HSC 2014 MX2 Marathon ADVANCED (archive) (1 Viewer)

Sy123 · Jul 29, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Another one, I can post my solution to the last question if requested:

$\\ $Find, with proof, all polynomials with real coefficients$ \ P(x) \ $such that$ \\ P(P(x)) = (P(x))^k$

glittergal96 · Jul 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
Whoops, my first inequality is wrong . I still get the feeling that there is a short solution from that first expression though, I will try again after netball.

No, I wish!

I solved it, but it's a pretty dirty solution lol. Anyway, better than nothing.

By the sine rule (this isn't always stated as part of the sine rule, but it is easy to prove with HSC circle geometry):

$\frac{a}{\sin(\alpha)}=\frac{b}{\sin(\beta)}=\frac{c}{\sin(\gamma)}=2R.$

So, making the substitutions: $s=\sin(\alpha),t=\sin(\beta),u=\cos(\alpha),v=\cos(\beta)$ we get

$RHS=\frac{a^2+b^2}{2\sqrt{a^2+b^2+2ab\cos(\gamma)}}=R\cdot\frac{s^2+t^2}{\sqrt{s^2+t^2+2st\cos(\pi-(\alpha+\beta))}}$

$=R\cdot\frac{s^2+t^2}{\sqrt{s^2+t^2-2st(uv-st)}}.$

So it is sufficient to show that the second factor in this last line is smaller that 1.

$(su-tv)^2\geq 0$

$\Rightarrow s^4+t^4+2s^2t^2 \leq s^2+t^2+2s^2t^2-2stuv$

$\Rightarrow \frac{s^2+t^2}{\sqrt{s^2+t^2-2st(uv-st)}}\leq 1.$

glittergal96 · Jul 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
Another one, I can post my solution to the last question if requested:

$\\ $Find, with proof, all polynomials with real coefficients$ \ P(x) \ $such that$ \\ P(P(x)) = (P(x))^k$

Comparing degrees we get deg(P)^2 = k*deg(P).

So solutions are either constant or of degree k.

For k=0 P=1 is trivially the only solution. For k=1, any constant is a solution. For k > 1, the only constant solutions are 0 and 1.

We can now assume k > 0 and just look for the non-constant solutions (which must be of degree k).

Write

$P(x)=ax^k+Q(x)$

with

$\deg(Q)<k .$

Putting this in the original equation we get

$(a-1)P(x)^k+Q(P(x))=0.$

By degree considerations, this forces a=1 (if the first term is nonzero, nothing in the second is of high enough degree to cancel it out) and then Q=0.

So the only such non-constant solution is $P(x)=x^k.$

integral95 · Jul 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Yeah lol so like no one seems to be responding to the integration marathon so i thought i might post it here.

$$Show that$ \int_{0}^{1}x^m(1-x)^n dx = \frac{m!n!}{(m+n+1)!}\\ \\ Find \int \frac{x^2-1}{x^2+1}\frac{1}{\sqrt{1+x^4}} dx$

For the second integral there's a hint that the substitution $\ \ \ \ \ u^2 = x^2 + \frac{1}{x^2}$ would help, but you could use another method if you can come up with one

Sy123 · Jul 31, 2014

Re: HSC 2014 4U Marathon - Advanced Level

integral95 said:
Yeah lol so like no one seems to be responding to the integration marathon so i thought i might post it here.

$$Show that$ \int_{0}^{1}x^m(1-x)^n dx = \frac{m!n!}{(m+n+1)!}\\ \\ Find \int \frac{x^2-1}{x^2+1}\frac{1}{\sqrt{1+x^4}} dx$

For the second integral there's a hint that the substitution $\ \ \ \ \ u^2 = x^2 + \frac{1}{x^2}$ would help, but you could use another method if you can come up with one

Don't reveal the substitution, that takes all the fun away :/

Sy123 · Jul 31, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
I solved it, but it's a pretty dirty solution lol. Anyway, better than nothing.

