HSC 2012 MX1 Marathon #2 (archive) (1 Viewer)

RealiseNothing · Jul 31, 2012

Re: HSC 2012 Marathon

seanieg89 said:
$Let $f(x)$:=\begin{cases}0&\mbox{if } x\leq 0\\e^{-1/x}&\mbox{if }x>0.\end{cases}.\\ Sketch $f$. Prove that $f$ is $k$ times differentiable at $0$ with $\frac{d^kf}{dx^k}(0)=0$ for each non-negative integer $k$.$

(This is an example of a smooth function which is zero and has all derivatives zero at a certain point, but is NOT the zero function.)

At x=0, both functions are equal to 0.

Now we show that the derivatives are also equal as the functions approach x=0.

It is trivial that the derivative of 0 is 0 as x approaches 0.

Now $\frac{d}{dx} e^{\frac{-1}{x}} = \frac{e^{\frac{-1}{x}}}{x^2}$

As x approaches 0, the 'e' will dominate the 'x'. So by considering the 'e' as x approaches 0:

$e^{\frac{-1}{0}} = \frac{1}{e^\infty} = 0$

So the functions have equal derivatives as x approaches 0.

Hence the piece-wise function is differentiable as x=0 with the derivative being 0. It then follows that it must be 'k' times differentiable at x=0 with the k'th derivative = 0 for each non-negative integer 'k'.

seanieg89 · Aug 1, 2012

Re: HSC 2012 Marathon

You have certainly given a correct argument for why f'(0)=0. It is not obvious though why EVERY derivative of f vanishes at x=0.

Carrotsticks · Aug 1, 2012

Re: HSC 2012 Marathon

Here's a question I thought of a while ago whilst I was trying to make a ramp with a sheet of cardboard and a tissue box:

$\\ $A stick has a fixed length $ L $ such that $ L \geq 2 \sqrt 2. \\\\ $A unit square is placed against the corner of a wall, and the stick is 'leaned' against the wall such that it is in contact at 3 points. The three points are the floor, the vertex of the square, and the wall.$ \\\\ $Find all possible values of $ \theta $ in terms of $ L $ that the stick can be leaned at, such that it is in contact with all 3 points.$$

Sy123 · Aug 8, 2012

Re: HSC 2012 Marathon

I have finally found a solution to the question carrot posted, well here we go:

Take the triangle in Carrot's diagram, within respect to theta:

$\cos {\theta}=\frac{x}{L} \\ \sin {\theta}=\frac{y}{L} \\ \\ \therefore x=L\cos{\theta} \ \ y=L\sin{\theta}$

Where y is the vertical side, and x is the horizontal side. Now lets take 3 points:

$(L\cos{\theta}, 0) \ \ (0, L\sin \theta} \ \ (1,1)$

Lets create a linear function which satisfies the x and y-intercepts:

$y=mx+b \\ \\ y=-x\tan \theta + L\sin \theta$

But function goes through (1,1)

$\therefore 1+\tan \theta=L \sin \theta \\ \\ (1+\tan \theta)^2=L^2 \sin^2{\theta} \\ \\ 1+\tan^2 \theta +\frac{2\sin \theta}{\cos \theta}=L^2 \sin ^2 \theta \\ \\ \frac{1}{\cos^2 \theta}+\frac{2\sin \theta}{\cos \theta}=L^2 \sin^2 \theta \\ \\ 1+2\sin \theta \cos \theta=L^2 \sin^2 \theta \cos^2 \theta \\ \\ 1+\sin{2\theta}=\frac{L^2}{4} \sin^2 {2\theta}$

Now by use of quadratic formula

$\sin {2\theta}=\frac{1\pm\sqrt{1-4 \frac{L^2}{4}(-1)}}{2\frac{L^2}{4}} \\ \\ \sin{2\theta}=\frac{2\pm2\sqrt{1+L^2}}{L^2} \\ \\ \therefore \theta = \frac{1}{2} \sin^{-1} \left(\frac{2\pm 2\sqrt{1+L^2}}{L^2} \right) \\ \\ \\ \square$

Complete (checked with Carrotsticks)

Now for my question:

The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

Fakeuser · Aug 9, 2012

Re: HSC 2012 Marathon

Edit: Accidental post.

