- Messages
- 681
- Reaction score
- 438
- Points
- 73
We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
Thanks for the help I get it !! May Allah bless you!I need help in another question it is May/June 2012 Paper 32 Q10.Vectors part iii.I am unable to form the modular form can u please tell me how to solve it step by step here is the link:http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_32.pdfok so
all u do is
pie/4 times 40
so
i'm gonna refer to the shaded are as a triangle which it roughly is
area of the first triangle is A
the triangle right next to it (under the x-axis)
area of those two would be 2A and area of 3 triangles would be 3A and so on............
and then the x-coordinates will be pie/4 + pie/4+...................
now that we want 40A
we do pie/4 times 40 which is equal to 10 pie so k=10
i hope you get it
Anyone with Further? or planning on taking further next year?
no more sketching plz... @dragonninja Dug littlecloud11
10ii(b) From 10(iia) the roots for z =-2, -2 +(2√3)i and -2 -(2√3)i.
So now we have z^2 = -2 or z^2 = -2+(2√3)i or z^2 = -2 -(2√3)i.
For the first equation-
z^2 = -2
so z = +/- √-2, z= √2i or -√2i
For the second equation assume z as the complex z= a+ bi
use the formula for finding the square root of a complex number to find the value of a and b. Like-
For z^2 = -2 +(2√3)i
(a + bi)^2 = -2 + (2 √3)i
so a^2 +2aib - b^2= -2 +(2 √3)i
then we have a^2 -b^2 = -2 -------1 and 2ab= 2√3 --------2
Rearrange eq 1 to give a = √3 / b
substitute for a in the second equation
(√3/b)^2 -b^2 = - 2
3 -b^4 = -2b^2
b^4 - 2b^2 - 3 = 0
(b^2 - 3 ) (b^2 + 1 ) = 0
b^2 is not equal to -1 since b is real and so b^2 = 3
so b=√ 3 or - √ 3, the corresponding values for a are -1 and 1 obtained by substituting in (2).
Hence z = a + bi = 1 + (√3)i or -1 - (√3)i.
Use the same method to solve for z^2 = -2 -(2√3)
The answers will be the same except with when b= √3 a=1 and when b=-√3 a=1
So the roots for p(z^2) =0 are √2i , -√2i , 1 + (√3)i , -1 - (√3)i, -1 + (√3)i and 1 - (√3)i.
thx you bro u really helped i got half of this thx you man
Here!
You doing Further right now? .. How would you classify it .. do you find the subject concepts difficult? I am not planning on taking it.. but was thinking of at least studying some part of it..
And If I plan on taking it.. would I be able to cope..? cuz i only took P1 S1 in AS Level.. and will be having P3 and M1 next year...
For almost 10 years, the site XtremePapers has been trying very hard to serve its users.
However, we are now struggling to cover its operational costs due to unforeseen circumstances. If we helped you in any way, kindly contribute and be the part of this effort. No act of kindness, no matter how small, is ever wasted.
Click here to Donate Now