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Mathematics: Post your doubts here!

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Sorry to bother, but this is the last past paper I'm solving before the exam and I just can't get it right at all >.<
Darn complex numbers!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_33.pdf
It's question ten, usually i try as much as I can from complex number questions and let it be, but since this has eleven marks I'm really scared :(
Also, the document Akira shared didn't quite help me understand the solution, sorry..

Anyone?
Esme
littlecloud11

10i) u-w =4i ----1 and uw = 5--------2
u= 5/w
substitute this in eq 1
5/w -w =4i
5-w^2 = 4i
w^2 +4i -5 =0
w= -4i +/- √{(4i)^2 - 4*-5*1}
w= -41 +/- √(-16+ 20)
w = -4i +2/2 -4i-2/2
w= -2i+1 and -2i-1

u= 5/w
u= 5/1-2i
u= 5(1+2i) /4+1
u= 10i +5/5
u= 2i + 1
and it's conjugate u= 2i-1

ii) Sorry this is a little messy-Untitled.jpg
The red shaded region is the answer for part ii

iii) I labeled the length of the real axis that represents the max ReZ
You know that the argument of Z is π/4
Cos π/4 = max ReZ/ √8 + 2
ReZ = 1/√2 * (√8+2)
ReZ= 2+ 2/√2
 
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Anika Raisa and rosogolla993
Sorry for the late reply.. Net was too slow last night


Let Q be the closest point on the line to P... Then PQ is perpendicular to the direction of the line.....
as Q lies on the line, the position vector of Q
OQ=(1 3 -4) - t(2 1 3)= (1+2t 3+t -4+3t)

Now PQ=OQ-OP
=(1+2t 3+t -4+3t)- (-1 4 11)=(2+2t -1+t -15+3t)
Since PQ is perpendicular to the line
PQ* direction of l=0

(2+2t -1+t -15+3t)(2 1 3)=0
u get the value of t ie 3
sub value of t in PQ
(2+2t -1+t -15+3t)= (8 2 -6)
lPQl=sqroot(8^2+2^2+6^2)= 10.2

NO need 4 being sorry! The net prob is commoon so i knw!

Thank u very much!! :)
 
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10i) u-w =4i ----1 and uw = 5--------2
u= 5/w
substitute this in eq 1
5/w -w =4i
5-w^2 = 4i
w^2 +4i -5 =0
w= -4i +/- √{(4i)^2 - 4*-5*1}
w= -41 +/- √(-16+ 20)
w = -4i +2/2 -4i-2/2
w= -2i+1 and -2i-1

u= 5/w
u= 5/1-2i
u= 5(1+2i) /4+1
u= 10i +5/5
u= 2i + 1
and it's conjugate u= 2i-1

ii) Sorry this is a little messy-View attachment 26656
The red shaded region is the answer for part ii

iii) I labeled the length of the real axis that represents the max ReZ
You know that the argument of Z is π/4
Cos π/4 = max ReZ/ √8 + 2
ReZ = 1/√2 * (√8+2)
ReZ= 2+ 2/√2

Thank you so much!! This was very helpful!!
just a little question though, in part (iii) How did you get the length sqrt(8) + 2 ?
I know it's a silly question but thanks a heap :$
God bless you!
 
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argz.JPG Please tell me if i am doing it right or not?
Q9 iii. argz> arg u => arg z>Pye/3 hence region greater than pye/3 will be shaded?
 
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Thank you so much!! This was very helpful!!
just a little question though, in part (iii) How did you get the length sqrt(8) + 2 ?
I know it's a silly question but thanks a heap :$
God bless you!

you know that the center of the circle is 2+2i
form a right angled triangle for which the hypotenuse is from the origin to the center of the circle. to find the length of this hypotenuse its sqrt(2^2 +2^2)= sqrt(8)
2 is the radius of the circle. sqrt(8) + 2 should give you the hypotenuse of the larger triangle which you can use to find the ReZ. The adjacent of this triangle which I marked in the diagram is the ReZ.

Feel free to ask questions if you're still confused.
 
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=[x+2lnx-1/x - 3/2 ln(2x-1)] (limits, i didn't mention) , i just wanna know how to arrive at that! :p
Good..now see, with 1 you will get 0
and with limit 2

you will get
2 + 2ln2 - 1/2 - 3/2ln3
3/2 +1/2(4ln2 = 3ln3)
3/2 + 1/2 ln( 16 /27)
 
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7i)
ok here goes to find the limits of this curve y= sin cube 2x cos cube 2x we will put y=0 as on the x axis y=o
0= cos cube2x sin cube 2x o=sin cube 2x
sin cube 2x = 0 0 = cos cube 2x
x=0 2x=pi/2

x= pi/4

so the limits be pi/4 and 0

by substituting u= sin2x
du/dx = 2cos2x
du= 2cos2x dx

now substituting

(cos cube 2x)( u^3) (1/ 2cos2x) <----------- where this is du
one cos would be cut by the cos in the denominator
u^3 (cos square 2x)/2
remember that cos square x = 1-sin square x
so cos square x = 1-sin square x
so
cos square 2x = 1-sin square 2x
and sin 2x =u
so
cos square 2x= 1- u^2
then the total thing wud be
1/2 integral of u^3(1-u^2)
1/2 integral of u^3 - u^5
1/2 (( u^4)/4 - (u^6)/6)
now open u as sin2x
1/2 (( sin 2x^4)/4 - (sin 2x^6)/6)
1/2 (1/4- 1/6)
1/2 (1/12)
1/24

Rutzaba solved this before.

ii)
the curves on the i will be taking them as triangles
okay so as given in the question area of one triangle is A with upper limit pi/4
the area of two triangles is 2A with upper limit as pi/4 + pi/4
so then the area of 40 triangles which is 40A with upper limit as 40 x pi/4 so its 10 pi so k=10

Andddd this was done by applepie1996
 
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I dnt get from the second step?! :(
okay
2 - 1/2 + 2ln2 - 3/2ln3
3/2 + 2ln2 - 3/2ln3
3/2 + 1/2 (4ln2 - 3ln3) [i took 1/2 as common factor]
3/2 +1/2 (ln 2^4 - ln3^3)
3/2 + 1/2 (ln 2^4/3^3)
3/2 + 1/2 (ln 16/27)
 
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