# Edexcel P2 Maths

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

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#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**username:alevel

password:mrhughes

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#5

(Original post by

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

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

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#6

(Original post by

neone?

**TheWolf**)neone?

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

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#7

(Original post by

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

**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

ahh cool - you can use any starting value right?

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#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?

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#9

(Original post by

how did you make that iteration formula?

ahh cool - you can use any starting value right?

**TheWolf**)how did you make that iteration formula?

ahh cool - you can use any starting value right?

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#10

(Original post by

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...

**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...

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#11

(Original post by

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?

**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?

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

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#12

(Original post by

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

**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

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#13

(Original post by

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

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

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#14

(Original post by

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

**lgs98jonee**)no cos 6a is... Show that the x co-ord of A lie in interval (0.5,1)

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#15

(Original post by

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

**TheWolf**)yea..ln0.5 = a negative answer..and

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#16

(Original post by

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

**TheWolf**)yea..ln0.5 = a negative answer..and

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#17

(Original post by

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

**lgs98jonee**)but 2x0.5-10ln0.5-3=-2+10ln2which is +ve

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#18

(Original post by

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

**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

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

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#19

(Original post by

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

**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

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#20

**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?

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

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