HSC 2012 Marathon :)

**RealiseNothing** · 17 Jun 2012, 10:50 PM

Originally Posted by Carrotsticks

Oh right yes yes I remember now.

Here's an interesting question:

A and B are two positive integers such that A > B and A =/= B.

Now consider A^B and B^A.

Which one is larger? Do any conditions need to be imposed to satisfy each inequality?

The bolded bit seems a bit redundant given A > B lol.

But I'll give it a go now.

**Carrotsticks** · 17 Jun 2012, 10:50 PM

Oh woops haha! Not thinking straight.

**Sanjeet** · 24 Jun 2012, 10:40 PM

Originally Posted by Kingportable

Lol plox someone help me solve this question, its driving me crazy!

John Fitzpatrick 3U Mathematics 25(c) Question 3

A particle movies in a straight line. At time t seconds, its displacement x cm from a fixed poin O in the line is given by x=5sin((pi/2)T + pi/6). Express the acceleration in terms of x only and hence show that the motion is simple harmonic. Find:

iii) The speed when x=-2 1/2 or x = - 5/2
iv) the acceleration when x = -2 1/2 or x = - 5/2

$x = 5sin \left ( \frac{\pi}{2}t + \frac{\pi}{6} \right ) \\ \dot{x} = \frac{5\pi}{2}cos\left ( \frac{\pi}{2}t + \frac{\pi}{6} \right ) \\ \ddot{x} = \frac{-5\pi^2}{4}sin \left ( \frac{\pi}{2}t + \frac{\pi}{6} \right ) \\ \ddot{x} = a = -\frac{\pi^2}{4}x \therefore $ It is in SHM as acceleration is proportional to displacement$ \\ \\ (iii)$ When $x = \frac{-5}{2},\\ \frac{-5}{2} = 5sin \left ( \frac{\pi}{2}t + \frac{\pi}{6} \right ) \\ \\ \frac{-1}{2} = sin \left ( \frac{\pi}{2}t + \frac{\pi}{6} \right ) \\ \\ \therefore \frac{\pi}{2}t + \frac{\pi}{6} = \frac{\pi}{6} + \pi = \frac{7\pi}{6}$ (ASTC)$ \\ \dot{x} = \frac{5\pi}{2}cos \left \frac{7\pi}{6} \right \Rightarrow -\frac{5\sqrt{3}\pi}{4} \\ \therefore speed = \frac{5\sqrt{3}\pi}{4}cms^{-1} \\$

(iv) Not even gonna bother to do this.

**Timske** · 24 Jun 2012, 10:49 PM

$x = -2\frac{1}{2} ~ and~\ddot{} x=-\frac{\pi^{2}}{4}x \\ Sub ~x ~in~,~\ddot{} x=-\frac{\pi^{2}}{4} *-2\frac{1}{2} = \frac{5\pi^{2}}{8} ~cm/s^2$

**barbernator** · 24 Jun 2012, 11:04 PM

if sally the problem gambler has 6 dice, all of which are 50% biased to a different number (i.e. for one die, chance of rolling a 1 - 50%, 2,3,4,5,6 - each 10%)

if she chooses 2 die at random and rolls them, what is the chance that she rolls a 1 and a 6?

If she wanted fair odds on the bet, what would the odds be?

**RealiseNothing** · 25 Jun 2012, 12:17 AM

Originally Posted by barbernator

if sally the problem gambler has 6 dice, all of which are 50% biased to a different number (i.e. for one die, chance of rolling a 1 - 50%, 2,3,4,5,6 - each 10%)

if she chooses 2 die at random and rolls them, what is the chance that she rolls a 1 and a 6?

If she wanted fair odds on the bet, what would the odds be?

Just having a quick look at the question, but wouldn't the fact that the even distribution of bias cancels out mean that the question is just the probability of rolling a 1 and 6 from 2 dice, which is 1/18?

**barbernator** · 25 Jun 2012, 5:23 PM

Originally Posted by RealiseNothing

Just having a quick look at the question, but wouldn't the fact that the even distribution of bias cancels out mean that the question is just the probability of rolling a 1 and 6 from 2 dice, which is 1/18?

yep lol

**Fus Ro Dah** · 25 Jun 2012, 5:24 PM

More questions please =) I'm bored..

