Mathematics: Post your doubts here!

[Esme]

Aoa wr wb'
Could someone please urgently post some good short notes on vectors??
JazakAllah khair!

Wa laikum as salaam
Might be of help

 

[littlecloud11]

And qn 10 ii of this paper please

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_31.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_ms_31.pdf

littlecloud11 PhyZac and everyone else doing p3

For the summer 05
10ii) You know that one of the points on line l is 4i -2j +2k
Let this point be X.
Find the direction vector of AX
AX =(4 -2 2) -(2 2 1) =(2 -4 1)
As both A and X lie on the plane the normal of the plane is also normal to this direction vector.
You already know the direction vector of line l.
find the cross product of these two direction vectors to find the normal of the plane.
(2 -4 1) x (1 2 1)
=-6i -j +8k
Now,
r.n = a.n
(x y z). (-6 -1 8) = (-6 -1 8) (2 2 1)
-6x -y +8z =-6
or
6x + y -8z =6

 

[Esme]

And qn 10 ii of this paper please

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_31.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_ms_31.pdf

littlecloud11 PhyZac Esme and everyone else doing p3

Equation of line AB:
r=3i-2j+4k + t(-i+j+3k)
OC = i+5j-35
Any point on the line AB will have the position vector: (3-t)i + (-2+t)j + (4+3t)k <--- let this point be OD
CD=OD-OC = (2-t)i + (3+t)j + (7+3t)
Now you have two direction vectors: CD and AB
CD.AB=0 because they're perpendicular
-(2-t)+(3+t)+3(7+3t)=0
11t=-22
t=-2

Substituting t in CD
CD=4i+j+k
|CD|=3(2)^1/2= 4.24

 

[abbey789]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_31.pdf
Question 5 PLEASE

 

[iKhaled]

how to integrate the first part of question 10..10(i)

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_33.pdf

 

[Esme]

how to integrate the first part of question 10..10(i)

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_33.pdf

Posted by applepie1996

I = ⌡(tan^(n+2) x + tan^n x) dx
take tan^n out as a common factor
I = ⌡[tan^n x(tan^2 x + 1)] dx
u = tan x
dx = du/sec^2 x
I = ⌡[tan^n x(tan^2 x + 1)] du/sec^2 x
tan^2 x + 1=sec^x so you can cancel sec^2 x
I = ⌡u^n du
I = u^n+1/(n + 1)
put in the new limits
u = tan x
tan(π/4) = 1
tan(0) = 0
I = 1/(n + 1)

 

[littlecloud11]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_31.pdf
Question 5 PLEASE

5) xy dy/dy = y^2 + 4
⌡(y/ y^2 +4) dy = ⌡1/x dx
Diffrentiating y^2+ 4 = 2y
so 1/2⌡2y/y^2 +4 dy = ln x + c
1/2 ln |y^2+4| = lnx +c
when y=0, x= 1
1/2 ln4 = ln1 +c
2*1/2 ln 2 = c
c= ln 2

1/2ln |y^2+4| = lnx +ln2
ln (y^2+4)^1/2 = ln (2x)
cancel ln from both sides
(y^2+4)^1/2 =2x
y^2 = 4x^2 - 4

 

[LMGD33]

Sorry to bother, but this is the last past paper I'm solving before the exam and I just can't get it right at all >.<
Darn complex numbers!
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_33.pdf
It's question ten, usually i try as much as I can from complex number questions and let it be, but since this has eleven marks I'm really scared
Also, the document Akira shared didn't quite help me understand the solution, sorry..

Anyone?
Esme
littlecloud11

 

[abbey789]

i have done only part (i) which I think yu must be stuck in ... if you want othrs tell me ...
yu should know standard Formulas of S.area and Volume of sphere

Thank you so much.

