Sure all of you here are awesome!Just a second, it's a long one
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Sure all of you here are awesome!Just a second, it's a long one
Definitely inshaAllah! Please do the same for meYou're welcome Remember me in your prayers
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
Esme Soldier313 any1 ?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) ?
WaitingJust a second, it's a long one
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
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
From where did 'they' get e^-1 in the end?!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
Someone please help? :'(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
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!
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!Did you try solving it? tell me which step u reached
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