HSC 2014 MX2 Marathon ADVANCED (archive) (5 Viewers)

VBN2470 · Apr 8, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$z^2 - z(x+y) + (x^2+y^2-xy - 8) = 0 \ \ \triangle \geq 0 \ \ $since$ \ x,y,z \ $real$ \\ \\ (x+y)^2 - 4(x^2+y^2-xy-8) \geq 0 \\ 32 - 3(x-y)^2 \geq 0 \Rightarrow \ \sqrt{\frac{32}{3}} \geq |x -y|$

Also if you're question if Q16 difficulty or beyond, please post it in the Advanced Level marathon. This marathon is for easy questions

Nice work

. Did you also manage to work out the answer for the other question posted above (finding $m+n$ )?

Sy123 · Apr 8, 2014

Re: HSC 2014 4U Marathon

VBN2470 said:
If $m$ and $n$ are positive integers such that $\frac{m+n}{m^2+mn+n^2}=\frac{4}{49}$ find the value of $m+n$ .

Its a little brute force but this is what I came up with:

http://imgur.com/5t9AOOB

Davo_01 · Apr 8, 2014

Re: HSC 2014 4U Marathon

This is what i thought of:
We know that m+n and mn is an integer and hence (m+n)^2-mn (the denominator) is an integer
So the expression can be written in the form p/q where p and q are integers.
Now p/q=4/49=8/98=12/147=16/196... (multiplying top and bottom by 1,2,3,4...)
Suppose p=m+n= 4 or 8, then mn would be negative (which it isnt),
if m+n=12 then mn=3 which do not have integer solutions
if p=m+n = 16 then mn=60 which has integer solutions m=10 and n=6 (or other way around), hence m+n=16

Davo_01 · Apr 8, 2014

Re: HSC 2014 4U Marathon

a bit trial and error but all i could think of

braintic · Apr 8, 2014

Re: HSC 2014 4U Marathon

VBN2470 said:
If $m$ and $n$ are positive integers such that $\frac{m+n}{m^2+mn+n^2}=\frac{4}{49}$ find the value of $m+n$ .

Could there be a neat way of doing this by multiplying the numerator and denominator each by (m-n) to get
(m^2 - n^2) / (m^3 - n^3) ?

Davo_01 · Apr 8, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Find lim n-> -infinity (x^2+4x+7)^0.5+x

RealiseNothing · Apr 8, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\frac{m+n}{m^2+mn+n^2} = \frac{4}{49}$

So $m+n = 4k$ for some integer $k$

Flipping the equation gives:

$\frac{m^2+mn+n^2}{m+n} = \frac{49}{4}$

Adding and subtracting $mn$ from the numerator allows us to decompose the fraction into:

$m+n-\frac{mn}{m+n} = \frac{49}{4}$

Since $m, n > 0$ we can deduce that:

$\frac{mn}{m+n} > 0$

And hence:

$m+n > \frac{49}{4} = 12.25$

Thus:

$m+n > 12.25$

Call this the lower bound of $m+n$

Now we will find an upper bound:

$\frac{m^2+mn+n^2}{m+n} = \frac{49}{4}$

$4(m^2+mn+n^2) = 49(m+n)$

$m^2+n^2 = \frac{49(m+n)}{4} - mn$

But we know that $m+n > 12$ which implies that $mn > 11$

So we can deduce that:

$m^2+n^2 < \frac{49(m+n)}{4}-11$

Now recalling the QM-AM inequality:

$\sqrt{2}\sqrt{m^2+n^2} > m+n$

Using what we just found:

$\sqrt{2}\sqrt{\frac{49(m+n)}{4}-11} > \sqrt{2}\sqrt{m^2+n^2} > m+n$

$\sqrt{\frac{49(m+n)}{2}-22} > m+n$

Squaring both sides gives:

$\frac{49}{2}(m+n) - 22 > (m+n)^2$

$0 > (m+n)^2 - \frac{49}{2}(m+n) + 22$

Using the quadratic formula we get:

$m+n = \frac{49}{4} \pm \frac{\sqrt{2049}}{4}$

$m+n = 17.91$ or $m+n = 6.59$

I've just used decimal answers as we are looking for a upper bound, not an exact answer.

