Mathematics: Post your doubts here!

[anonymous123]

Could some explain me how to do question 9 part ii? i dont understand why do we first have find the volume of the cylinder (as given in the marksheet)?


http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_13.pdf

Marksheet: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_ms_13.pdf

Thanks alot!

they just named it lyk dat bcz whn u revolve the region under y=5 it forms a cylinder..its the usual thing find volume under y=5 and then under curve the diff bw them is the req ans

 

[XPF MASTER]

can anyone please help me in finding the intersection of two planes? is there any specific method? please help

 

[XPFMember]

can anyone please help me in finding the intersection of two planes? is there any specific method? please help

Assalamoalaikum wr wb!
I came here to ask the same thing Although I do know a methid, yet ms gives so many, and I wanted to know how to do it...

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_32.pdf

Q:9 ii

 

[sea_princess]

Assalamoalaikum wr wb!
I came here to ask the same thing Although I do know a methid, yet ms gives so many, and I wanted to know how to do it...

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_32.pdf

Q:9 ii

I use this method of finding the directional vector of the line through the cross product of the planes normals, then I put either the x/y /z equal zero in both equations and solve them simultaneously to get the position vector of any point.
but i also want to know the other methods

 

[badrobot14]

can anyone please help me in finding the intersection of two planes? is there any specific method? please help

Assalamoalaikum wr wb!
I came here to ask the same thing Although I do know a methid, yet ms gives so many, and I wanted to know how to do it...

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_32.pdf

Q:9 ii

That's easy...
two planes intersect when they aren't parallel, so step 0: Check that their normal vectors aren't parallel (i.e. they don't are a scaler multiple of each other...)

now if they intersect, the logical thing is that their intersection wd be a line....
to get eqn of a line you need a point (pos vector on line) and it's direction vector.
x + 2y − 2z = 7 and 2x + y + 3z = 5.
now we need x, y and z that satisfy both eqnz..(that'd be our pt, da pos vector o line)
we have 2 eqnz n 3 variables.. so let's assume x=0
we get solving eqn simultaneously y = 31/8, z = 3/8
so our pos vactr on line is (0, 31/8, 3/8)

now for the direction vector of line:
extract out the normal vector of both planes... take their cross product to fet a vector perp to both of em.... i.e. d=N1 x N2 = [1,2,-2] x [2,1,3]
we get d= (8, -7, -3)

last step add all the ingredients and bake for i mins.... meaning: eqn of line is r = (0, 31/8, 3/8) + t(8, -7, -3)

 

[badrobot14]

^ if any1's wondering, I made computer do all the dirty calculations....

 

[Hammad Siddiqi]

for part two just suppose that a point q is on the line l so find the coordinates of that point
i.e x = 4 +2t
y = 2=t
z = -1-2t these are the coordinates of any point online l (obtained throgh the eq. of line)

now considering this point Q , make an eq. of line PQ. u already know the coordinate of p and coordinates of Q are now here in terms of "t"

the eq. of line PQ wil be r= ( 0 2 4 ) +s(4+2t -t -5 -2t)

noe take the direction vector of this line and the drection vector of the line l obtain vector product of them equating it to zero as they are perpendicular! (stated in question)

U wil have the value of t = -2

substitute the value in the cordinates of q obtained at the begining ( u wil have exact coordinates of Q )

subtract p - q and then calculate the magnitude

Nice..
Thnx

 

[Nikesh]

Could some explain me how to do question 9 part ii? i dont understand why do we first have find the volume of the cylinder (as given in the marksheet)?


http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_13.pdf

Marksheet: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_ms_13.pdf

Thanks alot!

It is because the shaded region is above the curve and if you try to find the volume directly by integrating square of given equation about x-axis it gives the volume of substance formed when the area bounded by the curve and x-axis (which is below the curve) is rotated about x-axis.
So first you have to rotate a rectangle formed by line AB with x-axis so that a cylinder is formed then find its volume and finally subtract the volume of curve rotated about x-axis which will give the volume of required region...
GOT IT????

