Mathematics: Post your doubts here!

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freezingfires

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applepie1996 said:
ok 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 :p
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 :)
Click to expand...
Thanks for the help I get it :D !! 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.pdf
 
littlecloud11

littlecloud11

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Rutzaba said:
no more sketching plz... @dragonninja Dug littlecloud11
Click to expand...

arg.PNG

When you see equations in the form |z-a| = |z-b| the locus for this would be the perpendicular bisector of the points (a, o) and (b,0)
similarly, |z-(a+bi)| =|z-(c+di)| would also be a perpendicular bisector, but of the points (a, b) and (c,d)

For a circle the equation will be in the form |z-a| = x, where x is any real number. Here, a= the centre of the circle and x= the radius
(I can explain how each equation comes to be but that might seem a little confusing, also, it's not restricted to the P3 syllabus so I avoided that but ask me if you want to know.)

In this question, |z| < |z-2| follows the pattern of the first equation, so the locus for this would be the perpendicular bisector of the points (0,0) and (2,0), i.e. the line x=1
for the next part 0<arg(z-u) </π/4, you have to consider the argument of Z from u and not the origin. Here, 'u' would be the origin and you assume a horizontal line through 'u' as the real-axis. Then label the argument of z as π/4. As the argument is less than π/4 but greater than o the inner region would be shaded. Also, |z| is < |z-2| so the region to the left of the bisector would be shaded.

Hope this help!
 
littlecloud11

littlecloud11

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Alice123 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_33.pdf
Q10 iib........... need help littlecloud11 Dug
Click to expand...

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.
 
Alice123

Alice123

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littlecloud11 said:
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.
Click to expand...



THANKS :)
 
syed1995

syed1995

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littlecloud11 said:
Here!
Click to expand...

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...
 
zain786

zain786

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how to solve these small functions :(

integrate:

cosx^2 and cotx^2 etc

can you please explain how its integrated?
thanks :)
 
littlecloud11

littlecloud11

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syed1995 said:
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...
Click to expand...

Yes, I'm writing my paper this May. Difficulty is a relative term. The subject content is essentially an extension of the whatever you cover in P1, P3, M1/S1 and m2/S2 with a few new concepts. It's more difficult if you don't have prior S2/M2 knowledge in the sense that there is more subject matter that you have never previously dealt with. If you have a solid base in P1, P3 then the further pure math part shouldn't cause you much trouble. Personally I think the further mechanics is the easiest section. Since you won't have M2 you'll have to put in more work because you practically end up covering the M2 syllabus as well. I have the most trouble with the further statistic section but that's because I never really liked S1. In short, what I can say is that you really have to put in time and a lot of work if you want to get everything done as the syllabus is pretty lengthy. It's just not a subject you can put off for the last minute.
Also, you get to understand how difficult the paper is once you start the question paper. I won't sugar coat it but the first time I did mine I almost cried. The only consolation- like everything else, it gets easier with time.
 
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