By the sine rule (this isn't always stated as part of the sine rule, but it is easy to prove with HSC circle geometry):

$\frac{a}{\sin(\alpha)}=\frac{b}{\sin(\beta)}=\frac{c}{\sin(\gamma)}=2R.$

So, making the substitutions: $s=\sin(\alpha),t=\sin(\beta),u=\cos(\alpha),v=\cos(\beta)$ we get

$RHS=\frac{a^2+b^2}{2\sqrt{a^2+b^2+2ab\cos(\gamma)}}=R\cdot\frac{s^2+t^2}{\sqrt{s^2+t^2+2st\cos(\pi-(\alpha+\beta))}}$

$=R\cdot\frac{s^2+t^2}{\sqrt{s^2+t^2-2st(uv-st)}}.$

So it is sufficient to show that the second factor in this last line is smaller that 1.

$(su-tv)^2\geq 0$

$\Rightarrow s^4+t^4+2s^2t^2 \leq s^2+t^2+2s^2t^2-2stuv$

$\Rightarrow \frac{s^2+t^2}{\sqrt{s^2+t^2-2st(uv-st)}}\leq 1.$

Here was my solution:

$\\ $It is needed to prove$ \ \frac{a^2+b^2}{2\sqrt{a^2+b^2 + 2ab\cos C}} \leq R \\ \\ $This translates to$ \\ \\ 4R^2(a^2+b^2+2ab\cos C) \geq (a^2+b^2)^2 \\ \\ $By dividing the triangle into 3 by its circumcentre, and looking at the isocoles triangle with angle$ \ 2C \ $we get$ \ c = 2R \sin C \\ \\ $So$ \ 4R^2 = \frac{a^2+b^2 - 2ab\cos C}{1 - \cos^2 C} \\ \\ $So it is needed to prove$ \\ \\ (a^2+b^2 + 2ab\cos C)(a^2+b^2-2ab\cos C) \geq (a^2+b^2)^2 - \cos^2 C (a^2+b^2)^2 \\ \\ $Expanding out we get$ \\ \\ (a^2+b^2)^2 \geq 4a^2b^2 \Rightarrow \ (a^2-b^2)^2 \geq 0 \ \square$

glittergal96 said:
Comparing degrees we get deg(P)^2 = k*deg(P).

So solutions are either constant or of degree k.

For k=0 P=1 is trivially the only solution. For k=1, any constant is a solution. For k > 1, the only constant solutions are 0 and 1.

We can now assume k > 0 and just look for the non-constant solutions (which must be of degree k).

Write

$P(x)=ax^k+Q(x)$

with

$\deg(Q)<k .$

Putting this in the original equation we get

$(a-1)P(x)^k+Q(P(x))=0.$

By degree considerations, this forces a=1 (if the first term is nonzero, nothing in the second is of high enough degree to cancel it out) and then Q=0.

So the only such non-constant solution is $P(x)=x^k.$

Yep exactly my solution

Sy123 · Aug 8, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $Show that only the constant function$ \ f \ $satisfies$ \ f(x) = f(x^2) \ $for all real$ \ x$

glittergal96 · Aug 8, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Show that only the constant function$ \ f \ $satisfies$ \ f(x) = f(x^2) \ $for all real$ \ x$

Isn't the function that is 1 at 0 and 0 elsewhere a non-constant solution?

What I can prove though, is that the only continuous solutions are constant.

Because if we let c:=f(0), then for x in (-1,1) we have

$f(x)=f(x^{2^n})$

for any positive integer n. But for such x, $x^{2^n}\rightarrow 0$ which means by continuity that the RHS tends to c. As the LHS is constant in n, we are forced to conclude that this constant is c.

So we know that f(x)=c on the interval (-1,1). This also implies by continuity that f(1)=c.

We can do something similar with square roots to show that for x > 1, we have $f(x)=f(x^{1/2^n})\rightarrow f(1)= c$ , so f(x) = c for all x >1 (so we have now shown that f(x) = c at all positive reals).

Finally, for negative x, x^2 is positive. So f(x)=f(x^2)=c.

$\Box$

Sy123 · Aug 8, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
Isn't the function that is 1 at 0 and 0 elsewhere a non-constant solution?

What I can prove though, is that the only continuous solutions are constant.

