HSC 2016 MX2 Marathon (archive) (1 Viewer)

leehuan · Jan 5, 2016

Re: HSC 2016 4U Marathon

Drsoccerball said:
are you sure it can even have a maximum value? Its a min parabola ?

If I try straightforwardly using the triangle inequality directly I get a complex modulus. This is mathematically invalid.

There has to be a corollary involved.

KingOfActing · Jan 5, 2016

Re: HSC 2016 4U Marathon

Drsoccerball said:
are you sure it can even have a maximum value? Its a min parabola ?

I'm pretty sure I have the answer, let me type it up.

dan964 · Jan 5, 2016

Re: HSC 2016 4U Marathon

yes it is a 1/2 sorry mistake in working.

dan964 · Jan 5, 2016

Re: HSC 2016 4U Marathon

never mind...

Paradoxica · Jan 5, 2016

Re: HSC 2016 4U Marathon

dan964 said:
Max occurs when equal
$|z|+\frac{2}{|z|}=|z+\frac{2}{z}|$

$|z|+\frac{2}{|z|}= 2$
$|z|^2+2 = |z|$ , we can do this because |z|>0, and |z|=0 is an irrelevant solution.
$|z|^2-|z|+2 = 0$
letting a = |z| just coz.
$a^2-a+2=0$
Finding the derivative
$y' = 2a-1$
When $y' = 0, a=0.5$
Since |z|>0, second derivative = a which is positive; hence max T.P. at |z|=1/2

Try again. The answer is not a rational number.

KingOfActing · Jan 5, 2016

Re: HSC 2016 4U Marathon

$\begin{align*} \text{We begin with the triangle inequality:}\\|(z -(z+\frac{2}{z})) + (z+\frac{2}{z})| &\leq |z -(z+\frac{2}{z})) |+| (z+\frac{2}{z})| \\|z| &\leq |-\frac{2}{z}|+2\\|z|^2 &\leq 2+2|z|\\|z|^2 - 2|z| - 2 &\leq 0\\1 - \sqrt3 \leq\, &|z| \leq 1+ \sqrt3\\\text{Hence, the maximum value of }|z|\text{ is }1 + \sqrt3\end{align*}$

dan964 · Jan 5, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
Try again. The answer is not a rational number.

just wait. nvm

Paradoxica · Jan 5, 2016

Re: HSC 2016 4U Marathon

dan964 said:
just wait. nvm

What made you think the maximum occurs when both sides are equal? As far as I'm concerned, there are almost no inequalities that work that way.

braintic · Jan 5, 2016

Re: HSC 2016 4U Marathon

I've done this without the triangle inequality. Confirming the answer is 1 + sqrt 3.

It also has a minimum value: sqrt 3 - 1
So there is something wrong with the sign in that working.

dan964 · Jan 5, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
What made you think the maximum occurs when both sides are equal? As far as I'm concerned, there are almost no inequalities that work that way.

I dk

KingOfActing · Jan 5, 2016

Re: HSC 2016 4U Marathon

braintic said:
I've done this without the triangle inequality. Confirming the answer is 1 + sqrt 3.

It also has a minimum value: sqrt 3 - 1
So there is something wrong with the sign in that working.

Oh yeah, damn, modulus can't be negative. So the correct solution would've read $|1-\sqrt3|\leq|z|\leq|1+\sqrt3|$

braintic · Jan 5, 2016

Re: HSC 2016 4U Marathon

KingOfActing said:
Oh yeah, damn, modulus can't be negative. So the correct solution would've read $|1-\sqrt3|\leq|z|\leq|1+\sqrt3|$

Actually, that reasoning will not do.
If that was correct working, you would have to interpret your answer as 0 <= |z| <= .....
I haven't checked the working, but there must be an error there somewhere.

KingOfActing · Jan 5, 2016

Re: HSC 2016 4U Marathon

braintic said:
Actually, that reasoning will not do.
If that was correct working, you would have to interpret your answer as 0 <= |z| <= .....
I haven't checked the working, but there must be an error there somewhere.

No, I got that. I know you can't just throw absolute value signs everywhere

I meant that if I worked it out properly, that would've been equivalent to the solution.

KingOfActing · Jan 5, 2016

Re: HSC 2016 4U Marathon

$NEXT QUESTION$\\$When a train travels around a curve of radius 400m at 36km/hr, it provides the same downwards lateral force as it does upwards when it travels at 72km/hr. Determine the angle of inclination of the railway lines.$

braintic · Jan 5, 2016

Re: HSC 2016 4U Marathon

Alternative solution without directly using the triangle inequality:

Let z = r cis theta and substitute.

