HSC 2013 Maths Marathon (archive) (2 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.

HSC 2013 Maths Marathon (archive) (2 Viewers)

Member

This too shall pass

Member

This too shall pass

Member

This too shall pass

This too shall pass

Member

This too shall pass

Member

This too shall pass

Member

This too shall pass

Well-Known Member

Member

Well-Known Member

This too shall pass

Member

Well-Known Member

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)