Because if we let c:=f(0), then for x in (-1,1) we have

$f(x)=f(x^{2^n})$

for any positive integer n. But for such x, $x^{2^n}\rightarrow 0$ which means by continuity that the RHS tends to c. As the LHS is constant in n, we are forced to conclude that this constant is c.

So we know that f(x)=c on the interval (-1,1). This also implies by continuity that f(1)=c.

We can do something similar with square roots to show that for x > 1, we have $f(x)=f(x^{1/2^n})\rightarrow f(1)= c$ , so f(x) = c for all x >1 (so we have now shown that f(x) = c at all positive reals).

Finally, for negative x, x^2 is positive. So f(x)=f(x^2)=c.

$\Box$

Ah yes I meant continuous
And yes you got it (can you please post a problem)

glittergal96 · Aug 8, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
Ah yes I meant continuous
And yes you got it (can you please post a problem)

Sure. Show that:

$\frac{1}{n}+\frac{1}{n+1}+\ldots+\frac{1}{2n-1}>n(\sqrt[n]{2}-1).$

Sy123 · Aug 9, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
Sure. Show that:

$\frac{1}{n}+\frac{1}{n+1}+\ldots+\frac{1}{2n-1}>n(\sqrt[n]{2}-1).$

$\\ $Re-arrange, the problem now becomes to prove$ \\ \\ n\sqrt[n]{2} < \frac{n+1}{n} + \frac{n+2}{n+1} + \dots + \frac{2n}{2n-1}$

$\\ $Using the general AM-GM inequality, we get$ \\ \\ \sum_{k=0}^{n-1} \frac{n+k+1}{n+k} \geq n\sqrt[n]{\frac{n+1}{n} \frac{n+2}{n+1} \dots \frac{2n}{2n-1}} = n\sqrt[n]{2} \ \square$

Sy123 · Aug 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $The angles of a triangle$ \ \alpha, \beta, \gamma \ $are opposite sides$ \ a,b,c \ $respectively$ \\ \\ $This triangle is such that$ \ a^2+b^2 = kc^2 \ $for some constant$ \ k>1 \\ \\ $Find$ \ \frac{\cot \gamma}{\cot \alpha + \cot \beta}$

dunjaaa · Aug 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

α+β+γ=180 -> cot(α+β)=-cot(γ)
[cot(α)cot(β)-1]/[cot(α)+cot(β)] = -cot(γ)
cot(γ)/[cot(α)+cot(β)] = cot^2(γ)/[cot(α)cot(β)-1] = [-sin(α)sin(β)cos^2(γ)]/[sin^2(γ)cos(α+β)] = [sin(α)sin(β)cos(γ)]/sin^2(γ)
But, cos(γ) = [a^2+b^2-c^2]/2ab = [c^2(k-1)]/2ab (Given a^2+b^2=kc^2)
∴ cot(γ)/[cot(α)+cot(β)] = [c^2(k-1)sin(α)sin(β)]/[2absin^2(γ)]
Since sin(α)/a = sin(β)/a = sin(γ)/c;
cot(γ)/[cot(α)+cot(β)] = [c^2(k-1)]/2 x sin(γ)/c x sin(γ)/c x 1/sin^2(γ) = (k-1)/2

Sy123 · Aug 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

dunjaaa said:
α+β+γ=180 -> cot(α+β)=-cot(γ)
[cot(α)cot(β)-1]/[cot(α)+cot(β)] = -cot(γ)
cot(γ)/[cot(α)+cot(β)] = cot^2(γ)/[cot(α)cot(β)-1] = [-sin(α)sin(β)cos^2(γ)]/[sin^2(γ)cos(α+β)] = [sin(α)sin(β)cos(γ)]/sin^2(γ)
But, cos(γ) = [a^2+b^2-c^2]/2ab = [c^2(k-1)]/2ab (Given a^2+b^2=kc^2)
∴ cot(γ)/[cot(α)+cot(β)] = [c^2(k-1)sin(α)sin(β)]/[2absin^2(γ)]
Since sin(α)/a = sin(β)/a = sin(γ)/c;
cot(γ)/[cot(α)+cot(β)] = [c^2(k-1)]/2 x sin(γ)/c x sin(γ)/c x 1/sin^2(γ) = (k-1)/2