**barbernator** · 25 Jun 2012, 5:34 PM

ok hmmm.

a projectile is fired at an angle of 40 degrees to the positive x axis. If the particle just clears a 4m high wall, 5.5m away from the firing spot on its ascent, what initial velocity V is it fired at, and at what distance away would it just clear a 2m high wall on its decline

**Fus Ro Dah** · 25 Jun 2012, 5:46 PM

Maybe a 'cooler' one that isn't just calculations and number crunching please? I like proofs more :3

**barbernator** · 25 Jun 2012, 5:48 PM

ok ill dig through a past cssa paper=

**barbernator** · 25 Jun 2012, 5:54 PM

a lower half of a hemisphere is obtained by rotating x^2+y^2-20y=0 around the x axis, and cutting it through the centre. Water is poured in, Find equations for the volume of the water, and the surface area of the water. lol thats super simple haha ill find a better one

ok, heres one.

find the limiting value of 1 + 1 + 3/4 + 1/2 + 5/16 + 3/16 + .... and i won't give you the general term because you want a challenge :P

**Fus Ro Dah** · 25 Jun 2012, 6:06 PM

We can break the above series into two sub-series and use the limiting sum for each one.

**barbernator** · 25 Jun 2012, 6:37 PM

Originally Posted by Fus Ro Dah

We can break the above series into two sub-series and use the limiting sum for each one.

which are? i will add a few more terms if you want

**Trebla** · 26 Jun 2012, 5:26 PM

The second term of a geometric series is 1/4 and its limiting sum is -1/3. Find the value of the fifth term of the series.

**barbernator** · 26 Jun 2012, 5:34 PM

-1/32. Does anyone have an answer to the one above?

**Aesytic** · 26 Jun 2012, 5:56 PM

i got -1/32 as well

**Kingportable** · 30 Jun 2012, 2:15 PM

Simple Harmonic Motion Fitzpatrick 3U Chapter 25(C) Q4(a)
The displacement x at time t of a point moving in a straight line is given by x=asin(nt+epsiton). Find the form wich this expression takes if initially:
b) x=0 and the velocity is negative

-------
Hey i'm trying to this question and currently working on question "a"
a) x'=0 and x=-5

**Kingportable** · 30 Jun 2012, 2:54 PM

Forget my previous question... Lol this one is an even more annoying one

This question's answer was completely different to mine.

Question 7 (CHAPTER 25C)
Solve the differential equation d^2x/dt^2 + 16x = 0 subject to the conditions x=3 and dx/dt=16 when t=0. Find the maximum displacement and the maximum speed if x metres is the displacement of a particle moving in a straight line at time t seconds.

what i did
i said that x=asin(nt+alpha)
and that d^2/dt^2 +16x=0
is the same as x''+16x=0
so x''=-16x <-- SHM where n=4

Since this is a sin function at t=0, x=0
so asin(alpha)=0
alpha=Sin^-1(0)
so, alpha=0 at t=0

since v=16 at t=0
i got that v=ancos(nt+alpha)=16
since at t=0, alpha =0
ancos(0)=16
an=16
since n=4
4a=16
a=4
so... x=4sin(4t)
however the back of John Fitzpatrick 3U says
x=4sin(4t) +3cos4t

the 4sin(4t) part makes total sense.... BUT WHERE THE HELL DID THAT 3cos4t COME FROM!!!!!!!!....WTTTFFFFF??????
So yeah im stuck on finding the displacement, is the axiliary methods in play in here?

