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so in about 15mins or so i'll attempt it
the whole thing sounded gibberish to me....then i realized its paper 3Can any one please help me with this sum's part iv below please!(I could do till iii but got badly stuck with part iv)Q.The complex number −2 + i is denoted by u.(i) Given that u is a root of the equation x3 − 11x − k = 0, where k is real, find the value of k. [3](ii) Write down the other complex root of this equation. [1](iii) Find the modulus and argument of u. [2](iv) Sketch an Argand diagram showing the point representing u. Shade the region whose pointsrepresent the complex numbers z satisfying both the inequalities|z| < |z − 2| and 0 < arg(z − u) < 1/4π. Please help asap am at door! And please draw n upload the diagram here too...pleaaase i m really confused abt the diagram dat i should draw for part ivThank u! May u be blessed for helping!BESIDES here is the mark scheme below... to me it sounded completely gibberish so i couldnot work out the last part of the sum!(iv) Show point representing u in relatively correct position in an Argand diagram B1Show vertical line through z = 1 B1Show the correct half-lines from u of gradient zero and 1 B1Shade the relevant region B1 [4][SR: For parts (i) and (ii) allow the following alternative method:
State that the other complex root is –2 – i B1
State quadratic factor x2 + 4x + 5 B1
Divide cubic by 3-term quadratic, equate remainder to zero and solve for k, or, using
3-term quadratic, factorise cubic and obtain k M1
Obtain k = 20 A1]
AND Also :
Can you please help me with 9709/paper 33 year:October/November 2012 question 10 b (ii). Please help badly stuck with it! Thank you
Link to Question: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_33.pdf
Link to Mark Scheme: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_ms_33.pdf
Anika Raisa said: ↑
Can you please help me with 9709/paper 33 year:October/November 2012 question 10 b (ii). Please help badly stuck with it! Thank you
Pleaase help!
First of all made equations for the dist covered by p and q :
Dist of p= o.5*1.75*(2+t)^2 (s=ut+0.5at^2) (u=0)
dist of Q = o.5*1.75*t^2
the difference was 4.9 => dist of P - Dist of Q
o.5*1.75*(2+t)^2 - o.5*1.75*t^2 = 4.9 (took o.5*1.75 common and divided it by 4.9 )
(2+t)^2 - t^2 =5.6
expanded the eq 4 +4t+t^2 -T^2 =5.6 ( T^2 is cancelled )
t =0.4
Haha...sorry i should have mentioned above!the whole thing sounded gibberish to me....then i realized its paper 3
well they both started from rest with same conditions (driving force resistance etc...)
i like how you emphasized that it was paper 3 xD
what year is this? pls when posting question post the paper so i can check the mark scheme after solving it to make sure i am not giving u the wrong answer!
sorry for the delay but here you go:http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w06_qp_4.pdf
Pie-man
or anyone ....... how to do 7(iii) ?
Thank you
i like how you emphasized that it was paper 3 xD
the people that take paper 3 in this forum (of which i know about) will be online soon so solve more equations until then. good luck
The values of mod Z is a line through the origin passing through the centre of the circle.So the greatest and least values of mod z is the length that corresponds to the two points that the line touches the circle with.and since the diameter is four,you know that the greatest mod is four more than the vaule for the min mod Z
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