Sy123 · Sep 17, 2012

Re: HSC 2012 Marathon

The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

Difficulty: Moderate

Lets revive this thread

s8891 · Sep 28, 2012

Re: HSC 2012 Marathon

What are the chances this years 3U hsc exam will have harder 2U questions (such as financial repayments etc/sine+cosine rule)?

I've notice they often appear in the 1990s exams, but very rare in the 2000s papers.

Shadowdude · Sep 28, 2012

Re: HSC 2012 Marathon

Well you can find an observed probability by going through the past papers and making a tally yourself.

twinklegal19 · Sep 30, 2012

Re: HSC 2012 Marathon

Sy123 said:
The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

Difficulty: Moderate

Lets revive this thread

cbb to align the equal signs, sorry!

$Let the side length of the square be $x$ and the radius of the circle be $r$. The perimeter $k$ is given by$\\k=4x+2\pi r\qquad (1)\\$and the area\\ $A=x^2+\pi r^2\qquad (2)\\$Rearranging (1) gives$\\r=\frac{k-4x}{2\pi}\qquad (3)\\$sub into (2)$\\A=x^2+\pi \left ( \frac{k-4x}{2\pi} \right )^2\\=x^2+\frac{1}{4\pi}\left ( k^2-8kx+16x^2 \right )\\\frac{dA}{dx}=2x+\frac{1}{4\pi}(32x-8k)\\$For stationary points, $\frac{dA}{dx}=0\\2x+\frac{1}{4\pi}(32x-8k)=0\\x=\frac{k}{\pi+4}$ which must give a minimum value as $A$ is a quadratic with $a> 0\\$sub into (3)$\\r=\frac{k-4\left ( \frac{k}{\pi +4} \right )}{2\pi}\\=\frac{1}{2}\left ( \frac{k}{\pi +4} \right )\\=\frac{1}{2}x\\\therefore $The area is least when the side of square is double the radius of the circle.$

twinklegal19 · Sep 30, 2012

Re: HSC 2012 Marathon

New question to revive this thread

$$Evaluate $\int \sqrt{1+\sqrt x}\; dx$ using the substitution $u^2=1+\sqrt{x}$

asianese · Sep 30, 2012

Re: HSC 2012 Marathon

$\int \sqrt{1+\sqrt{x}} \;dx \\ u^2=1+ \sqrt{x} \Rightarrow 2udu=\frac{dx}{2 \sqrt{x}} \therefore dx=4u\sqrt{x} dx\\I=\int u \cdot 2u \cdot 2\sqrt{x} dx\\=4\int u^2(u^2-1)du \\ =4\int u^4-u^2 du\\=4 \left[\frac{(1+\sqrt{x})^{\frac{5}{2}}}{5} -\frac{(1+\sqrt{x})^{\frac{3}{2}}}{3}\right]+C$

Sy123 · Sep 30, 2012

Re: HSC 2012 Marathon

Here is a neat question that I just made (and solved, so it is solvable)
It is a 5 part question, so here we go:

Difficulty: Easy -> Hard

$$Given that the angle PO makes with the x-axis is$ \ \ \theta$

$$i) Find the cartesian locus of point P where$ \ \ P(\cos \theta, \sin \theta) \ \ \ \ [1] \\ \\ $ii)State restrictions on the y value which make this locus a function, and hence give the resultant restriction on$ \ \ \theta \ \ \ \ [1]$

$$iii) Show that the equation of the tangent of the locus at P is$ \ \\ \\ y=\frac{1-x \cos \theta}{\sin \theta} \ \ \ \ [2]$

$$iv) This tangent intersects the y-axis at Q, show that the equation of the normal at Q is$ \ \ \\ \\ y=x\tan \theta + \frac{1}{\sin \theta} \ \ \ \ [3]$

$$v) Find the value(s) of$ \ \ \theta \ \ $such that the normal at Q is a tangent to the locus of P. Give reasons for your answer$ \ \ \ [3]$

