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FP1 June 11 Edexcel - Paper and Model answers in the FIRST post.

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Original post by TheMagicMan

Original post by TheMagicMan
As I have just edited my post to say, this seems very roundabout for what can only be 1 or 2 marks of a 4 mark question.
And there is the problem of + or - with this method which gets even more long-winded


I agree, but the mark schemes penalise you if you just quote dy/dx = 1/t
Original post by Soapy Molloy
I agree, but the mark schemes penalise you if you just quote dy/dx = 1/t


Lucky I didn't do that then; I differentiated implicitly wrt t & x and then chain ruled
Reply 382
Original post by Groat
I shouldn't lose any marks for leaving my answer to 2c as below, right?

z2,z3=5±i3 z_2, z_3 = 5 \pm i\sqrt{3}


should be fine
Reply 383
Original post by Groat
Phew, I was starting to get a bit worried even though I meant exactly what Arsey had written.

Although it does say answers in the question! Arsey, any opinions?


yeah but by giving +- you're giving 2 answers
Reply 384
Original post by Arsey
yes

you have changed 72k17^{2k-1} to 72k7^{2k} in the third line

and the last line of working is all kinds of wrong


Oh yeah :|. I didn't see that ergghhh :frown:

Okay so it should be:

7(7^2k) - 7^-1(7^2k)
= 7(7^2k) - 1/7(7^2k)
=48/7(7^2k)

and then how do I show thats divisible by 12?

So would I get any method marks for that question based on what I've done?
Original post by Kiran03
Oh yeah :|. I didn't see that ergghhh :frown:

Okay so it should be:

7(7^2k) - 7^-1(7^2k)
= 7(7^2k) - 1/7(7^2k)
=48/7(7^2k)

and then how do I show thats divisible by 12?

So would I get any method marks for that question based on what I've done?


48/7 (7^2k)=48(7^2k-1)
Feeling confident about 100 in this paper.
Original post by Xyonzz
Hopefully 100 :biggrin: this makes me feel marginally better about the dream-crusher that was C4


I had to mentally prepare myself about not getting into my firm after C4... :frown: On the bright side, this was such a lovely exam! Thoroughly enjoyed it, possibly my favourite module of Maths! :smile:
(edited 12 years ago)
Reply 388
Original post by Kiran03
Oh yeah :|. I didn't see that ergghhh :frown:

Okay so it should be:

7(7^2k) - 7^-1(7^2k)
= 7(7^2k) - 1/7(7^2k)
=48/7(7^2k)

and then how do I show thats divisible by 12?

So would I get any method marks for that question based on what I've done?


I can not remember what you did up to that point but you could say

48×71(72k)48 \times 7^{-1}(7^{2k})

12×[4(72k1)]12 \times [4(7^{2k-1})]

hence divisible by 12
(edited 12 years ago)
Reply 389
Original post by Soapy Molloy
y^2 = 48x
y = root(48)x^0.5
dy/dx = 0.5root(48)x^-0.5
dy/dx = 2root(3)/root(x)
then sub x = 12t^2


Original post by TheMagicMan
You can awlog that Im(z2)>Im(z3)

Also the solution to 9b) should read f(k+1)=49f(k) - 20X12 rather than + 20X12

You should then really show that 49f(k)>240 but probably unnecessary as f(n) is clearly increasing. I do induction a different way anyway.

I'm also wondering how you do 8 b) well without implicitly differentiating as that surely is not part of the FP1 syllabus. Maybe you can just assume that for a parabola dy/dx = 1/t. lOf course you can rearracnge for y, differentiate wrt x and then substitute t but that seems very roundabout


MagicMan - I have no idea what you mean by

You can awlog that Im(z2)>Im(z3)


I realise I made a transcription error putting a + instead of a - in the final line of working, it has been mentioned quite a few times :wink:

The method given by Soapy is the intended method for this question but implicit is a lot quicker.
Reply 390
Original post by Arsey
I can not remember what you did up to that point but you could say

48×71(72k)48 \times 7^{-1}(7^{2k})

12×[4(72k1)]12 \times [4(7^{2k-1})]

hence divisible by 12


thts wt i did :colone:
Original post by Arsey
MagicMan - I have no idea what you mean by


Sorry. Maths forum shorthand for 'assume without loss of generality'.
Reply 392
Original post by Arsey
MagicMan - I have no idea what you mean by



I realise I made a transcription error putting a + instead of a - in the final line of working, it has been mentioned quite a few times :wink:

The method given by Soapy is the intended method for this question but implicit is a lot quicker.


how the **** do you access the first post
Reply 393
f(k+1) = 7^2k+1 + 5
f(k+1) = 7 * 7^-1 * 7^2k-1 + 5
f(k+1) = 7^(2k-1) + 5

which is divisible by 12

i somehow ended up doing this after trying f(k+1) - f(k) about 6 times and it not working and i'm 99.9% sure this is wrong but can anyone tell me why?
Reply 394
Will I lose any marks on the proof by induction question, for the one where you prove it's divisible by 12.

I stupidly left it as just 48 [ can't remember ]

Should I have changed it to 12 [ 4 x can't remember ] ?
Reply 395
Original post by iAGP
Will I lose any marks on the proof by induction question, for the one where you prove it's divisible by 12.

I stupidly left it as just 48 [ can't remember ]

Should I have changed it to 12 [ 4 x can't remember ] ?


That should be fine as long as you explicitly said it is divisible by 12 (and the bit you can't remember was correct).
Reply 396
Original post by dave2381
how the **** do you access the first post


:s-smilie:

click on the first page and the first post is the first post on the first page
I mustve accidently wrote -48 byt did divide it by 12 so got -4 (which is still divisible by 12 i guess). Am i looking at lost marks?
hey guys
Im gonna do fp1 from now on but i dont have the book. could somebody pls scan the chapters 2 and 3 pls and attach it. I would be very grateful. I really want to stand out from the others because I want to do it in January. Otherwise I have to do it in june. Only good students are allowed and i dont get the book before september.
Reply 399
After looking at the paper again, I think I kinda messed up Q7b. I got the 1/3n ( 36n^2 - 1 ) - ( 4n^2 - 1 ), it's just I don't remember getting the 3 anywhere inside the brackets. Really quite annoyed at that.

And thanks to whoever replied to me earlier :smile:

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