Mathematics: Post your doubts here!

[daredevil]

I've solved this but the working is pretty lengthy.
Ill just tell you in short, you gotta break the 1/x^2(2x+1) into partial fractions, and then integrate. youre gonna get -2ln(x) -1/x + 2ln(2x+1)= lny +c
Then use (1,1) to find c. Then youre gonna get too many ln's. just put them together in a single ln function (divide the ones being subtracted, multiply the ones being added) and then cancel the ln from both sides. THEN, substitute 2 in x and youre gonna get 25e/36e^(1/2) = 25e^(1/2) / 36

oohh okay thankss... my approach was wrong i guess...

 

[daredevil]

i did this question sooooooo long ago i forgot how i did it am searching for the working . may take a while

yeaah okay thankss.....

i'll start another paper till then and just HOPE for better results than this one.... i didnt even do 1 SINGLE question by myself!! >_<

 

[asd]

l

For q(2).....

Could you do the "show" part in the same question as well?
Well, Ik that |z-2i| = 4
is gonna go to:
a^2 + b^2 -4b +4 = 16
a^2 + b^2 -4b -12 = 0 which is same as zz* - 2iz* + 2iz -12= o. But is this enough to show what theyre asking for?

 

[sumeru]

Could you do the "show" part in the same question as well?
Well, Ik that |z-2i| = 4
is gonna go to:
a^2 + b^2 -4b +4 = 16
a^2 + b^2 -4b -12 = 0 which is same as zz* - 2iz* + 2iz -12= o. But is this enough to show what theyre asking for?

I think the question asks to show /z-2i/=4 from the equation that is obtained from above.....so maybe if we reverse the process.like
zz* - 2iz* + 2iz =12.....(Add 4 on both sides, which gives,
zz* - 2iz* + 2iz +4=16
(z - 2i)(z* + 2i)= 16
(z - 2i)(z - 2i)*=16
/z - 2i/^2=16..(square root on both sides)
/z - 2i/=4

 

[Namehere]

I think the question asks to show /z-2i/=4 from the equation that is obtained from above.....so maybe if we reverse the process.like
zz* - 2iz* + 2iz =12.....(Add 4 on both sides, which gives,
zz* - 2iz* + 2iz +4=16
(z - 2i)(z* + 2i)= 16
(z - 2i)(z - 2i)*=16
/z - 2i/^2=16..(square root on both sides)
/z - 2i/=4

Wow! Please could you explain me your strategy here. Plus, explain what you did from (z - 2i)(z* + 2i)= 16 downwards, and why did you decide to factorise? Would really appreciate if you answer all of this for me.

Thank you very much in advance.

 

[mmmmmm]

plz tell me if my ans. is correct:

yea, u r right. Thanks man.

 

[daredevil]

someone plz explain the trapezium rule to me


here's a question if u need it for reference

 

[sumeru]

Wow! Please could you explain me your strategy here. Plus, explain what you did from (z - 2i)(z* + 2i)= 16 downwards, and why did you decide to factorise? Would really appreciate if you answer all of this for me.

Thank you very much in advance.

I think the question wants us to show that /z-2i/=4 from the very equation that is obtained above......if you add 4 the equation becomes zz* - 2iz* + 2iz +4=16........and now if u multiply (z - 2i)(z* + 2i) , it is same as zz* - 2iz* + 2iz +4.........and (z* + 2i) is same as (z - 2i)*......so, (z - 2i)(z - 2i)* means /z - 2i/^2=16.....and putting sq.root on both sides gives /z - 2i/=4

 

[sitooon]

Here is Q8(i):

I´ll upload Q8(ii) soon.

Lets suppose next time they asked greatest value , then which side we need to find ??

 

[asma tareen]

Do we hv to subtract the area under curve from area of trapezium :/??

someone plz explain the trapezium rule to me
View attachment 39361

here's a question if u need it for reference

W

 

[asd]

I think the question asks to show /z-2i/=4 from the equation that is obtained from above.....so maybe if we reverse the process.like
zz* - 2iz* + 2iz =12.....(Add 4 on both sides, which gives,
zz* - 2iz* + 2iz +4=16
(z - 2i)(z* + 2i)= 16
(z - 2i)(z - 2i)*=16
/z - 2i/^2=16..(square root on both sides)
/z - 2i/=4

So basically, what I did is correct as well? cause its just the reverse of what you did.

 

[Angelina_25]

Oct nov 13 varient 32 question 3 and 4 :/ Aly Emran

 

[Talha Irfan]

someone plz explain the trapezium rule to me
View attachment 39361

here's a question if u need it for reference

https://www.khanacademy.org/math/in...trapezoidal-approximation-of-area-under-curve

perfect explanation by Khan

 

[periyasamy]

Hai guys need some help.

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s13_qp_31.pdf

Guys,i dont get the question 7(b).Thank u guys.

 

[A star]

Oct nov 13 varient 32 question 3 and 4 :/ Aly Emran View attachment 39374

I got the answer of three how to do Question no 2? how did you get the second value –0.369

 

[periyasamy]

-----Hei guys,have a doubt here.In question 6(11),why does the r is equal to zero?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_31.pdf

-----Guys n how do u make 32ln4-8ln2=56ln2.Thanks in advance


----Last question guys.The question 8(b) from the same paper.Thanks guys.Seriously have no idea.Sorry for asking many questions,but i need help from u guys.Thanks a lot.

 

[daredevil]

solve ii for me plz...

 

[SitiPutri]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_62.pdf
Q.no.3...(ii)....Any idea anyone?

Sorry if it's too late lol

 

[sumeru]

-----Hei guys,have a doubt here.In question 6(11),why does the r is equal to zero?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_31.pdf

-----Guys n how do u make 32ln4-8ln2=56ln2.Thanks in advance


----Last question guys.The question 8(b) from the same paper.Thanks guys.Seriously have no idea.Sorry for asking many questions,but i need help from u guys.Thanks a lot.

32ln4-8ln2
8 into 4 ln4-8ln2
8ln256-8ln2
8(ln256/2)
8ln128
8ln2*7
56ln2

 

[sumeru]

-----Hei guys,have a doubt here.In question 6(11),why does the r is equal to zero?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_31.pdf

-----Guys n how do u make 32ln4-8ln2=56ln2.Thanks in advance


----Last question guys.The question 8(b) from the same paper.Thanks guys.Seriously have no idea.Sorry for asking many questions,but i need help from u guys.Thanks a lot.