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Mathematics: Post your doubts here!

Rutzaba

Rutzaba

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19islandprincess96 said:
Please help! This paper http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_3.pdf
Question 5 part 2!
And question 8?
How do we do them?
Click to expand...
∫ 2/ u(4-u)
by partial fractions
A/ u + B/ 4-u
((((A(4-u) +B(u)))) / u(4-u)
4A -Au +Bu
compare with 2/ u(4-u)
the denominator eliminates... and we have two equations
a-b=0 since there is no term in u---i
4A =2 the term without variable
A= 1/2
b=1/2
then ∫ 1/2(u) + 1/2(4-u) for the limits 2 and 1
1/2 ln u + 1/2 ln (4-u)
1/2 ln ( u +(4-u)(-1)) note that the -1 is the differential of 4-u
minus means to divide in log
so 1/2 ln ( u/ (4-u))
out limits 2 and 1
1/2 ln (2/4-2) -1/2 ln (1/4-1)
1/2ln 1 - 1/2 ln 1/3 note that ln 1=0
0 -1/2 ln 1/3
-1/2 ln (3^-1) by log propery the power would come down
-(-1/2 ) ln 3
1/2 ln 3
:) sorry abt the complex numbers ive completely forgotten em ... dun frgt to pray fr me
:)
 
Rutzaba

Rutzaba

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like i dint do 75 percent of the explaining i did here... i mean no alphabets except for variables... plus here two t0pics are involved one is integration and other is partial fractions
 
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Fatima18

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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s03_qp_6.pdf
Hey guys...Can anyone explain well number 5ii and 5 iii?????Thanx![/quote]
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_62.pdf
Question 5ii and iii.
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_62.pdf
number 6 i and iii?
I know the answers since they are given in the marking scheme..but I don't understand how to get them.
 
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19islandprincess96

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Rutzaba said:
∫ 2/ u(4-u)
by partial fractions
A/ u + B/ 4-u
((((A(4-u) +B(u)))) / u(4-u)
4A -Au +Bu
compare with 2/ u(4-u)
the denominator eliminates... and we have two equations
a-b=0 since there is no term in u---i
4A =2 the term without variable
A= 1/2
b=1/2
then ∫ 1/2(u) + 1/2(4-u) for the limits 2 and 1
1/2 ln u + 1/2 ln (4-u)
1/2 ln ( u +(4-u)(-1)) note that the -1 is the differential of 4-u
minus means to divide in log
so 1/2 ln ( u/ (4-u))
out limits 2 and 1
1/2 ln (2/4-2) -1/2 ln (1/4-1)
1/2ln 1 - 1/2 ln 1/3 note that ln 1=0
0 -1/2 ln 1/3
-1/2 ln (3^-1) by log propery the power would come down
-(-1/2 ) ln 3
1/2 ln 3
:) sorry abt the complex numbers ive completely forgotten em ... dun frgt to pray fr me
:)
Click to expand...
Thank you so much for ur help- but, umm... u solved the wrong question. :/
I don't know how to do the integration of question 5.
 
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19islandprincess96

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syed1995 said:
HEY YOU ARE TOO AWESOME .. you are just being modest :p

Well I hope P3 is easy.. cuz next year Physics is going to kill me.. I found AS physics hard.. what will A2 Physics do to me -,-
Click to expand...
You'll like A2 Physics (hopefully)... me and my friend found As hard too and this year, the As part seems so easy and A2 is interesting too, except the applications part. Insha'Allah you'll find it good and easy!
 
syed1995

syed1995

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19islandprincess96 said:
You'll like A2 Physics (hopefully)... me and my friend found As hard too and this year, the As part seems so easy and A2 is interesting too, except the applications part. Insha'Allah you'll find it good and easy!
Click to expand...

Thanks for that.. That relieves me a little :) I hope so too.. Insha`Allah :)
 
syed1995

syed1995

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Rutzaba said:
p3 if you get a good teacher 6months are enuff fr prep... obv u go out of practice very easily in maths :/
Click to expand...

That's the problem.. Going out of practice in maths is v. v. easy.. I mean i left stats for 6-7 days and i have totally forgotten Permutation and Combination in Statistics :(

Again the teacher.. We don't have a 'very good' teacher at our school.. who'd you recommend to study maths from in Karachi?

I have heard that Sir Zeeshan is pretty impressive for maths. Any suggestions? I'd only need the teacher for P3.. I can manage M1 on my own.
 
Esme

Esme

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19islandprincess96 said:
I cant get the argument of u^2. And yeah part 2. Thank you! :)
Click to expand...

Argument of u^2 is found by multiplying the arg(u) by 2.

The picture is a bit unclear
But you have to mark the points u and u^2.. then make a perpendicular bisector of the line joining these two points.
And make a circle with centre (0,0) and radius 2.
Shade the part to the right of the bisector, and inside the circle
 

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19islandprincess96

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Esme said:
Argument of u^2 is found by multiplying the arg(u) by 2.

The picture is a bit unclear
But you have to mark the points u and u^2.. then make a perpendicular bisector of the line joining these two points.
And make a circle with centre (0,0) and radius 2.
Shade the part to the right of the bisector, and inside the circle
Click to expand...
The argument of u is -3π/4 in the ms and the argument of u^2 is π/2...

Thank You for the drawing! May Allah bless you with awesomeness in this life and in the hereafter!
 
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