P3 Urgent Help

[yasser37]

HI Everyone
So I went through the past papers and here are the questions I couldn't do... if anyone can help me I'll be thankful

November 04 , Question 10, Part (ii)
June 05 , Question 4, Part (i)
June 05, Question 10, Part (ii) ---> The ms says a vector (2,-4,1) is used. where did that come from?
November 05, Question 10, Part (iii)
November 08 , Question 10 Part (iv)
November 09, Paper 31, Question 7, part (iv)
June 09, Paper31, Question 9, Part (ii)

 

[WellWIshER]

guys does anyone know hw to integrate 1/(sqrt( x^3-x^4))

 

[yasser37]

use chain rule mate
still waiting for my questions....

 

[WellWIshER]

wt chain rule???

 

[abuhantash]

Could you be here after about 4 hrs? I would have posted some answers by then. I wanna finish my last O/N/10 paper and then go through my doubts and yours. If I had doubts of my own would you help as well?? THE EXAM IS SO CLOSE!

 

[WellWIshER]

sure btw i have didnt start da past papers do u think i will finish them all??? :no: :Search: :x

 

[ahmed t]

maybe if ur not studying anything else

 

[yasser37]

ahmed I am here wen u'tr back
tyt

 

[ahmed t]

plzz i wanna help but i have slow interent and it takes alot to open a page so pls post the links otherwise it will take along time to open each paper thanks

 

[abuhantash]

November 04 Q10 (ii)
x-3/0.005x dx = dt
to make it easier do : x/0.005x - 3/0.005x to get : 200 - 600/x
Now integrate
200 - 600/x dx = dt
200x - 600ln(x) = t +c when x=3 t=0
c= 600 - 600ln(3)
t= 200x - 600 [ln(X) + 1 -ln(3)] Tada! hope you got it


June 05 Q4
y=thita!!! (1-tan^2y).(sec^2y dy) / (sec^2) ^2
x=tany
dx=sec^2 y dy Now here cancel out the sec^2 and do the rest

Q10 (ii)
To find the plane you need 2 direstions
1) from the line (1,2,1)
2) direction AL between A(2,2,1) and point on L (4,-2,2) >>>> this will be (2,-4,1) or (-2,4,-1)

november 05 Q10 (iii)
N is the perpindicular foot of A to p.
find line equation of PN r=(2,2,1) + t(1,-2,2)
so you get x = s+t
y = 2-2t
z = 1+2t replace x, y, z in the plane equation. You will get t=6/9

Find position vector N ( x,y,z) which is gonna be (8/3, 2/3, 7/3)
then find the distance I assume you know the formula XD

AAAhh this was tiring, i will try to answer the rest. But i, gonna post my questions later if you dont mind PLEASE answer em =]

 

[yasser37]

ok mate sure

November 04 , Question 10, Part (ii)
http://www.xtremepapers.me/CIE/index.ph ... 4_ms_3.pdf
June 05 , Question 4, Part (i)
http://www.xtremepapers.me/CIE/index.ph ... 5_ms_3.pdf
June 05, Question 10, Part (ii) ---> The ms says a vector (2,-4,1) is used. where did that come from?
http://www.xtremepapers.me/CIE/index.ph ... 5_ms_3.pdf
November 05, Question 10, Part (iii)
http://www.xtremepapers.me/CIE/index.ph ... 5_ms_3.pdf
November 08 , Question 10 Part (iv)
http://www.xtremepapers.me/CIE/index.ph ... 8_ms_3.pdf
November 09, Paper 31, Question 7, part (iv)
http://www.xtremepapers.me/CIE/index.ph ... 9_ms_3.pdf
June 10, Paper31, Question 9, Part (ii)
http://www.xtremepapers.me/CIE/Internat ... _ms_31.pdf

 

[yasser37]

November 04 Q10 (ii)
x-3/0.005x dx = dt
to make it easier do : x/0.005x - 3/0.005x to get : 200 - 600/x
Now integrate
200 - 600/x dx = dt
200x - 600ln(x) = t +c when x=3 t=0
c= 600 - 600ln(3)
t= 200x - 600 [ln(X) + 1 -ln(3)] Tada! hope you got it


