geq 2x_k x_j \ $(by the AM-GM inequality)$$

$\\ \Rightarrow \sum_{k=1}^n \sum_{j=1}^n (x_k^2 + x_j^2) \geq \sum_{k=1}^n \sum_{j=1}^n (2x_kx_j)$

$\\ \Rightarrow \sum_{k=1}^n \left( (nx_k^2) + \left(\sum_{j=1}^n x_j^2 \right) \right) \geq 2 \left(\sum_{i=1}^n x_i \right)^2$

$\\ \Rightarrow n \sum_{k=1}^n x_k^2 + n \left(\sum_{j=1}^n x_j^2 \right) \geq 2 \left(\sum_{i=1}^n x_i \right)^2$

$\\ \Rightarrow 2n \sum_{i=1}^n x_i^2 \geq 2 \left(\sum_{i=1}^n x_i \right)^2$

$\\ \Rightarrow n \sum_{i=1}^n x_i^2 \geq \left(\sum_{i=1}^n x_i \right)^2$

InteGrand · Jun 18, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $i)$ \ \left( \sum_{i=1}^n x_i \right)^2 = \left(\sum_{i=1}^n x_i \right) \left(\sum_{i=1}^n x_i \right) \\ = x_1 \left(\sum_{i=1}^n x_i \right) + x_2 \left(\sum_{i=1}^n x_i \right) + \dots + x_n \left(\sum_{i=1}^n x_i \right) \\ = \sum_{k=1}^n x_k \left(\sum_{j=1}^n x_j \right) \\ = \sum_{k=1}^{n} \sum_{j=1}^n x_j x_k$

$\\ $ii)$ \ x_k^2 + x_j^2 \geq 2x_k x_j \ $(by the AM-GM inequality)$$

$\\ \Rightarrow \sum_{k=1}^n \sum_{j=1}^n (x_k^2 + x_j^2) \geq \sum_{k=1}^n \sum_{j=1}^n (2x_kx_j)$

$\\ \Rightarrow \sum_{k=1}^n \left( (nx_k^2) + \left(\sum_{j=1}^n x_j^2 \right) \right) \geq 2 \left(\sum_{i=1}^n x_i \right)^2$

$\\ \Rightarrow n \sum_{k=1}^n x_k^2 + n \left(\sum_{j=1}^n x_j^2 \right) \geq 2 \left(\sum_{i=1}^n x_i \right)^2$

$\\ \Rightarrow 2n \sum_{i=1}^n x_i^2 \geq 2 \left(\sum_{i=1}^n x_i \right)^2$

$\\ \Rightarrow n \sum_{i=1}^n x_i^2 \geq \left(\sum_{i=1}^n x_i \right)^2$

Part (i) is also basically via distributivity (as in one line answer: "distributivity" (although obviously a risk to leave it at that)).

Sy123 · Jun 18, 2015

Re: HSC 2015 4U Marathon

$\\ $i) Prove that for$ \ x \ $sufficiently close to$ \ 0 \ $that$ \ \cos x < \frac{\sin x}{x} < 1$

$\\ $ii) Hence show that$ \ \lim_{x \to 0} \frac{\sin x}{x} = 1$

HSC 2015 MX2 Marathon (archive) (8 Viewers)

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