HSC 2012 MX1 Marathon #2 (archive) (2 Viewers)

nightweaver066 · Jun 17, 2012

Re: HSC 2012 Marathon

barbernator said:
thats not too bad, typing up working now. ok i wont lol

haha

RealiseNothing said:
Depends which way you do it tbh. If you weren't given the second part to prove it would be a lot harder. ie - Find $I_n$ in terms of (n-1) only, so that no other term appaears in the expression but some form of (n-1).

True. Should have gave it without the result!

deswa1 · Jun 17, 2012

Re: HSC 2012 Marathon

I like these sorts of questions so hopefully you guys will find it fun as well:

P(x) is a cubic polynomial with real coefficients. One zero of P(x) is 1+2i, the constant term is -15 and P(2)=5. Find the equation of P(x)

nightweaver066 · Jun 17, 2012

Re: HSC 2012 Marathon

deswa1 said:
I like these sorts of questions so hopefully you guys will find it fun as well:

P(x) is a cubic polynomial with real coefficients. One zero of P(x) is 1+2i, the constant term is -15 and P(2)=5. Find the equation of P(x)

This is MX1 thread lol

Nevertheless...

As all the coefficients are real, by the Conjugate Root Theorem, 1 - 2i is also a root.
By factor theorem, (x - 1 + 2i)(x - 1 - 2i), i.e. (x^2 - 2x + 5) is a factor of P(x).

Let P(x) = (x^2 - 2x + 5)(ax + b)
As constant = -15, 5b = -15, b = -3
P(2) = (4 - 4 + 5)(2a - 3) = 5
2a - 3 = 1
a = 2

Therefore P(x) = (2x - 3)(x^2 - 2x + 5)
= 2x^3 -7x^2 + 16x - 15

deswa1 · Jun 17, 2012

Re: HSC 2012 Marathon

nightweaver066 said:
This is MX1 thread lol

Oh really? Whoops, should have been paying attention...

Though I guess the only 4U knowledge you need is that complex roots occur in conjugate pairs and the definition of i.

For 3U students: Add to my question, both 1+2i and 1-2i are roots, where i is the square root of -1 (i.e. i^2=-1). This might not make sense though...

SpiralFlex · Jun 17, 2012

Re: HSC 2012 Marathon

So tempted to join you guys, but must resist. Tomorrow is my big day!

Kingportable · Jun 17, 2012

Re: HSC 2012 Marathon

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:

i) The period;
ii) Amplitude
iii) The speed when x=-2+1/2
iv) the acceleration when x = -2 + 1/2

Period was easy but then i got to amplitude, i thought i had to use this thing that this dude thought me Amplitude=Endpoints - centre but i realised that the question does specify values at certain points (eg v=7, x=4, x''=-7 at t=0). So i would imagine we would use the knowledge of the value of n=pi/2 and and that the period is 4 in order to work this question out. Lol i need better notes on SHM anyone got links? My text book or maths online is really not working for me.

nightweaver066 · Jun 17, 2012

Re: HSC 2012 Marathon

deswa1 said:
Oh really? Whoops, should have been paying attention...

Though I guess the only 4U knowledge you need is that complex roots occur in conjugate pairs and the definition of i.

For 3U students: Add to my question, both 1+2i and 1-2i are roots, where i is the square root of -1 (i.e. i^2=-1). This might not make sense though...

