Edexcel P2 Maths

nikhil_misri

any1 here no of any websites with edexcel past papers, or edexcel questions for P2 maths? ne help appreciated :smile:

Reply 1

TheWolf

nikhil_misri

any1 here no of any websites with edexcel past papers, or edexcel questions for P2 maths? ne help appreciated :smile:

http://www.mrhughes.net/Maths/Hidden%20Pages/login.htm

username:alevel
password:mrhughes

Reply 2

IntegralAnomaly

www.rdoh.com\exams
user is exams and password is ukl.

Reply 3

TheWolf

guys an anyone do pg 190 heinemann p2 examination style paper 6b?

Reply 4

TheWolf

guys an anyone do pg 190 heinemann p2 examination style paper 6b?

neone?

Reply 5

lgs98jonee

TheWolf

neone?

i used an iteration
rearranged to x=(10lnx+3)/2

with a starting value of 6 and got between 15 and 16

i suppose u could use trial and error until u see a sign change but iteration possibly best...intereseting to see how it should be done

Reply 6

TheWolf

lgs98jonee

how did you make that iteration formula?

ahh cool - you can use any starting value right?

Reply 7

TheWolf

and by using the iteration formula - how do you rearrange it so that you make sure youre not getting the x value of A?

Reply 8

lgs98jonee

TheWolf

how did you make that iteration formula?

ahh cool - you can use any starting value right?

u cant always..here u seem to have to be able to use any value above that of the other intersection (i.e about 1)...i am not quite sure about the exact rules for when the iteration coverges to certain values...

Reply 9

TheWolf

lgs98jonee

i figured out the iteration too, but there is point a and point b,when you put why = 0, but why the hell does the iteration formula only converge to point b's value?

Reply 10

lgs98jonee

TheWolf

i figured out the iteration too, but there is point a and point b,when you put why = 0, but why the hell does the iteration formula only converge to point b's value?

well if x<1 lnx is -ve and 10lnx+3<0 generally, so when x2 is worked out (which is negative) u can not continue as u can not do ln of a negative number

if u put x>=1, then u get a larger +ve number which increases until 10lnx+3 is constant

Reply 11

TheWolf

lgs98jonee

but both xs are bigger than one in no.6? soz im sure youre right - but im slow at understanding :eek:

Reply 12

lgs98jonee

TheWolf

but both xs are bigger than one in no.6? soz im sure youre right - but im slow at understanding :eek:

no cos 6a is... Show that the x co-ord of A lie in interval (0.5,1)

Reply 13

TheWolf

lgs98jonee

no cos 6a is... Show that the x co-ord of A lie in interval (0.5,1)

yea..ln0.5 = a negative answer..and

Reply 14

lgs98jonee

TheWolf

yea..ln0.5 = a negative answer..and

ln1=0 so when x=1, it equals -1

Reply 15

lgs98jonee

TheWolf

yea..ln0.5 = a negative answer..and

but 2x0.5-10ln0.5-3=-2+10ln2which is +ve

Reply 16

TheWolf

lgs98jonee

but 2x0.5-10ln0.5-3=-2+10ln2which is +ve

mate thanks, but i dont have a clue how this means the iterative formula only shows the other solution - ill figure a way to understand it myself

Reply 17

lgs98jonee

TheWolf

mate thanks, but i dont have a clue how this means the iterative formula only shows the other solution - ill figure a way to understand it myself

well yeah it is a bit ambiguous...but u know there r only 2 solutions and u know one is between 0.5 and 1 so u know the other is the only other one. look at it this way

if x<1
then lnx is -ve
and then when this value is put back in to the iteration thing u get an error becos u can not do ln of a -ve number

but if u do ln of x where x is >=1 u get 10lnx-3 which is bigger than x so this increases until u reach the x value which u r looking for

Reply 18

TheWolf

lgs98jonee

ahh lol ofcourse - thanks for clearing this up mate

Reply 19

Fermat

TheWolf

i figured out the iteration too, but there is point a and point b,when you put why = 0, but why the hell does the iteration formula only converge to point b's value?

In general. if you have a simple iteration formula,

x_(r+1) = g(x_r)

then there wil be convergence, to a point a if |g'(a)| < 1 and g' is continuous at a.

when g(x) = (10lnx + 3)/2
g'(x) = 5/x
and
|g(x)| < 1 for |x| > 5

So the iteration formula will converge for x=15.063 (approx) (5/15.063 < 1 => converges) but trying to converge for x=0.884 won't work 'cos 5/0.884 >> 1 => non-convergence :frown: