Mathematics: Post your doubts here!

[Silent Hunter]

you got it or you want me to finish the whole thing

nope ....... its fine ....... and thank you very much .....

have some more

and thanks again

 

[Rutzaba]

y
yeap its pure
i just hate ....hate this chapter -_-
i never get the answer

we have tto find value of w?
and then how wud i sketch :0 lemme try

 

[applepie1996]

we have tto find value of w?
and then how wud i sketch :0 lemme try

oops sorry
just do b part 2

 

[applepie1996]

nope ....... its fine ....... and thank you very much .....

have some more

and thanks again

you're welcome
anytime !

 

[tanmaydube]

Please help thank you

 

[Silent Hunter]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_33.pdf

applepie1996 or Rutzaba

its question 7 (i) ....i so hate this topic and the complex one

 

[Rutzaba]

oops sorry
just do b part 2

i dont remember sorry

 

[applepie1996]

i dont remember sorry

its okay

 

[applepie1996]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_33.pdf

applepie1996 or Rutzaba

its question 7 (i) ....i so hate this topic and the complex one

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

Silent Hunter

 

[Rutzaba]

and now ii say that this post has been used by more than three ppl that my effort dfint waste

 

[Binyamine]

View attachment 23754

Please help thank you

Here you are. Compliment from Miss Rutaba.

 

[tanmaydube]

Thanks a lot Binyamine !!!!

 

[Silent Hunter]

Rutzaba and applepie1996

thank you very much ..... ( there maybe more questions on the way )

and thanks again

 

[applepie1996]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_33.pdf
question 6 part 2
Rutzaba

 

[applepie1996]

Rutzaba and applepie1996

thank you very much ..... ( there maybe more questions on the way )

and thanks again

lucky us

 

[Rutzaba]

intelligent you !

 

[Rutzaba]

lyt jane wali in 2 mins... rat tak post krdungi ans insha Allah

 

[applepie1996]

lyt jane wali in 2 mins... rat tak post krdungi ans insha Allah

awwwww
its okay
post zaroor kurnah

 

[freezingfires]

Can anyone please solve winter 2012 Question 7 part ii of this winter 2012 paper 33 question. The value of k=10 .Can someone please explain how do we get it?