June 05 Q4
y=thita!!! (1-tan^2y).(sec^2y dy) / (sec^2) ^2
x=tany
dx=sec^2 y dy Now here cancel out the sec^2 and do the rest

Q10 (ii)
To find the plane you need 2 direstions
1) from the line (1,2,1)
2) direction AL between A(2,2,1) and point on L (4,-2,2) >>>> this will be (2,-4,1) or (-2,4,-1)

november 05 Q10 (iii)
N is the perpindicular foot of A to p.
find line equation of PN r=(2,2,1) + t(1,-2,2)
so you get x = s+t
y = 2-2t
z = 1+2t replace x, y, z in the plane equation. You will get t=6/9

Find position vector N ( x,y,z) which is gonna be (8/3, 2/3, 7/3)
then find the distance I assume you know the formula XD

AAAhh this was tiring, i will try to answer the rest. But i, gonna post my questions later if you dont mind PLEASE answer em =]

THANKS MATE
will read the explanation now
sure mate I'll be glad to help

 

[ahmed t]

u posted the markschemes

 

[yasser37]

my bad

November 04 , Question 10, Part (ii)
http://www.xtremepapers.me/CIE/index.ph ... 4_qp_3.pdf
June 05 , Question 4, Part (i)
http://www.xtremepapers.me/CIE/index.ph ... 5_qp_3.pdf
June 05, Question 10, Part (ii) ---> The ms says a vector (2,-4,1) is used. where did that come from?
http://www.xtremepapers.me/CIE/index.ph ... 5_qp_3.pdf
November 05, Question 10, Part (iii)
http://www.xtremepapers.me/CIE/index.ph ... 5_qp_3.pdf
November 08 , Question 10 Part (iv)
http://www.xtremepapers.me/CIE/index.ph ... 8_qp_3.pdf
November 09, Paper 31, Question 7, part (iv)
http://www.xtremepapers.me/CIE/index.ph ... 9_qp_3.pdf
June 10, Paper31, Question 9, Part (ii)
http://www.xtremepapers.me/CIE/Internat ... _qp_31.pdf

 

[ahmed t]

ok november 08
i have no clue
for november 09 its not opening up for some reason

june 10
look wen u differientiate y and quote to zero it gives u a maximum value of y or minimum
so if u diff dy/dx and quote to zero it gives u the maximum and min value of dy/dx
so all u do is diff the anser of i and quote to zero

 

[ahmed t]

for w08
in the markscheme they take 4+2i as z and find out zw and z/w as in the previous question z,zw,z/w are verticies of a equalatiral triangle
and part iii)of the same question i understand the distance but i dont understand the subtended arc thing
can u pleas explain this point

 

[Newbie]

dO visit this site .... it has short tutorials for fast revision .... thank me later :d
http://adf.ly/k621

 

[yasser37]

For June 10 I keep getting x=1 which is clearly wrong :S
can someone post their differentiation process ?
thanks

 

[yasser37]

bump

 

[infinity00]

@ Well Wisher

Integral of 1/Sqrt(x^3-x^4), Follow The Steps

1/Sqrt(x^3-x^4) = 1/sqrt (x^3(1-x))

Let x = sin^2u, Then dx = 2sinucosu du Substituting, We'll Have

1/Sqrt(sin^6u(1-sin^2u)) dx.

dx = 2sinucosu du, so

1/Sqrt(sin^6u(1-sin^2u)) * 2sinu cosu

taking Squareroot

: 1/(Sin^3u(Cosu)) * 2sinucosu

You'll Get

Integral 2/sin^2u du

Now, Multiply The Numerator And Denominator Both With Sec^2u, You'll Get

Integral 2sec^2u/Tan^2u

Integrating, You'll Have

-2Cotu + c

Substitute Back The Value of u In Terms of x

= -2Cot(Arcsin(sqrtx)) + c, Which According To the Identity 1 + Cot^2x = Cosec^2x Equals

-2Sqrt ((1/x) - 1) + c