Mathematics: Post your doubts here!

[Alice123]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_33.pdf
help needed in Q1o...Dug thanks in advance

 

[applepie1996]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_32.pdf
question 6 part 3
Rutzaba

 

[applepie1996]

Someone help!!!!! http://www.novapapers.com/images/PDF/CIE/A-LEVEL/maths/2011nov/9709_w11_qp_33.pdf
P3 q6 part 2. Its an argand diagram question! Please help

you first plot points A and B on the diagram and i am assuming you know how to do that
next you find the mid point between A and B
you get -1/2-1/2i
this will be the centre of the circle
no you have to find the radius of the circle
find the distance between the two points using sqrt[(x2-x1)^2+(y2-y1)^2 and you get an answer
square root of answer will give you the radius

 

[applepie1996]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_33.pdf
help needed in Q1o...Dug thanks in advance

I = ⌡(tan^(n+2) x + tan^n x) dx
take tan^n out as a common factor
I = ⌡[tan^n x(tan^2 x + 1)] dx
u = tan x
dx = du/sec^2 x
I = ⌡[tan^n x(tan^2 x + 1)] du/sec^2 x
tan^2 x + 1=sec^x so you can cancel sec^2 x
I = ⌡u^n du
I = u^n+1/(n + 1)
put in the new limits
u = tan x
tan(π/4) = 1
tan(0) = 0
I = 1/(n + 1)

 

[sagystu]

can someone please help , w09_qp_51 question no 1 , i got the X of 0.05 and multiplied it by 2 due to the 2 forces acting on both sides why didin't the answer do that ?!! http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w09_qp_51.pdf

 

[applepie1996]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_33.pdf
help needed in Q1o...Dug thanks in advance

part 2 is easy i hope you can do it now
part 3 ) (tan9 + tan7) + (tan5+tan3) + 4(tan7+tan5)
tan n + tan (n+2) = 1/(n+1)
so for tan9 + tan7 it will be 1/(7+1)
for tan3 + tan5 = 1/(3+1)
for tan5 + tan7 = 1/(5+1)
1/(7+1) + 1/(3+1) + 4*1/(5+1)

 

[GorgeousEyes]

please pure 1 jun2010 v11 no.10 and n0.8 iii .

 

[syed1995]

a better dialogue p3 p3 every where xD. anyway in s1 heres a good one there are 9 pies to be divided between 3 people so that each person gets a odd no of pies. in how many no of ways can they be divided. i know the answerso i am posting it so u guys can solve it and others can get help.

Ways to be divided
3 5 1 x 3!
3 3 3 x 3!/3!
1 7 1 x 3!/2!

9C3 * 6C5 * 1C1 * 3! = 3024
9C3 * 6C3 * 3C3 * 3!/3! = 1680
9C1 * 8C7 * 1C1 * 3!/2! = 216

Total No. Of Ways = 4920..

If the pies are identical then it would be a whole different story..

 

[sagystu]

hey , if u had a spring and u have 2 tensions acting on both sides how would u calculate tension

ex

100>>>spring <<< 70 N

thanks

 

[syed1995]

please pure 1 jun2010 v11 no.10 and n0.8 iii .

i ) OB = OA + OC
OB = 4i + 2j + 4k
Unit Vector of OB = (4i + 2j + 4k) / (√4^2+2^2+4^2)
= (4i + 2j + 4k)/6

ii) AC = OC - OA
AC = 2i -4j - 2k

Now use the scalar product formulae
OB = 4i + 2j + 4k
AC = 2i -4j - 2k
AC * OB = |AC| * |OB| * Cos(Theta)

(4*2)+(2*-4)+(4*-2) = 2√6 * 6 * Cos(Theta)
-8/12√6 = Cos(Theta)
Theta = Cos^-1(-8/12√6)
Theta = 105.8 Degrees
They asked for acute angle = 180 - Ans
= 74.2 Degrees Answer

iii) 2 * (Magnitude of OA + Magnitude of OC)
2 * (|OA| + |OC|)
2* (√19 + √11)
= 15.35 Answer

 

[syed1995]

please pure 1 jun2010 v11 no.10 and n0.8 iii .

8iii) The perpendicular bisector of AB meets BC at D.

Perpendicular bisector of a line is always from the middle of the line (Midpoint of the line) and is 90 Degrees to the line.

So find the co-ordinates of midpoint.. then find the equation of normal on the midpoint (normal gradient = -m/2 as [2m * grad = -1])

Then simultaneously solve equation of Normal (of AB) and equation of BC.. to get the co-ordinates of D.

 

[GorgeousEyes]

i ) OB = OA + OC
OB = 4i + 2j + 4k
Unit Vector of OB = (4i + 2j + 4k) / (√4^2+2^2+4^2)
= (4i + 2j + 4k)/6

ii) AC = OC - OA
AC = 2i -4j - 2k

Now use the scalar product formulae
OB = 4i + 2j + 4k
AC = 2i -4j - 2k
AC * OB = |AC| * |OB| * Cos(Theta)

(4*2)+(2*-4)+(4*-2) = 2√6 * 6 * Cos(Theta)
-8/12√6 = Cos(Theta)
Theta = Cos^-1(-8/12√6)
Theta = 105.8 Degrees
They asked for acute angle = 180 - Ans
= 74.2 Degrees Answer

iii) 2 * (Magnitude of OA + Magnitude of OC)
2 * (|OA| + |OC|)
2* (√19 + √11)
= 15.35 Answer

i can't get why did we do this --> OB = OA + OC

 

[GorgeousEyes]

8iii) The perpendicular bisector of AB meets BC at D.

Perpendicular bisector of a line is always from the middle of the line (Midpoint of the line) and is 90 Degrees to the line.

So find the co-ordinates of midpoint.. then find the equation of normal on the midpoint (normal gradient = -m/2 as [2m * grad = -1])

Then simultaneously solve equation of Normal (of AB) and equation of BC.. to get the co-ordinates of D.

THANK YOU

 

[Alice123]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_32.pdf
10 b, cant find the greatest value of arg z Rutzaba
and also 10iii http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_32.pdf

 

[MoonShock]

Someone HELP!
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_31.pdf
Question no. 10.
I'm getting the partial fractions value wrong! Getting A= 1/1800 and B= 1/1800... but they're each 2. HOW?

 

[salvatore]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_63.pdf
Could anyone please help me with qn no. 6 (a)? I don't understand the marking scheme.
Thanks

PhyZac Dug

 

[ahmed abdulla]

ahmed abdulla syed1995 A star

This of pure 1, no Mechanics nor Physics knowledge needed.

as you did,
2(1-0.75^n)/1-0.75

but we also have to multiply with 2, since when a bounces it goes up and down so it travel the distance twice not once, except for the first where it only go down, then bounces..therefore

{[2(1-0.75^n)/1-0.75] x 2} -2
[8-8(0.75^n)] x 2 -2
16 - 16(0.75^n) - 2
14 - 16(o.75^n)


(b) sum of infinity
(2/1-0.75) x 2 -2
8 x 2 -2
16 -2
14.

bro .... can i now why the .... -2 after multiplying with2
syed1995 A star

 

[GorgeousEyes]

Please Help jun2010 v.12 pure 1 no.7 (a)

 

[falcon678]

Please Help jun2010 v.12 pure 1 no.7 (a)

link please?

 

[PhyZac]

bro .... can i now why the .... -2 after multiplying with2

the ball travel the first time once...then twice after ward...so u multiply with 2 but u minus the first 2 meters