HSC 2013 MX2 Marathon (archive) (1 Viewer)

themoonspretty · Sep 1, 2013

Re: HSC 2013 4U Marathon

Prove the volume of sphere

asianese · Sep 1, 2013

Trebla said:
ftfy

O yep heh

asianese · Sep 2, 2013

Re: HSC 2013 4U Marathon

$P(x) = k(x-a_1)(x-a_2)\cdots(x-a_n) \\ P(-x) = k(-x-a_1)(-x-a_2)\cdots(-x-a_n)$

Any takers?

HeroicPandas · Sep 2, 2013

Re: HSC 2013 4U Marathon

asianese said:
$P(x) = k(x-a_1)(x-a_2)\cdots(x-a_n) \\ P(-x) = k(-x-a_1)(-x-a_2)\cdots(-x-a_n)$

Any takers?

Multiply them?

Spring Chicken · Sep 2, 2013

Re: HSC 2013 4U Marathon

Do you guys have the book in which you got these questions from? Thanks.

Sy123 · Sep 2, 2013

Re: HSC 2013 4U Marathon

$Prove using differentiation by first principles the following$

$$i)$ \ \frac{d}{dx} \ e^x = e^x$

$$ii)$ \ \frac{d}{dx} \ \sin x = \cos x$

$$Using the fact that$ \ e^x =\lim_{n \to \infty}\left( 1 + \frac{x}{n} \right )^n$

$$iii)$ \ \frac{d}{dx} \ \ln x = \frac{1}{x}$

Sy123 · Sep 2, 2013

Re: HSC 2013 4U Marathon

Spring Chicken said:
Do you guys have the book in which you got these questions from? Thanks.

A lot of the inequalities I post here I take directly from the books:
- Mathematical Circles (Russian Experience)
- The Art and Craft of Problem Solving - Paul Zeitz

Some other questions I 'make', by copying a Wikipedia article
Some of the sums that I post come from the book:
- Summation of Series - Jolley

These books have many non-syllabus topics though

Otherwise they are taken from trials and modified sometimes to make it harder or genuinely I sometimes make my own questions. (such as some of the binomials in the 3U marathon and many questions in the 2U marathon)

study1234 · Sep 2, 2013

Re: HSC 2013 4U Marathon

ABCD is a quadrilateral inscribed in a circle. BA and CD when produced meet at P. O the centre of the circle PAC. Prove that BD is perpendicular to OP. (Patel Ex 10C Question 5)

JJ345 · Sep 3, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$i) Prove$ \ \ \frac{\sin x}{x} = \cos \left( \frac{x}{2} \right ) \cdot \cos \left( \frac{x}{4} \right) \cdot \cos \left( \frac{x}{8} \right ) \cdot \cos \left( \frac{x}{16} \right ) \cdot \dots$

$ii) Hence prove that$

$\pi = 2\cdot \frac{2}{\sqrt{2}} \cdot \frac{2}{\sqrt{2+\sqrt{2}}} \cdot \frac{2}{\sqrt{2 + \sqrt{2+\sqrt{2}}}} \cdot \frac{2}{\sqrt{2+ \sqrt{2 + \sqrt{2+ \sqrt{2}}}}} \cdot \dots$

Using double angle formula:
sin(x)=2cos(x/2)sin(x/2)
=2cos(x/2)[2sin(x/4)cos(x/4)]
=..........
This process is repeated till the nth term so
sin(x)=2^n [cos(x/2)cos(x/4)....cos(x/2^n)]sin(x/2^n)

Divide both sides by x
sin(x)/x=[cos(x/2)cos(x/4)....cos(x/2^n)]sin(x/2^n)/x/2^n

As n-->infinity sin(x/2^n)/x/2^n-->1
So we get
sin(x)/x=cos(x/2)cos(x/4)....
For lim n-->infinity
ii) I'm guessing we sub x=pi/2 then play around with it?

Sy123 · Sep 3, 2013

Re: HSC 2013 4U Marathon

JJ345 said:
Using double angle formula:
sin(x)=2cos(x/2)sin(x/2)
=2cos(x/2)[2sin(x/4)cos(x/4)]
=..........
This process is repeated till the nth term so
sin(x)=2^n [cos(x/2)cos(x/4)....cos(x/2^n)]sin(x/2^n)

Divide both sides by x
sin(x)/x=[cos(x/2)cos(x/4)....cos(x/2^n)]sin(x/2^n)/x/2^n

As n-->infinity sin(x/2^n)/x/2^n-->1
So we get
sin(x)/x=cos(x/2)cos(x/4)....
For lim n-->infinity
ii) I'm guessing we sub x=pi/2 then play around with it?

Yep that's it, well done.