Please answer this as well
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_31.pdf
Question 5 PLEASE

 

[farhan143]

CAN SOMEONE PLEASE EXPLAIN ME THE ANSWER TO THIS QUESTION




IT IS 2012 MAY-JUNE P61 STATISTICS - 9709
QP: http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_61.pdf

MS: http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_ms_61.pdf

 

[LMGD33]

CAN SOMEONE PLEASE EXPLAIN ME THE ANSWER TO THIS QUESTION

View attachment 26652


IT IS 2012 MAY-JUNE P61 STATISTICS - 9709
QP: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_61.pdf

MS: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_ms_61.pdf

Although you'd find better luck on the Statistics forum, I'll attempt to solve it tomorrow

 

[Esme]

CAN SOMEONE PLEASE EXPLAIN ME THE ANSWER TO THIS QUESTION

View attachment 26652


IT IS 2012 MAY-JUNE P61 STATISTICS - 9709
QP: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_61.pdf

MS: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_ms_61.pdf

I suggest you post in this thread as everyone here is focusing on P3 which is tomorrow
https://www.xtremepapers.com/commun...tics-doubt-post-your-doubts-here.25911/page-9

 

[abbey789]

5) xy dy/dy = y^2 + 4
⌡(y/ y^2 +4) dy = ⌡1/x dx
Diffrentiating y^2+ 4 = 2y
so 1/2⌡2y/y^2 +4 dy = ln x + c
1/2 ln |y^2+4| = lnx +c
when y=0, x= 1
1/2 ln4 = ln1 +c
2*1/2 ln 2 = c
c= ln 2

1/2ln |y^2+4| = lnx +ln2
ln (y^2+4)^1/2 = ln (2x)
cancel ln from both sides
(y^2+4)^1/2 =2x
y^2 = 4x^2 - 4

I didn't get the part where you diffrentiate y^2+4. PLEASE HELP ME! I'm panicking. I can't get anything through my brain!

 

[Esme]

I didn't get the part where you diffrentiate y^2+4. PLEASE HELP ME! I'm panicking. I can't get anything through my brain!

You don't have to differentiate. She shifted the terms so that you have y on one side and x on the other. Then you integrate them both.

 

[biba]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_33.pdf
can anybody help me with qs 7 (iii) ?

 

[abbey789]

I G

5) xy dy/dy = y^2 + 4
⌡(y/ y^2 +4) dy = ⌡1/x dx
Diffrentiating y^2+ 4 = 2y
so 1/2⌡2y/y^2 +4 dy = ln x + c
1/2 ln |y^2+4| = lnx +c
when y=0, x= 1
1/2 ln4 = ln1 +c
2*1/2 ln 2 = c
c= ln 2

1/2ln |y^2+4| = lnx +ln2
ln (y^2+4)^1/2 = ln (2x)
cancel ln from both sides
(y^2+4)^1/2 =2x
y^2 = 4x^2 - 4

I GOT IT! THANK YOU SO MUCH! You are my saviour!!

 

[abbey789]

You don't have to differentiate. She shifted the terms so that you have y on one side and x on the other. Then you integrate them both.

We have to differentiate it so we have the numerator as the differentiation of the denominator. so, now we can write it as ln(y^2+4), as when this is differentiated it gives us numerator 2y. And (1/2) is there to balance it!

 

[abbey789]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_31.pdf
Question 9! HELP ME!

 

[LMGD33]

I didn't get the part where you diffrentiate y^2+4. PLEASE HELP ME! I'm panicking. I can't get anything through my brain!

The rule he used here is this one :
S is integral
S f ' (x)/ f(x) dx = ln |f(x)| + c
So he realized that if you diff y^2 + 4 you get 2y , but that's not the case in the numerator, to make it the case though, we can simply multiply inside the integral by two AND divide outside the integral by half , that way it's like we didn't change any value yet we were able to use the rule to get 1/2 ln (Y^2 + 4)
Yeah?

 

[LMGD33]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_31.pdf
Question 9! HELP ME!

Just a second, it's a long one