HSC 2014 MX2 Marathon ADVANCED (archive) (3 Viewers)

glittergal96 · Aug 17, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
A little hard for the lower level marathon but a little easier for this level:

$\\ 0 < u < 1 \\ $Let$ \ u_{n+1} = \frac{1}{u_n} + u \ $where$ \ u_1 = 1 + u \\ \\ $Show that$ \ u_n > 1 \ $for all integral$ \ n \geq 1$

If

$1<u_n\leq 1+u$

then

$1< \frac{1+u+u^2}{1+u}=\frac{1}{1+u}+u\leq \frac{1}{u_n}+u \leq 1+u.$

Noting that the penultimate term is just $u_{n+1}$ , induction does the trick.

TL1998 · Aug 18, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Find:

lim ((2n)! / (n!n^n))^(1/n)
n->infinity

glittergal96 · Aug 18, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Let

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

Then by log laws we have

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

But the limit of this expression as $n\rightarrow \infty$ is exactly the definition of the integral of log(x) between 1 and 2!

So since the exponential function is continuous, we get

$x_n\rightarrow \exp\left(\int_1^2 \log(u) \, du\right)=\exp(2\log(2)-1)=\frac{4}{e}.$

TL1998 · Aug 18, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
Let

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

Then by log laws we have

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

But the limit of this expression as $n\rightarrow \infty$ is exactly the definition of the integral of log(x) between 1 and 2!

So since the exponential function is continuous, we get

$x_n\rightarrow \exp\left(\int_1^2 \log(u) \, du\right)=\exp(2\log(2)-1)=\frac{4}{e}.$

that's what i did. This was question's clearly too easy for the advanced thread. Here's another one which is a little hard for the normal thread but easier than the rest of the questions here.

A function f(x) is integral e --> 1 abs(t-x) e^t dt (0<=x<=1) . Find f(x) and the x value which gives its minimum value.

glittergal96 · Aug 20, 2014

Re: HSC 2014 4U Marathon - Advanced Level

TL1998 said:
that's what i did. This was question's clearly too easy for the advanced thread. Here's another one which is a little hard for the normal thread but easier than the rest of the questions here.

A function f(x) is integral e --> 1 abs(t-x) e^t dt (0<=x<=1) . Find f(x) and the x value which gives its minimum value.

Am a bit confused by your notation here, you should use LaTeX!

Anyway, presuming my guess at what you mean is right (and that the absolute values are irrelevant because x =< 1):

$f(x)=\int_1^e (t-x)e^t\, dt=\int_1^e te^t\, dt -x\int_1^e e^t \, dt=e^{e+1}-e^e -x(e^e-e)$

which is minimised at x=1, because the slope of this line is negative. (Unless you meant integrating from e to 1, in which case change sign of the above and x=0 gives the minimum).

This seems too easy though, so let me know if you meant something else.

TL1998 · Aug 20, 2014

Re: HSC 2014 4U Marathon - Advanced Level

when i say abs(t-x), i'm talking the absolute value of it. so |t-x|. and sorry i meant integrate 1 to 0, with the 1 on top. Don't know why i put an e.

also, contrary to what you had stated, the absolute values do play a role.

To help you, here's the value of f(x): 2e^x -(e+1)x-1

glittergal96 · Aug 20, 2014

Re: HSC 2014 4U Marathon - Advanced Level

TL1998 said:
when i say abs(t-x), i'm talking the absolute value of it. so |t-x|. and sorry i meant integrate 1 to 0, with the 1 on top. Don't know why i put an e.

also, contrary to what you had stated, the absolute values do play a role.

To help you, here's the value of f(x): 2e^x -(e+1)x-1

In the question you originally posted, absolute values didn't play a role. t went between 1 and e and x was smaller than 1. So t-x was always positive.

After you changed the interval of integration, of course the absolute values matter. Now we do the following:

$f(x)=\int_0^x (x-t)e^t \, dt +\int_1^x (x-t)e^t \, dt = 2F(x)-F(1)-F(0)$

$=2e^x-(e+1)x-1$

where

$F(t)=(x+1-t)e^t$

is a primitive of the integrand.

Then by differentiating f, we find that it is stationary at

$\log\left(\frac{1+e}{2}\right).$

And this is the minimum, because f'' > 0 for 0 < x < 1.

