Mathematics: Post your doubts here!

[Copy Cat]

He had the one I have. He left the XPC though.

so if i have any doubt can i post it over here since we dont know the answers can anyone give it a shot.

 

[The Sarcastic Retard]

so if i have any doubt can i post it over here since we dont know the answers can anyone give it a shot.

Will try.

 

[Copy Cat]

Will try.

thanks

 

[Sarosh Jameel]

Thanks a lot !!!

Let theta be x,
Part i)
sin^4x - cos^4x
= (sin^2x)(sin^2x) - (cos^2x)(cos^2x)
= (sin^2x)^2 - (cos^2x)^2
We know a^2 - b^2 = (a - b)(a + b)
= (sin^2x - cos^2x)(sin^2x + cos^2x)
We know sin^2x + cos^2x = 1
= (sin^2x - cos^2x)(1)
= sin^2x - cos^2x
sin^2x - cos^2x
= sin^2x - (1 - sin^2x)
= 2sin^2x - 1
Part ii)
sin^4x - cos^4x = 2sin^2x - 1
2sin^2x - 1 = 1/2
2sin^2x = 3/2
sin^2x = 3/4
sin(x) = (Roof of 3)/2
x = 60
As x lies between 0 to 360, there will be more 3 values of x.
x = 60 ; (180 - x) = (180 -60) = 120
x = 60 ; (x + 180) = (60 + 180) = 240
x = 120 ; (x +180) = (120 + 180) = 300
Therefore, x = 60, 120, 240, 300.

ks

 

[Sarosh Jameel]

PLZ SOME HELP TO ME TO SOLVE THIS QUESTION !!

 

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w10_qp_41.pdf
Q3 Please someone explain

 

[The Chill Master]

Help asap q2
https://docs.google.com/file/d/0B4puqUpdi6_SS0VfU3RDWXJXaFk/edit?usp=docslist_api

 

[Sarosh Jameel]

Help asap q2
https://docs.google.com/file/d/0B4puqUpdi6_SS0VfU3RDWXJXaFk/edit?usp=docslist_api

I got answer .. x=1.9 . but not sure if i am correct ...

 

[The Chill Master]

I got answer .. x=1.9 . but not sure if i am correct ...

well can u please post solution

 

[The Sarcastic Retard]

well can u please post solution

in^-1(x - 1) = tan^-1(3)
sin^-1(x - 1) = 1.23
Take sin at both the sides,
x - 1 = sin(1.23)
x = 0.9 + 1
x = 1.9

 

[The Chill Master]

in^-1(x - 1) = tan^-1(3)
sin^-1(x - 1) = 1.23
Take sin at both the sides,
x - 1 = sin(1.23)
x = 0.9 + 1
x = 1.9

THanx a bunch

 

[The Sarcastic Retard]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_41.pdf
Q3 Please someone explain

"use lami's rule to get the angle opposite to P1 then get the angle AP1X using subtraction from 360 degrees, then lami again to get the value of W."

 

[The Sarcastic Retard]

PLZ SOME HELP TO ME TO SOLVE THIS QUESTION !!

WHICH QUESTION!!

 

[Sarosh Jameel]

WHICH QUESTION!!

Question 8.

 

[The Chill Master]

P1 SOLVED
https://www.xtremepapers.com/community/threads/maths-p1-solved.36373/

 

[The Chill Master]

http://i.imgur.com/3y1SlVh.png?1
help asap

 

[The Chill Master]

 

[Sarosh Jameel]

http://i.imgur.com/3y1SlVh.png?1
help asap

PART ( i ) : Differentiate y = ( 4x + 1 ) ^ 1/2
dy/dx = 2 ( 4x + 1 ) ^ -1/2
put x = 2 to get gradient of 1st tangent
gradient of tangent = 2 ( 4 * 2 + 1 )^ -1/2
gradient of tangent = 2/3

Differentiate y = ( 1/2) X^2 +1
dy/dx = x
put x = 2 in above equation to get gradient of 2nd tangent
gradient of 2nd tangent = 2

let theta be x
Tan x = ( M2 - M1 ) / 1 + M2M1
Tan x = ( 2 - 2/3 ) / 1 + 4/3
Tan x = 4/7
theta = 29.7 degree

Part ( ii ) : find area under the 1st curve using 2 and 0 as limits .
find area under 2nd curve using 2 and 0 as limits .
subtract both areas to get shaded area ... final answer is 1 unit square ..

 

[Sarosh Jameel]

View attachment 49875

PART ( I ) : Sum to infinity of progression P is a / ( 1 - r )
a= 2 r= 1/2
Sum to infinity of P is 2 / ( 1 - 1/2 ) = 4

Sum to infinity of progression Q is a / ( 1 - r )
a=3 r= 1/3
Sum to infinity of Q is 3 / ( 1 - 1/3 ) = 9/2

Arithmetic progression is 4 , 9/2 , 5

Sum to infinity of R is 5


PART ( II ) : Sum to infinity of progression R is a / ( 1 - r )

R = 5 , a =4 and r = ?

5 = 4 / ( 1 - r )

r= 1/5

sum of first 3 terms is a ( 1 - r^n ) / 1 - r
put the corresponding values of a , r and n

sum of first 3 terms is 4.96

 

[The Chill Master]

PART ( i ) : Differentiate y = ( 4x + 1 ) ^ 1/2
dy/dx = 2 ( 4x + 1 ) ^ -1/2
put x = 2 to get gradient of 1st tangent

PART ( I ) : Sum to infinity of progression P is a / ( 1 - r )
a= 2 r= 1/2
Sum to infinity of P is 2 / ( 1 - 1/2 ) = 4

Sum to infinity of progression Q is a / ( 1 - r )
a=3 r= 1/3
Sum to infinity of Q is 3 / ( 1 - 1/3 ) = 9/2

Arithmetic progression is 4 , 9/2 , 5

Sum to infinity of R is 5


PART ( II ) : Sum to infinity of progression R is a / ( 1 - r )

R = 5 , a =4 and r = ?

5 = 4 / ( 1 - r )

r= 1/5

sum of first 3 terms is a ( 1 - r^n ) / 1 - r
put the corresponding values of a , r and n

sum of first 3 terms is 4.96

gradient of tangent = 2 ( 4 * 2 + 1 )^ -1/2
gradient of tangent = 2/3

Differentiate y = ( 1/2) X^2 +1
dy/dx = x
put x = 2 in above equation to get gradient of 2nd tangent
gradient of 2nd tangent = 2

let theta be x
Tan x = ( M2 - M1 ) / 1 + M2M1
Tan x = ( 2 - 2/3 ) / 1 + 4/3
Tan x = 4/7
theta = 29.7 degree

Part ( ii ) : find area under the 1st curve using 2 and 0 as limits .
find area under 2nd curve using 2 and 0 as limits .
subtract both areas to get shaded area ... final answer is 1 unit square ..

Thanx a bunch i was rele worried hv exam tom thanx a bunch