Mega A Level Maths Thread MK II

Original post by thers

This has got to be the worst a level maths question I have seen lmfao. I scored 1/14 on it.

http://filestore.aqa.org.uk/subjects/AQA-MFP2-QP-JAN12.PDF

Question 8. I can't make any progress past part (a). Can anyone explain what to do at all?

I've not actually checked, but I imagine that the roots of

\ z^5 - 1 =0

are the same as the roots of

z^4+z^3+z^2+z+1=0

given that it says hence :smile:

Reply 1782

Original post by joostan

I've not actually checked, but I imagine that the roots of

\ z^5 - 1 =0

are the same as the roots of

z^4+z^3+z^2+z+1=0

given that it says hence :smile:

They are apparently but why?

Reply 1783

Original post by thers

They are apparently but why?

Because

\dfrac{z^5-1}{z-1} = z^4 + z^3+z^2+z+1

Reply 1784

GCSE-help

6

Original post by justinawe

No.

\log (28x) - \log (9) = \log \left( \dfrac{28x}{9} \right)

, using basic log rules.

So definitely not the same thing.

Yeah silly me :colondollar:

Thanks a lot :smile:

Reply 1785

Original post by joostan

Because

\dfrac{z^5-1}{z-1} = z^4 + z^3+z^2+z+1

Right, do you have any ideas for the next 2 parts?

Reply 1786

Original post by joostan

Because

\dfrac{z^5-1}{z-1} = z^4 + z^3+z^2+z+1

Actually I've figured out how to do (d) by using the given answer in (c). But can't do (c)

Reply 1787

Original post by thers

Right, do you have any ideas for the next 2 parts?

You know (by de Moivre) that

e^{i\theta} + e^{-i\theta} = 2cos(\theta)

How does this help, do you think?

Reply 1788

L'Evil Fish

OP

18

Lost 3 marks on Jan 2013 on this...

I think I know why, but the mark scheme has no workings, just the values :redface:

Do I work out the gradient and make it negative to find a?

Reply 1789

username937760

16

M5, June 2009. 99 UMS.

This is what my life has become.

Reply 1790

Original post by joostan

You know (by de Moivre) that

e^{i\theta} + e^{-i\theta} = 2cos(\theta)

How does this help, do you think?

So I've got z^2 + z^-2 + 2cos2pi/5 + 1 = 0
What next?

Reply 1791

Boy_wonder_95

15

Original post by DJMayes

M5, June 2009. 99 UMS.

This is what my life has become.

Arithmetic errors? :tongue:

Reply 1792

Original post by L'Evil Fish

Lost 3 marks on Jan 2013 on this...

I think I know why, but the mark scheme has no workings, just the values :redface:

Do I work out the gradient and make it negative to find a?

Yep

Reply 1793

L'Evil Fish

OP

18

Original post by joostan

Yep

Darn it!

Lets hope I can smash Friday's exam now :cool:

lets see what UMS I'd have had..

2012: 100 UMS
2013: 97 UMS

I want 97 (birth year :cool:

)

Reply 1794

Original post by thers

So I've got z^2 + z^-2 + 2cos2pi/5 + 1 = 0
What next?

What you're trying to do is factorise the quartic into 2 quadratics, then divide by z^2 to get the desired result :smile:

Reply 1795

Original post by L'Evil Fish

Darn it!

Lets hope I can smash Friday's exam now :cool:

lets see what UMS I'd have had..

2012: 100 UMS
2013: 97 UMS

I want 97 (birth year :cool:

)

100 is a bit showy :wink:

- same goes for you DJ :tongue:

Reply 1796

L'Evil Fish

OP

18

Original post by joostan

100 is a bit showy :wink:

- same goes for you DJ :tongue:

Showy? I want 97/98... So it looks nice... I'd love a 100 though, in one of C3/4 but not gonna happen :mmm:

Reply 1797

username937760

16

Original post by Boy_wonder_95

Arithmetic errors? :tongue:

Not even arithmetic, even worse. On a vectors question on systems of forces I forgot to multiply by -1 so forces would cancel out, -2 accuracy marks. Then, on a moments of inertia question with finding the force on a hinge, an absolute monster worth 9 marks, I did it all completely correctly, managed to avoid the 1001 different places I could have made an arithmetical error, but forgot to add an mg on the end of 1 of two things when expanding them out from brackets in the final line, so minus another mark. :facepalm:

Reply 1798

Original post by joostan

What you're trying to do is factorise the quartic into 2 quadratics, then divide by z^2 to get the desired result :smile:

Right so I've got a quartic

z^4 + (2cos(2pi/5) + 1 )z^2 + 1 =0

How am I supposed to factorize this?

Reply 1799