# Edexcel P2 Maths

Watch
Announcements
This discussion is closed.
#1
any1 here no of any websites with edexcel past papers, or edexcel questions for P2 maths? ne help appreciated 0
15 years ago
#2
(Original post by nikhil_misri)
any1 here no of any websites with edexcel past papers, or edexcel questions for P2 maths? ne help appreciated 0
15 years ago
#3
www.rdoh.com\exams
user is exams and password is ukl.
0
15 years ago
#4
guys an anyone do pg 190 heinemann p2 examination style paper 6b?
0
15 years ago
#5
(Original post by TheWolf)
guys an anyone do pg 190 heinemann p2 examination style paper 6b?
neone?
0
15 years ago
#6
(Original post by 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
0
15 years ago
#7
(Original post by lgs98jonee)
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
how did you make that iteration formula?

ahh cool - you can use any starting value right?
0
15 years ago
#8
and by using the iteration formula - how do you rearrange it so that you make sure youre not getting the x value of A?
0
15 years ago
#9
(Original post by 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...
0
15 years ago
#10
(Original post by lgs98jonee)
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...
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?
0
15 years ago
#11
(Original post by 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
0
15 years ago
#12
(Original post by lgs98jonee)
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
but both xs are bigger than one in no.6? soz im sure youre right - but im slow at understanding 0
15 years ago
#13
(Original post by TheWolf)
but both xs are bigger than one in no.6? soz im sure youre right - but im slow at understanding no cos 6a is... Show that the x co-ord of A lie in interval (0.5,1)
0
15 years ago
#14
(Original post by lgs98jonee)
no cos 6a is... Show that the x co-ord of A lie in interval (0.5,1)
0
15 years ago
#15
(Original post by TheWolf)
ln1=0 so when x=1, it equals -1
0
15 years ago
#16
(Original post by TheWolf)
but 2x0.5-10ln0.5-3=-2+10ln2which is +ve
0
15 years ago
#17
(Original post by 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
0
15 years ago
#18
(Original post by 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
0
15 years ago
#19
(Original post by lgs98jonee)
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
ahh lol ofcourse - thanks for clearing this up mate
0
15 years ago
#20
(Original post by 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 0
X
new posts Back
to top

view all
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

Yes (117)
55.19%
Yes, but I want to swap them (16)
7.55%
No, but I know who I want to choose (19)
8.96%
No, I still don't know who I want to choose (50)
23.58%
I have decided I don't want to go to uni anymore and will not be choosing (10)
4.72%