Mathematics: Post your doubts here!

[Esme]

yeah what u said is right because now the question makes sense..thank u!!

You're welcome. Actually i got confused cuz you said the curve is parallel to the x-axis, which isn't possible. So you probably were talking about the tangent

 

[iKhaled]

You're welcome. Actually i got confused cuz you said the curve is parallel to the x-axis, which isn't possible. So you probably were talking about the tangent

oh yeah sorry my bad i meant tangent :$

 

[iKhaled]

can someone please solve this differential equation..its question 6

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_31.pdf

 

[PhyZac]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_61.pdf
Please help me asap. PhyZac Q3 ii) Why do we take for short interval X<a , and for long interval X>a...Why can't we do vice versa.
And also, Q5 a (ii)
Jazak Allah khair!
Cheers!

See..for Q3 ii) check the upload pic to imagine it

for 5 a (ii) syed1995 answered in this post.

 

[AlishaK]

See..for Q3 ii) check the upload pic to imagine it

for 5 a (ii) syed1995 answered in this post.

Jazak Allah Khair!! Thanks a ton. May allah grant you success!

 

[iKhaled]

question 2 also of the same paper arghhh !!

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_31.pdf

 

[AlishaK]

This is not technically a permutation question since it won't require the P...

5000 and 6000 .. meaning it's a four digit number which starts with a 5.

so first number is 5.. that's a given.

5 _ _ _

Now since repetition is allowed..

the next 3 numbers can be anything from 1,2,3,4,5,6

meaning it can be 5555 5432 or 5646 .. all are possible.

so 6 choices for all the 3 other spaces..

meaning it becomes

1 * 6 * 6 * 6... = 1*6^3

a bit absurd question it is, but why do u multiply a '1' in there?!
Jazak Allah khair !

 

[iKhaled]

question 2 also of the same paper arghhh !!

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

nevermind anymore about question 2..i only need question 6 the diff equation

 

[AlishaK]

http://www.examsolutions.net/maths-revision/syllabuses/CIE/period-1/specification.php
This is an amazing and hell useful website! (guaranteed) So, if anyone of u need help with understanding or anything!! this is it. I assure u. I only used this website to prepare for my AS math...and past papers of course...nothing else...no books! some topics arent covered yet thou. but use whatever u can!! Best of luck!
Jazak Allah khair! Cheers!

 

[AlishaK]

nevermind anymore about question 2..i only need question 6 the diff equation

xdy/dx=1-y^2
rearrange so that all the x terms are at one side and y terms at the other.
so.... 1/(1-y^2) dy = 1/x dx
Now integrate both the sides,
$-sign for integration
$ 1/(1-y^2) dy = $1/x dx
P.S (1-y^2) is in the form a^2 - b^2 which is nothing but (a+b)(a-b)(remember??!)
so in order to integrate expressions like $1/a^2 - b^2 = [ 1/2a* ln(a+b/a-b)] (memorise this)
Therefore,
continuing with the integration,
$ 1/(1-y^2) dy = $1/x dx
[1/2*1 * ln(1+y/1-y)] = lnx +K where K is the common constant!
You're given: x=2, y=0
so put those in the eqn:
1/2 * ln (1+0/1-0) = ln2 +K
ln(1) is '0'.
therefore, K+ln2 = 0
K= -ln2 D
Now get the eqn
1/2 * ln(1+y/1-y) =lnx -ln2
1/2 * ln(1+y/1-y) -lnx + ln2 = 0

There u gooo!!
I hope u got it!! it can be easy...if u didnt understand then refer to the website i posted! u'll get it forever stuck in ur mind!!

Cheers!

 

[syed1995]

a bit absurd question it is, but why do u multiply a '1' in there?!
Jazak Allah khair !

Well just to make it a bit clearer..

We have 5 to choose from one five... so it will be 1C1 which is 1 for first number.. then we can choose any one number from the 6 numbers.. for each of the next three numbers (1,2,3,4,5,6) since repetition is allowed...