For (ii) that is pretty much what you do

===

$$Prove for any real$ \ x,y$

$\frac{x^6}{y^2} + \frac{y^6}{x^2} \geq x^4 + y^4$

RealiseNothing · Sep 3, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$Prove for any real$ \ x,y$

$\frac{x^6}{y^2} + \frac{y^6}{x^2} \geq x^4 + y^4$

We know that:

$a+b \geq 2\sqrt{ab}$

Now let $a=x^2y^6$ and $b=x^6y^2$

$x^2y^6 + x^6y^2 \geq 2\sqrt{x^8y^8}$

$x^2y^6 + x^6y^2 \geq 2x^4y^4$

Multiply both sides by 3 to establish the inequality:

$3x^2y^6 + 3x^6y^2 \geq 6x^4y^4$

Now consider:

$(x^2-y^2)^4 \geq 0$

$x^8 - 4x^6y^2 + 6x^4y^4 - 4x^2y^6 + y^8 \geq 0$

$x^8 + y^8 + 6x^4y^4 \geq 4x^6y^2 + 4x^2y^6$

Split the RHS up into the following:

$x^8 + y^8 + 6x^4y^4 \geq x^6y^2 + x^2y^6 + 3x^6y^2 + 3x^2y^6$

$x^8 + y^8 + 6x^4y^4 - 3x^6y^2 - 3x^2y^6 \geq x^6y^2 + x^2y^6$

$x^8 + y^8 - (3x^6y^2 + 3x^2y^6 - 6x^4y^4) \geq x^6y^2 + x^2y^6$

Now using the inequality we establish before, we get:

$3x^6y^2 + 3x^2y^6 - 6x^4y^4 \geq 0$

Let this be A:

$x^8 + y^8 - A \geq x^6y^2 + x^6y^2$

Hence since $A \geq 0$

$x^8+y^8 \geq x^6y^2 + x^2y^6$

$x^8+y^8 \geq x^2y^2(x^4 + y^4)$

$\frac{x^8+y^8}{x^2y^2} \geq x^4 + y^4$

$\frac{x^6}{y^2} + \frac{y^6}{x^2} \geq x^4 + y^4$

obliviousninja · Sep 3, 2013

Re: HSC 2013 4U Marathon

Holy shit, would have never gotten that. RealiseNothing, can you discuss you logic as to how you approach inequality questions.

RealiseNothing · Sep 3, 2013

Re: HSC 2013 4U Marathon

obliviousninja said:
Holy shit, would have never gotten that. RealiseNothing, can you discuss you logic as to how you approach inequality questions.

The biggest part is really just intuition lol. The fact that all the powers are even ( $x^2~x^6~x^8$ ) combined with the symmetry of everything made it look like some sort of binomial expansion.

Like we had $x^6y^2$ and $x^2y^6$ which just screams out binomial expansion.

Sy123 · Sep 3, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
We know that:

$a+b \geq 2\sqrt{ab}$

Now let $a=x^2y^6$ and $b=x^6y^2$

$x^2y^6 + x^6y^2 \geq 2\sqrt{x^8y^8}$

$x^2y^6 + x^6y^2 \geq 2x^4y^4$

Multiply both sides by 3 to establish the inequality:

$3x^2y^6 + 3x^6y^2 \geq 6x^4y^4$

Now consider:

$(x^2-y^2)^4 \geq 0$

$x^8 - 4x^6y^2 + 6x^4y^4 - 4x^2y^6 + y^8 \geq 0$

$x^8 + y^8 + 6x^4y^4 \geq 4x^6y^2 + 4x^2y^6$

Split the RHS up into the following:

$x^8 + y^8 + 6x^4y^4 \geq x^6y^2 + x^2y^6 + 3x^6y^2 + 3x^2y^6$

$x^8 + y^8 + 6x^4y^4 - 3x^6y^2 - 3x^2y^6 \geq x^6y^2 + x^2y^6$

$x^8 + y^8 - (3x^6y^2 + 3x^2y^6 - 6x^4y^4) \geq x^6y^2 + x^2y^6$

Now using the inequality we establish before, we get:

$3x^6y^2 + 3x^2y^6 - 6x^4y^4 \geq 0$

Let this be A:

$x^8 + y^8 - A \geq x^6y^2 + x^6y^2$

Hence since $A \geq 0$

$x^8+y^8 \geq x^6y^2 + x^2y^6$

$x^8+y^8 \geq x^2y^2(x^4 + y^4)$

$\frac{x^8+y^8}{x^2y^2} \geq x^4 + y^4$

$\frac{x^6}{y^2} + \frac{y^6}{x^2} \geq x^4 + y^4$

Well done.