Diagram of the Question.

asianese · Sep 30, 2012

Re: HSC 2012 Marathon

i) $x^2+y^2=1$

ii) $y\geq0 \therefore y=\sqrt{1-x^2} \Rightarrow 0\leq \theta \leq \pi$

iii) $\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{-\cos\theta}{\sin\theta}\\y-\sin\theta=\frac{-\cos\theta(x-\cos\theta)}{\sin\theta}\\y=\frac{\cos^2\theta-x\cos\theta}{\sin\theta}+\sin\theta\\=\frac{1-x\cos\theta}{\sin\theta}$

iv) $x=0, y=\frac{1}{\sin\theta} \Rightarrow Q(0,\frac{1}{\sin\theta})\\ m_{\mathrm{normal}}=\frac{\sin\theta}{\cos\theta}=\tan\theta\\y-\frac{1}{\sin\theta}=\tan\theta x\\y=x\tan\theta+\frac{1}{\sin\theta}$

Not sure about v)

Sy123 · Sep 30, 2012

Re: HSC 2012 Marathon

asianese said:
i) $x^2+y^2=1$

ii) $y\geq0 \therefore y=\sqrt{1-x^2} \Rightarrow 0\leq \theta \leq \pi$

iii) $\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{-\cos\theta}{\sin\theta}\\y-\sin\theta=\frac{-\cos\theta(x-\cos\theta)}{\sin\theta}\\y=\frac{\cos^2\theta-x\cos\theta}{\sin\theta}+\sin\theta\\=\frac{1-x\cos\theta}{\sin\theta}$

iv) $x=0, y=\frac{1}{\sin\theta} \Rightarrow Q(0,\frac{1}{\sin\theta})\\ m_{\mathrm{normal}}=\frac{\sin\theta}{\cos\theta}=\tan\theta\\y-\frac{1}{\sin\theta}=\tan\theta x\\y=x\tan\theta+\frac{1}{\sin\theta}$

Not sure about v)

$\frac{1-x\cos\theta}{\sin\theta}=x\tan\theta+\frac{1}{\sin\theta}\\1-x\cos\theta=x\sin\theta\tan\theta+1\\-\cos\theta=\sin\theta\frac{\sin\theta}{\cos\theta}\\-\cos^2\theta=\sin^2\theta \\\sin^2\theta+\cos^2\theta=1 \Rightarrow $All values of $ 0\leq\theta\leq\pi $ are such that the normal at Q is also a tangent at P???$$

Correct! Except for part v. Use the fact that theta is the inclined angle to x-axis and circle properties to find theta such that it is
true (thats how I did it)

Shadowdude · Sep 30, 2012

Re: HSC 2012 Marathon

For part v.. you equated the equations of the tangent and normal?

From skimming, would you not differentiate the equations and solve for theta in the expression m_1 m_2 = -1? That was my first thought. That might not work but... you know how you always think of one way to solve it at the very beginning - which might be right or wrong.

asianese · Sep 30, 2012

Re: HSC 2012 Marathon

I'm still kinda not understanding v.

Are you asking when does (a) normal through Q 'equal' or 'is' the tangent at P?

Sy123 · Sep 30, 2012

Re: HSC 2012 Marathon

It is asking for what values of theta does it make the normal at Q a tangent to the circle

Sy123 · Oct 20, 2012

Re: HSC 2012 Marathon

Here is the solution for just v, or at least how I did it:

http://i981.photobucket.com/albums/ae294/sy08071996/Image10.jpg

I dont want to leave this question unanswered lol. Will make another question if the demand is high enough
(Keep dyeing marathon alive!)

barbernator · Oct 20, 2012

Re: HSC 2012 Marathon

what unanswered questions are there? im ready to rock for half an hour.

Sy123 · Oct 20, 2012

Re: HSC 2012 Marathon

Ah no one answered v which made the thread die, asianese answered the other ones but you can do them, its a good excercise I reckon

HSC 2012 MX1 Marathon #2 (archive) (1 Viewer)

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