 

[farhan143]

question 5:
i- assume that the probability that the spinner lands on the blue side is x
that will make P( landing on red)= 4x , P(landing on green)= 3x , the probability can never exceed 1
so now you have x+4x+3x = 1
8x=1 >>>>> x= 1/8 ( this is the probability it lands on the blue side)
ii- the spinner lands on a different colour each time , so that makes the possible arrangements of colours ( RGB, RBG, BRG, BGR, GRB, GBR) so now the probability is ( 1/8 X 4/8 X 3/8) X 6 = 9/64 ( this is because you'll only put the probabilities in different order while multiplying them for 6 times ,so as a shortcut you multiply them once and multiply the result by 6)
iii- from (i) the probability of landing on blue is 1/8 , so ( 1-p)= 7/8, and the n is given in the question to be 136
so first get the mean = np= 136 X 1/8 =17 , variance = np(1-p)= 136 X 1/8 X 7/8 =14.875
now P (it will land on blue less than 20 times )= P( Z < (19.5 - 17 )/ square root 14.875)= P ( Z < 0.648) , get it from the normal distribution table, it will equal 0.7415 = 0.742

I'll do the rest later

i stil am waiting for the others
just to remind you!!

 

[XPFMember]

That's easy...

two planes intersect when they aren't parallel, so step 0: Check that their normal vectors aren't parallel (i.e. they don't are a scaler multiple of each other...)

now if they intersect, the logical thing is that their intersection wd be a line....
to get eqn of a line you need a point (pos vector on line) and it's direction vector.
x + 2y − 2z = 7 and 2x + y + 3z = 5.
now we need x, y and z that satisfy both eqnz..(that'd be our pt, da pos vector o line)
we have 2 eqnz n 3 variables.. so let's assume x=0
we get solving eqn simultaneously y = 31/8, z = 3/8
so our pos vactr on line is (0, 31/8, 3/8)

now for the direction vector of line:
extract out the normal vector of both planes... take their cross product to fet a vector perp to both of em.... i.e. d=N1 x N2 = [1,2,-2] x [2,1,3]
we get d= (8, -7, -3)

last step add all the ingredients and bake for i mins.... meaning: eqn of line is r = (0, 31/8, 3/8) + t(8, -7, -3)

JazakAllah khairen!
What we were told by our teacher was to find two points OA and OB by substituting x, y or z as zero respectively. Then take one of them as the point, eg OA and find AB which is the direction of the line.

When I was solving this question, forgetting what we learnt, I just found it sensible to find the direction using the way you told. *feel good* But, I couldn’t get how to find a point on the line, so I turned back to check my notes.
Anyway your method is nice I guess. I feel this one better. Thanks a lot. JazakAllah Khairen.

 

[aaakhtar19]

assalam.u.alaikum every0ne

can s0meone solve questi0n Q6 mechanics 1 n0v 2003 (9706_w03_qp04)

 

[ousamah112]

plz help with may/june 08 p3 q 6...

 

[sea_princess]

i stil am waiting for the others
just to remind you!!

yeah right I'll start on them right now
btw which year was it and the session too , please?

 

[sea_princess]

This is so hard i cant understnd a thing...... can any 1 explain me this im in need of serious help!!!!!

View attachment 7397

Please help me solve this question .....
THIS IS 2011 OCT/NOVEMBER
STATISTICS PAPER 62

question 6:
i) you will only consider boys here : P( music)= 0.15, P( other )= 0.85
thus the probability that fewer than 3 boys chose music = 6C0 X(0.85)^6 + 6C1 X(0.15) X (0.85)^5 + 6C2 X (0.15)^2 X(0.85)^4 = 0.953
ii) here first get the probability of all the drama students in the school = 0.60 X 0.10 + 0.40 X 0.55 = 0.28
then out of these drama students , the probability of having a boy = (0.60 X 0.10)/ 0.28= 3/14 = 0.214
then use binomial distribution again :
probability that at least one is a boy = 1- probability of no boys at all = 1 - 5C0 x(0.7857)^5= 0.7005= 0.701

 

[sea_princess]

This is so hard i cant understnd a thing...... can any 1 explain me this im in need of serious help!!!!!

