Mathematics: Post your doubts here!

[Rutzaba]

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

guys q4,
i am getting
1-2ax+3a^2x^2-4a^3x^3
so
3a^2=-4a^3
a=-3/4
but answer 1/square root 2
and pls explain part ii as well

the ans is 1/ sqrt 2

read the question carefully
it talks abt the coeff of x and xcube
not x square and x cube...

 

[Rutzaba]

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

guys q4,
i am getting
1-2ax+3a^2x^2-4a^3x^3
so
3a^2=-4a^3
a=-3/4
but answer 1/square root 2
and pls explain part ii as well

simple that u put this value of a bak into the terms
simplify and that will be ur answer.

 

[David Hussey]

the ans is 1/ sqrt 2

read the question carefully
it talks abt the coeff of x and xcube
not x square and x cube...

*facepalm*!!!
Jazkallahu khair!

 

[sitooon]

ummm... okay well i don't know... we did study it in A1 for our AS...

I'm talking about the quotient and product rules in differentiation....

Wasn't it in P1 A star asd syed1995 ??

I'm sorry if it was a false alarm but i think it was included...

No it wans't

 

[sitooon]

Second Part

 

[David Hussey]

how to integrate 3y^2/4y^3+1
is it by substituition ?
how do i know which method of integration i should use? really confusing at times

 

[A star]

how to integrate 3y^2/4y^3+1
is it by substituition ?
how do i know which method of integration i should use? really confusing at times

lol its from nov 2013 cause i just solved this question heres goes the logic:
the differential of the function in denominator is ( 12y^3) it is present in the numerator only 4 is missing so multiply/divide by four forming:
1/4 (ln(4y^3+1) )+c

 

[Haya Ahmed]

Some one help me please

9709_w07_qp_6 Q6 (iii)

 

[SitiPutri]

Some one help me please

9709_w07_qp_6 Q6 (iii)

(i) E(x) = np > (or equals to) 8
p = 0.65.

Thus, n x 0.65 > (or equals to) 8
n = 12.3
thus minimum trial is 12.3, make it to 13.

 

[Haya Ahmed]

(i) E(x) = np > (or equals to) 8
p = 0.65.

Thus, n x 0.65 > (or equals to) 8
n = 12.3
thus minimum trial is 12.3, make it to 13.

Ohh crap .. I missed the word expected .. anyways thanks you are awesome .. stay awesome

 

[Haya Ahmed]

Please someone help me

9709_w05_qp_6 - Q5
9709_s06_qp_6 - Q6 (iii) (why do we add 5th match and 2 teams again) :?

 

[SitiPutri]

Please someone help me

9709_w05_qp_6 - Q5
9709_s06_qp_6 - Q6 (iii) (why do we add 5th match and 2 teams again) :?

Q6 (iii)
Match 1: 32 teams play
Match 2: 16 teams play
Match 3: 8 teams play
Match 4: 4 teams play
Match 5: 2 teams play

Now look, 2 teams ONLY play 4 matches (Match 1, match 2, match 3, match 4)
And 2 teams play 5 matches (Match 1, match 2, match 3, match 4, match 5)

Do you get it? Ask me if you're still confused.

For Q5 which part are you confused with?

 

[Haya Ahmed]

Q6 (iii)
Match 1: 32 teams play
Match 2: 16 teams play
Match 3: 8 teams play
Match 4: 4 teams play
Match 5: 2 teams play

Now look, 2 teams ONLY play 4 matches (Match 1, match 2, match 3, match 4)
And 2 teams play 5 matches (Match 1, match 2, match 3, match 4, match 5)

Do you get it? Ask me if you're still confused.

For Q5 which part are you confused with?

I'm sad to say that I didn't get it .. and for Q5 .. I'll be thankfull If you did it fully because It depends on each other

 

[daredevil]

no it isnt its confusing to go so downgraded in P1 you tend to confuse it with addmaths



I am guessing you guys dont have addmaths and yeah its not wrong to learn it it saves alot of time you only have simple differential equations or rate of change questions

Ohh i didnt take add maths either :`)
I guess our teacher just made us do it prematurely bcz i did it in AS .... sorry for the hitch people I thot i was helping xD

 

[princeali97]

can anyone solve this one? :/ im confused
http://maxpapers.com/wp-content/uploads/2012/11/9709_w13_qp_1.pdf
component 3 q8(iii)

 

[omarzamzam97]

Al salamo 3alayko
I have 2 questions if any anyone could help me
1 ) The first term of an arithmetic progression is 8 and the common difference is d, where d ≠ 0. The first
term, the fifth term and the eighth term of this arithmetic progression are the first term, the second
term and the third term, respectively, of a geometric progression whose common ratio is r.
(i) Write down two equations connecting d and r. Hence show that r = 3/4
and find the value of d.
.
.
Question 2)
g : x → 2x-1/x+3
, x ∈ >, x ≠ −3.

(ii) Express g−1(x) in terms of x.
(iii) Show that the equation g−1(x) = x has no solutions.

 

[ZaqZainab]

can anyone solve this one? :/ im confused
http://maxpapers.com/wp-content/uploads/2012/11/9709_w13_qp_1.pdf
component 3 q8(iii)

came in my exam

 

[Thought blocker]

psychiatrist HOPE YOU CAN READ THE SOLUTIONS
ZaqZainab help me and him/her about the questions I had not attempt.
Question 11 part ii) ----> Like this 4-3sinx graph plot it

 

[ZaqZainab]

did u guys have addmaths?

nope

 

[ZaqZainab]

psychiatrist HOPE YOU CAN READ THE SOLUTIONS
ZaqZainab help me and him/her about the questions I had not attempt.
Question 11 part ii) ----> Like this 4-3sinx graph plot it

which paper