Mathematics: Post your doubts here!

[PhyZac]

=[x+2lnx-1/x - 3/2 ln(2x-1)] (limits, i didn't mention) , i just wanna know how to arrive at that!

Good..now see, with 1 you will get 0
and with limit 2

you will get
2 + 2ln2 - 1/2 - 3/2ln3
3/2 +1/2(4ln2 = 3ln3)
3/2 + 1/2 ln( 16 /27)

 

[littlecloud11]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_33.pdf
question 10 n 7 please

Alice123
Esme
KnowITAll
SararaIH
LMGD33
littlecloud11
PhyZac .... any1??
Please help!! May Allah bless u! Thank u!!

7i)

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

Rutzaba solved this before.

ii)

the curves on the i will be taking them as triangles
okay so as given in the question area of one triangle is A with upper limit pi/4
the area of two triangles is 2A with upper limit as pi/4 + pi/4
so then the area of 40 triangles which is 40A with upper limit as 40 x pi/4 so its 10 pi so k=10

Andddd this was done by applepie1996

 

[AlishaK]

Good..now see, with 1 you will get 0
and with limit 2

you will get
2 + 2ln2 - 1/2 - 3/2ln3
3/2 +1/2(4ln2 = 3ln3)
3/2 + 1/2 ln( 16 /27)

I dnt get from the second step?!

 

[PhyZac]

I dnt get from the second step?!

okay
2 - 1/2 + 2ln2 - 3/2ln3
3/2 + 2ln2 - 3/2ln3
3/2 + 1/2 (4ln2 - 3ln3) [i took 1/2 as common factor]
3/2 +1/2 (ln 2^4 - ln3^3)
3/2 + 1/2 (ln 2^4/3^3)
3/2 + 1/2 (ln 16/27)

 

[AlishaK]

okay
2 - 1/2 + 2ln2 - 3/2ln3
3/2 + 2ln2 - 3/2ln3
3/2 + 1/2 (4ln2 - 3ln3) [i took 1/2 as common factor]
3/2 +1/2 (ln 2^4 - ln3^3)
3/2 + 1/2 (ln 2^4/3^3)
3/2 + 1/2 (ln 16/27)

Im so sorry, I did such a dumb mistake and u had to write this all! :/
Jazak Allah khair! May Allah bless u ! :')

 

[LMGD33]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_33.pdf
question 10 n 7 please

Alice123
Esme
KnowITAll
SararaIH
LMGD33
littlecloud11
PhyZac .... any1??
Please help!! May Allah bless u! Thank u!!

Tried ten to the best of my abilities!! I hope you can do part two now that one is cleared up

 

[LMGD33]

you know that the center of the circle is 2+2i
form a right angled triangle for which the hypotenuse is from the origin to the center of the circle. to find the length of this hypotenuse its sqrt(2^2 +2^2)= sqrt(8)
2 is the radius of the circle. sqrt(8) + 2 should give you the hypotenuse of the larger triangle which you can use to find the ReZ. The adjacent of this triangle which I marked in the diagram is the ReZ.

Feel free to ask questions if you're still confused.

Perfect sense! can't believe I was such an idiot D: Thank you!!

 

[waver525]

hey can any one prove this.....
1/(sinx cosx) ==> ( secx)whole sqre /tanx
pls some try this thnkx

 

[littlecloud11]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_33.pdf
question 10 n 7 please

Alice123
Esme
KnowITAll
SararaIH
LMGD33
littlecloud11
PhyZac .... any1??
Please help!! May Allah bless u! Thank u!!

10i) iw^2= (2-2i)^2
iw^2 = 4 - 2*2*2i + (2i)^2
iw^2 = 4- 8i - 4
iw^2 = - 8i
w^2 = -8
w= sqrt(8)i and -sqrt(8)i

ii) http://imgur.com/gVGylCL

I drew the same diagram twice. The first is for the argument.
The green angle reprsents the smallest possible value and the blue represents the largest possible value.
First find angle 'a' (the black one)
Sin a = 2/√32
a= .362 rad
The green angle is therefore pi/4 - .361 =.424
and the blue angle is pi/4 +.361 = 1.15

In the second diagram, the blue line is the least modulus and the blue+ red line is the greatest possible modulus.
For the blue line, it's √32- 2 = 3.66
and for the red line it's √32 +2 +2 = 7.66

LMGD33 here.

 

[AlishaK]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_33.pdf
Q5 ii)
Also, how do we generally deal with these approaching questions. :/
PhyZac

 

[PhyZac]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_33.pdf
Q5 ii)
Also, how do we generally deal with these approaching questions. :/
PhyZac

This question is asked so many times!

Anyway..make the equation in calculator...put a big x value..u will notice what happens!

 

[AsadZaidi]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_32.pdf
would anyone help me with 10(iii).. i don't get how its ans is 1

 

[AsadZaidi]

y= lnx/x^2...
we have to use integration by parts ..
u = lnx so du/dx=(1/x)
dv/dx=(1/x^2) so, integral of v=(-1/x)
(i am using ! as the sign of integral)
the formula is: !u(dv/dx) = uv - !v(du/dx)
= lnx(-1/x) - ! (-1/x)(1/x)
= -lnx/x + ! (1/x^2)
obtain the second integral:
= (-lnx/x) - (1/x)
the limits are x=e and x = 1
now u can put in the limits and obtain the answer !

thanks ....u r a life saver...

 

[biba]

hey can any one prove this.....
1/(sinx cosx) ==> ( secx)whole sqre /tanx
pls some try this thnkx

i am taking R.H.S:
(secx)^2 = 1/(cosx)^2
so,
[1/(cosx)^2] / tanx..
[1/(cosx)^2] x (1/tanx)
[1/(cosx)^2] x (cosx/sinx)
it becomes : (cosx)/[(cosx)^2 sinx]
simplify u get : 1/cosx sinx

 

[biba]

thanks ....u r a life saver...

no problem

 

[Anika Raisa]

7i)


Rutzaba solved this before.

ii)


Andddd this was done by applepie1996

Thank u very much to u all for helping out.
# littlecloud11
LMGD33
applepie1996
Rutzaba!!!

May Allah Bless u!

 

[Anika Raisa]

Hey any one know how to do no. 7iii here :
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_31.pdf .... (Cud do i n ii bt not iii )

Here is wat i hv done in iii:



Please Help!! Thank u!!!

littlecloud11 Rutzaba PhyZac
VampBhums or any1?

 

[D0cEngi]

take a=1+2i n b=1-3i

nw u=a/b

theres this formula dat arg(a/b)=arg a - arg b

n u knw arg=tan(inverse) imaginary/real

so here
arg a= tan(inverse) 2/1

arg b= tan(inverse) -3/1


Hope it helped!! BOL!

Thanks a lot.

 

[Anika Raisa]

Thanks a lot.

N p! Just remember me in your prayers!

 

[LMGD33]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_32.pdf
would anyone help me with 10(iii).. i don't get how its ans is 1

I'm so sorry, my brain isn't working right now, but I checked my copybook, and for some reason I equated dy/dx to x
So I got : +x^3 - 2x^2 - x = 0
Hence x = 1
Why though I equated it to x, I have no clue :O If you figure it out let me know!!