Alternatively:

Re-arranging:

$x^4 \left( \frac{x^2}{y^2} - 1 \right ) + y^4 \left( \frac{y^2}{x^2} - 1\right)$

$(x^2-y^2) \left(\frac{x^6-y^6}{x^2y^2} \right )$

If x > y, then the both brackets are positive, multiplied together to become positive
If x < y, then both brackets are negative, multiplied together to become positive.

Therefore always positive.

Sy123 · Sep 3, 2013

Re: HSC 2013 4U Marathon

$Find$

$\sum_{k=1}^n \frac{1}{k(k+1)(k+2)}$

asianese · Sep 3, 2013

Re: HSC 2013 4U Marathon

asianese said:
$P(x) = k(x-a_1)(x-a_2)\cdots(x-a_n) \\ P(-x) = k(-x-a_1)(-x-a_2)\cdots(-x-a_n)$

Any takers?

HeroicPandas said:
Multiply them?

Yep. So it'd just be $-P(x)P(-x)$ .

I hope this is what sy was asking for..

Clearly that function is even, so we've checked that both plus or minus a_i is accounted for.

Sy123 · Sep 3, 2013

Re: HSC 2013 4U Marathon

asianese said:
Yep. So it'd just be $-P(x)P(-x)$ .

I hope this is what sy was asking for..

Clearly that function is even, so we've checked that both plus or minus a_i is accounted for.

Yeah that's what I was going for

This inequality is a good one, it had me stumped for a while:

$Show that$

$\frac{1}{\sqrt{4n}} \leq \left(\frac{1}{2} \right ) \left(\frac{3}{4} \right ) \dots \left(\frac{2n-1}{2n} \right ) < \frac{1}{\sqrt{2n}}$

Sy123 · Sep 4, 2013

Re: HSC 2013 4U Marathon

$Using the squeeze theorem$

$$i) Prove that$ \ \ \lim_{x \to \infty} \frac{\ln x}{x} = 0$

$ii) Prove then that$

$\lim_{n \to \infty} \frac{n}{\sqrt[n]{n!}} = e$

largarithmic · Sep 4, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Yeah that's what I was going for

This inequality is a good one, it had me stumped for a while:

$Show that$

$\frac{1}{\sqrt{4n}} \leq \left(\frac{1}{2} \right ) \left(\frac{3}{4} \right ) \dots \left(\frac{2n-1}{2n} \right ) < \frac{1}{\sqrt{2n}}$

Can be solved by squaring all sides and doing tricks like changing a $\frac{k-1}{k}$ to $\frac{k}{k+1}$ to make it bigger/smaller as appropriate.

Something you mightn't have noticed about the expression in the middle is that it is actually equal to

$\frac{(2n)!}{\left((2)(4)...(2n)\right)^2} = \frac{1}{2^{2n}} \frac{(2n)!}{n! n!}$

which is actually a binomial coefficient (sorry i dont know the tex for them!). However also, $2^{2n}$ is the total number of subsets of $\{1,2,...,2n\}$ , while the binomial coefficient is the total number of $n$ element subsets. In other words, the expression in the middle is the probability a random subset of a 2n-element set has size exactly n.

The result just proven in the question says that... as n gets larger, this probability is bounded between constant multiples of $\frac{1}{\sqrt{n}}$ . I.e. we have an *asymptotic growth estimate*.

So here's a new question which might be great to think about: can you prove using combinatorial / counting / probability estimates (and not savvy algebraic manipulation) that this is true?

Sy123 · Sep 4, 2013

Re: HSC 2013 4U Marathon

largarithmic said:
Can be solved by squaring all sides and doing tricks like changing a $\frac{k-1}{k}$ to $\frac{k}{k+1}$ to make it bigger/smaller as appropriate.

Something you mightn't have noticed about the expression in the middle is that it is actually equal to

$\frac{(2n)!}{\left((2)(4)...(2n)\right)^2} = \frac{1}{2^{2n}} \frac{(2n)!}{n! n!}$

which is actually a binomial coefficient (sorry i dont know the tex for them!). However also, $2^{2n}$ is the total number of subsets of $\{1,2,...,2n\}$ , while the binomial coefficient is the total number of $n$ element subsets. In other words, the expression in the middle is the probability a random subset of a 2n-element set has size exactly n.

The result just proven in the question says that... as n gets larger, this probability is bounded between constant multiples of $\frac{1}{\sqrt{n}}$ . I.e. we have an *asymptotic growth estimate*.

So here's a new question which might be great to think about: can you prove using combinatorial / counting / probability estimates (and not savvy algebraic manipulation) that this is true?

Yep that is the method that I had done.
I did notice the binomial co-efficient in the middle but I'm afraid my permutations/combinations argument skills aren't very good so I couldn't do anything with it. Also you can tex in binomials using the \binom{}{} command.
I'll leave it to others to try and solve that and I'll keep in my comfortable domain of algebra

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