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Mathematics: Post your doubts here!

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Hello! following the rule where S denotes integration
S f ' (x) / f (x) = ln |f(x) | + c
You realize that 4 + x^2 differentiated is 2x
Now to make the numerator 2x, you simply divide by 2 so 4x/2 = 2x
Now you can integrate to : 1/2 ln (4 + x^2)

actuali may be i typed wrong its 4x/(4+x^2) !! Sorry!!! Thnk u!
 
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Hey any one know how to do no. 7iii here :
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_31.pdf .... (Cud do i n ii bt not iii :( )

Here is wat i hv done in iii:

View attachment 26713

Please Help!! Thank u!!!

littlecloud11 Rutzaba PhyZac
VampBhums or any1?

First we would make the eq of line AB
We find b1 = OB-OA
(3,4,0) - (1,2,2) = (2,2,-2)
Line of eq AB = r +tb1
Where t is the constant given as alpha in the question
And r is any one point on the line. Lets take OA
(1,2,2) +t (2, 2, -2) ---------- eq of line AB
Any point on line ab will have points
(1+2t), (2+2t),(2-2t) we shall take this to be the point P
Now to find cos AOP we will have two points OP and OA
(OA) . (OP) = | OA | |OP | cos AOB
Where mod of OA= root of ( 1^2 +2^2+2^2) = 3
And mod of OP= root of ((1+2t) ^2 + (2+2t) ^2+(2-2t) ^2) = (√ 12t^2 +4t+9)
We have (1,2,2). ((1+2t), (2+2t),(2-2t) ) = (√9) (√ 12t^2 +4t+9) cos AOB
1+2t + 4+4t +4-4t = 3 (√ 12t^2 +4t+9) cos AOB
9+2t = 3(√ 12t^2 +4t+9) cos AOB
9+2t / 3(√ 12t^2 +4t+9) = cos AOB ---- I
Then we come to BOP
(OB) .(OP) = | OB | |OP | cos BOP
(3,4,0) . ((1+2t), (2+2t),(2-2t) ) = (√25) (√ 12t^2 +4t+9) cos BOP
14t +11 /5 (√ 12t^2 +4t+9) =cos BOP
Both (√ 12t^2 +4t+9) cancelled
We have
14t +11/ 5 = 9+2t / 3
42t +33 = 45 +10t
32t= 12
T= 3/8
 
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the rule is that if fx = function
and f' x = its differential
then the integral of f'x/fx = ln fx+c
the differential of 4+x^2 is 2x
so we take 2 common and make the eq
2 integral of 2x/ 4 +x^2
2 ln 4+x^2
hope u got it :)
Thank you so much! :D I was confused with this as well!
 
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Hey everyone,
I know that there is a saying that you shouldn't work too much or put too much pressure on yourself before the day of your exam. But can anyone give me some tips and advice about what I should do before the day of the exam? Like revision methods or revision styles that should follow? Or what papers I should solve before the day of the exam?
 
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Hey everyone,
I know that there is a saying that you shouldn't work too much or put too much pressure on yourself before the day of your exam. But can anyone give me some tips and advice about what I should do before the day of the exam? Like revision methods or revision styles that should follow? Or what papers I should solve before the day of the exam?

try the 2012 ones they will be gud practice..
 
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First we would make the eq of line AB
We find b1 = OB-OA
(3,4,0) - (1,2,2) = (2,2,-2)
Line of eq AB = r +tb1
Where t is the constant given as alpha in the question
And r is any one point on the line. Lets take OA
(1,2,2) +t (2, 2, -2) ---------- eq of line AB
Any point on line ab will have points
(1+2t), (2+2t),(2-2t) we shall take this to be the point P
Now to find cos AOP we will have two points OP and OA
(OA) . (OP) = | OA | |OP | cos AOB
Where mod of OA= root of ( 1^2 +2^2+2^2) = 3
And mod of OP= root of ((1+2t) ^2 + (2+2t) ^2+(2-2t) ^2) = (√ 12t^2 +4t+9)
We have (1,2,2). ((1+2t), (2+2t),(2-2t) ) = (√9) (√ 12t^2 +4t+9) cos AOB
1+2t + 4+4t +4-4t = 3 (√ 12t^2 +4t+9) cos AOB
9+2t = 3(√ 12t^2 +4t+9) cos AOB
9+2t / 3(√ 12t^2 +4t+9) = cos AOB ---- I
Then we come to BOP
(OB) .(OP) = | OB | |OP | cos BOP
(3,4,0) . ((1+2t), (2+2t),(2-2t) ) = (√25) (√ 12t^2 +4t+9) cos BOP
14t +11 /5 (√ 12t^2 +4t+9) =cos BOP
Both (√ 12t^2 +4t+9) cancelled
We have
14t +11/ 5 = 9+2t / 3
42t +33 = 45 +10t
32t= 12
T= 3/8

Thank u!
But 7 iii has only 1 mark do we need dat much detail! Besides ur ans doesnt prove :(
cz AP: OB=3:5 !!!
its just OA:OB!!! :(
 
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y dun u open the question... i havent even done part three xD
this is part two

oh ryt u did ii ! Sorry i m really lost 2day preperin 4 chem wid P3! Btwn i cud do dat bt not part iii Cud u pls help with dat part please!!!
 
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This question got to everyone. No one ever posted a solution in this thread! If someone could PLEASE solve it, I would be so grateful.
What I would recommend, however, is to draw it on a graph paper, and then use trigonometry.
i can only help with a half answer.. So sorry
0
 
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Hey can someone please tell me how to find the gradient by dy/dx in the following cases:

1) When the tangent passes through the origin

2) When the tangent is parallel to the y-axis

3) When the tangent is parallel to the x-axis
 
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