Mathematics: Post your doubts here!

[Edison Weeho]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_33.pdf

Please help, 33/m/j/13 question 7 i), ii) and iii)

 

[Edison Weeho]

Video solution :

Thank for your help, can you help me this 33/m/j/13 question 7 i), ii) and iii) ?

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s13_qp_33.pdf

 

[Wolfgangs]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s06_qp_3.pdf

Q3 please

 

[Nazree]

papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s13_ms_32.pdf

Help me with question 10(ii) thankyou so much!!

 

[Wenxin]

9709 w10 qp33

10 The polynomial p(ß) is defined by
p(ß) = ß3 + mß2 + 24ß + 32,
where m is a constant. It is given that (ß + 2) is a factor of p(ß).
(i) Find the value of m. [2]
(ii) Hence, showing all your working, find
(a) the three roots of the equation p(ß) = 0, [5]
(b) the six roots of the equation p(ß2) = 0. [6]

(ii)how to solve?anyone can solve it??thanks

 

[Thought blocker]

Thank for your help, can you help me this 33/m/j/13 question 7 i), ii) and iii) ?

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s13_qp_33.pdf

Thank for your help, can you help me this 33/m/j/13 question 7 i), ii) and iii) ?

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s13_qp_33.pdf

I hope you can do part (i) because that's just understanding what a complex conjugate is.

(ii)
|z - 10i| = 2|z - 4i|
You are told to square both sides. Question (i) tells you how to do this: |z - ki|^2 = (z - ki)(z* + ki).

(z - 10i)(z* + 10i) = 4(z - 4i)(z* + 4i)

Simplify to get 0 = zz* - 2iz* + 2iz - 12.

Meanwhile, |z - 2i|^2 = zz* - 2iz* + 2iz + 4. So the original equation becomes
|z - 2i|^2 - 16 = 0
| |z - 2i| | = 4
|z - 2i| = 4


(iii) The graph of |z| = k is a circle of radius k centered at the origin. What does subtracting 2i from z do to that graph?
Everything summarized. Hope you can understand. Am in a bit hurry.
If you don't get inbox me.

 

[Thought blocker]

Well RoOkaYya G is blocked. She forgot her password. So its just me to solve this piles of doubts. So I will take time.

 

[Dynamite]

guys help me in http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w08_qp_3.pdf
question 7 part 2 !!!!!

 

[Lily9605]

Thank you..could you help with qn 7 part 2

 

[Emma23]

Could someone help me with question 10? I did part one but I can't figure out the other parts.

 

[Thought blocker]

guys help me in http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w08_qp_3.pdf
question 7 part 2 !!!!!

he normal to each plane is

<2, -1, -3> and <1, 2, 2> respectively.

To get the angle between the planes, consider the angle between the normals.

To get the equation of the line of intersection, consider the cross product of the normals. This will give you the direction numbers for the line of intersection.

That's all I can say.

 

[Thought blocker]

Could someone help me with question 10? I did part one but I can't figure out the other parts.

(ii)
|wz| = |w||z| = (1)(R) = R
arg|wz| = arg|w| + arg|z| = 2π/3 + θ

|z/w| = |z|/|w| = R/1 = R
arg|z/w| = arg|z| - arg|w| = θ - 2π/3

(iii)
Modulus of all three are same meaning their lengths are equal ( = R)
All angles subtended are π/3

(iv)
z = 4 +2i
The other two vertices are zw and z/w

zw = (4 + 2i)(-0.5 +i√3/2)

zw = - (2 + √3) + (2√3 -1)i

z/w = (4 + 2i)/(-0.5 +i√3/2)
Rationalise the denominator to get
z/w = - (√3 + 2) + (2√3 + 1)i

 

[Thought blocker]

Thank you..could you help with qn 7 part 2

its simple OP is actually the normal to the required plane.. coaz it says that AB is on the plane and OP is perpendicular to the plane.

simply use
r.n = a.n ( take OA or OB in place of a)
(x,y,z) x (2/3,5/3,7/3) = (1,2,2) x (2/3,5/3,7/3)
2x + 5y + 7z = 26

 

[Lily9605]

its simple OP is actually the normal to the required plane.. coaz it says that AB is on the plane and OP is perpendicular to the plane.

simply use
r.n = a.n ( take OA or OB in place of a)
(x,y,z) x (2/3,5/3,7/3) = (1,2,2) x (2/3,5/3,7/3)
2x + 5y + 7z = 26

Not part 3, I was asking about part 2..how do you get (2/3,5/3,7/3)?

 

[Dynamite]

he normal to each plane is

<2, -1, -3> and <1, 2, 2> respectively.

To get the angle between the planes, consider the angle between the normals.

To get the equation of the line of intersection, consider the cross product of the normals. This will give you the direction numbers for the line of intersection.

That's all I can say.

Thanks bro for the explanation I got it now

 

[Edison Weeho]

I hope you can do part (i) because that's just understanding what a complex conjugate is.

(ii)
|z - 10i| = 2|z - 4i|
You are told to square both sides. Question (i) tells you how to do this: |z - ki|^2 = (z - ki)(z* + ki).

(z - 10i)(z* + 10i) = 4(z - 4i)(z* + 4i)

Simplify to get 0 = zz* - 2iz* + 2iz - 12.

Meanwhile, |z - 2i|^2 = zz* - 2iz* + 2iz + 4. So the original equation becomes
|z - 2i|^2 - 16 = 0
| |z - 2i| | = 4
|z - 2i| = 4


(iii) The graph of |z| = k is a circle of radius k centered at the origin. What does subtracting 2i from z do to that graph?
Everything summarized. Hope you can understand. Am in a bit hurry.
If you don't get inbox me.

I got it, Thank for your help =), can you help me this ? 32/0/n/11 question 7) iii)
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_ms_32.pdf

 

[Edison Weeho]

who can help me this ? 32/0/n/11 question 7) iii)
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_ms_32.pdf

 

[snowbrood]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_qp_3.pdf

Q3 please

here you go http://sh.st/y4i3I

 

[Lily9605]

question

(ii)
|wz| = |w||z| = (1)(R) = R
arg|wz| = arg|w| + arg|z| = 2π/3 + θ

|z/w| = |z|/|w| = R/1 = R
arg|z/w| = arg|z| - arg|w| = θ - 2π/3

(iii)
Modulus of all three are same meaning their lengths are equal ( = R)
All angles subtended are π/3

(iv)
z = 4 +2i
The other two vertices are zw and z/w

zw = (4 + 2i)(-0.5 +i√3/2)

zw = - (2 + √3) + (2√3 -1)i

z/w = (4 + 2i)/(-0.5 +i√3/2)
Rationalise the denominator to get
z/w = - (√3 + 2) + (2√3 + 1)i

Thank you for explaining. Could you help with question 6 part 2?

 

[Haya Ahmed]

Can you guys differentiate with respect to x ..
Q=
Sinx + Cosx
Sinx - Coxs

the answer is supposed to be 2 / Sin2x - 1


RoOkaYya G