TL1998 · Aug 20, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
In the question you originally posted, absolute values didn't play a role. t went between 1 and e and x was smaller than 1. So t-x was always positive.

After you changed the interval of integration, of course the absolute values matter. Now we do the following:

$f(x)=\int_0^x (x-t)e^t \, dt +\int_1^x (x-t)e^t \, dt = 2F(x)-F(1)-F(0)$

$=2e^x-(e+1)x-1$

where

$F(t)=(x+1-t)e^t$

is a primitive of the integrand.

Then by differentiating f, we find that it is stationary at

$\log\left(\frac{1+e}{2}\right).$

And this is the minimum, because f'' > 0 for 0 < x < 1.

Our methods are different, but our answers are the same. Since Sy123 seems to be busy, i'll just post more questions.

Given x>1 or <1/2 and n is an integer, prove the following

(1+ x/(x-1) )^n >= 1+nx/(x-1)

glittergal96 · Aug 20, 2014

Re: HSC 2014 4U Marathon - Advanced Level

TL1998 said:
Our methods are different, but our answers are the same. Since Sy123 seems to be busy, i'll just post more questions.

Given x>1 or <1/2 and n is an integer, prove the following

(1+ x/(x-1) )^n >= 1+nx/(x-1)

Thanks for the questions!

It's easy to show that

$\frac{x}{x-1}>-1 \Leftrightarrow x< 0.5 \textrm{ or } x>1$

by multiplying out and solving the quadratic inequality.

So I only need to show that

$(1+y)^n \geq ny$

for y > -1 and all integers n.

This comes from observing that $x^n$ is a concave up function for $x>0$ (from graphs we have drawn for a long time, or by taking second derivatives), and so the tangent line to this curve at $x=1$ lies below this curve for all $x>0$ .

Since the derivative of $x^n$ at 1 is $n$ , the above fact translates to

$(1+y)^n\geq 1+ny$

for all $y>-1$ as required.

TL1998 · Aug 20, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Yep, that also works. note that equality holds iff x=0.

given 0<x<y, determine which of (1+x)^y or (1+y)^x is larger

Sy123 · Aug 21, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $If$ \ a+b+c+d = 1 \\ \\ $Prove$ \ \ \frac{a^2}{a+b} + \frac{b^2}{b+c} + \frac{c^2}{c+d} + \frac{d^2}{d+a} \geq \frac{1}{2}$

glittergal96 · Aug 21, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $If$ \ a+b+c+d = 1 \\ \\ $Prove$ \ \ \frac{a^2}{a+b} + \frac{b^2}{b+c} + \frac{c^2}{c+d} + \frac{d^2}{d+a} \geq \frac{1}{2}$

I think this is only necessarily true if a,b,c,d are positive so I will make that assumption.

For real $a_j,b_j$ , the polynomial

$\sum_{k=1}^n (a_kx-b_k)^2=\left(\sum_{k=1}^n a_k^2\right)x^2-2\left(\sum_{k=1}^n a_kb_k\right)x+\sum_{k=1}^n b_k^2$

is non-negative and so must have non-positive discriminant.

This implies that

$\left(\sum_{k=1}^n a_k^2\right)\left(\sum_{k=1}^n b_k^2\right)\geq \left(\sum_{k=1}^n a_kb_k\right)^2$

upon simplification. (This is called the Cauchy-Schwartz inequality and is very famous).

Now a simple application of Cauchy-Schwartz gives us

$\left( \left(\frac{a}{\sqrt{a+b}}\right)^2+\left(\frac{b}{\sqrt{b+c}}\right)^2+\left(\frac{c}{\sqrt{c+d}} \right)^2+\left(\frac{d}{\sqrt{d+a}}\right)^2 \right) (\sqrt{a+b}^2+\sqrt{b+c}^2+\sqrt{c+d}^2+\sqrt{d+a}^2)\geq (a+b+c+d)^2$

but our assumption that a+b+c+d=1 implies that the second factor on the LHS is 2 and that the RHS is 1.

So we are left with

$\frac{a^2}{a+b} + \frac{b^2}{b+c} + \frac{c^2}{c+d} + \frac{d^2}{d+a} \geq \frac{1}{2}.$

panda15 · Aug 21, 2014

Re: HSC 2014 4U Marathon - Advanced Level

TL1998 said:
Please stick to 4u concepts.

