Mathematics: Post your doubts here!

[Jiyad Ahsan]

PhyZac Jiyad Ahsan
Here... They are photocopied notes so they might not be very neat sorry.

thanx alot !!

 

[PhyZac]

PhyZac Jiyad Ahsan
Here... They are photocopied notes so they might not be very neat sorry.

Thank you so much so much so much!!!!!!!!!!!!!!!!!!! Jazaki Allah Khairan!!! Thanks ALOOOOOOOOOT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! May Allah S.w.t grant you and your family with Jannatul firdous, and shower His blessing on you. Aameeeeeeeeeeeeeeeeennn!!!!!!!!!! Thank you SO much!
May Allah S.w.t grant you with best results Aaameeen!

 

[PhyZac]

PhyZac

could you help me understand how to know what loci to draw on argand diagrams?i cant seem to grasp the concept..

I learned it from this file

 

[Esme]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w09_qp_31.pdf
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w09_ms_31.pdf

Q7iv) How do we show the shading for this: 0 < arg(z − u) < 1/4 π ?

 

[Esme]

Thank you so much so much so much!!!!!!!!!!!!!!!!!!! Jazaki Allah Khairan!!! Thanks ALOOOOOOOOOT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! May Allah S.w.t grant you and your family with Jannatul firdous, and shower His blessing on you. Aameeeeeeeeeeeeeeeeennn!!!!!!!!!! Thank you SO much!
May Allah S.w.t grant you with best results Aaameeen!

You're welcome.. I'm glad I could help after all the help you've given me
And thank you for all the duas..
May Allah grant you success too and shower his blessings on you Ameen

 

[haroon740]

june 2012 p3 9707 last question .

 

[Esme]

june 2012 p3 9707 last question .

Which variant ?

 

[haroon740]

32

 

[Esme]

32

This is only the last part, I think:

applepie1996
We have to find the point p where the perpendicular distance to the two planes is same.

first we will for an equation the way the paper gave.
therefore
for plane m
|x+2y-2z-1|/sqrt((1^2) + (2^2) + (-2^2) = |x+2y-2z-1|/3 [P.S, sqrt is square root]
for plane n
|2x-2y+z-7|/sqrt((2^2)+(-2^2)+(1^2) = |2x-2y+z-7|/3


since you are finding a point where both distance is same , therefore.
|x+2y-2z-1|/3 = |2x-2y+z-7|/3 (3 can be cancelled both sides)
=
|x+2y-2z-1| = |2x-2y+z-7|

now we will sub the line values in the equation.
the x component (1+2t) [P.S, t is lamda ]
the y comp. (1+t)
the z comp. (-1+2t)

therefore.
| (1+2t) +2 ( 1+t) -2(-1+2t) -1 | = |2(1+2t) - 2(1+t) + (-1+2t) - 7 |
simplifying it you get
|4| = |-8+4t|
now to solve modulus, we do squaring method
16 = 64-64t+16t^2
simplify
2 = 8 - 8t +2t^2
2t^2 - 8t + 6 = 0
solve it and get
t = 3 or t= 1
when t=3 the position of point is [P.S, u do this by sub t value in line equation)
(7, 4, 5) => OA
when t = 1 the position of point is
(3, 2, 1) => OB

BA (or AB, same thing) = OA - OB
= (4 , 2, 4)

now find the mod
sqrt(4^2 + 2^2 + 4^2)
= 6

I dint do every step in detail, (very long) if u dont get any step, just ask. [Pray for me please.]

 

[haroon740]

This is only the last part, I think:

Thanks genious InshaAllah u'll ace the exams with huge success XD

 

[Esme]

Thanks genious InshaAllah u'll ace the exams with huge success XD

Insha'Allah
But errmm I didn't solve that..
Phyzac did it in a previous post so I quoted it for you

 

[PhyZac]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_31.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_ms_31.pdf

Q7iv) How do we show the shading for this: 0 < arg(z − u) < 1/4 π ?

What i got is this (doesnt seem correct though)
littlecloud11 or Dug please check!

 

[Esme]

What i got is this (doesnt seem correct though)
littlecloud11 or Dug please check!

Well that's how I did it too but I'm not sure either, so I wanted to check....

 

[Esme]

What i got is this (doesnt seem correct though)
littlecloud11 or Dug please check!

Actually no hold on... the perpendicular bisector is at x=1, so we have to shade to the left of it... you shaded to the left of x=2

 

[PhyZac]

Actually no hold on... the perpendicular bisector is at x=1, so we have to shade to the left of it... you shaded to the left of x=2

Now that is where I got wrong ! Thanks again! Yup, you said it correct! Now if I made it from 1, the answer will correct. (i think) Jazaki Allah khairan!

 

[Esme]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_32.pdf

PhyZac Could you tell me what you did in Q8ii) ?
I didn't exactly understand the question

 

[PhyZac]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_32.pdf

PhyZac Could you tell me what you did in Q8ii) ?
I didn't exactly understand the question

check this

α = √ln(4 + 8α²)
α² = ln(4 + 8α²)
e^α² = 4 + 8α²
¼ e^α² = 1 + 2α²
Sq root b.s:
½ e^½α² = √(1 + 2α²)
½ = (e^-½α²)√(1 + 2α²)

WAIT all credits goes Dug ! Jazahu Allah khairan!!!

 

[Esme]

check this

α = √ln(4 + 8α²)
α² = ln(4 + 8α²)
e^α² = 4 + 8α²
¼ e^α² = 1 + 2α²
Sq root b.s:
½ e^½α² = √(1 + 2α²)
½ = (e^-½α²)√(1 + 2α²)

WAIT all credits goes Dug ! Jazahu Allah khairan!!!

Thanks a lot PhyZac and Dug

 

[karim dakroury]

Help how to know wats coming of the exam

 

[PhyZac]

Help how to know wats coming of the exam

Check this! This is the syllabus, go to Pure 3, and you will find the stuff that might come in exam!
http://papers.xtremepapers.com/CIE/...d AS Level/Mathematics (9709)/9709_y13_sy.pdf