# Ekman's Compilation of MX2 Questions (1 Viewer)

#### Ekman

##### Well-Known Member
Hey everyone, I decided to take the liberty of collecting all the hard MX2 questions that I solved during my HSC year and develop a more organised compilation for students that study MX2.

https://www.dropbox.com/sh/6t6f4tcz...hL7Hgca/Compilation of All Questions.pdf?dl=0

Feel free to point out any errors, and I will hopefully fix them up as soon as possible

Also a big thanks to Integrand, Sy123 and a few others for helping me validate the answers as well

-Ekman

#### Drsoccerball

##### Well-Known Member
Hey everyone, I decided to take the liberty of collecting all the hard MX2 questions that I solved during my HSC year and develop a more organised compilation for students that study MX2.

https://www.dropbox.com/sh/6t6f4tcz...hL7Hgca/Compilation of All Questions.pdf?dl=0

Feel free to point out any errors, and I will hopefully fix them up as soon as possible

Also a big thanks to Integrand, Sy123 and a few others for helping me validate the answers as well

-Ekman
No way. This file was to good to release props to Ekman.

#### leehuan

##### Well-Known Member
I suck. And I'm in holiday mode. So I'm to lazy to do these.

But dayum

#### gsinclair

##### New Member
Thanks Ekman; that's awesome.

#### KingOfActing

##### lukewarm mess
I am officially your best friend ever I am going to do them all these holidays yessss

Wow!

#### SammyT123

##### Active Member
Hey everyone, I decided to take the liberty of collecting all the hard MX2 questions that I solved during my HSC year and develop a more organised compilation for students that study MX2.

https://www.dropbox.com/sh/6t6f4tcz...hL7Hgca/Compilation of All Questions.pdf?dl=0

Feel free to point out any errors, and I will hopefully fix them up as soon as possible

Also a big thanks to Integrand, Sy123 and a few others for helping me validate the answers as well

-Ekman
Do you happen to have worked solutions to these ?
Regardless thankyou so much for taking the time to make these

#### Drsoccerball

##### Well-Known Member
Do you happen to have worked solutions to these ?
Regardless thankyou so much for taking the time to make these
I made these for integration, complex numbers, volumes and partially mechanics but I threw them in the bin ahaha.

EDIT: After final's I could make it again? ...

#### InteGrand

##### Well-Known Member
I made these for integration, complex numbers, volumes and partially mechanics but I threw them in the bin ahaha.

EDIT: After final's I could make it again? ...
Why did you throw them away?

#### Drsoccerball

##### Well-Known Member
Why did you throw them away?
I was cleaning and it was in the way and I had no where to put it LOL.

#### Green Yoda

##### Hi Φ
this is nice..gonna be useful next year

#### marxman

##### Member
Hi all, as Drsoccerball requested i'll just post the questions on here from now on, these are all pretty hard.

Could I also get some help with Q10 from the polynomial set? Is it similar logic to Q3?

Resticky?

#### wu345

##### Member
Isn't the solution to Q15 of the polynomials section, 8, not 5?

#### Drsoccerball

##### Well-Known Member
Isn't the solution to Q15 of the polynomials section, 8, not 5?
5 is correct.

$\bg_white (\alpha + \beta + \gamma - \gamma)(\alpha + \beta + \gamma - \alpha)(\alpha + \beta + \gamma - \beta)$

$\bg_white = (2 - \alpha)(2 - \beta)(2 - \gamma)$

$\bg_white Which is the same thing as saying P(2). \therefore P(2) = 5.$

#### InteGrand

##### Well-Known Member
5 is correct.

$\bg_white (\alpha + \beta + \gamma - \gamma)(\alpha + \beta + \gamma - \alpha)(\alpha + \beta + \gamma - \beta)$

$\bg_white = (2 - \alpha)(2 - \beta)(2 - \gamma)$

$\bg_white Which is the same thing as saying P(2). \therefore P(2) = 5.$
$\bg_white \noindent If this was the intention of the solution method, maybe the Q. was typed up incorrectly? The Q. asks to find \left(\alpha+\beta - \gamma \right)\left(\beta + \gamma -\alpha \right)\left(\gamma +\alpha - \beta \right), where \alpha,\beta,\gamma are the roots of x^3 -2x^2 + 3x-1=0. So we can't just do what you did above; we could only generally do that if we were adding and subtracting something in each term in the product, which we're not doing for the Q. as asked. If we wanted to do what you did, the Q. should've asked for the value of something like \left(\alpha+\beta \right)\left(\beta + \gamma \right)\left(\gamma +\alpha \right).$

#### Drsoccerball

##### Well-Known Member
$\bg_white \noindent If this was the intention of the solution method, maybe the Q. was typed up incorrectly? The Q. asks to find \left(\alpha+\beta - \gamma \right)\left(\beta + \gamma -\alpha \right)\left(\gamma +\alpha - \beta \right), where \alpha,\beta,\gamma are the roots of x^3 -2x^2 + 3x-1=0. So we can't just do what you did above; we could only generally do that if we were adding and subtracting something in each term in the product, which we're not doing for the Q. as asked. If we wanted to do what you did, the Q. should've asked for the value of something like \left(\alpha+\beta \right)\left(\beta + \gamma \right)\left(\gamma +\alpha \right).$
Sorry I just activated auto mode
You can still do it that way though.
$\bg_white (2-2\gamma)(2-2\beta)(2-2\alpha)$

$\bg_white 8(1 - \gamma)(1 - \beta)(1 - \alpha)$

$\bg_white 8P(1)$

#### InteGrand

##### Well-Known Member
Sorry I just activated auto mode
You can still do it that way though.
$\bg_white (2-2\gamma)(2-2\beta)(2-2\alpha)$

$\bg_white 8(1 - \gamma)(1 - \beta)(1 - \alpha)$

$\bg_white 8P(1)$
Yep, and P(1) = 1 – 2 + 3 – 1 = 1, so the answer is indeed 8 as wu345 said.

#### wu345

##### Member
Can I please have a hint for Q18 of integration? I simplified the integrand a bit by using difference of cubes and cancelling some common factors but after that im pretty stuck. Chucking into wolfram Alpha and integral calculator resulted in both timing out

#### ssukarieh1

##### New Member
Are these from HSC papers ?