Mathematics: Post your doubts here!

[$~SauD~$]

the area inside the circle if i'm not mistaken
because u'll have |z - (3 + 4i)| =< 5
$~SauD~$ m i right?

yes, you are right

btw, guys.. stupid doubt..
how is 3+4i as 5 on the right side?

 

[Kumkum]

yes, you are right

btw, guys.. stupid doubt..
how is 3+4i as 5 on the right side?

do u mean like 5 =< |z - (3 + 4i)| ?

 

[$~SauD~$]

do u mean like 5 =< |z - (3 + 4i)| ?

it's w^2 on the right, right?
so should it be 3+4i just like on the left?

 

[Kumkum]

it's w^2 on the right, right?
so should it be 3+4i just like on the left?

yes...
on the right is the modulus of 3 + 4i and the left is the complex number itself
not sure if this is what ur saying

 

[geminisign]

Hi there
May I please get help with question 7b?
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_32.pdf
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_ms_32.pdf

If possible, could you please show me a diagram of the correct answer and an explanation?

Thank you!

 

[$~SauD~$]

yes...
on the right is the modulus of 3 + 4i and the left is the complex number itself
not sure if this is what ur saying

yes modulus of 3+4i is 5? :s how so?

 

[$~SauD~$]

yes...
on the right is the modulus of 3 + 4i and the left is the complex number itself
not sure if this is what ur saying

yes modulus of 3+4i is 5? :s how so?

 

[Kumkum]

yes modulus of 3+4i is 5? :s how so?

modulus = sq.rt ( 3^2 + 4^2)
=sq.rt( 9 + 16)
=5
it's ok now?

 

[$~SauD~$]

Hi there
May I please get help with question 7b?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_32.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_ms_32.pdf

If possible, could you please show me a diagram of the correct answer and an explanation?

Thank you!

https://skydrive.live.com/redir?resid=8E431BE676F5BD8!1202&authkey=!AH77uTDQqhkEH5E&v=3

Just donno where to shade.. |z| < 2 shows to shade whole circle.. But then i am confused with the other inequality..
Sorry.

 

[Maz]

https://skydrive.live.com/redir?resid=8E431BE676F5BD8!1202&authkey=!AH77uTDQqhkEH5E&v=3

Just donno where to shade.. |z| < 2 shows to shade whole circle.. But then i am confused with the other inequality..
Sorry.

Just put in the value of a point to check which side of the line gives smaller value of |z| than |z-2-2i| eg.z=i.

 

[6Astarstudent]

https://skydrive.live.com/redir?resid=8E431BE676F5BD8!1202&authkey=!AH77uTDQqhkEH5E&v=3

Just donno where to shade.. |z| < 2 shows to shade whole circle.. But then i am confused with the other inequality..
Sorry.

that is correct, except you use dotted lines for both the circle and the y=2-x line because its < not smaller or equal to.
You shade the bottom part of the circle, that is the quarter circle occupied in 2, 3 and 4 quadrant and shade the area below the y=2-x line (the triangle) in the first quadrant.

 

[$~SauD~$]

Just put in the value of a point to check which side of the line gives smaller value of |z| than |z-2-2i| eg.z=i.

Thanks bud

that is correct, except you use dotted lines for both the circle and the y=2-x line because its < not smaller or equal to.
You shade the bottom part of the circle, that is the quarter circle occupied in 2, 3 and 4 quadrant and shade the area below the y=2-x line (the triangle) in the first quadrant.

it is smaller dude.. :/
Thanks..

 

[6Astarstudent]

Thanks bud



it is smaller dude.. :/
Thanks..

it's the big one, substitute any point eg. -1-i
|z| < |z − 2 − 2i|
|-1-i| < |-1-i -2-2i|
|-1-i|< |-3-3i|
square root 2 < 3 square root 2
and -1-i is below the bisector, so you shade the big one.

unless I'm doing something really wrong

 

[$~SauD~$]

it's the big one, substitute any point eg. -1-i
|z| < |z − 2 − 2i|
|-1-i| < |-1-i -2-2i|
|-1-i|< |-3-3i|
square root 2 < 3 square root 2
and -1-i is below the bisector, so you shade the big one.

unless I'm doing something really wrong

nah, you are right brah.. Thanks bud

 

[SudhaPS]

9709/31/M/J/10 Questin no. 7 iii How to find the least arg Z and its corresponding mag z. Kindly help.

 

[auk]

I have a question please.
A geometric series has first term a and common ratio r,where absolute value of r<1.Sum of infinity of this series is 8.The sum of infinity of this series obtained by adding all the odd-numbered terms(i.e. 1st term+3rd term+5th term+....) is 6.Find the value of r

 

[Hemdon]

im facing problems in 9709/06/M/J/04 question no: 3......sorry cant paste the link cause the site is down!!! please do help

 

[saadgujjar]

solve these questions
inequality (x+1)/(x-1) <4
find stationary point y=2x^2 + e^-x^4 .....y=x^n-lnx^n for n>= 1
solve the integration of 1/2x dx
plz explain

 

[$~SauD~$]

solve these questions
inequality (x+1)/(x-1) <4
find stationary point y=2x^2 + e^-x^4 .....y=x^n-lnx^n for n>= 1
solve the integration of 1/2x dx
plz explain

(x+1) < 4(x-1) ==> x+1 < 4x - 4 ==> +1 + 4 < 4x -x ==> 5< 3x so x > 5/3

dy/dx = 4x - e^-x WAit what? :s e^-x^4? da fuq is this? :s

1/2x dx ==> ~ is integration sign for example..

~ 1/2x dx ==> differentiation of 2x is 2.. so numerator should be 2..
1/2 ~ 2/2x dx

1/2 ln 2x + c

 

[saadgujjar]

(x+1) < 4(x-1) ==> x+1 < 4x - 4 ==> +1 + 4 < 4x -x ==> 5< 3x so x > 5/3

dy/dx = 4x - e^-x WAit what? :s e^-x^4? da fuq is this? :s

1/2x dx ==> ~ is integration sign for example..

~ 1/2x dx ==> differentiation of 2x is 2.. so numerator should be 2..
1/2 ~ 2/2x dx

1/2 ln 2x + c

power of e is -x raise to power 4.......plz explain this question and the integration