[BobGreggary]

I noticed that there was no thread open for the Further Pure 1 examination coming up this June (or atleast not one that I can find :P)!

How's everyone feeling about it, and how are you revising?

I've currently made a document containing all the FP1 past papers and organised them into chapters and probably going to make my way through that.

Probably the most difficult topic for me is Complex Loci, but hopefully everything else should be fine!

Good luck everyone!

[Steve0606]

I've just got to finish the last two chapters in the book and I'm done with my revision.

Tbh, there's nothing really that difficult except the induction questions can sometimes be a bit dodgy, but most of the time even they're (I dare say it) easy!

[Snowdog94]

(Original post by Steve0606)
I've just got to finish the last two chapters in the book and I'm done with my revision.

Tbh, there's nothing really that difficult except the induction questions can sometimes be a bit dodgy, but most of the time even they're (I dare say it) easy!

I fully agree with this, induction is the thing which catched me out of one or two past papers. (That 3^n-1 thing!) My method for revising is to try and do one paper a day until the exam (no other exams) and try and be quite relaxed about it.

[Nathdragon5]

Feeling pretty good too, just doing past papers mainly. The annoying thing about FP1 is that everyone who does it is pretty good at maths, so your raw percentage will often be higher than your UMS... means you just have to try to get 100 otherwise there's big drops.... I've seen it where 90% raw was the A boundary!

And yeah, doing past papers is the best

[BobGreggary]

(Original post by Nathdragon5)
Feeling pretty good too, just doing past papers mainly. The annoying thing about FP1 is that everyone who does it is pretty good at maths, so your raw percentage will often be higher than your UMS... means you just have to try to get 100 otherwise there's big drops.... I've seen it where 90% raw was the A boundary!

And yeah, doing past papers is the best

Oh god don't tell me this! XD

If you like I'll send you this file I've made which basically has all the FP1 papers from 2012 Jan to 2005 June with all the questions organised into chapters (excluding Ch.1 which I've kind of mixed with Chapter 10).

Good luck with your studying regardless

[Nathdragon5]

(Original post by BobGreggary)
Oh god don't tell me this! XD

If you like I'll send you this file I've made which basically has all the FP1 papers from 2012 Jan to 2005 June with all the questions organised into chapters (excluding Ch.1 which I've kind of mixed with Chapter 10).

Good luck with your studying regardless

Hahahaha, I know I wish it wasn't true, but take a look at the grade boundaries from Jan 2012 :P maybe June 11 as well :P

And wow, thanks, that would be amazing... do you want to put it as a link on here or send it to me, and I'll upload it (if that's okay) obviously not claiming it was me but if you cba to do it

And yeah, you too, any particular areas you don't like? I personally just keep making little silly errors which really bugs me.

[Snowdog94]

A difficult induction brings the boundaries down thankfully :P June 2009 is an example of VERY low boundaries.

[Nathdragon5]

(Original post by Snowdog94)
A difficult induction brings the boundaries down thankfully :P June 2009 is an example of VERY low boundaries.

Ahh well then here's hoping for one of those.... so long as you're good at those you should be fine. I personally get confused on the ones with sequences, and which bit to prove. :P

[KoalaKim]

I just hope there isn't a question about cubic roots like the one on the jan 12 paper

[ninuzu]

Have any of you guys got the Jan 2012 paper and mark scheme that they wouldn't mind sharing? thanks

[Nathdragon5]

(Original post by ninuzu)
Have any of you guys got the Jan 2012 paper and mark scheme that they wouldn't mind sharing? thanks

Yeah I do, unfortunately I'm out at the moment, if nobody has posted it by the time I get back I'll do it

[ninuzu]

(Original post by Nathdragon5)
Yeah I do, unfortunately I'm out at the moment, if nobody has posted it by the time I get back I'll do it

Thanks

[Nathdragon5]

(Original post by ninuzu)
Thanks

Okay, here you go! Unfortunately, I don't have a copy of the mark scheme, but I've done this test and had it marked, so I'm going to write down the final answers you should get.... if anybody wants me to go through a question, just quote me

Unfortunately the file size is too large, so I've put it in dropbox. Don't worry, its still free to get and you don't have to sign up or anything

here's the link: http://db.tt/2XH60P7X

Hope that helps!

