Ekman's Compilation of MX2 Questions (2 Viewers)

Ekman · Nov 27, 2015

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 · Nov 27, 2015

Ekman said:
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 · Nov 28, 2015

I suck. And I'm in holiday mode. So I'm to lazy to do these.

But dayum

gsinclair · Dec 21, 2015

Thanks Ekman; that's awesome.

KingOfActing · Dec 21, 2015

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

ze- · Jan 4, 2016

SammyT123 · May 19, 2016

Ekman said:
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 · May 19, 2016

SammyT123 said:
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 · May 19, 2016

Drsoccerball said:
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 · May 19, 2016

InteGrand said:
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 · May 19, 2016

this is nice..gonna be useful next year

marxman · May 23, 2016

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?

dan964 · May 24, 2016

Resticky?

wu345 · May 31, 2016

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

Drsoccerball · May 31, 2016

wu345 said:
Isn't the solution to Q15 of the polynomials section, 8, not 5?

5 is correct.

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

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

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

InteGrand · May 31, 2016

Drsoccerball said:
5 is correct.

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

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

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

$\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 · May 31, 2016

InteGrand said:
$\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.
$(2-2\gamma)(2-2\beta)(2-2\alpha)$

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

$8P(1)$

InteGrand · May 31, 2016

Drsoccerball said:
Sorry I just activated auto mode
You can still do it that way though.
$(2-2\gamma)(2-2\beta)(2-2\alpha)$

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

$8P(1)$

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

wu345 · Jun 4, 2016

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 · Oct 3, 2016

Are these from HSC papers ?

Ekman's Compilation of MX2 Questions (2 Viewers)

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