HSC 2016 MX2 Marathon (archive) (4 Viewers)

Paradoxica · Apr 27, 2016

Re: HSC 2016 4U Marathon

Using the Integral test, prove the following limit:

$\lim_{n \to \infty} \sum_{r=2^n-1}^{2^{n+1}-1} \frac{1}{r} = \log{2}$

You may assume uniform convergence and the squeeze theorem.

KingOfActing · Apr 27, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
$\pi \int_0^{\frac{\pi}{4}} \cos^2{x} - \sin^2{x} $d$x = \pi \int_0^{\frac{\pi}{4}} \cos{2x} $d$x = \frac{\pi}{2}\left[\sin{2x}\right]_0^{\frac{\pi}{2}} = \frac{\pi}{2} u^3$

Isn't the integral $\pi\left(\int_{0}^{\frac{1}{\sqrt 2}} \arcsin^2{y}\,dy + \int_{\frac{1}{\sqrt 2}}^{1} \arccos^2{y}\,dy \right)$

Paradoxica · Apr 27, 2016

Re: HSC 2016 4U Marathon

KingOfActing said:
Isn't the integral $\pi\left(\int_{0}^{\frac{1}{\sqrt 2}} \arcsin^2{y}\,dy + \int_{\frac{1}{\sqrt 2}}^{1} \arccos^2{y}\,dy \right)$

jks

when you don't read the question properly

*weep*

hedgehog_7 · Apr 27, 2016

Re: HSC 2016 4U Marathon

uhh the provided answer is 1/2 (pi(pi root 2 -1))

hedgehog_7 · Apr 27, 2016

Re: HSC 2016 4U Marathon

KingOfActing said:
Isn't the integral $\pi\left(\int_{0}^{\frac{1}{\sqrt 2}} \arcsin^2{y}\,dy + \int_{\frac{1}{\sqrt 2}}^{1} \arccos^2{y}\,dy \right)$

wait how did you end up with this integral? the area i chose was just between 0 and 1/root 2, why does it include between 1/root 2 and 1?

Paradoxica · Apr 27, 2016

Re: HSC 2016 4U Marathon

hedgehog_7 said:
wait how did you end up with this integral? the area i chose was just between 0 and 1/root 2, why does it include between 1/root 2 and 1?

the area bounded by the two curves are bounded by two different curves.

so you have to consider separate cases.

KingOfActing · Apr 27, 2016

Re: HSC 2016 4U Marathon

hedgehog_7 said:
wait how did you end up with this integral? the area i chose was just between 0 and 1/root 2, why does it include between 1/root 2 and 1?

Just crudely graphing it suffices (yes, my pro paint skills are back

)

Paradoxica · Apr 27, 2016

Re: HSC 2016 4U Marathon

hedgehog_7 said:
uhh the provided answer is 1/2 (pi(pi root 2 -1))

According to Mathematica, the exact value of the integral is

$\pi \left(\frac{\pi}{\sqrt{2}}-2\right)$

A numerical approximation using Simpson's Rule confirms the result.

I guess this integral is just tedious bashing with IBP.

math man · Apr 27, 2016

Re: HSC 2016 4U Marathon

KingOfActing said:
View attachment 33155

If anyone wants to do that, that's the question that made me rethink all of my life choices during the 4u half yearly

I hate circular motion with a passion

The beauty of circular motion is its ease, it's so simple the math behind it

Paradoxica · Apr 28, 2016

Re: HSC 2016 4U Marathon

math man said:
The beauty of circular motion is its ease, it's so simple the math behind it

go back to appliedville

actually, this is going to be entertaining...

*continues to scrawl random stuff*

math man · Apr 28, 2016

Re: HSC 2016 4U Marathon

Appliedville kicked me out, I'm just strolling these streets to find a town that will accept me

