Mathematics: Post your doubts here!

[Zash Riyash]

http://maxpapers.com/wp-content/uploads/2012/11/9709_w14_qp_12.pdf
MS ---> http://maxpapers.com/wp-content/uploads/2012/11/9709_w14_ms_12.pdf
Q11(iii)
Mark scheme says there is an inflexion at ( π ) in the graph, how am i supposed to know there was an inflexion at that point??? I mean if you draw the graph it looks like a straight line.
Please Helppp!

http://derivative-functions.cours-de-math.eu/advanced-math-english11.php

Go through the web page, you might understand what exactly is an inflexion. Also, the graph isn't exactly a straight line!

 

[Turki AbdulAziz]

http://derivative-functions.cours-de-math.eu/advanced-math-english11.php

Go through the web page, you might understand what exactly is an inflexion. Also, the graph isn't exactly a straight line!

I went throught it, but im still not sure how the shape of my graph would look like?can you please draw the shape?

 

[Furqan Azam]

I need a bit of you people's help on this question: Oct/nov 2014/12 P1, Q 7 part i
Also, It will be nice if someone can come up with the graph asked in the last question. Thanks.

 

[Turki AbdulAziz]

I need a bit of you people's help on this question: Oct/nov 2014/12 P1, Q 7 part i
Also, It will be nice if someone can come up with the graph asked in the last question. Thanks.

AM = OM - OA
OM= 1/2 OX , OM= 1/2 (4i + 4j + 10k) , OM= (2i + 2j + 5k)
OA= 8i
AM= (2i + 2j + 5k) - (8i) = (-6i +2j + 5k)

AC will simply be: ( -8i + 8j )

 

[Zash Riyash]

I went throught it, but im still not sure how the shape of my graph would look like?can you please draw the shape?

https://www.icloud.com/photostream/#A7Grq0zwwjrd

 

[The Sarcastic Retard]

Specimen papers 2015?

 

[BhaiArshad]

I need a bit of you people's help on this question: Oct/nov 2014/12 P1, Q 7 part i
Also, It will be nice if someone can come up with the graph asked in the last question. Thanks.

http://maxpapers.com/wp-content/uploads/2012/11/9709_w14_qp_12.pdf
MS ---> http://maxpapers.com/wp-content/uploads/2012/11/9709_w14_ms_12.pdf
Q11(iii)
Mark scheme says there is an inflexion at ( π ) in the graph, how am i supposed to know there was an inflexion at that point??? I mean if you draw the graph it looks like a straight line.
Please Helppp!

Animation of Point of Inflexion:

Graph of the curve given in Q11:

 

[ashcull14]

 

[Aina]

plz help me out with question 2

 

[sara kamal]

z=9 e^i(1/3)

z=9(e^0(cos(1/3)+i sin(1/3))

square root of z=(9 (e^0(cos(1/3)+i sin(1/3)) )^1/2

=9^(1/2)( cos(0.5*(1/3 π))+i sin(0.5*(1/3 π)) )

=+-3(cos(1/6 π) + i sin(1/6 π))

one square root is =3(cos(1/6 π) + i sin(1/6 π))
= (3√3/2)+i (3/2) (open the bracket)
r=3
θ=1/6 π
so 3 e^i (1/6) π

other root = - 3(cos(1/6 π) + i sin(1/6 π))
= - (3√3/2)- i (3/2)
r=3
θ = (1/6 π)
θ= (1/6 π)-π = -(5/6)π

so 3 e^i (-5/6) π

thank you so much.....

 

[TheJDOG]

plz help me out with question 2

Here you go
Is it correct ?

 

[manya]

Can someone tell me how to draw the graph

 

[sara kamal]

z=9 e^i(1/3)

z=9(e^0(cos(1/3)+i sin(1/3))

square root of z=(9 (e^0(cos(1/3)+i sin(1/3)) )^1/2

=9^(1/2)( cos(0.5*(1/3 π))+i sin(0.5*(1/3 π)) )

=+-3(cos(1/6 π) + i sin(1/6 π))

one square root is =3(cos(1/6 π) + i sin(1/6 π))
= (3√3/2)+i (3/2) (open the bracket)
r=3
θ=1/6 π
so 3 e^i (1/6) π

other root = - 3(cos(1/6 π) + i sin(1/6 π))
= - (3√3/2)- i (3/2)
r=3
θ = (1/6 π)
θ= (1/6 π)-π = -(5/6)π
so 3 e^i (-5/6) π

Please Ahmed,explain how to represent Rez>1 on graph where Rez denotes real part of z

 

[TheJDOG]

http://studentbounty.com/pastpapers/Cambridge International Examinations (CIE)/International AS and A Level/Mathematics (9709)/2012 Jun/9709_s12_ms_33.pdf
Q 7 (i) Q9..please anyone...

Here's 9

 

[TheJDOG]

Please Ahmed,explain how to represent Rez>1 on graph where Rez denotes real part of z

I think that would be a straight vertical line at 1 on an Argand diagram, since it says Real of z greater than 1, if you need to shade you'll have to shade on the area more than 1, that would be to the right of the vertical line. I might be wrong though, if I am correct me please.

 

[BhaiArshad]

View attachment 52927

 

[BhaiArshad]

Can someone tell me how to draw the graph

Restrictions Applied: (0 < x < 2pi)

 

[manya]

Restrictions Applied: (0 < x < 2pi)

oh ok. thanks alot

 

[papajohn]

Q:8 iii Please anyone
http://maxpapers.com/wp-content/uploads/2012/11/9709_s13_qp_11.pdf
MS------> http://maxpapers.com/wp-content/uploads/2012/11/9709_s13_ms_11.pdf

 

[The Sarcastic Retard]

Q:8 iii Please anyone
http://maxpapers.com/wp-content/uploads/2012/11/9709_s13_qp_11.pdf
MS------> http://maxpapers.com/wp-content/uploads/2012/11/9709_s13_ms_11.pdf

f(x) = 2x^2-12x+13
2(x^2-6x+13/2)
2((x-3)^2-5/2)
2(x-3)^2 -5
x>=7 so use x as 7 range is y>=27.