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Mathematics: Post your doubts here!

Anika Raisa

Anika Raisa

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LMGD33 said:
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)
Click to expand...

actuali may be i typed wrong its 4x/(4+x^2) !! Sorry!!! Thnk u!
 
Rutzaba

Rutzaba

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Anika Raisa said:
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?
Click to expand...

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
 
SararaIH

SararaIH

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Rutzaba said:
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 :)
Click to expand...
Thank you so much! :D I was confused with this as well!
 
SararaIH

SararaIH

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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?
 
W

waver525

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SararaIH said:
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?
Click to expand...

try the 2012 ones they will be gud practice..
 
Anika Raisa

Anika Raisa

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Rutzaba said:
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
Click to expand...

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!!! :(
 
Anika Raisa

Anika Raisa

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Rutzaba said:
y dun u open the question... i havent even done part three xD
this is part two
Click to expand...

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!!!
 
LMGD33

LMGD33

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Ashique said:
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.
Click to expand...
i can only help with a half answer.. So sorry
0
 
SararaIH

SararaIH

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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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