Mathematics: Post your doubts here!

[LindaKim123]

Hope you get this:
View attachment 26738

Oh ok!! I stuck at ln(1-y^2) = 2ln(x/2) before but I got it now!!!

Thank you very much!

 

[Ashique]

xc

View attachment 26735

@LindaKim123
Oh I finally got it!!!! Excuse the untidiness. I work in a messy way, when I get excited:

 

[abbey789]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_31.pdf

question 6 part ii PLEASE!

 

[kiara15]

are we suppose to attempt all varient papers?

 

[abbey789]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_31.pdf

question 6 part ii PLEASE! SOMEONE PLEASE!

 

[LindaKim123]

Hey, you switched your axis! You put the imaginary one in the real one!

dis is great!!! Thank you!!

umm, can you do
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_31.pdf
question 7 ??

 

[Ashique]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_31.pdf

question 6 part ii PLEASE!

Let x= 2theta
So you get: cosx + 3 sinx
Which is equal to Sqrt10 (x-71.57)
We know x= 2 theta
sqrt 10 cos (2 theta -71.57) =2
2theta-71.57= cos invers( 2/sqrt 10)
2 theta-71.57= 50.77 AND 2theta-71.57= -50.77 [Since cos theta= -cos theta)


Solving two equations, you get theta= 61.17 and 10.4

 

[Barney 909]

Re: Maths help available here!!! Stuck somewhere?? Ask here!

Assalamoalaikum!!

UPDATE: Link to Sequences Help by destined007 added!

i need help

 

[Ashique]

dis is great!!! Thank you!!

umm, can you do
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_31.pdf
question 7 ??

When the tangent is parallel to the y axis, the dy/dx is infinity. To express infinity= 1/0
So we equate the dy/dx=1/0
y/[x(3y^3-1)] = 1/0
cross multiplying:
x(3y^3-1)=o

3y^3-1=0
y= cube root of (1/3)
y= 0.693

Substitue this value into x(3y^3-1)=o to find the x coordinate.

 

[Jaf]

Question 7 part ii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_33.pdf

Please! The er says something about symmetry. Donno what!!
Ashique bro?

 

[LindaKim123]

When the tangent is parallel to the y axis, the dy/dx is infinity. To express infinity= 1/0
So we equate the dy/dx=1/0
y/[x(3y^3-1)] = 1/0
cross multiplying:
x(3y^3-1)=o

3y^3-1=0
y= cube root of (1/3)
y= 0.693

Substitue this value into x(3y^3-1)=o to find the x coordinate.

how did u do i) part? I did but I didn't get the answer

 

[WayneRooney10]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_31.pdf
Q9 part ii. Please. Thanks!

ANYONE?

 

[Jaf]

When the tangent is parallel to the y axis, the dy/dx is infinity. To express infinity= 1/0
So we equate the dy/dx=1/0
y/[x(3y^3-1)] = 1/0
cross multiplying:
x(3y^3-1)=o

3y^3-1=0
y= cube root of (1/3)
y= 0.693

Substitue this value into x(3y^3-1)=o to find the x coordinate.

The issue I had with the answer to this is that instead of rounding off to 0.693 before substitution into the equation, I continue to use cuberoot of 1/3, and end up getting a different answer for the x coordinate (the difference is quite a bit actually). Confusing. :S

 

[Ashique]

Question 7 part ii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_33.pdf

Please! The er says something about symmetry. Donno what!!
Ashique bro?

You found your limit in part i, which was 1/4 π. Now as you can see, before the area under the curve was denoted by A. Now they equated the integral to 40A, which means the area under the curve is 40 times bigger. Hence the limits are to be multiplied by 40.
kπ= 1/4π *40
k=10

 

[Jaf]

ANYONE?

Dude, just multiply your two answers from part i) to get one quadratic factor and find the other one by (Ax^2 + Bx + C) inspection.
That is, the roots you get in i) are 1 - √2 i and 1 + √2 i. So the factors become (x-1+√2 i)(x-1-√2 i). You multiply the brackets and you'll see everything in terms of 'i' gets cancelled to give you a quadratic factor of x^2 - 2x +3.

If you still don't get it, let me know I'll post a screenshot.

 

[Ashique]

ANYONE?

Never really understood what they did. I would want some help in this question as well!

 

[minie23]

Never really understood what they did. I would want some help in this question as well!

November 2012 P3, was the hardest paper which CIE released since 2002 !
Lets hope we get one which will be much easier tomorrow so that we all can ace Insha Allah

 

[applepie1996]

November 2012 P3, was the hardest paper which CIE released since 2002 !
Lets hope we get one which will be much easier tomorrow so that we all can ace Insha Allah

InshaAllah

 

[iKhaled]

t

November 2012 P3, was the hardest paper which CIE released since 2002 !
Lets hope we get one which will be much easier tomorrow so that we all can ace Insha Allah

the easier the harder to ace to harder the easier to ace

 

[Jaf]

You found your limit in part i, which was 1/4 π. Now as you can see, before the area under the curve was denoted by A. Now they equated the integral to 40A, which means the area under the curve is 40 times bigger. Hence the limits are to be multiplied by 40.
kπ= 1/4π *40
k=10

I see. Thank you so much.