Mathematics: Post your doubts here!

[syed1995]

syed1995, a lil help pls....

Don't have S2. Currently only Have P1,S1

 

[gary221]

Don't have S2. Currently only Have P1,S1

neways thnxx

 

[syed1995]

neways thnxx

Sorry

 

[gary221]

Sorry

np yaar, just leave it..
btw do u know of any1 who does take S2??

 

[freezingfires]

 

[A star]

np yaar, just leave it..
btw do u know of any1 who does take S2??

no way m1 is much easier s1 is hard enough

On the 7th May afternoon.

i guess we will have to siscuss all the papers by friday with three consecutive papers no time will be there. half grade decided by 14th may and still going on to 11th june cie logic -_-

 

[Silent Hunter]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_31.pdf

its question 9 .... problem is in the first few steps...... how come C = - 3 ? its coming 6 on myside (when doing the partial fractions) and finding A = 2 and B = 1 ?

applepie1996

or any one ? thanks alot

 

[mominzahid]

Hi.. Can someone please help me solve this problem? its from w10 qp 33.
this is thje question. any help would be really appreciated. Thanks

 

[syed1995]

no way m1 is much easier s1 is hard enough


i guess we will have to siscuss all the papers by friday with three consecutive papers no time will be there. half grade decided by 14th may and still going on to 11th june cie logic -_-

I find S1 to be easy.. but then again I haven't studied M1 at all .. the only main thing in S1 which is difficult is Probability/Permutation .. but since my PnC has been good from add maths itself that doesn't seem that difficult to me.

Yaar sahi kar rahay hain .. jab pahlay do paperz mein bunda ghaltian karay ga.. tu he will need atleast 15 days each (phy,chem) to get atleast 30-35 in MCQs taakay kuch tu izzat reh jaye

Honestly, 7,8,9 are deadly for me. I have to attempt 4 papers in 3 days. from then onwards it is a breeze until 10-11 June.

 

[syed1995]

no way m1 is much easier s1 is hard enough


i guess we will have to siscuss all the papers by friday with three consecutive papers no time will be there. half grade decided by 14th may and still going on to 11th june cie logic -_-

I will try and complete my chemistry physics by this Monday .. and then concentrate on Maths P1.. I haven't done even a single Maths P1 yet :\

My Chem Practical is on the 23rd..

 

[daviruss]

plzz help in number 6 can any one solve it for me plzz http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s08_qp_3.pdf

 

[A star]

I will try and complete my chemistry physics by this Monday .. and then concentrate on Maths P1.. I haven't done even a single Maths P1 yet :\

My Chem Practical is on the 23rd..

i dont even know when my practical is
well i dont have gp so have three papers in three days

 

[syed1995]

i dont even know when my practical is
well i dont have gp so have three papers in three days

haha .. GP is compulsory at our school.. honestly as long as i give Chemistry AL first and then GP it will be all good.. else i am screwed.. :\ lol both in morning

in P1.. Functions is where i lose most marks.. I don't know why but I lose marks in this.

 

[applepie1996]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_31.pdf

its question 9 .... problem is in the first few steps...... how come C = - 3 ? its coming 6 on myside (when doing the partial fractions) and finding A = 2 and B = 1 ?

applepie1996

or any one ? thanks alot

okay
i will try to do it

 

[applepie1996]

View attachment 23718

ok here goes to find the limits of this curve y= sin cube 2x cos cube 2x we will put y=0 as on the x axis y=o
0= cos cube2x sin cube 2x o=sin cube 2x
sin cube 2x = 0 0 = cos cube 2x
x=0 2x=pi/2

x= pi/4

so the limits be pi/4 and 0

by substituting u= sin2x
du/dx = 2cos2x
du= 2cos2x dx

now substituting

(cos cube 2x)( u^3) (1/ 2cos2x) <----------- where this is du
one cos would be cut by the cos in the denominator
u^3 (cos square 2x)/2
remember that cos square x = 1-sin square x
so cos square x = 1-sin square x
so
cos square 2x = 1-sin square 2x
and sin 2x =u
so
cos square 2x= 1- u^2
then the total thing wud be
1/2 integral of u^3(1-u^2)
1/2 integral of u^3 - u^5
1/2 (( u^4)/4 - (u^6)/6)
now open u as sin2x
1/2 (( sin 2x^4)/4 - (sin 2x^6)/6)
1/2 (1/4- 1/6)
1/2 (1/12)
1/24

No need to thank me
because i didn't solve it
Rutzaba solved it for me

 

[Rutzaba]

ok here goes to find the limits of this curve y= sin cube 2x cos cube 2x we will put y=0 as on the x axis y=o
0= cos cube2x sin cube 2x o=sin cube 2x
sin cube 2x = 0 0 = cos cube 2x
x=0 2x=pi/2

x= pi/4

so the limits be pi/4 and 0

by substituting u= sin2x
du/dx = 2cos2x
du= 2cos2x dx

now substituting

(cos cube 2x)( u^3) (1/ 2cos2x) <----------- where this is du
one cos would be cut by the cos in the denominator
u^3 (cos square 2x)/2
remember that cos square x = 1-sin square x
so cos square x = 1-sin square x
so
cos square 2x = 1-sin square 2x
and sin 2x =u
so
cos square 2x= 1- u^2
then the total thing wud be
1/2 integral of u^3(1-u^2)
1/2 integral of u^3 - u^5
1/2 (( u^4)/4 - (u^6)/6)
now open u as sin2x
1/2 (( sin 2x^4)/4 - (sin 2x^6)/6)
1/2 (1/4- 1/6)
1/2 (1/12)
1/24

No need to thank me
because i didn't solve it
Rutzaba solved it for me

lawl u dint have to mention my name

 

[applepie1996]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_31.pdf

its question 9 .... problem is in the first few steps...... how come C = - 3 ? its coming 6 on myside (when doing the partial fractions) and finding A = 2 and B = 1 ?

applepie1996

or any one ? thanks alot

okay i think since the powers of the x's of the top half and the bottom half of the equations are the same you divide the two equations like how we do it in the polynomials chapter yeah then after you divide the two quadratic equation you get the quotient as "2" which will be A now to find B and C you have to factorize the bottom half of the equation which i am hoping you know how to do so the factors are (2x+1) and (x+2) sooooooo.......
now A + B/2x+1 + C/x+2 = 4x^2+5x+3 and you know A is 2 so find B and C . (which i am assuming you can do )

 

[applepie1996]

lawl u dint have to mention my name

nahhhh its fine
i didn't deserve the credit anyways
except maybe for the copy paste part

 

[applepie1996]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_31.pdf

its question 9 .... problem is in the first few steps...... how come C = - 3 ? its coming 6 on myside (when doing the partial fractions) and finding A = 2 and B = 1 ?

applepie1996

or any one ? thanks alot

if you don't get it tell me okay ??

 

[Rutzaba]

nahhhh its fine
i didn't deserve the credit anyways
except maybe for the copy paste part

oh come on u desrve equal amount of credit for the intention of helping^_^