HSC 2013 Maths Marathon (archive) (4 Viewers)

andybandy · Oct 22, 2013

Re: HSC 2013 2U Marathon

vilst3r said:
God, they should mention that in the solutions, I got so confused LOL

haha they should!

vilst3r · Oct 22, 2013

Re: HSC 2013 2U Marathon

Alright good luck on 2 unit tomorrow, thanks for your help man

andybandy · Oct 22, 2013

Re: HSC 2013 2U Marathon

vilst3r said:
Alright good luck on 2 unit tomorrow, thanks for your help man

you to

spanky125 · Oct 27, 2013

Re: HSC 2013 2U Marathon

Solve $2sinx=1$

Shazer2 · Oct 27, 2013

Re: HSC 2013 2U Marathon

$2sinx = 1 $ for $ 0 \le x \le 2\pi \\ \\ \indent sinx = \frac {1}{2} \\ \\ \indent \therefore x = \frac {\pi}{3}, \frac {2\pi}{3}$

Ikki · Nov 17, 2013

Re: HSC 2013 2U Marathon

Shazer2 said:
$2sinx = 1 $ for $ 0 \le x \le 2\pi \\ \\ \indent sinx = \frac {1}{2} \\ \\ \indent \therefore x = \frac {\pi}{3}, \frac {2\pi}{3}$

Correction:
$2sinx = 1 $ for $ 0 \le x \le 2\pi \\ \\ \indent sinx = \frac {1}{2} \\ \\ \indent \therefore x = \frac {\pi}{6}, \frac {5\pi}{6}$

asianese · Nov 28, 2013

Re: HSC 2013 2U Marathon

$$Consider the function $y(x) = \dfrac{u(x)}{v(x)}.$ Using only the product rule and the chain rule for differentiation, show that $\dfrac{dy(x)}{dx} = \dfrac{v\frac{du(x)}{dx} - u\frac{dv(x)}{dx}}{v^2(x)}.$

asianese · Jan 15, 2014

Re: HSC 2013 2U Marathon

$Consider the curve $ y=\dfrac{\ln x}{x} $ which is shown. Suppose a certain tangent of the curve goes through the origin. $\\ $a) By considering the equation of the tangent, find the co-ordinates of the point $P. $(3 marks)$ \\ $b) Write down the $ x-$intercept of the curve. (1 mark)$

$The curve $ y=\dfrac{1}{2x} $ is now shown on in the plane.$\\ $c) Find the equation of the tangent to the hyperbola at the point $P. $(2 marks)$\\ $d) Show that $\triangle POQ $ is isosceles. (2 marks) $ \\ $e) Find, in exact form, $\angle QPO. $(2 marks)$

RealiseNothing · Jan 15, 2014

Re: HSC 2013 2U Marathon

andybandy said:
yeah i would love if someone helped with that, but what im thinking is, since its a cubic, only one side will continously increase, and so that root might be above the x axis?

A bit late but:

If you differentiate a cubic you get it's gradient function, in this case a quadratic. The function is always increasing if the gradient is never 0 (and it increases for at least one value of x and is continuous - just minor subtleties). So you find the value of 'k' such that the gradient is never 0, which is when the discriminant of the gradient function is negative, meaning it has no roots, thus it is never 0.

photastic · May 26, 2014

Re: HSC 2013 2U Marathon

So no 2014 marathon?

Smile12345 · May 26, 2014

Re: HSC 2013 2U Marathon

We need a 2U 2014 Thread...

Mods, could we/ I start a 2014 Thread??

photastic · Aug 8, 2014

Re: HSC 2013 2U Marathon

The most hardest q16 question I have ever seen. Nah jokes, the biggest troll of a question to be placed near the end of the exam. Guys who sat the cssa this year, you would know this.

Carrotsticks · Aug 9, 2014

Re: HSC 2013 2U Marathon

Just Pythagoras that thing up haha. Indeed a troll question if thrown at the end of the paper. I'm sure many student would have drawn a line from A to B, then called the intersection with the horizontal X or something.

Carrotsticks · Aug 9, 2014

Re: HSC 2013 2U Marathon

Unless that was the solution that they were looking for, which would be very silly.

photastic · Aug 9, 2014

Re: HSC 2013 2U Marathon

Another troll question. Show that 5 cents is half a dollar

Trebla · Aug 9, 2014

Re: HSC 2013 2U Marathon

I will start a 2U Marathon for 2014.

HSC 2013 Maths Marathon (archive) (4 Viewers)

andybandy

Member

vilst3r

Member

andybandy

Member

spanky125

Just waitin' for a mate

Shazer2

Member

Ikki

Active Member

asianese

Σ

asianese

Σ

RealiseNothing

what is that?It is Cowpea

photastic

Well-Known Member

Smile12345

Active Member

photastic

Well-Known Member

Carrotsticks

Retired

Carrotsticks

Retired

photastic

Well-Known Member

Trebla

Administrator

Users Who Are Viewing This Thread (Users: 0, Guests: 4)