P2 - Edexcel nxt Friday, eak! Watch

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byb3
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#21
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#21
hi, i've had a go at doing some answers for the P2 exam in january since nobody seems to have the official ones.

You can find them here:

http://www.boddingtonbaby.pwp.blueyo...003ANSWERS.pdf

Can't guarantee if they are all right, but I'm fairly sure the 'vast' majority are

The paper is here for those who don't have it yet.

http://www.boddingtonbaby.pwp.blueyo...exam/Jan03.pdf

Regards,
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#22
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#22
tryn 2 get a look at that exam paper and it says page doesnt exsist . . . . .
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#23
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#23
Byb3 made a typo in the address. The correct address is:

http://www.boddingtonbaby.pwp.blueyo...2003/Jan03.pdf

And yeah, proof is the hardest thing about P2 imo
byb3
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#24
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lol, yeah I changed everything about sorry about that.

Anyways, for your help, if it does help, you can visit

www.boddingtonbaby.pwp.blueyonder.co.uk for a page linked to the exams online at the moment.

Regards,
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#25
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#25
Hey mate, you're being the most useful member on this thread and both your sites are down.
Please don't lead me down dark alleys I'm already desperate.
byb3
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#26
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sorry mate, but its working on this side.

maybe if you try putting index.htm on the end.

this is the sites exact URL.

http://www.boddingtonbaby.pwp.blueyo...o.uk/index.htm

lemme know if you can access it now.

regards,
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mensandan
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#27
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Hello guys.

If you are after any particular P2 solutions let me know as I have done all of them in prep for Friday. They are not electronic but I can tell you the answers, working and the question if needed. All the papers since the syllabus changed are here. They have all been graded and corrected. Let me know if you are interested. All the best for Friday, any other P2 questions let me know as I am keen to test myself, Dan
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byb3
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#28
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hey mate,

do you have answers to question 6 on p2 in jan 2003?

It is mainly part b which I haven't quite understood.

Regards,
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#29
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#29
Got the paper, thanku!! Was the one i did soooooo crap in this Jan, im gunna hav a go l8a and let u kno how i go, better than a D i hope.
Ne chance of Physics papers???? I got xams 2mroo! lol
Xenon
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#30
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hi, on question 4 on the P2 paper of jan 2003, I got an answer of pie(5 + ln4), and the answer is supposed to be pie(5 + 1/2ln2).

Here is the working which I did. Could sum1 please tell me where I went wrong, or if it's just a case of juggling around with the answer I got?

y=1+1/(2x^2)
=1+ 1/2 (x^-1/2)
= pie(integral(1 + 1/2 (x^-1/2))^2)
=pie (integral(1+ x^-1/2+1/4(x)^-1))dx
=pie [x + 2x^1/2 + lnx] with limits 1 and 4
=pie[(4 + (2*(4^1/2)) + ln4) - (1+2+ln1)]
=pie [(4+4+ln4)-(1+2+0)]
=pie (5 + ln4)

Thanks
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#31
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#31
i get the right answer...
heres my working

y = 1 + 1/(2 root x)
therfore y squared = (1 + 1/(2 root x))^2

expand and simplifying that giving
1 + 1/root x + 1/(4 root x)

integrate with limits 4 and 1 giving
pi [x + x^-1/2 + 1/(4x)] between 4 and 1

this can be written as:
pi [1 + x^-1/2 + 1/4 x^-1 ] between 4 and 1

integrate giving:
pi [x + 2x^1/2 + 1/4 lnx] between 4 and 1

sub in 4 and 1 giving
pi [(4 + 4 + 1/4 ln4) - (1 + 2 + 1/4 ln1)]

simplify to give
pi [5 + 1/4 ln4 - 1/4 ln1]

simplify more giving
pi (5+ 1/4 ln4)

taking the 1/4 ln4 ===> ln4 ^1/4

4^1/4 is the same as 2^1/2

so ln4 ^1/4 can be written as ln2 ^1/2

taking the 1/2 down in front of the ln giving 1/2 ln2

sub that in to replace 1/4 ln4

answer is then pi (5+ 1/2 ln2)

therefore shown
Xenon
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#32
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Thanks
I see where I went wrong. I thought that the integral of 1/4x would just be ln x. But I guess you separate it out to 1/4 and 1/x. Then put 1/4 outside and integrate 1/x. This is what you always do right? Phew, at least I realised this before the exams.
Oh and byb3, could you please just post the answers on the website, because for some reason (probably due to my computer), I can't open the paper or the answers. Much appreciated if you could do so.
Thanks again
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mensandan
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#33
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A hideous question, for the wording!!

dy/dx = 1/x

So for normal gradient is: -1/(dy/dx) = -x

Now at x= q, gradient = -q

y = ln 3q from sub in original

So equation of line is:

y - ln3q = -q(x - q)

Now at x= 0 y=0 since it passes through the origin:

So -ln3q = -q^2

So: q^2 - ln3q = 0

Now since x=q:

x^2 - ln3x = 0

QED

Dan
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#34
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#34
[QUOTE][i]

4^1/4 is the same as 2^1/2

[QUOTE]

how did you know that? sorry, i don't get how its the same!
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#35
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[QUOTE]Originally posted by Unregistered
[QUOTE][i]

4^1/4 is the same as 2^1/2


how did you know that? sorry, i don't get how its the same!
4 = 2^2 so 4^1/4 = (2^2)^1/4.

Multiply the powers, giving 2^1/2.
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