Anyways the other answers are: 5m;20m/s

**Sy123** · 2 Jul 2012, 11:39 PM

$\begin{align*}\ddot{x}&=-16x \\ \left(\frac{1}{2}v^2\right)\frac{d}{dx}&=-16x \\ \frac{v^2}{2}&=-8x^2+C \\ \\ x=3 \Rightarrow v=16 \\ \frac{16^2}{2}&=-8(3)^2+C \\ C=200 \\ \frac{v^2}{2}&=200-8x^2 \\ v^2&=400-16x^2 \\ v&=+\sqrt{16(25-x^2)} \\ v&=+4\sqrt{25-x^2} \\ \\ \frac{dt}{dx}&=\frac{1}{4\sqrt{25-x^2}} \\ \\ t&=\frac{1}{4}\sin^{-1}{\frac{x}{5}}+C \end{align*}$

$t=0 \Rightarrow x=3 \\ \\ t&=\frac{1}{4}\sin^{-1}{\frac{3}{5}}+C \\ \\ t&=\frac{1}{4}\sin^{-1}{\frac{x}{5}}-\frac{1}{4}\sin^{-1}{\frac{3}{5}} \\ \\ x&=5\sin{4t+\sin^{-1}{\frac{3}{5}}} \\ \\ x_{max}=5 \\ \\ v_{max} \Rightarrow x=0 (C.O.O) \\ \\ v_{max}=4\sqrt{25-0}=20m/s$

And if you notice, if you expand my x in terms of t, using the triangle to evaluate the inverse sine of 3/5

$x&=5\sin{4t+\sin^{-1}{\frac{3}{5}}} \\ \\ x&=5\left(\sin{4t} \cos{\sin^{-1}{\frac{3}{5}}}+\cos{4t} \sin{\sin^{-1}{\frac{3}{5}}}\right)$

$x&=5\left(\sin{4t}\left(\frac{4}{5}\right)+\cos{4t} \left (\frac{3}{5}\right) \right ) \\ \\ x&=4\sin{4t}+3\cos{4t}$

Which is the answer in the Fitzpatrick book. My way of solving this question is alot cleaner imo. (Also first time using alignment in latex, which turned out to be a semi failure since it was making my latexing invalid for some reason)

I will post my question up here very shortly

**Sy123** · 3 Jul 2012, 12:13 AM

Show that
$\sin{\cos^{-1}{x}}=\sqrt{1-x^2} \ \ \text{For all x} -1\leq x\leq 1$

Difficulty: Easy

I will have a better one tommorow.

**lolcakes52** · 7 Jul 2012, 3:24 PM

Originally Posted by barbernator

-1/32. Does anyone have an answer to the one above?

no, can you add more terms. is the initial 1 part of the series.

**OMGITzJustin** · 9 Jul 2012, 6:19 PM

A particle moves with simple harmonic motion. At the extremities of the motion the absolute value of the acceleration is 1cm^-2, and when the particle is 3cm from the centre of motion the speed is 2*squareroot2 cm^-1.

Find period and amplitude

**Sy123** · 9 Jul 2012, 10:46 PM

Originally Posted by OMGITzJustin

A particle moves with simple harmonic motion. At the extremities of the motion the absolute value of the acceleration is 1cm^-2, and when the particle is 3cm from the centre of motion the speed is 2*squareroot2 cm^-1.

Find period and amplitude

$\ddot{x} =-n^2 x \\ \frac{1}{2}v^2=\frac{-n^2 x^2}{2}+C \\ \\ x=a \rightarrow v=0 (\text{Where a is extremety}) \\ \\ \therefore C=\frac{n^2 a^2}{2} \\ v^2=-n^2 x^2+n^2 a^2$

$(2\sqrt{2})^2=-9n^2+n^2 a^2 \\ \\ 8=-9n^2+n^2 a^2 \\ \\ \Abs{\ddot{x}}=1 \ \ \ when \ \ \ x=a \\ \\ if \ \ n>0 \ \ \ a>0 \therefore n^2 a=1 \\ \\ \therefore 8=-9n^2+a \\ n^2=\frac{1}{a} \\ \\ 8=\frac{-9}{a}+a \\ \\ a^2-8a-9=0 \\ a=9 \ \ a=-1 \ \ \ a>0 \\ \therefore amplitude=9 \\ \\ n^2 a=1 \\ \therefore n=\frac{1}{3} \\ \therefore Period=6\pi$

I have a feeling it might be wrong though but at least I attempted.

**Sy123** · 9 Jul 2012, 11:08 PM

$$Prove that for all integers$ \ \ n>0 \\ \\ x^n \neq \ln{x}$

Its a question entirely possible within 3u boundaries (depending on how you do it you will have to use 3u knowledge)

I made the question up after I proved that result. So enjoy :3

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