Mathematics: Post your doubts here!

[hamza.k143]

Can someone please tell me how to do May/June 2010 P31 Q9?
QP: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_31.pdf
MS: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_ms_31.pdf

It is easy you just take out the gradient by quoitent rule and the reverse it for the gradient of normal and then put the point(x,y) u wuill get it

 

[Rahma Abdelrahman]

It is easy you just take out the gradient by quoitent rule and the reverse it for the gradient of normal and then put the point(x,y) u wuill get it

Part i is okay, I couldn't do part (ii).

 

[Aamir Bilal Syed]

The points A and B have position vectors, relative to the origin O, given by
−−→OA = i+2j+3k and −−→OB = 2i+j+3k.
The line l has vector equation
r = (1−2t)i+ (5+t)j+ (2−t)k.
(i) Show that l does not intersect the line passing through A and B.
(ii) The point P lies on l and is such that angle PAB is equal to 60◦
. Given that the position vector
of P is (1 − 2t)i + (5 + t)j + (2 − t)k, show that 3t
2 + 7t + 2 = 0. Hence find the only possible
position vector of P.

I solved the question but for part (ii) I got t = -2 and t=-1/3. What I don't understand is why do we reject t = -1/3 and use t=-2 to find the position vector of P?

 

[sudeep1097]

I can't solve 8 (b). Can anyone help me with it. If you are curious then this is from w_13_qp_31. I don't know how to find least value of |z-w|

 

[sumeru]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_62.pdf
Q.no.3...(ii)....anyone any idea?

 

[Aishvarya.s]

Hi can someone help with this question?
dy/dx=y-(4-y_.This question is from m/j 2005 p3 q8

Thanx in advance

 

[marimin]

please explain question 6(1) http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w07_qp_3.pdf

 

[sudeep1097]

please explain question 6(1) http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w07_qp_3.pdf

It means you have to draw a graph for the equation and show that it intersects with x-axis at only one point.

 

[AgentBlank000]

Agent blank p3 p1?

Paper 4 actually, mechanics M1

 

[Hassan Ali Abid]

Is there anyone who knows about any good maths book which should contains some difficult questions of complex no. ...just like the questions which are in mayjune 13 papers.??????????

 

[mak kings]

plzz sm explain my doubt!!

 

[periyasamy]

Hi guys.Anyone can help me with the diff of this question.Thanks a lot.....
Its question 8 from oct nov 12(31)

 

[midha.ch]

Hi guys.Anyone can help me with the diff of this question.Thanks a lot.....
Its question 8 from oct nov 12(31)

 

[periyasamy]

Thank u..........

 

[Namehere]

I can't solve 8 (b). Can anyone help me with it. If you are curious then this is from w_13_qp_31. I don't know how to find least value of |z-w|

Yes please someone explain this question with explanations for every step. I know how to do the diagram, im just having trouble finding the least value for z-w.

 

[sitooon]

Try this midha.ch
Obtain A =1
Obtain B = −1, C = 2, D = 0

 

[salvatore]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s08_qp_3.pdf
I have some problems with qn no. 5
In (i), how do I sketch the locus in an argand diagram?
(ii) I just can't solve this question. Please help me understand the working

I'll be grateful for any help provided.
Thanks

 

[sudeep1097]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s08_qp_3.pdf
I have some problems with qn no. 5
In (i), how do I sketch the locus in an argand diagram?
(ii) I just can't solve this question. Please help me understand the working

I'll be grateful for any help provided.
Thanks

5(i)
From what i have learned. for |z-i|=2 draw circle with center at 0,i. z is at o,o and -i means that now center point is 0,i if it was +i then center point would be 0,-i just opposite. Same for real numbers i.e |z-1| would mean starting point is 0,1. Now the 2 is radius. SO |z-i|=2 means draw circle with radius 2 and center 0,i. modulus of imaginary number is a circle.

 

[salvatore]

5(i)
From what i have learned. for |z-i|=2 draw circle with center at 0,i. z is at o,o and -i means that now center point is 0,i if it was +i then center point would be 0,-i just opposite. Same for real numbers i.e |z-1| would mean starting point is 0,1. Now the 2 is radius. SO |z-i|=2 means draw circle with radius 2 and center 0,i. modulus of imaginary number is a circle.

Thanks a lot
umm.. what about part ii?

 

[Namehere]

Please someone solve Q8b, just how to find the least value for z-w with explanations of why you are doing each step pleasee!