Cauchy-Schwarz is really just a fancy version of the triangle inequality.

Sy123 · Aug 21, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $If$ \ a+b+c+d = 1 \\ \\ $Prove$ \ \ \frac{a^2}{a+b} + \frac{b^2}{b+c} + \frac{c^2}{c+d} + \frac{d^2}{d+a} \geq \frac{1}{2}$

Here is my method which is more 4U level:

$\\ \frac{a^2}{a+b} + \frac{b^2}{b+c} + \frac{c^2}{c+d} + \frac{d^2}{d+a} \\ \\ = \frac{a(a+b-b)}{a+b} + \frac{b(b+c-c)}{b+c} + \frac{c(c+d-d)}{c+d} + \frac{d(d+a-a)}{d+a}\\ \\ = (a+b+c+d) - \frac{ab}{a+b} - \frac{bc}{b+c} - \frac{cd}{c+d} - \frac{da}{d+a} \\ \\ $Since$ \ (x+y)^2 \geq 4xy \ $then$ \ \frac{x+y}{4} \geq \frac{xy}{x+y} \ \Rightarrow \ -\frac{xy}{x+y} \geq -\frac{x+y}{4} \\ \\ \therefore \ (a+b+c+d) - \frac{ab}{a+b} - \frac{bc}{b+c} - \frac{cd}{c+d} - \frac{da}{d+a} \\ \\ \geq (a+b+c+d) - \frac{2a + 2b + 2c + 2d}{4} = \frac{a+b+c+d}{2} = \frac{1}{2}$

glittergal96 · Aug 21, 2014

Re: HSC 2014 4U Marathon - Advanced Level

I didn't assume anything from beyond the course though did I? I just mentioned the name of an intermediate thing I proved.

I proved everything using that squares are non-negative and that non-negative quadratics have non-positive discriminant...this is surely okay?

Sy123 · Aug 22, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
I didn't assume anything from beyond the course though did I? I just mentioned the name of an intermediate thing I proved.

I proved everything using that squares are non-negative and that non-negative quadratics have non-positive discriminant...this is surely okay?

I think what TL1998 was going for is that if you knew only what the 4U course would teach you, how would you do it. Cauchy Schwarz although can be proven with 4U methods is not taught nor is it an inequality that is entirely obvious

dunjaaa · Aug 23, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Does anyone know a book that has deeper complex numbers/polynomial theoretical styled problems like the 2011 Extension 2 HSC Exam Q8 (c). If so, please let us know

.

Carrotsticks · Aug 23, 2014

dunjaaa said:
Does anyone know a book that has deeper complex numbers/polynomial theoretical styled problems like the 2011 Extension 2 HSC Exam Q8 (c). If so, please let us know .

There are many books on polynomials containing those kinds of properties (bounding locations of roots, it's actually something I personally find fascinating). However, a lot of them need some understanding and knowledge of higher powered machinery such as Rouche's Theorem, which is one of the core theorems in the refinement of polynomial root locations.

Some fairly loose bounds on polynomial roots can be established without those tools (like 2011 HSC) but they are usually too insignificant to have a dedicated problem set for.

If you look up in google "Integer Polynomials", there is a paper that has a whole lot of these properties (including the 2011 HSC one I believe) but like I said, there are few such books out there with totally elementary proofs. If I recall correctly, there is a book on Polynomials by Barbeau that may have some of these problems.

In 2012 BOS Trials, the very last question establishes a fairly tight bound on the roots of polynomials.

glittergal96 · Aug 23, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
I think what TL1998 was going for is that if you knew only what the 4U course would teach you, how would you do it. Cauchy Schwarz although can be proven with 4U methods is not taught nor is it an inequality that is entirely obvious

Ah right, I see what you mean. (Our teacher did show us it though as a useful inequality to know and know how to prove, like AM-GM-HM.)

Do you know if I would lose any marks for having a solution like that? fleshed out a bit in an actual exam.

Sy123 · Aug 24, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
Ah right, I see what you mean. (Our teacher did show us it though as a useful inequality to know and know how to prove, like AM-GM-HM.)

Do you know if I would lose any marks for having a solution like that? fleshed out a bit in an actual exam.

I do not think you would lose any marks

I would ask Carrotsticks for confirmation though

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