Testing $m+n = 10$ gives $100 - 295 + 22 = -173$ . So we deduce that:

$6.59 < m+n < 17.91$

But incorporating our lower bound from earlier:

$12.25 < m+n < 17.91$

However we know from earlier that $m+n = 4k$ for some integer $k$ .

Hence $m+n=16$ is the only solution that satisfies the given inequality.

RealiseNothing · Apr 8, 2014

Re: HSC 2014 4U Marathon - Advanced Level

That's the only method I could come up with that completely gets rid of any trial and error.

Br0 · Apr 8, 2014

Re: HSC 2014 4U Marathon - Advanced Level

RealiseNothing said:
That's the only method I could come up with that completely gets rid of any trial and error.

i have pm'd regarding other methods. pls repond

hit patel · Apr 9, 2014

Re: HSC 2014 4U Marathon - Advanced Level

gahyunkk said:
Yea i did but i cant get those value with solving 16x^5-20x^3+5x-1=0 especially bcuz of -1 at the end

Hmm... Why do I have the feeling that you are talking sum of roots as 20/16 and not 0? This may justify the negative one possibly?

Davo_01 · Apr 9, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Managed to get rid of trial and error to:
Since m and n are integers m+n=4k (1) and (m+n)^2-mn=49k (2) (k>0)
Subbing (1) into (2):
16k^2-m(4k-m)=49k
M^2-4mk+16k^2-49k=0
M=2k+(49k-12k^2)^0.5
N=2k-(49k-12k^2)^0.5
Discriminant must be positive (and perfect square):
K<49/12
Also n>0
So 2k>(49k-12k^2)^0.5
K>49/16
Hence k=4
Which gives discriminat=4 (perfect square) and m=10 n=6
M+n=16

seanieg89 · Apr 9, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$Let $P,Q,R$ be real polynomials and let $S(x)=P(x^3)+xQ(x^3)+x^2R(x^3)$.\\ \\If $S$ is of degree $n$ and has all roots distinct, find a polynomial of degree $n$ whose roots are the cubes of the roots of $S$. (In terms of $P,Q,R$).$

Davo_01 · Apr 9, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
$Let $P,Q,R$ be real polynomials and let $S(x)=P(x^3)+xQ(x^3)+x^2R(x^3)$.\\ \\If $S$ is of degree $n$ and has all roots distinct, find a polynomial of degree $n$ whose roots are the cubes of the roots of $S$. (In terms of $P,Q,R$).$

View attachment 30255

Here is my attempt, its a bit long, im sure you have a quicker method.

Sy123 · Apr 10, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $Find$ \ f(x) = \lim_{n \to \infty} \frac{\sin^{n+2}x + \cos^{n+2}x}{\sin^nx + \cos^nx}$

Someone else post a question please

seanieg89 · Apr 10, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Davo_01 said:
View attachment 30255

Here is my attempt, its a bit long, im sure you have a quicker method.

Mine involved the cube roots of unity but is roughly the same length.

Note that it is important that the degrees of P(x^3),xQ(x^3) and x^2R(x^3) are distinct mod 3 in order to conclude deg(T)=n, otherwise good solution.

Davo_01 · Apr 10, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Find$ \ f(x) = \lim_{n \to \infty} \frac{\sin^{n+2}x + \cos^{n+2}x}{\sin^nx + \cos^nx}$

Someone else post a question please

I got (1+lcos2xl)/2?

Davo_01 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Find$ \ f(x) = \lim_{n \to \infty} \frac{\sin^{n+2}x + \cos^{n+2}x}{\sin^nx + \cos^nx}$

Someone else post a question please

Here is my attempt

edit: damn just realized there is an error on like 3rd last line, how do i prove l is the positive case?

Davo_01 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
Mine involved the cube roots of unity but is roughly the same length.

Note that it is important that the degrees of P(x^3),xQ(x^3) and x^2R(x^3) are distinct mod 3 in order to conclude deg(T)=n, otherwise good solution.

Hmm what do you mean by "distinct mod 3"? and do you mind showing me your method, sounds interesting

Sy123 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Davo_01 said:
I got (1+lcos2xl)/2?

Correct, the original question had $0 \leq x \leq \frac{\pi}{2}$ , and I forgot to write that.

Sy123 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $Find all polynomials$ \ f \ $such that$ \ \ f(x) + f(1-x) = 1 \ $for all$ \ x$

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