I'll post the answers in a bit, I'll just write them out

ANSWERS:

Spoiler:

Show

1)12+5i and 0.395 radians

2)q=-2,p=3

3)Roots are (root6) + (root3)i and -(root6) - (root3)i

4)0.25n(n+1)(n+3)(n-2)

5)
(0 -1)
(-1 0)

bi)stretch in the y direction, stretch factor 4.

bii) ¦C¦=4, the transformation has enlarged the shape by 4 (ie area scale factor)

6) Circle centre (root3, 1), radius 2, and arg z = pi/6

7i)M^4 =
(81 0)
(80 1)

ii)M^n =
(3^n 0)
((3^n)-1 1)

iii)proof by induction, you know what you need to get to...

8i) 1/[r(r+1)]

ii) n/(n+1)

iii)1-(n/(n+1))

9i)-a^2 - 9a + 10

ii) a=1 or a=-10

iii)this is going to look confusing, sorry :P *will represent spaces to lay it out nicer.
_____1______ (determinant)
-a^2 - 9a + 10

(**-a***2****6-9a***) (matrix)
(**5**-a-9**-3a+18**)
(*-a****2***(a^2)-4*)

10i)a+b+y=3, ab+by+ya=2, aby=-2/3

ii)a=-5, b=8, c=-4/9

Hope that helped, as mentioned above, quote me if you want me to help with anything

[ninuzu]

(Original post by Nathdragon5)
Okay, here you go! Unfortunately, I don't have a copy of the mark scheme, but I've done this test and had it marked, so I'm going to write down the final answers you should get.... if anybody wants me to go through a question, just quote me

Unfortunately the file size is too large, so I've put it in dropbox. Don't worry, its still free to get and you don't have to sign up or anything

here's the link: http://db.tt/2XH60P7X

Hope that helps!

I'll post the answers in a bit, I'll just write them out

ANSWERS:

Spoiler:

Show

1)12+5i and 0.395 radians

2)q=-2,p=3

3)Roots are (root6) + (root3)i and -(root6) - (root3)i

4)0.25n(n+1)(n+3)(n-2)

5)
(0 -1)
(-1 0)

bi)stretch in the y direction, stretch factor 4.

bii) ¦C¦=4, the transformation has enlarged the shape by 4 (ie area scale factor)

6) Circle centre (root3, 1), radius 2, and arg z = pi/6

7i)M^4 =
(81 0)
(80 1)

ii)M^n =
(3^n 0)
((3^n)-1 1)

iii)proof by induction, you know what you need to get to...

8i) 1/[r(r+1)]

ii) n/(n+1)

iii)1-(n/(n+1))

9i)-a^2 - 9a + 10

ii) a=1 or a=-10

iii)this is going to look confusing, sorry :P *will represent spaces to lay it out nicer.
_____1______ (determinant)
a^2 + 9a - 10

(**-a***2****6-9a***) (matrix)
(**5**-a-9**-3a+18**)
(*-a****2***(a^2)+4*)

10i)a+b+y=3, ab+by+ya=2, aby=-2/3

ii)a=-5, b=8, c=-4/9

Hope that helped, as mentioned above, quote me if you want me to help with anything

Thank you

[ninuzu]

(Original post by Nathdragon5)
Okay, here you go! Unfortunately, I don't have a copy of the mark scheme, but I've done this test and had it marked, so I'm going to write down the final answers you should get.... if anybody wants me to go through a question, just quote me

Unfortunately the file size is too large, so I've put it in dropbox. Don't worry, its still free to get and you don't have to sign up or anything

here's the link: http://db.tt/2XH60P7X

Hope that helps!

I'll post the answers in a bit, I'll just write them out

ANSWERS:

Spoiler:

Show

1)12+5i and 0.395 radians

2)q=-2,p=3

3)Roots are (root6) + (root3)i and -(root6) - (root3)i

4)0.25n(n+1)(n+3)(n-2)

5)
(0 -1)
(-1 0)

bi)stretch in the y direction, stretch factor 4.

bii) ¦C¦=4, the transformation has enlarged the shape by 4 (ie area scale factor)

6) Circle centre (root3, 1), radius 2, and arg z = pi/6

7i)M^4 =
(81 0)
(80 1)

ii)M^n =
(3^n 0)
((3^n)-1 1)

iii)proof by induction, you know what you need to get to...

