Mathematics: Post your doubts here!

[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.

 

[sumeru]

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

 

[periyasamy]

Wow.What a great working man .Thank u.

 

[periyasamy]

View attachment 39408

solve ii for me plz...

Whats the answer for both parts?

 

[A star]

can you also explain Q2 of the same year

 

[A star]

 

[periyasamy]

Do u mean this que 2(3^x-1)=3^x
If that is the one,expand the mod n u will get 2(3^x)-2(1)=3^x
then 2(3^x)-1(3^x)=2
(3^x)=2
then u can get the x.
I hope it helps.

 

[daredevil]

Whats the answer for both parts?

 

[periyasamy]

View attachment 39424

Hi ,this is the way.Ok .
from part 1 u get=>>-tant
given point is(acos^3t,asin^3t)----->These r the coordinates
y-y1=m(x-x1)
y-asin^3t=-tant(x-acos^3t)
U continue expanding it n u will get it....

 

[periyasamy]

sumeru,Bro thanks for the solution today man.But i dont get a slight part in it.I underlined it here.Do u mind to explain it to me man.It can't be integration by parts.So whats it?Thanks man.Sorry man,very dumb in maths.

 

[Angelina_25]

Thanks alot bro

 

[sitooon]

Hi ,this is the way.Ok .
from part 1 u get=>>-tant
given point is(acos^3t,asin^3t)----->These r the coordinates
y-y1=m(x-x1)
y-asin^3t=-tant(x-acos^3t)
U continue expanding it n u will get it....

Can you solve part iii ,

 

[A star]

Do u mean this que 2(3^x-1)=3^x
If that is the one,expand the mod n u will get 2(3^x)-2(1)=3^x
then 2(3^x)-1(3^x)=2
(3^x)=2
then u can get the x.
I hope it helps.

yes this gives us only one value and the ms says twqo values :/

 

[periyasamy]

Can you solve part ii & iii , i cant get it even with expanding

oo.Sorry for that eqn.It actually is the same eqn like y=mx+c but only differnt in form.I continue from what i said above
y-asin^3t=-sint/cost(x-acos^3t)
so bring the cos the otherside,,,,ycost-asin^3tcost=-xsint+asintcos^3t
simplify the rhs eqn with a sintcost
so u get finally xsint+ycost=asintcost

For the 3rd part.use the eqn of part 2
when yintecept,x=o so x=acost
when x intercept,y=o so y=asint
so for finding the length,u use the formula of distance------>(xy)=sqr root of( (acost)^2+(asint)^2)

 

[periyasamy]

yes this gives us only one value and the ms says twqo values :/

whats the other value

 

[periyasamy]

whats the other value

I think i know why.The first line of eqn i wrote should exist in 2 forms.1st is the +3^x n the other form i guess is-3^x.
Try n see .I think u wll get the other answer.

 

[sumeru]

sumeru,Bro thanks for the solution today man.But i dont get a slight part in it.I underlined it here.Do u mind to explain it to me man.It can't be integration by parts.So whats it?Thanks man.Sorry man,very dumb in maths.

 

[daredevil]

what is Sin^3(t) + Cos^3(t)?

 

[periyasamy]

what is Sin^3(t) + Cos^3(t)?

Sorry for that it means sin power of 3then u write a t beside it.Dont get confused its not sin power of 3t.