HSC 2013 Maths Marathon (archive) (8 Viewers)

anthony_wheels · Sep 4, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
$$i) Sketch the line$ \ \ y=A-x \ \ $for$ \ x\geq 0 \ , \ y \geq 0 \ , \ A>0$

$ii) The sum of 2 non-negative numbers is$ \ \ A \ \ $what is the maximum$

$and the minimum value of the sum of the squares of these numbers$

Solutions to the question above

Sy123 · Sep 5, 2013

Re: HSC 2013 2U Marathon

anthony_wheels said:
View attachment 28600 Solutions to the question above

Yes that is true but you haven't explained why (a=0) gives maximum, I've asked for the sketch in the first part for a reason.

$Prove that$

$x^{\frac{1}{x}} \leq e^{\frac{1}{e}}$

$$For$ \ x > 0$

anthony_wheels · Sep 6, 2013

Re: HSC 2013 2U Marathon

Sorry about that.

However, the question didn't really ask for an explanation. Anyway, I tried doing the question above and my proof is a little bit 50/50, but I couldn't think of any other way to do it.

I hope it is sufficient

Cheers!

Sy123 · Sep 6, 2013

Re: HSC 2013 2U Marathon

anthony_wheels said:
Sorry about that. However, the question didn't really ask for an explanation. Anyway, I tried doing the question above and my proof is a little bit 50/50, but I couldn't think of any other way to do it. View attachment 28606 I hope it is sufficient Cheers!

That is essentially what I was aiming for, well done.

$By multiplying the numerator and denominator by the same function (essentially multiplying by 1 in a clever way), Find$

$\int \frac{e^{2x} - 1}{e^{2x} + 1} \ dx$

anthony_wheels · Sep 6, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
That is essentially what I was aiming for, well done.

$By multiplying the numerator and denominator by the same function (essentially multiplying by 1 in a clever way), Find$

$\int \frac{e^{2x} - 1}{e^{2x} + 1} \ dx$

Sy123 · Sep 6, 2013

Re: HSC 2013 2U Marathon

anthony_wheels said:
View attachment 28610

Yep well done.

$$Consider the functions$ \ \ C(x) \ \ $and$ \ \ S(x)$

$Which are such that$

$C(\cos x) = x$

$S(\sin x) = x$

$i) By differentiating, show that$

$S'(z) + C'(z) = 0$

$$For$ \ -1 \leq z \leq 1$

$$Consider another function$ \ \ f(z) = S(z) + C(z)$

$$ii) Show that$ \ \ f(z) = \frac{\pi}{2}$

Sy123 · Sep 8, 2013

Re: HSC 2013 2U Marathon

$$The curves$ \ y=\cos x \ \ $and$ \ y=\tan x \ \ $intersect at a point$ \ P \ $with$ \ x$-coordinate of$ \ \alpha$

$0 \leq \alpha < \frac{\pi}{2}$

$$i) Show that the curves intersect at right angles to each other at$ \ P$

$ii) Show that$

$\sec^2 \alpha = \frac{1+\sqrt{5}}{2}$

anthony_wheels · Sep 8, 2013

Re: HSC 2013 2U Marathon

Good question... Took a lot of working out (which probably could have been avoided but oh well)

... Also, I wouldn't say my proof is very good but that's how it is haha

Sy123 · Sep 8, 2013

Re: HSC 2013 2U Marathon

Yes well done, it was the long way however.

$\cos \alpha = \tan \alpha$

$\therefore \frac{\sin \alpha}{\cos^2 \alpha} = 1 \ \ \fbox{1}$

Now, differentiate the curves, and find the gradients of each curve at P

$-\sin \alpha = m_1$

$m_2 = \sec^2 \alpha$

$m_1 m_2 = -\frac{\sin \alpha}{\cos^2 \alpha} = -1 \ \ $due to$ \ \fbox{1}$

===

By the way, that was from HSC 4U Q7 =)
But you can (as you ahve shown) can do it at 2U level, and its good that you were able to solve it.

anthony_wheels · Sep 8, 2013

Re: HSC 2013 2U Marathon

Ah sick! ahah... Yeah, I went the very very long way but I eventually got there... So much quicker ^

Sy123 · Sep 9, 2013

Re: HSC 2013 2U Marathon

$$Consider the right angle triangles$ \ \ \triangle ABC \ , \ \triangle ACD \ $and$ \ \triangle AFE$

$$Let$ \ \angle BAC = \angle CAD = x$

$AC = 1$

$$Let$ \ t=\cos x$

$$i) Prove that$ \ \ DE = \frac{1}{t^2} \sqrt{1-t^2} \ \ \ \fbox{2}$

$$ii) Prove that$ \ \ \triangle AFE \parallel \parallel \triangle ABC \ \ \ \fbox{1}$

$$iii) Hence prove that$ \ \ \cos(2x) = 2\cos^2 x - 1 \ \ \ \ \fbox{3}$

$$iv) Hence find$ \ \ \int_0^{\frac{\pi}{2}} \cos^2 x \ dx \ \ \ \ \fbox{2}$

anthony_wheels · Sep 9, 2013

Re: HSC 2013 2U Marathon

Biggest trek of my life due to a range of mistakes I made initially, but I finally got the answers (I hope

) lol

Sy123 · Sep 9, 2013

Re: HSC 2013 2U Marathon

Well done, though I was intending that you didn't refer to the actual trigonometry and just used 't' only, but its all good.

=======

$The graph$ \ \ f(x)=\frac{Ax}{1+x^2} \ \ $is sketched below$

$$A rectangle$ \ \ ABCD \ \ $is drawn, with$ \ \ B(a,0)$

$$i) Prove$ \ \ f(x) = f \left( \frac{1}{x} \right ) \ \ \ \fbox{1}$

$ii) Show that the area of the rectangle$ \ \ ABCD \ \ $increases as$ \ a \ $increases$

$$and the area of the rectangle has no maximum value$ \ \ \ \fbox{3}$

$$iii) Find the limiting value of the area of the rectangle$ \ \ \ \fbox{2}$

seanieg89 · Sep 17, 2013

Re: HSC 2013 2U Marathon

$A man runs n laps of a circular racetrack. He runs the first lap at $v_1\, ms^{-1}$, the second at $v_2\, ms^{-1}$, and so on. What is his average speed over his entire run?$

anthony_wheels · Sep 18, 2013

Re: HSC 2013 2U Marathon

seanieg89 said:
$A man runs n laps of a circular racetrack. He runs the first lap at $v_1\, ms^{-1}$, the second at $v_2\, ms^{-1}$, and so on. What is his average speed over his entire run?$

Hope it's correct

seanieg89 · Sep 18, 2013

Re: HSC 2013 2U Marathon

anthony_wheels said:
View attachment 28671 Hope it's correct

Yep, well done

.

Sy123 · Sep 26, 2013

Re: HSC 2013 2U Marathon

$Solve the equation$

$x^2 -6x + \frac{3}{x^2-6x+3} = -1$

HSC2014 · Sep 27, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
$Solve the equation$

$x^2 -6x + \frac{3}{x^2-6x+3} = -1$

Wolfram says no real solution; I couldn't do it

SpiralFlex · Sep 27, 2013

Re: HSC 2013 2U Marathon

Change the minus 1 to 1 and there will be real solutions then.

HSC2014 · Sep 27, 2013

Re: HSC 2013 2U Marathon

Oh does that work hahaha... I assumed it was a typo but i didn't know which part.

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