InteGrand · May 7, 2016

Re: HSC 2016 4U Marathon

$\noindent [Bernoulli's Inequality]$

$\noindent Let $x>-1$.$

$\noindent i) Show that $\left(1+x\right)^{r} \geq 1+rx$ if $r > 1$ or $r<0$.$

$\noindent ii) Show that $\left(1+x\right)^{r} \leq 1+rx$ if $0\leq r \leq 1$.$

Paradoxica · May 7, 2016

Re: HSC 2016 4U Marathon

$$\noindent Let $\xi$ be a complex $n^{\text{th}}$ root of unity.$\\\\$a) i) Show $\sum_{r=1}^{n-1} \xi^r = -1 \\\\ \indent $ii) Hence, show $\sum_{r=1}^{n-1} (1-\xi^r) = n \\\\ \indent $iii) Hence, or otherwise, evaluate $\sum_{r=1}^{n-1} r\xi^{n-r-1}\\\\ $b) i) Prove the following extension of the triangle inequality using induction:$ \\\\ \indent |\pm z_1 \pm z_2 \pm \cdots \pm z_{n-1} \pm z_n| \leq |z_1| + |z_2| + \cdots + |z_{n-1}| + |z_n| \\\\ \indent $You may assume The Triangle Inequality: $ |\pm x \pm y| \leq |x|+|y| \\\\ \indent $ii) By using the result(s) from part a), prove $|1-\xi| > \frac{2}{n-1} \\\\ \indent $iii) Hence prove $\left|\sin{\frac{k\pi}{n}}\right| > \frac{1}{n-1}$

hedgehog_7 · May 9, 2016

Re: HSC 2016 4U Marathon

A particle is thrown vertically upwards where the air resistance is given by R = 0.01mv^2. If the velocity of projection is 60m/s find the time to reach the highest point. Answer is 3.43 s

Drsoccerball · May 10, 2016

Re: HSC 2016 4U Marathon

hedgehog_7 said:
A particle is thrown vertically upwards where the air resistance is given by R = 0.01mv^2. If the velocity of projection is 60m/s find the time to reach the highest point. Answer is 3.43 s

$\ddot{x} = 0.01v^2$

$\frac{dv}{dt}= 0.01v^2$

$\frac{dt}{dv} =\frac{1}{0.01v^2}$

$T = \int_u^0{\frac{1}{0.01v^2}}dv$

hedgehog_7 · May 10, 2016

Re: HSC 2016 4U Marathon

a projectile is fired vertically upwards from the earths surface with velocity U m/s. The retardation due to gravity is given by the law k/x^2 where x is the distance of the projectile from the centre of the earth, and k is a constant. The acceleration due to gravity on the earths surface is g. The earths radius is R. Neglecting the air resistance show that if U^2 = gR, then the projectile reaches the height R above the earths surface. What is the time for this journey?

having trouble with getting the U^2 = gR bit

InteGrand · May 10, 2016

Re: HSC 2016 4U Marathon

hedgehog_7 said:
a projectile is fired vertically upwards from the earths surface with velocity U m/s. The retardation due to gravity is given by the law k/x^2 where x is the distance of the projectile from the centre of the earth, and k is a constant. The acceleration due to gravity on the earths surface is g. The earths radius is R. Neglecting the air resistance show that if U^2 = gR, then the projectile reaches the height R above the earths surface. What is the time for this journey?

having trouble with getting the U^2 = gR bit

The question as asked requires us to assume U² = gR, and then prove from this assumption that the projectile reaches the height R above the Earth's surface.

hedgehog_7 · May 10, 2016

Re: HSC 2016 4U Marathon

oh i see... T_T

ze- · May 25, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
$$\noindent Let $\xi$ be a complex $n^{\text{th}}$ root of unity.$\\\\$a) i) Show $\sum_{r=1}^{n-1} \xi^r = -1 \\\\ \indent $ii) Hence, show $\sum_{r=1}^{n-1} (1-\xi^r) = n \\\\ \indent $iii) Hence, or otherwise, evaluate $\sum_{r=1}^{n-1} r\xi^{n-r-1}\\\\ $b) i) Prove the following extension of the triangle inequality using induction:$ \\\\ \indent |\pm z_1 \pm z_2 \pm \cdots \pm z_{n-1} \pm z_n| \leq |z_1| + |z_2| + \cdots + |z_{n-1}| + |z_n| \\\\ \indent $You may assume The Triangle Inequality: $ |\pm x \pm y| \leq |x|+|y| \\\\ \indent $ii) By using the result(s) from part a), prove $|1-\xi| > \frac{2}{n-1} \\\\ \indent $iii) Hence prove $\left|\sin{\frac{k\pi}{n}}\right| > \frac{1}{n-1}$

This is a great question

Paradoxica · May 25, 2016

Re: HSC 2016 4U Marathon

ze- said:
This is a great question

so great I don't recall it existed?

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