HSC 2014 MX2 Marathon (archive) (3 Viewers)

dunjaaa · Apr 30, 2014

Re: HSC 2014 4U Marathon

(i) Can be proven using vectors, I've answered this before so I'll let someone else have a go
(ii) Basic circle geometry; you get a(Ɵ)=sqrt[2(1-cos(Ɵ))] and b(Ɵ)=sqrt[2(1+cos(x))], hence a(Ɵ)/b(Ɵ) turns out to be tan(Ɵ/2). Integrating we obtain the required result
(iii)We can deduce that the arg(z+1)=Ɵ/2 (ext. angle of a triangle is = to the sum of the two opposite int. angles). Therefore the locus of arg(z+1)=Ɵ/2 is a ray with equation y=tan(Ɵ/2)(x+1) for x>-1 and y>0. Thus, P(ω)=itan(Ɵ/2).

dunjaaa · Apr 30, 2014

Re: HSC 2014 4U Marathon

Screen shot 2014-04-29 at 8.04.30 PM.png

(iii) Let u=tan(x) and you get the required result
(iv) Hint for this part
(v) Im guessing as n approaches infinity the integral approaches 0 with the use of part (iv)

Davo_01 · May 1, 2014

Re: HSC 2014 4U Marathon

Prove that

$A_n=8+88+888+...+888...8=\dfrac{8}{81} (10^{n+1}-9n-10)$

Where $A_n$ contains n terms, with the term $T_k$ having k digits, each of them being 8.

dunjaaa · May 1, 2014

Re: HSC 2014 4U Marathon

Screen shot 2014-05-01 at 9.55.49 PM.png

Davo_01 · May 1, 2014

Re: HSC 2014 4U Marathon

dunjaaa said:
View attachment 30378

Nice work

VBN2470 · May 1, 2014

Re: HSC 2014 4U Marathon

VBN2470 said:
The function $f(x)= \frac{|x-2|}{x^2-5x+6}$ is not defined for $f(2)$ . Is there a value that could be given for $f(2)$ that would make the function continuous at 2? Give reasons for your answer.

Bump!

emilios · May 1, 2014

Re: HSC 2014 4U Marathon

God damn dunja some of your solutions are crazy. Are you aiming for a state rank? I think you've got a good chance.

Sy123 · May 3, 2014

Re: HSC 2014 4U Marathon

$\\ a < b < c \\ \\ $Find$ \ \ \max \left \{a^2(b^2+c^2), \ b^2(a^2+c^2), \ c^2(a^2+b^2) \right \}$

$Note: $ \ \max \{\dots \} \ \ $refers to the maximum value in the set inside the$ \ \max \ $function$

Kurosaki · May 3, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ a < b < c \\ \\ $Find$ \ \ \max \left \{a^2(b^2+c^2), \ b^2(a^2+c^2), \ c^2(a^2+b^2) \right \}$

$Note: $ \ \max \{\dots \} \ \ $refers to the maximum value in the set inside the$ \ \max \ $function$

Do we assume they are (a,b,c) all greater than 0? Just checking.

Sy123 · May 4, 2014

Re: HSC 2014 4U Marathon

Kurosaki said:
Do we assume they are (a,b,c) all greater than 0? Just checking.

Ah yes sorry

Kurosaki · May 4, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ a < b < c \\ \\ $Find$ \ \ \max \left \{a^2(b^2+c^2), \ b^2(a^2+c^2), \ c^2(a^2+b^2) \right \}$

$Note: $ \ \max \{\dots \} \ \ $refers to the maximum value in the set inside the$ \ \max \ $function$

$\\ a <b < c \\ \\ $We multiply by a,b and c$ \\ \\$We arrive at $ a^2 < ab < ac, ab < b^2 < bc, ac<bc<c^2 \\ \\$Now, we can quickly see that $ ab < ac < bc \\ \\ $ Squaring, we find that a^2b^2 < a^2c^2 < b^2c^2 (1)$
$\\ \\$Now, expanding the elements of the set, we get (a^2b^2+a^2c^2), (a^2b^2+b^2c^2), (a^2c^2+b^2c^2)$. \\ \\ $ Using (1), we see that the maximum value is clearly c^2(a^2+b^2), as $ bc > ac > ab $$
Not too sure about this haha.

Sy123 · May 4, 2014

Re: HSC 2014 4U Marathon

Kurosaki said:
$\\ a <b < c \\ \\ $We multiply by a,b and c$ \\ \\$We arrive at $ a^2 < ab < ac, ab < b^2 < bc, ac<bc<c^2 \\ \\$Now, we can quickly see that $ ab < ac < bc \\ \\ $ Squaring, we find that a^2b^2 < a^2c^2 < b^2c^2 (1)$
$\\ \\$Now, expanding the elements of the set, we get (a^2b^2+a^2c^2), (a^2b^2+b^2c^2), (a^2c^2+b^2c^2)$. \\ \\ $ Using (1), we see that the maximum value is clearly c^2(a^2+b^2), as $ bc > ac > ab $$
Not too sure about this haha.

Yea that's pretty much it, not too hard

dunjaaa · May 4, 2014

Re: HSC 2014 4U Marathon

(i) Prove that for any point along the y-axis, there is exactly two normals that can drawn to the hyperbola xy=c^2
(ii) Hence, explain why it is impossible to have 3 normals drawn from an arbitrary point to the hyperbola in the x-y plane

dunjaaa · May 4, 2014

Re: HSC 2014 4U Marathon

Highly doubt it, there are many people who are better than me. All you can do is work hard I guess

TL1998 · May 8, 2014

Re: HSC 2014 4U Marathon

Axio said:
View attachment 30177
The altitudes AP and BQ in an acute-angled triangle ABC meet at H. AP produced cuts the circle through A, B and C at K.
Prove that HP = PK.

I don't think anyone answered this so i'll answer it.

Angle BKP = Angle BCA (subtended by same arc)
Angle BCA = Angle BHK (exterior angle of cyclic quad)
BP is common, and Angle BPH and Angle BPK are both 90

So triangle BPK is congruent to triangle BHP.

Thus HP = PK (corresponding sides on congruent triangles)

not difficult but can't leave a question unanswered.

Parvee · May 12, 2014

Re: HSC 2014 4U Marathon

$Show~ \int_{-\pi }^{\pi}cos(nx)cos(mx)dx \\ a)=0,~if~m\neq~n \\ b)=\pi,~if~m=n$

dunjaaa · May 13, 2014

Re: HSC 2014 4U Marathon

Screen shot 2014-05-13 at 8.29.44 AM.png

Sy123 · May 13, 2014

Re: HSC 2014 4U Marathon

$\\ $Prove that$ \ \ 1 - \frac{2}{\pi} x \leq \cos x \leq 1 - \frac{x^2}{\pi} , \ $for$ \ 0 \leq x \leq \frac{\pi}{2}$

dunjaaa · May 16, 2014

Re: HSC 2014 4U Marathon

Part 1

Screen shot 2014-05-16 at 12.23.04 AM.png

and Part 2

Screen shot 2014-05-16 at 12.24.33 AM.png

, I suck at inequality proving, but this is what I manage to salvage

mathing · May 16, 2014

Re: HSC 2014 4U Marathon

dunjaaa not sure if what you wrote is correct, it's a bit difficult for me to follow all your steps.

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