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I'm thinking of the same lol... P1 on 7th, and I barely started preparing. I do that by attempting to solve the questions posted hereNow where to start preparations for P1... I have no idea
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I'm thinking of the same lol... P1 on 7th, and I barely started preparing. I do that by attempting to solve the questions posted hereNow where to start preparations for P1... I have no idea
Thanks!!PANDA-
Okay, see.
From first part u get 150 as well as 30
Now they said smallest value now is 10 meaning
sin (n(10))= 1/2
10n = 30
n = 3
Okay now to find the largest value...first change the range
from 0 < q < 360
by multiplying with 3 (since now 3q)
to
0< q<1080
that is a three rounds in the quadrants [hope u get this point]
I drew a sketch, every round is different colour. So the largest value is on the third round, on the blue point, from the sketch u find, 2 rounds plus 150
(2 rounds, 360 x 2 = 720
3q = 720 + 150 = 870
q = 290
maybe do past papers? ever heard of that?
I'm thinking of the same lol... P1 on 7th, and I barely started preparing. I do that by attempting to solve the questions posted here
Wow... same problem I'm having, I got a bit rusty in functions because I didn't practice for a long time.haha that's all the practice that I have done.. Honestly I don't know functions one bit!
got bored of solving them and you finish before time by 45 mins means you r well prepared..am i wrong ?Well I have heard of them.. but get bored while solving them... 1hr 45min on each paper isn't fair when doing at home .. It should be like chem and physics P2.. 1 hour and you're done
got bored of solving them and you finish before time by 45 mins means you r well prepared..am i wrong ?
Okay, from first part you get that dy/dx = -sint/costi need help with question 6(ii) may june 2009 paper 3
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s09_qp_3.pdf
Honestly for P1 1hr 45 is plenty.naw.. I was talking about physics and chemistry.. they are 1 hour each for theory.. It takes 1 hour to solve them.. so one can solve like 3 or 4 papers in a day.. but with maths P1 I can't even solve 2 papers a day :\
And no one can solve the paper in 45 minutes! .. takes me about 1hr 30-40 minutes to solve a complete paper..
Wait, you mean phi of statistic.PhyZac or anyone else..
Know of any place (online or via calculator) where one can get inverse of phi? without referring to the normal tables I mean...
there is a calculator function but i tried but didnt knoe its use properlyPhyZac or anyone else..
Know of any place (online or via calculator) where one can get inverse of phi? without referring to the normal tables I mean...
thanks man that really helped!Okay, from first part you get that dy/dx = -sint/cost
Now they want the equation of tangent when t is t!!
Equation is in form
(y-y1) = m (x-x1)
M is the gradient, and when t is t, the gradient is -sint/cost ( the dy/dx)
the y1 is asin^3 (t)
and the x1 is acos^3(t)
Sub those value, and do some algebra and then you get the answer! [P.S, i am sure you know how to do this step, if you get any difficulty ask]
OKaay..see..!http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_31.pdf
Please help for no. 7 (i)
OKaay..see..!
First we will do the limits.
in the first one it is 2 and 0
so, the sub is x = t^2 + 1
when t is 2, x is 5
when t is 0, x is 1
So now new limit are 5 and 1
Now to main thing!
First thing I do is find the dy/dx of substitution.
not of this >> x = t^2 + 1
make t the subject first.
t^2 = x - 1
t = (x - 1)^1/2 [power 1/2 is square root]
okay now, take dy/dx, but to be precise dt/dx
dt/dx = 1/2 ( x - 1)^(-1/2)
Now next thing!
We now should substitute the t's of formula.
since x=t^2 +1
then this
4t^3 ln(t^2 + 1)
is
4t^3 lnx
One more t is left! what is t alone?
x = t^2 +1
t = (x - 1)^1/2
so sub this and get
4((x-1)^1/2)^3 lnx
4(x-1)^3/2 ln x
Now add the differential to the top equation
4(x-1)^3/2 ln x 1/2 ( x - 1)^(-1/2)
4/2 (x-1) ^(3/2 -1/2) lnx
2 (x-1) ^1 ln x
2x - 2 ln x , with limits 5 to 1
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