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Mathematics: Post your doubts here!

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CIE PURE MATH 3:
Can 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 x
3 − 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 points
represent 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 iv​
Thank 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 B1​
Show vertical line through z = 1 B1​
Show the correct half-lines from u of gradient zero and 1 B1​
Shade 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!
 
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Can 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 x
3 − 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 points
represent 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 iv​
Thank 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 B1​
Show vertical line through z = 1 B1​
Show the correct half-lines from u of gradient zero and 1 B1​
Shade 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!
the whole thing sounded gibberish to me....then i realized its paper 3 :D
 
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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 :D

how did you know that the both particles have same acceleration?
 
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CIE Puremath3: Some help here please!​
Please help me with 9709/paper 33 year:October/November 2012 question 10 b (ii). Please help badly stuck with it! 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 :)
 
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help guys in solving this .. (2x-1)ln5=ln2 + xln3 .. i have difficulities in this sort of questions
 
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View attachment 25695

Plz help me solve this math.....i will be really grateful...:)

Here is the answer:


Direction of plane=n= direction of line
n=(a b c)
Direction of line:
AB =OB-OA
=r⃗ .n⃗ =a⃗ .n
=(2 -2 11)-(-1 2 5)
=(3 -4 6)
= (a b c)

Therefore equation of plane= 3x-4y+6z=d

Let r= (x y z) and n=(3 -4 6)

Therefore, (x y z). (3 -4 6)=(2 -2 11).(3 -3 4)
ð 3x-4y+6z= 6+8+66=80

Now Solution to part 2:

At y axis xand z is equal to zero
Therefore, Subsituting values In the plane equation:
We get… 4y=80
ð y=20

Now using dot product, we have…
(0 20 0).(3 -4 6)= |(0 20 0)|.|(3 -4 6)|.cos O
=> -80=(Ö202).(Ö32+42+62) .cos 0
=> cos O = -80/(20Ö61)
Or a=cos-1(80/(20Ö61))=59.193 o (Look that – sign has been ignored)
As there is a minus sign so we have to consider 2nd quadrant (Remember Trigonometry! ; ))
so we get, O =120.807 o
Now O is the angle between the normal of plane and y
To get require angle we need to subtract 90o from it n so we have
the angle=30.81

(Note that the Os with two dots above them are square roots n O with a strikethrough is theta!Weird it turns out dis waay wen i copy n paste it here frm ms word!)
Hope it helped!
 
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@iKhaled
Cud u help wid this...
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
Thank you
@iKhaled
Cud u help wid this...
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
Thank you
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
Hope this helps you.
 
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