Mathematics: Post your doubts here!

[littlecloud11]

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-
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

 

[biba]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_33.pdf
can anybody help me with qs 7 (iii) ?

Esme Soldier313 any1 ?

 

[abbey789]

Wai

Just a second, it's a long one

Waiting

 

[LMGD33]

Sure all of you here are awesome!

You're awesomee!! There you go hope it's clear!

 

[abbey789]

Thank You so much

 

[Anika Raisa]

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!!

 

[LindaKim123]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_33.pdf

question 10 b)ii) please!!

 

[LMGD33]

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!

 

[AlishaK]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_32.pdf
Q10 (ii) Pleaseeee!!
PhyZac

 

[Anika Raisa]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_33.pdf
question 10 n 7 please

Alice123
Esme
KnowITAll
SararaIH
LMGD33
littlecloud11
PhyZac .... any1??
Please help!! May Allah bless u! Thank u!!

 

[AlishaK]

View attachment 26611View attachment 26611

NOT DONE BY ME! FOUND ON A WEBSITE!

Answers to all questions up in the zip file above in post #9196

From where did 'they' get e^-1 in the end?!

 

[Gordious]

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?

 

[Gordious]

Can someone help me with these?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_qp_3.pdfQ2. I am getting two different answers by doing two different methods: 1) Squaring both sides 2) 2x> x-1 and -2x<x-1

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_32.pdfQ6ii; what will be the shape of the curve?

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s03_qp_3.pdfQ8ii

Someone please help? :'(

 

[littlecloud11]

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.

 

[PhyZac]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_32.pdf
Q10 (ii) Pleaseeee!!
PhyZac

Did you try solving it? tell me which step u reached

 

[AlishaK]

Did you try solving it? tell me which step u reached

=[x+2lnx-1/x - 3/2 ln(2x-1)] (limits, i didn't mention) , i just wanna know how to arrive at that!

 

[PhyZac]

=[x+2lnx-1/x - 3/2 ln(2x-1)] (limits, i didn't mention) , i just wanna know how to arrive at that!

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)

 

[littlecloud11]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_33.pdf
question 10 n 7 please

Alice123
Esme
KnowITAll
SararaIH
LMGD33
littlecloud11
PhyZac .... any1??
Please help!! May Allah bless u! Thank u!!

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

 

[AlishaK]

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)

I dnt get from the second step?!

 

[PhyZac]

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)