View attachment 7397

Please help me solve this question .....
THIS IS 2011 OCT/NOVEMBER
STATISTICS PAPER 62

question 7:
i) that's normal distribution , u have the mean = 8 and standard deviation = square root 24
then P( 7 <X<12) = P ( 7-8 / root 24 <Z< 12- 8/ root 24) = P (-0.204 <Z<0.816) = phi(0.816) + phi(0.204) - 1 = 0.7926 + 0.5808 -1 = 0.3734 = 0.373
ii) here you will calculate normally since the two mews will be cancelled together:
P(X <0) = P( Z< 0- mew/ 2 mew) = P(Z< - 0.5)= 1 - phi ( o.5) = 1- 0.6915 = 0.3085 = 0.309
iii)three times the mean will equal to 3 mew
so P(X>3 mew) =P(Z> 3mew - mew/ 2 mew) =P(Z> 1) = 1- phi (1) = 1- 0.8413 = 0.1587
so the number of days = 70 X0.1587 = 11.1 days ( you can round it to 11 days)
iv) P(X> 6)= P(Z> 6- mew/ 2mew) = 0.0735 >>>>> P(X<6) = 0.9265
get z from the table , it will equal 1.45
so 6 - mew/ 2 mew = 1.45
6 - mew = 2.9 mew
3.9 mew = 6
mew =1.54

 

[ppaayas]

Hello!
My question is not really pedagogical but bear with me. I am sitting AS Math in a week's time. Am doing P1 and S1, but my real query is, which papers can I sit next year? I want to do S2 and P3, is that combination possible? Help please.

 

[aaakhtar19]

plz help with may/june 08 p3 q 6...

sory dnt have super script 0pti0n in my m0bile so (x^2)y means x square.y

xy(x+y) = 2(a^3)
(x^2)y + x(y^2) = 2(a^3)
taking d/dx on both sides
d/dx[(x^2)y] + d/dx[x(y^2)]=d/dx[2(a^3)]
using product rule
2xy + (x^2).dy/dx + (y^2) + 2xy.dy/dx = 0
[(x^2)+2xy]dy/dx = -(y^2) -2xy
dy/dx = [-(y^2)-2xy]/[(x^2)+2xy]
take dy/dx=0
u’ll get -(y^2)-2xy=0
y=-2x
substitute this val of y in the real eq u’ll get x=a
y =-2x
as x=a so y=-2a
so ans is (a,-2a)

 

[2pac]

sory dnt have super script 0pti0n in my m0bile so (x^2)y means x square.y

xy(x+y) = 2(a^3)
(x^2)y + x(y^2) = 2(a^3)
taking d/dx on both sides
d/dx[(x^2)y] + d/dx[x(y^2)]=d/dx[2(a^3)]
using product rule
2xy + (x^2).dy/dx + (y^2) + 2xy.dy/dx = 0
[(x^2)+2xy]dy/dx = -(y^2) -2xy
dy/dx = [-(y^2)-2xy]/[(x^2)+2xy]
take dy/dx=0
u’ll get -(y^2)-2xy=0
y=-2x
substitute this val of y in the real eq u’ll get x=a
y =-2x
as x=a so y=-2a
so ans is (a,-2a)

Could you please explain me q9 of may 2010 p31?
I have done half of the procedure but then can't figure what to do after the 2 gets cancelled out.
Thanks

 

[2pac]

Hello!
My question is not really pedagogical but bear with me. I am sitting AS Math in a week's time. Am doing P1 and S1, but my real query is, which papers can I sit next year? I want to do S2 and P3, is that combination possible? Help please.

s2 and p3 are the papers you must do in a2 if you have taken statistics and pure math in AS,so don't worry,you'll be doing s2 and p3 like everyone else who opted for stats and pure math.

 

[ousamah112]

sory dnt have super script 0pti0n in my m0bile so (x^2)y means x square.y

xy(x+y) = 2(a^3)
(x^2)y + x(y^2) = 2(a^3)
taking d/dx on both sides
d/dx[(x^2)y] + d/dx[x(y^2)]=d/dx[2(a^3)]
using product rule
2xy + (x^2).dy/dx + (y^2) + 2xy.dy/dx = 0
[(x^2)+2xy]dy/dx = -(y^2) -2xy
dy/dx = [-(y^2)-2xy]/[(x^2)+2xy]
take dy/dx=0
u’ll get -(y^2)-2xy=0
y=-2x
substitute this val of y in the real eq u’ll get x=a
y =-2x
as x=a so y=-2a
so ans is (a,-2a)

thankyou...