8i) 1/[r(r+1)]

ii) n/(n+1)

iii)1-(n/(n+1))

9i)-a^2 - 9a + 10

ii) a=1 or a=-10

iii)this is going to look confusing, sorry :P *will represent spaces to lay it out nicer.
_____1______ (determinant)
a^2 + 9a - 10

(**-a***2****6-9a***) (matrix)
(**5**-a-9**-3a+18**)
(*-a****2***(a^2)+4*)

10i)a+b+y=3, ab+by+ya=2, aby=-2/3

ii)a=-5, b=8, c=-4/9

Hope that helped, as mentioned above, quote me if you want me to help with anything

Quite a nice paper actually, and was very pleased with myself The grade boundary for an A in this paper was 65/72!!!
http://pdf.ocr.org.uk/download/admin...nd_jan_12.pdf?

[Nathdragon5]

(Original post by ninuzu)
Thank you

You're very welcome

(Original post by ninuzu)
Quite a nice paper actually, and was very pleased with myself The grade boundary for an A in this paper was 65/72!!!
http://pdf.ocr.org.uk/download/admin...nd_jan_12.pdf?

And here is the problem with fp1 high grade boundaries.. I got 68/72 and that wasn't even 95.... it's ridiculous, but understandable because the people that do fp1 do further maths and are therefore likely to score higher. The only way to beat the system is to get 100%, which always equals 100 UMS.... and damn, did you just do the paper now?!?!?!

[ninuzu]

(Original post by Nathdragon5)
You're very welcome



And here is the problem with fp1 high grade boundaries.. I got 68/72 and that wasn't even 95.... it's ridiculous, but understandable because the people that do fp1 do further maths and are therefore likely to score higher. The only way to beat the system is to get 100%, which always equals 100 UMS.... and damn, did you just do the paper now?!?!?!

Oh no, I didn't do it that quick! I found the paper elsewhere online, but not the markscheme.

[cameron95]

can someone help with question 10 from the june 2010 paper:

10 The complex number z, where 0 < arg z < pi/2 , is such that z^2 = 3 + 4i.
(i) Use an algebraic method to find z. [5]
(ii) Show that z^3 = 2 + 11i. [1]
The complex number w is the root of the equation
w^6 − 4w^3 + 125 = 0
for which -pi/2 < argw < 0.
(iii) Find w. [5]

i managed to do the question, but i don't understand why for part (i) z is just (2+i), not +/-(2+i).
Also, for part (iii), I got w^3 = (2+/-11i), but the mark scheme says you have to choose the negative sign (2-11i) and 'relate required value to conjugate of part (i)'.
if someone could explain this to me that would be great, thanks

[Nathdragon5]

(Original post by cameron95)
can someone help with question 10 from the june 2010 paper:

10 The complex number z, where 0 < arg z < pi/2 , is such that z^2 = 3 + 4i.
(i) Use an algebraic method to find z. [5]
(ii) Show that z^3 = 2 + 11i. [1]
The complex number w is the root of the equation
w^6 − 4w^3 + 125 = 0
for which -pi/2 < argw < 0.
(iii) Find w. [5]

i managed to do the question, but i don't understand why for part (i) z is just (2+i), not +/-(2+i).
Also, for part (iii), I got w^3 = (2+/-11i), but the mark scheme says you have to choose the negative sign (2-11i) and 'relate required value to conjugate of part (i)'.
if someone could explain this to me that would be great, thanks

Put very quickly:
1) the answer is only 2+i because of the arg being between 0 and pi/2. Only the 2+i will give an arg of this, -2-i will be in the bottom left of the graph, indicating that the angle is greater that pi/2.

2) for the last bit, the rule is as follows, if w^3 is the conjugate of z^3... ie if (z^3)* = w^3, then w = z*.

You have that w^3 is 2-11i, again relating to the arg of w being negative therefore 2-11i is the answer, not 2+11i as this would be in the positive region of the argann diagram.

So we can see that w^3 is the conjugate of z^3 and therefore w is the conjugate of z.

As z is 2+i, you use the conjugate of this to find out that w = 2-i

Hope that helped

[Nathdragon5]

(Original post by ninuzu)
Oh no, I didn't do it that quick! I found the paper elsewhere online, but not the markscheme.

Hahahaha xD I was gonna say, you're pretty quick :O