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Edexcel FP1 Thread - 20th May, 2016

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No I have not, I'm on my phone so dont have the link.

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Original post by economicss
Ah just found it, thanks so much for that :smile: I don't suppose you've done part b? Thanks :smile:


No I haven't as Im on my phone now so don't have the link

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Original post by Billsonbubbles
Can somone answer June 2010 Question 7a for me ?


I dont get the answers on the MS


Cheers


check attachment
Original post by kingaaran
What have you tried?

awww cute
Original post by Billsonbubbles
Can somone answer June 2010 Question 7a for me ?


I dont get the answers on the MS


Cheers


Hope this helps!
Original post by techfan42
check attachment


Cheers man,


How did you factorise it in part a ... I mean I get how you use the indice law in line 1 ... but then you went to 6( 2^k ... etc?


P,S there is a good exam on the physics and maths tutor website for the gold silver and bronze papers for FP1 i am sure you knew anyway ah
Original post by Brailey
Hope this helps!



Thanks, I dont get the factorisation on the second line of working.


Also I have taught myself AS further maths, do you have any tips that your teacher has mentioned , or you think is nice to know.


Like for me I noted that


Area of the new shape = Determinant of the transformation matrix x area of the old shape



And this works for quadrilaterals and triangles when you want to work out the area of shape .


Enjoy and thanks !
Original post by Billsonbubbles
Cheers man,


How did you factorise it in part a ... I mean I get how you use the indice law in line 1 ... but then you went to 6( 2^k ... etc?


P,S there is a good exam on the physics and maths tutor website for the gold silver and bronze papers for FP1 i am sure you knew anyway ah


You don't actually factorise, all you have to do is try and get the expression of f(k) multiplied by a number, in this case 6. Once you do that, you'll see that 6(2^k + 6^k) is basically 6f(k). However, remember that you can't change the equation. This is still an expression for f(k+1) Therefore, once you realise that, you see that 6f(k) gives you 4(2^k) extra. Therefore, you have to minus that in order to retain the original equation, giving you the required expression

p.s. yeah, thanks for the tip, but I've already finished them hahahaha
Original post by TomWeller
question: for the argument of z

i know you do tan^-1 (b/a) of a + bi

but do you keep a and b exactly how they are, or are they the mod of a and b?

example:
z = 1 - i

would it be tan-1(-1/1) or simply tan-1(1/1)... and why?


you dont take the mod. It depends on where it is on a argand diagram, as the angle for the argument is taken from the real numbers axis/x axis anticlockwise.

so for your example the angle would be minus tan-1(1/1) or 360/2pie - tan(1/1).

I always draw a diagram like a tiny sketch and see where it lies and then go from there!

Hope this helps
Original post by Billsonbubbles
Cheers man,


How did you factorise it in part a ... I mean I get how you use the indice law in line 1 ... but then you went to 6( 2^k ... etc?


P,S there is a good exam on the physics and maths tutor website for the gold silver and bronze papers for FP1 i am sure you knew anyway ah


post 983; all he did was use the law of indices 2(2^k) is basically 2^k+1 and vice versa

(p.s. I've also had to teach myself most of further maths, just started in jan :smile:
Original post by Billsonbubbles
Thanks, I dont get the factorisation on the second line of working.


Also I have taught myself AS further maths, do you have any tips that your teacher has mentioned , or you think is nice to know.


Like for me I noted that


Area of the new shape = Determinant of the transformation matrix x area of the old shape



And this works for quadrilaterals and triangles when you want to work out the area of shape .


Enjoy and thanks !


So for the second line of working the questions asks for 6f(k) you already have 6.6^k so by changing the 2.2^k to 6^k-4.2^k you still get 2.2K^2 and can factorise to get in the form that the question is asking, its a case of jigging it around to get what the question asks.,

In terms of tips I havnt learnt the transformations i prove them this takes maybe a minute longer but you save time if you cant remember instead of a guess!

Cheers!
Original post by techfan42
post 983; all he did was use the law of indices 2(2^k) is basically 2^k+1 and vice versa

(p.s. I've also had to teach myself most of further maths, just started in jan :smile:


Oh nice, cheers by the way.


I take it you have finished it then ?


I still have half of S2 to do aha FML ah.


Not to mention going over pretty much off all of A2 economics.

And A2 physics ah :wink:

get it I just need to go over it in my head again for the last two

Cheers
Original post by iMacJack
Yep - silly me.
Cheers


explain to me please!!!!!!! stuck on it for 20 mins (i did c2 2 years ago and got 100 yet i still cant remember how to do it loooool)
Original post by Billsonbubbles
Oh nice, cheers by the way.


I take it you have finished it then ?


I still have half of S2 to do aha FML ah.


Not to mention going over pretty much off all of A2 economics.

And A2 physics ah :wink:

get it I just need to go over it in my head again for the last two

Cheers


I'm actually doing it AS, so they're making us doing FP1, FP2 and D1. Which is messed up because FP2 is usually an A2 module and you need C3/C4 knowledge, but it's fine cause I went over the C3/C4 books too
https://gyazo.com/bb2a0cc0002c1fdf91f31d04facbb76a

part b help with the geometric series bit please
Original post by tazza ma razza
explain to me please!!!!!!! stuck on it for 20 mins (i did c2 2 years ago and got 100 yet i still cant remember how to do it loooool)


Split it up into the sum of r = 1 to 12 of (9r^2-4r) + k lots of the sum of r = 1 to 12 of 2^r

put r = 1 in to get your a, your n = 12, use the sum of a geometric series formula, youll get a value, add the bits together, equate it to the value given, rearrange for k
Original post by Billsonbubbles
Thanks, I dont get the factorisation on the second line of working.


Also I have taught myself AS further maths, do you have any tips that your teacher has mentioned , or you think is nice to know.


Like for me I noted that


Area of the new shape = Determinant of the transformation matrix x area of the old shape



And this works for quadrilaterals and triangles when you want to work out the area of shape .


Enjoy and thanks !


bruh it says that in the textbook
Original post by tazza ma razza
https://gyazo.com/bb2a0cc0002c1fdf91f31d04facbb76a

part b help with the geometric series bit please


sum of 2^r from r=0 to r=n is 2^(n+1) -1
Can someone please explain the answers to part (b) and (ii) of question 6 on the 2015 IAL paper
https://699d3c34b250207412778630ab10c15e39222eaf.googledrive.com/host/0B1ZiqBksUHNYd0pJSmxJSHozWVk/January%202015%20(IAL)%20MA%20-%20F1%20Edexcel.pdf

My answer to (b) was 150 clockwise? and in (ii) I don't understand how using
b^2 -4ac shows that det(M) is not equal to zero

Thank you!
Guys, can you do the f(k)=......=5m for an e.g divisible by 5 questions and then rearrange for one term and sub into f(k+1) for *every* divisibility induction question? I think it's my preferred method now. A reply would be much appreciated. :smile:

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