Mathematics: Post your doubts here!

AhmedNasser

AhmedNasser

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For 9709_s04_qp_1.pdf
I need help with Question Number 10 (ii)
Thanks in advance,
AHMED NASSER
 

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thnask
XPFMember said:
Assalamoalaikum!

Good news for you all!! :Yahoo!:

Stuck somewhere in Maths?? Post your queries here and destined007 (others are also welcome! ;) ) will answer it at the earliest, InshaAllah! 8)

P.S. I'm busy these days, so I can't promise to be there for help. I've done P1, M1, S1 so far and will be starting P3 in a couple of days, InshaAllah!

May Allah give us all success in this world as well as the HereAfter...Aameen!! :)

And oh yeah, let me put up some links for A level Maths Notes here.

Check this out! - Nice website, with video tutorials!

A level Maths Notes - Uploaded by 'destined007'

My P1 Notes! - Only few chapters available at the moment!

Maths Notes by destined007

Some Notes for P1 and P3 - shared by hamidali391

A LEVEL MATHS TOPIC WISE NOTES

MATHS A LEVEL LECTURES

compiled pastpapers p3 and p4 - by haseebriaz

Jazak Allah Khair!
XPFMember 8)
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ffaadyy

ffaadyy

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Khan_971 said:
MJ 2005. P3 Q8 (ii) :( http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s05_qp_3.pdf
Click to expand...

dy/dx = y(4-y)
[1/y(4-y)].dy = 1.dx

As we can't directly integrate [1/y(4-y)], we'll use its partial fractions form which we've found out in part (i) of this question.

[1/y(4-y)] dy = 1 dx

Replace '[1/y(4-y)]' with '(1/4)(1/y) - (1/4)[1/(4-y)] '

(1/4)(1/y) - (1/4)[1/(4-y)] dy = 1 dx

Now integrate both the sides.

(1/4)(ln y) - (1/4)[ ln (4-y) ] = x + c

Substitute y=1 and x=0 in the above equation to find the value of the constant 'c'.

- (1/4)( ln 3 ) = c

Put back the value of 'c' in the integrated equation.

(1/4)(ln y) - (1/4)[ ln (4-y) ] = x - (1/4)( ln 3 )

Next, obtain an expression of 'y' in terms of 'x'.

(1/4)(ln y) - (1/4)[ ln (4-y) ] = x - (1/4)( ln 3 )
(1/4)(ln y) - (1/4)[ ln (4-y) ] + (1/4)( ln 3 ) = x
(1/4)[ ln y - ln (4-y) + ln 3 ] = x
ln y + ln 3 - ln (4-y) = 4x
ln 3y - ln (4-y) = 4x
ln [ 3y / (4-y) ] = 4x
3y / (4-y) = e^4x
3y / (4-y) = 1/e^-4x

Cross multiply.

4 - y = 3ye^-4x
4 = 3ye^-4x + y
4 = y ( 3e^-4x + 1 )
4 / ( 3e^-4x + 1 ) = y

Therefore, the final answer is y = 4 / ( 3e^-4x + 1 ).
 
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unique840

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Silent Hunter said:
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s11_qp_61.pdf

Q7 (b)
Click to expand...
if ronnie has to win thn she must be the first to draw a yellow ball. julie draws the first ball which means julie hasto draw a green ball so that ronnie may draw a yellow one i.e P(gy). second possibility is that the first ball drawn by julie is green, the next by ronnie is also green and the next by julie is also green. the 4th by ronnie is yellow. i.e P(gggy). the last possibility is P(gggggy)
as it is without replacement, the probability will not be constant
 
zain786

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PLZ GUYZ I GOT A MOCK TOMORROW...IM DOING MATHS PAST PAPER 9709/01/MJ/08.....

im stuck in Qs 8 (ii)

and Q9 (i)... wich seems easy bt just cant get the 2 in the formula
]
ans ASAP
thnx:)
 
mhalvi

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Assalam o alikum
does any one know where the nov2011 session papers are available...??
I really need them.
I would be very thankful if someone post a link ...:)
 
ffaadyy

ffaadyy

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Hussnain said:
HELP NEEDED.
P3
COMPLEX NUMBERS
M/J/08
Q 5 (ii)
Please reply asap.
Click to expand...

This is how we'll do this question:

1/(z+2-i)

We are given that ' z = 2 cos θ + i ( 1 - 2 sin θ )' so we'll put in this value of 'z' in the above equation.

1/(z+2-i)
1/[2 cos θ + i ( 1 - 2 sin θ )+2-i]

Arrange the real numbers and the imaginary numbers.

