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# Edexcel FP1 Thread - 20th May, 2016 watch

1. (Original post by fpmaniac)
Can anyone help me on how to solve sigma r=0 to r=1. Someone explained it before but i couldnt fully get it. Maybe give me some examples with n=1,2,3 etc. Thanks
If you look at the screenshot I have put examples in.

All you do is add on what r=0 is to the the summation for your soultion.

Hope this helps!
Attached Images

2. (Original post by iMacJack)
Should be (12,12) not (27,12)
Thought it looked too easy

For some reason I used p=3 when I got the x coordinate, whoops
3. (Original post by Music With Rocks)
Just to check when rearranging matrices does the term always go to the front so if
AB=C

A=B-1C
and
B=A-1C

Or is this wrong? it looks wrong
It depends on what side your value is on, if we have

XA = B
and we have to isolate the 'X'
then:
XAA^-1 = BA^-1
XI = BA^-1
X = BA^-1

Basically you apply it to the same side of the term on either side
(Sorry for the bad explanation!)
4. (Original post by TheMoon)
I don't understand, isn't it like two right angled triangles which you can use 1/2 base times height with?

I tried drawing it, obviously I've drawn it very inaccurately (terrible paint skills) since it doesn't look like a normal but is it something like this? Might be wrong. Sorry if I'm completely wrong with how it looks.
This gets 225 anyway.
If p=2, sub that into (3p^2, 6p), you get (12, 12) not (27, 12)
I drew a diagram and it was pretty much irregular as far as I could tell
5. Would you get credit for using this in the exam? TeeEm, Zacken ?

(Original post by WhatIsSleep)
For that PQS question, I found an area formula which helped me work it out..

(I have attached it as an image but the source is http://www.mathopenref.com/coordtrianglearea.html)

So A is (3, 0)
B is (12, 12)
and C is (27, -18)

Ax=3, Ay=0
Bx=12, By=12
Cx=27, Cy=-18

Subbing those into the formula, I got 225 units^2

Honestly though, I tried drawing diagrams and everything...couldn't find a better way of doing it tbh
6. (Original post by Brailey)
If you look at the screenshot I have put examples in.

All you do is add on what r=0 is to the the summation for your soultion.

Hope this helps!
Where did the 2n come from in the end equation. If u add r=0 shouldnt it be just +2
7. (Original post by WhatIsSleep)
If p=2, sub that into (3p^2, 6p), you get (12, 12) not (27, 12)
I drew a diagram and it was pretty much irregular as far as I could tell
yeah sorry, I used p=3 for some reason when getting the x coordinate. not sure then

There has to be an easier way though.
8. (Original post by TrueDAN)
That is a beastly question haha!! Just done the first part now! How far have you got?
tbh I think that's fp3. They introduce Loci in FP2 so FP1 takers wouldn't know what a loci of points is
9. (Original post by iMacJack)
It depends on what side your value is on, if we have

XA = B
and we have to isolate the 'X'
then:
XAA^-1 = BA^-1
XI = BA^-1
X = BA^-1

Basically you apply it to the same side of the term on either side
(Sorry for the bad explanation!)
Oh I think I get what you mean so

AB=C
A=CB-1

AB=C
B=A-1C

Is this correct now?
10. (Original post by WhatIsSleep)
If p=2, sub that into (3p^2, 6p), you get (12, 12) not (27, 12)
I drew a diagram and it was pretty much irregular as far as I could tell
Just draw a diagram, find the gradients of SP and SQ and u will see that they are perpendicular, then its just half the area of the rectangle using length of a line formula to work out the lengths
11. (Original post by TheMoon)
yeah sorry, I used p=3 for some reason when getting the x coordinate. not sure then

There has to be an easier way though.
Maybe it's something that's not on the UK FP1 spec :/
12. (Original post by Music With Rocks)
Oh I think I get what you mean so

AB=C
A=CB-1

AB=C
B=A-1C

Is this correct now?
Correct
13. (Original post by iMacJack)
ii A is just finding the Det M, because the det is the scale factor, so ii (a) is 4, isn't it? Aren't you talking about part ii (b)?
Could you explain a bit more in depth please?
14. (Original post by iMacJack)
Correct
Thank you very much you explained it well
15. (Original post by TheRandomGenius)
6b anyone? (i and ii). I've gone around in circles and ended up in an algebraic mess.

https://7cba9babeb0db0ff9468853e0b2d...%20Edexcel.pdf
did you figure it out?
16. Nope.
(Original post by AmarPatel98)
did you figure it out?
17. (Original post by connorbarr)
Just draw a diagram, find the gradients of SP and SQ and u will see that they are perpendicular, then its just half the area of the rectangle using length of a line formula to work out the lengths
ahh seems you're right :P
I didn't think to find the gradients of the lines
18. (Original post by TheRandomGenius)
Nope.
I've done 6bi, i'm just doing ii. I'll post in a minute.
19. (Original post by TheRandomGenius)
Nope.

So you get the tangent at P from part a.

Similarly you get the tangent at Q.

The gradient of the tangent at P is 1/p, so similarly, the gradient at Q is 1/q. The two tangents are perpendicular, so you know their gradients must multiply to give -1. Form an eqn from this.

You are told the intersection point, so form another eqn from this.

Solve simultaneously to obtain p and q
20. (Original post by TheRandomGenius)
Nope.
I think you would make the gradient of p equal to the -1/gradient of q then you would substitute back into the equation to find p and q and substitute the R coordinates

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