Yes well done

$\\ $A base 2 representation of a number is using only 2 digits$ \ 1 \ $and$ \ 0 \ $to represent it, we commonly use decimal form, or base 10$ \\ \\ $That is, in normal numbers we have$ \ 123 = 1 \times 10^2 + 2 \times 10^1 + 3 \times 10^0 \\ \\ $However an example of a number in base 2 would be$ \\\\ 10101.01_2 = 1 \times 2^4 + 0 \times 2^3 + 1 \times 2^2 + 0\times 2^1 + 1 \times 2^0 + 0 \times 2^{-1} + 1 \times 2^{-2} \\ \\ $i) Show that any positive integer in base 10 can be represented in base 2 notation$ \\ \\ $ii) Show that$ \ 0.00011001100110011\dots_2 = \frac{1}{10}$

glittergal96 · Aug 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
Yes well done

$\\ $A base 2 representation of a number is using only 2 digits$ \ 1 \ $and$ \ 0 \ $to represent it, we commonly use decimal form, or base 10$ \\ \\ $That is, in normal numbers we have$ \ 123 = 1 \times 10^2 + 2 \times 10^1 + 3 \times 10^0 \\ \\ $However an example of a number in base 2 would be$ \\\\ 10101.01_2 = 1 \times 2^4 + 0 \times 2^3 + 1 \times 2^2 + 0\times 2^1 + 1 \times 2^0 + 0 \times 2^{-1} + 1 \times 2^{-2} \\ \\ $i) Show that any positive integer in base 10 can be represented in base 2 notation$ \\ \\ $ii) Show that$ \ 0.00011001100110011\dots_2 = \frac{1}{10}$

i) Suppose (for the sake of contradiction) that m is the least positive integer that cannot be written in base 2.

We have m=2n or m=2n+1 for some non-negative integer n < m.

But since n < m, n must have a binary representation. We append 0 or 1 to this binary representation to obtain one for m.

ii) $(2^5-2)x=11.\overline{0011}_2-0.\overline{0011}_2=11_2=3 \Rightarrow x=\frac{1}{10}.$

Sy123 · Aug 12, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $Let$ \ H_n = \sum_{k=1}^n \frac{1}{k} \\ \\ $Find$ \ \sum_{n=2}^{\infty} \frac{1}{nH_n H_{n-1}}$

aDimitri · Aug 13, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\text{Given that m & n are both positive integers, show that }\\ 17m^{2} + 4n^{2} \text{ and } 17n^{2} + 4m^{2} \text{ cannot both be perfect squares}$

SilentWaters · Aug 13, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Let$ \ H_n = \sum_{k=1}^n \frac{1}{k} \\ \\ $Find$ \ \sum_{n=2}^{\infty} \frac{1}{nH_n H_{n-1}}$

1.

2.

Stygian · Aug 13, 2014

Re: HSC 2014 4U Marathon - Advanced Level

aDimitri said:
$\text{Given that m & n are both positive integers, show that }\\ 17m^{2} + 4n^{2} \text{ and } 17n^{2} + 4m^{2} \text{ cannot both be perfect squares}$

from a brief look at this and a bit of brainstorming, would you use the property of the difference between two squares so like a^2-b^2=(a+b)(a-b) and then show that (17m^2+4n^2)-(17n^2+4m^2) cannot be expressed in that form?

i might have a better look at this later

aDimitri · Aug 13, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Stygian said:
from a brief look at this and a bit of brainstorming, would you use the property of the difference between two squares so like a^2-b^2=(a+b)(a-b) and then show that (17m^2+4n^2)-(17n^2+4m^2) cannot be expressed in that form?

i might have a better look at this later

It can be expressed in that form though.
$(\sqrt{17m^{2} + 4n^{2}} +\sqrt{17n^{2} + 4m^{2}} )(\sqrt{17m^{2} + 4n^{2}} - \sqrt{17n^{2} + 4m^{2}} )$

You would have to then show that a and b cant both be integers.

HSC 2014 MX2 Marathon ADVANCED (archive) (1 Viewer)

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