1/[(2 cos θ + 2) + i ( 1 - 2 sin θ -1 )]
1/[(2 cos θ + 2) - i ( 2 sin θ )]

Next, multiply the equation '1/[(2 cos θ + 2) - i ( 2 sin θ )]' by the conjugate of '[(2 cos θ + 2) - i ( 2 sin θ )]'

{1/[(2 cos θ + 2) - i ( 2 sin θ )]} x [(2 cos θ + 2) - i ( 2 sin θ )] / [(2 cos θ + 2) + i ( 2 sin θ )]
[2 cos θ + 2 + i (2 sin θ)] / [(2 cos θ + 2)^2 + (4 sin^2 θ)]

As the question has asked us to deal with the real part only, we'll remove 'i (2 sin θ)' which is imaginary.

(2 cos θ + 2) / [(2 cos θ + 2)^2 + (4 sin^2 θ)]
(2 cos θ + 2) / (4 cos^2 θ + 8 cos θ + 4 + 4 sin^2 θ)
(2 cos θ + 2) / (4 cos^2 θ + 4 sin^2 θ + 8 cos θ + 4)
(2 cos θ + 2) / [4 (cos^2 θ + sin^2 θ) + 8 cos θ + 4]

Recall the identity 'cos^2 θ + sin^2 θ=1'.

(2 cos θ + 2) / [4 (cos^2 θ + sin^2 θ) + 8 cos θ + 4]
(2 cos θ + 2) / [4 (1) + 8 cos θ + 4]
(2 cos θ + 2) / (8 + 8 cos θ)
[2 ( cos θ + 1 ) ] / [ 8 ( 1 + cos θ ) ]

'cos θ + 1' cancels out as its present in both the numerator and the denominator. We are left with '2/8' now which further simplifies to '1/4'. Therefore, '1/4' is our final answer.
 
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unique840

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zain786 said:
PLZ GUYZ I GOT A MOCK TOMORROW...IM DOING MATHS PAST PAPER 9709/01/MJ/08.....

im stuck in Qs 8 (ii)

and Q9 (i)... wich seems easy bt just cant get the 2 in the formula
]
ans ASAP
thnx:)
Click to expand...
ans9) integrate the derivative given.
dy = (-k/x^3) dx
dy = -k(x^-3) dx
integrate the whole equation
y = [-k(x^-2)] / -2 +c
y = k/(2x^2) +c
18 = k/ (2) +c
3 = k/(2*4^2) +c
solve both simultaneously
and keep the values of k and c in the integrated equation
 
Sajjad Hossain

Sajjad Hossain

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zain786 said:
PLZ GUYZ I GOT A MOCK TOMORROW...IM DOING MATHS PAST PAPER 9709/01/MJ/08.....

im stuck in Qs 8 (ii)

and Q9 (i)... wich seems easy bt just cant get the 2 in the formula
]
ans ASAP
thnx:)
Click to expand...
In the first part of Q8 you found K = 5 or K = -7 so wat u have to do in 8)ii) is substitue the k values in the equation (36/2-x)-2k=x and get the answers. Ans is x=-4 x=8 :)
In Q9)i) integrate -k/x^3 u will get y=k/x^2+c then they gave u the ponts (1,18) and (4,3) substitue this points in place of x and y u will get two equtaions then solve them simultaneously :D Hope it helped
 
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unique840

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Hussnain said:
HELP NEEDED.
P3
COMPLEX NUMBERS
M/J/08
Q 5 (ii)
Please reply asap.
Click to expand...
i m replacing theta with x
1/(2cosx +i - i2sinx +2 -i)
+i -i cancels out
1/(2cosx -i2sinx +2)
multiply and divide the whole fraction with the conjugate
[1/(2cosx + 2 - i2sinx)] * [(2cosx + 2 + i2sinx) / (2cosx + 2 + i2sinx)]
in the denominator: ( (2cosx + 2 + (2cosx + 2 + i2sinx) ) - i2sinx) * (2cosx + 2 + i2sinx)
taking (2cosx + 2) as a. and (i2sinx) as b. we apply the formula (a+b)(a-b) = a^2 - b^2; so
(2cosx + 2 + i2sinx) / {[(2cosx+2)^2] - [(i2sinx)^2]}
(2cosx + 2 + i2sinx) / {[4(cos^2)x + 8cosx + 4] + 4(sin^2)x}
-(i^2) is changed to +1 in the above step
4(cos^2)x + 4(sin^2)x is changed by identity [(cos^2)x +(sin^2)x = 1] to 4
(2cosx + 2 + i2sinx) / (8cosx + 8)
we jus have to prove the real part. so separate the real and imaginary part
(2cosx + 2) / (8cosx + 8)
(cosx + 1) / 4(cosx + 1)
cosx + 1 cancels out
1/4 is left which is constant for all values of theta
 
BeanZ 08

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

Assalamoalaikum!! :)

UPDATE: Link to Sequences Help by destined007 added!
Click to expand...
salam ....
can u plz tell me tat hw 2 expand a binomial having three terms? m really stuck hea !!! plz help me out
 
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