Rearrange to write cos^2 theta in terms of r.

Then using the fact that 0 <= cos^2 theta <= 1, solve the resulting inequality to find the range for r^2 and hence for r.

braintic · Jan 5, 2016

Re: HSC 2016 4U Marathon

KingOfActing said:
$NEXT QUESTION$\\$When a train travels around a curve of radius 400m at 36km/hr, it provides the same downwards lateral force as it does upwards when it travels at 72km/hr. Determine the angle of inclination of the railway lines.$

A downwards force is not a "lateral" force.

Edit - yes, I know what you mean now.

InteGrand · Jan 5, 2016

Re: HSC 2016 4U Marathon

braintic said:
A downwards force is not a "lateral" force.

Edit - yes, I know what you mean now.

It's referring to the force up and down the plane, right (like in the direction of friction forces in these Q's that have friction)?

omegadot · Jan 6, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
$$What is the maximum value of $|z|$, if $z$ satisfies $\left | z + \frac{2}{z} \right | = 2$

$$\noindent Geometrically $|z| = k$ where $k > 0$ is a circle in the Argand plane. So one is required to find the circle with the greatest radius subject to the constraint $\left |z + \frac{2}{z} \right | = 2$ (which is also just some curve, or be it rather complicated, in the Argand plane). So we are required to find the circle of greatest radius that still intersects the constraint curve.$

$$\noindent Let $z = x + i y$ where $x,y \in \mathbb{R}.$ So the equation for the circle becomes $x^2 + y^2 = k^2$ while the constraint curve can be written as$\\\left (x + \frac{2x}{x^2 + y^2} \right )^2 + \left (y - \frac{2y}{x^2 + y^2} \right )^2 = 4.$

$$\noindent If the two curves are to intersect, $x^2 + y^2 = k^2$ and the above equation reduces to $\\x^2 \left (1 + \frac{2}{k^2} \right )^2 + y^2 \left (1 - \frac{2}{k^2} \right )^2 = 4.$

$$\noindent By finding local maxima the maximum radius for the intersecting circle can be found. Implicitly differentiating the above equation with respect to $x$, we have$\\2x \left (1 + \frac{2}{k^2} \right )^2 + 2y \frac{dy}{dx} \left (1 - \frac{2}{k^2} \right )^2 = 0.$

$$\noindent Stationary points occur when $\frac{dy}{dx} = 0$. Thus $\\2x \left (1 + \frac{2}{k^2} \right )^2 = 0 \,\, \Rightarrow \,\, x = 0,$

$$\noindent giving $y = \pm \frac{2}{1 - 2/k^2}.$ Thus stationary points occur at $(x,y) = \left (0, \pm \frac{2}{1 - 2/k^2} \right ).$

$$\noindent Substituting these points into the equation of the circle $x^2 + y^2 = k^2$, gives $k^4 - 8k^2 + 4 = 0.$

$$\noindent After solving the biquadratic equation, since $k > 0$ one has $k = \sqrt{4 \pm 2 \sqrt{3}}.$

$$\noindent When the radius is greatest, $|z|_{\rm max} = \sqrt{4 + 2 \sqrt{3}} = \sqrt{3} + 1$. This occurs when $z = \pm (\sqrt{3} + 1)i.$

$$\noindent As a bonus, when the radius is least, $|z|_{\rm min} = \sqrt{4 - 2 \sqrt{3}} = \sqrt{3} - 1$. This occurs when $z = \pm (\sqrt{3} - 1)i.$

omegadot · Jan 6, 2016

Re: HSC 2016 4U Marathon

$$\noindent And here is what it looks like on an Argand diagram. The red curve corresponds to $\left |z + \frac{2}{z} \right | = 2$, the blue curve to the circle with minimum radius $\sqrt{3} - 1$, while the green curve to the circle with maximum radius of $\sqrt{3} + 1.$

hedgehog_7 · Jan 7, 2016

Re: HSC 2016 4U Marathon

Given P ( a cos theta , b sin theta ) and Q ( a cos phi , b sin phi ) lie on the ellipse x^2/a^2 + y^2/b^2 = 1

IF PQ subtends a right angle at (a,0) show that ( tan theta/2 ) X ( tan phi/2 ) = -b^2/a^2

How does b (sin theta - sin phi ) / a (cos theta - cos phi) = b/a * ( 2 sin ( (theta - phi) /2 ) ) ) * cos ( theta + phi /2 ) ) / - 2 sin ( (theta - phi) /2 ) ) sin (theta + phi /2 )

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