1C1*6C1*6C1*6C1 = 1*6*6*6 = Answer

 

[iKhaled]

xdy/dx=1-y^2
rearrange so that all the x terms are at one side and y terms at the other.
so.... 1/(1-y^2) dy = 1/x dx
Now integrate both the sides,
$-sign for integration
$ 1/(1-y^2) dy = $1/x dx
P.S (1-y^2) is in the form a^2 - b^2 which is nothing but (a+b)(a-b)(remember??!)
so in order to integrate expressions like $1/a^2 - b^2 = [ 1/2a* ln(a+b/a-b)] (memorise this)
Therefore,
continuing with the integration,
$ 1/(1-y^2) dy = $1/x dx
[1/2*1 * ln(1+y/1-y)] = lnx +K where K is the common constant!
You're given: x=2, y=0
so put those in the eqn:
1/2 * ln (1+0/1-0) = ln2 +K
ln(1) is '0'.
therefore, K+ln2 = 0
K= -ln2 D
Now get the eqn
1/2 * ln(1+y/1-y) =lnx -ln2
1/2 * ln(1+y/1-y) -lnx + ln2 = 0

There u gooo!!
I hope u got it!! it can be easy...if u didnt understand then refer to the website i posted! u'll get it forever stuck in ur mind!!

Cheers!

the key to the 8 marks is all in this point "P.S (1-y^2) is in the form a^2 - b^2 which is nothing but (a+b)(a-b)(remember??!)" -_- thanksss i got it now !!

 

[AlishaK]

the key to the 8 marks is all in this point "P.S (1-y^2) is in the form a^2 - b^2 which is nothing but (a+b)(a-b)(remember??!)" -_- thanksss i got it now !!

hahaah...u know that integration technique/point, i found it in some other book....it wasnt there in mine or maybe i did not check...hahah but learn that...i came across a similar question in o/n 2008... this was still simple!!

 

[khoshi]

I am having problem in 9709/62/O/N/09 q7 ii why are we adding the previous probablity it should be subtracted :/ can anyone help please

 

[kiara15]

thaank u so much specially permutation , i had great ambiguity here

 

[iKhaled]

god..winter exam 2012 paper 31 was RAAAPPEEEEE!!

 

[scouserlfc]

Can anyone solve this without using a calculator and giving the exact answer for x

(3x)^(lg3) = (4x)^(lg4)

I know its easy and even i solved it but can anyone tell me the answer in the exact form and if possible the steps u used to get them so i can compare to mine as i dont have the answer for this And a big thanks if u do solve it

 

[PhyZac]

Can anyone solve this without using a calculator and giving the exact answer for x

(3x)^(lg3) = (4x)^(lg4)

I know its easy and even i solved it but can anyone tell me the answer in the exact form and if possible the steps u used to get them so i can compare to mine as i dont have the answer for this And a big thanks if u do solve it

it is very hard

lg [(3x)^(lg3)] = lg [(4x)^(lg4)]
lg3 x lg3x = lg4 x lg4x
lg3 x (lg3 + lgx) = lg4 x (lg4 + lgx)
(lg3)^2 + lg3lgx = (lg4)^2 + lg4lgx
(lg3)^2 - (lg4)^2 = lg4lgx - lg3lgx
(lg3 - lg4) (lg3 + lg4) = lgx (lg4 - lg3)
-(lg4 -lg3) (lg3 + lg4) = lgx (lg4 - lg3)
-(lg3 + lg4) = lgx
-lg12 = lgx
lg(1/12) = lgx
x = 1/12

 

[iKhaled]

it is very hard

lg [(3x)^(lg3)] = lg [(4x)^(lg4)]
lg3 x lg3x = lg4 x lg4x
lg3 x (lg3 + lgx) = lg4 x (lg4 + lgx)
(lg3)^2 + lg3lgx = (lg4)^2 + lg4lgx
(lg3)^2 - (lg4)^2 = lg4lgx - lg3lgx
(lg3 - lg4) (lg3 + lg4) = lgx (lg4 - lg3)
-(lg4 -lg3) (lg3 + lg4) = lgx (lg4 - lg3)
-(lg3 + lg4) = lgx
-lg12 = lgx
lg(1/12) = lgx
x = 1/12

is this statistics or math? lg is log? :S:S:S

 

[PhyZac]

is this statistics or math? lg is log? :S:S:S

Pure 3
lg is log which is log to base 10