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Edexcel FP1 Thread - 20th May, 2016

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Original post by anndz3007
I put rotation 135 degree anticlockwise, cus when i put 135 back into the cos and sin formula i got the same matrix as the question
And 12.468 3dp


cheers, Im so annoyed seeing every one have it 0.001 different to mine as I checked it twice in the half an hour I had left :frown:
Original post by chewitt77
cheers, Im so annoyed seeing every one have it 0.001 different to mine as I checked it twice in the half an hour I had left :frown:


You'll almost definitely only drop 1 mark. Don't worry about it :tongue:
Original post by chewitt77
+-sqr2/2 (+-1/sqr2)
9/2*x^1/2 + 25/2*x^-3/2
12.467 (I double checked this however may still be wrong)
a=38 b=15 c=1
2-2i, -8i, p=-4 q=8
1/p
anti clockwise rotation 45 degree about (0,0) (this may be wrong)
p=-3 q=-9, matrix (0 1 (bottom line 1 0))
a^2 +2a -4 + 4i(a +1), a=-1, sqr5 and 2.03 rad, QR twice the length of OP and QR is parallel to OP
x=15sqr2/4


Wait, for the QR and OP I don't remember them being parallel. Wasn't R = -5 on the real axis, P = -1+2i. I don't remember what they were exactly

EDIT:
And Q = -3+4i? Not sure
(edited 7 years ago)
Original post by midgemeister7
Don't think it did


It definitely did not. Intersection between y=-x and the gradient was the condition to find the other point I think.
Original post by TheRandomGenius
Sorry yeah thats what I meant- T^2 = I. That would get the mark?


I dont kbow but that's what i put, i calculated TT and got I, so i put TT= I, TT^-1= I
=>T =T^-1
Original post by Chirstos Ioannou
Wait i just realized i'm talking about the IAL one, so the GCE was hard?

Is there a different blog for IAL as I did that paper as well
I got K=5
Wbu?
Someone please make a Hitler reacts video
Original post by Sallekmo
Lol dude they wernt 3 marks they were 5 marks each

Nah the very last two questions were 3 each.
anyone got a= -1 for z^2+2z ?
Original post by ws12758
No it said that the midpoint was on y=-x lol dumbdumb


That's a different question.
Original post by Vishna
Is there a different blog for IAL as I did that paper as well
I got K=5
Wbu?


Someone's made a thread for the IAL paper.

http://www.thestudentroom.co.uk/showthread.php?t=4102261
Original post by chewitt77
cheers, Im so annoyed seeing every one have it 0.001 different to mine as I checked it twice in the half an hour I had left :frown:


I was trying to be sure so i literally put all the digits in the calculation in the equation ._. Idk why everyone got .007
Does anybody remember how much were the two last sub questions in question 9 worth? 🙄
Original post by oShahpo
It definitely did not. Intersection between y=-x and the gradient was the condition to find the other point I think.


He made an account just to call someone dumb and got it wrong himself lol

Original post by annehappy
anyone got a= -1 for z^2+2z ?


yep
Original post by annehappy
anyone got a= -1 for z^2+2z ?


Yes
Original post by TheRandomGenius
Sorry yeah thats what I meant- T^2 = I. That would get the mark?


Yes.
Original post by anndz3007
I put rotation 135 degree anticlockwise, cus when i put 135 back into the cos and sin formula i got the same matrix as the question
And 12.468 3dp


agreed on 12.468.

Think I wrote mod z as root 5 but arg to 3sf. Looks like I lost one there
Original post by Sophie Wilczek
Does anybody remember how much were the two last sub questions in question 9 worth? 🙄


nein
Original post by Pyslocke
Wait, for the QR and OP I don't remember them being parallel. Wasn't R = -5 on the real axis, P = -1+2i. I don't remember what they were exactly


I can't rember exactly however if you look at OP as being the vector z (as that is what point P was)

Then travelling from the point z^2 two lots of the vector z means they have both travelled in the same direction, just one QR was two lots of z hence travelling twice as far
If anybody gets a hold of the paper, I don't mind doing model solutions.

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