tryn 2 get a look at that exam paper and it says page doesnt exsist . . . . .
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.
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
do you have answers to question 6 on p2 in jan 2003?
It is mainly part b which I haven't quite understood.
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
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?
=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 (5 + ln4)
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
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)
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.
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
4^1/4 is the same as 2^1/2
how did you know that? sorry, i don't get how its the same!