The thought of sqrt(-1) being defined is ridiculous

Timske · Jun 17, 2012

Re: HSC 2012 Marathon

Kingportable said:
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:

i) The period;
ii) Amplitude
iii) The speed when x=-2+1/2
iv) the acceleration when x = -2 + 1/2

Period was easy but then i got to amplitude, i thought i had to use this thing that this dude thought me Amplitude=Endpoints - centre but i realised that the question does specify values at certain points (eg v=7, x=4, x''=-7 at t=0). So i would imagine we would use the knowledge of the value of n=pi/2 and and that the period is 4 in order to work this question out. Lol i need better notes on SHM anyone got links? My text book or maths online is really not working for me.

amplitute is 5 .

deswa1 · Jun 17, 2012

Re: HSC 2012 Marathon

nightweaver066 said:
The thought of sqrt(-1) being defined is ridiculous

Yeah I know. I just hoped they would take it as fact (i.e. i+(-i)=0 and i^2=-1). Shouldn't have posted the question though- I'll try and find a nice 3U question

barbernator · Jun 17, 2012

Re: HSC 2012 Marathon

Kingportable said:
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:

i) The period;
ii) Amplitude
iii) The speed when x=-2+1/2
iv) the acceleration when x = -2 + 1/2

Period was easy but then i got to amplitude, i thought i had to use this thing that this dude thought me Amplitude=Endpoints - centre but i realised that the question does specify values at certain points (eg v=7, x=4, x''=-7 at t=0). So i would imagine we would use the knowledge of the value of n=pi/2 and and that the period is 4 in order to work this question out. Lol i need better notes on SHM anyone got links? My text book or maths online is really not working for me.

the amplitude is just 5. It is because the sine function on its own is maximum +- 1, but we are multiplying it by 5.

SHM is defined by x=asin(nt+e) where a is amplitude, period is 2pi/n and e is a constant.

NIGEL SIMMONS. look him up. he is the way forward.

Timske · Jun 17, 2012

Re: HSC 2012 Marathon

Requesting some hard integration questions

nightweaver066 · Jun 17, 2012

Re: HSC 2012 Marathon

Timske said:
Requesting some hard integration questions

$\int \frac{x^4}{1 + x^{10}}dx$

Timske · Jun 17, 2012

Re: HSC 2012 Marathon

nightweaver066 said:
$\int \frac{x^4}{1 + x^{10}}dx$

x^10?

deswa1 · Jun 17, 2012

Re: HSC 2012 Marathon

Timske said:
Requesting some hard integration questions

I'm about to start a 4U test that has integration as one of the topics. Any ones I see that don't require 4U techniques, I'll post for you.

Kingportable · Jun 17, 2012

Re: HSC 2012 Marathon

Zomg duhr lol i didn't realize i had the displace ment!! ARRHHHHHHHHHH!

Timske · Jun 17, 2012

Re: HSC 2012 Marathon

deswa1 said:
I'm about to start a 4U test that has integration as one of the topics. Any ones I see that don't require 4U techniques, I'll post for you.

Alright thanks,

nightweaver066 · Jun 17, 2012

Re: HSC 2012 Marathon

Timske said:
x^10?

I typed the question correctly.

barbernator · Jun 17, 2012

Re: HSC 2012 Marathon

1/5 tan^-1(x^5)+c?

nightweaver066 · Jun 17, 2012

Re: HSC 2012 Marathon

barbernator said:
1/5 tan^-1(x^5)+c?

Yep

Carrotsticks · Jun 17, 2012

Re: HSC 2012 Marathon

nightweaver066 said:
Mind if i go backwards? lol

$I_n = (n -1)I_{n - 1} + (n - 1)!$

$= (n - 1)[(n - 1)! - (n - 2)!] + (n - 1)!$

$= (n - 1)^2 (n - 2)! - (n - 1)! + (n - 1)!$

$= (n - 1)(n - 1)!$

$= \frac{(n - 1)n!}{n}$

$= n(n - 1)! - (n - 1)!$

$= n! - (n - 1)!$

Mind if I extend the question a bit?

$$Simplify $ \sum_{n=1}^{k} I_n$

HSC 2012 MX1 Marathon #2 (archive) (2 Viewers)

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Member

Well-Known Member

Sequential

Well-Known Member

Active Member

Sequential

Well-Known Member

Sequential

Well-Known Member

Member

Sequential

Well-Known Member

Active Member

Well-Known Member

Retired

Users Who Are Viewing This Thread (Users: 0, Guests: 2)