Mathematics: Post your doubts here!

[Jiyad Ahsan]

how many of u are giving s1 this year

meeeeee
im giving accel.

 

[A star]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w07_qp_1.pdf 10..i..

you can see than AXO is a right angle triangle so

Ax=rssinx
Ox=rcosx
area of OXA = 1/2 x rsinx x rcosx
area of sector = 1/2 x r^2 x x
hence are of shaded region = (r^2x)/2 - (r^2sinxcosx)/2 this makes r^2/2 x (x-sinxcosx)

 

[A star]

so me u daredevil asd mustafamotani asd and syed will be there to answer queries

 

[Jiyad Ahsan]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w07_qp_1.pdf 10..i..

oh well.. umm OP is 2i, OQ is 2j+4k, OR is 4i+2j+2k

PR is OR - OP
PR=2i+2j+2k

PQ is OQ - OP
PQ= -2i+2j+2k

 

[Jiyad Ahsan]

w-06-qp-1 no. 7. ii..help needed..

w-06-qp-1 no. 7. ii..help needed..

you integrate the equation with limits 1 and 0 to give you the area of the first region(R1), then you do it again with limits 2 and 1 this being the region (R2)
you'll have the same answers (1/4) so you can say R1=R2

 

[prekshya]

you integrate the equation with limits 1 and 0 to give you the area of the first region(R1), then you do it again with limits 2 and 1 this being the region (R2)
you'll have the same answers (1/4) so you can say R1=R2

you integrate the equation with limits 1 and 0 to give you the area of the first region(R1), then you do it again with limits 2 and 1 this being the region (R2)
you'll have the same answers (1/4) so you can say R1=R2

you integrate the equation with limits 1 and 0 to give you the area of the first region(R1), then you do it again with limits 2 and 1 this being the region (R2)
you'll have the same answers (1/4) so you can say R1=R2

thanks (Y)

 

[prekshya]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w08_qp_1.pdf 9 iii??

 

[Jiyad Ahsan]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w08_qp_1.pdf 9 iii??

umm putting it simply find the angles that the tangent makes with the x-axis, subtract the larger from the smaller and voila you have the acute angle

basically you differentiate the curve equation and first put in the x- coordinate of p then of q
what you'll have is (since the gradient y/x)
tan(theta)=the first gradient(2/3)
tan(alpha)= second(3/4)
larger angle(56.3) - samller angle(36.9)
answer (19.4)

the brackets are the answers..

 

[Your-Blood]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w08_qp_1.pdf 9 iii??

Differentiate the equation of the curve. you would get dy/dx i.e equation of then gradient of curve.
put the x-co-ordinates of P and Q in the gradient equation. You would get gradient of tangent at P and Q.
let the gradient of tangent at P= m1
let the gradient of tangent at Q=m2
Use formula to find angle
tan(x) = m1 - m2 / (1 + m1m2)
find x in degrees.

 

[Hamza1996]

MINE

mine is coming -4/15

wats urs..?

MINE WS ALSO -4/15

 

[syed1995]

so me u daredevil asd mustafamotani asd and syed will be there to answer queries

hmm.. S1 is well.. dry

 

[A star]

but this doesnt mean we wil not study it

 

[Buzzkiller]

Nov 12/13/Q 6
Help!!

 

[A star]

people we dont have time so pls give link with ur question

 

[Zarif009]

General Questions:
Should we underline our final answers with a double line? Some people say we should because it helps the examiner.

And when writing down degrees in radians... do we have to use the word 'radian' ? such as 1.21 radian

 

[A star]

i dont suggest it as some times it might make your answer messy and unreadable . ( if u have messy writting )

 

[Buzzkiller]

people we dont have time so pls give link with ur question

there you go:
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_13.pdf

 

[Buzzkiller]

question six in this

 

[Zarif009]

Okay thanks and what about the radian question??

 

[A star]

well in 6i) cos(x/2 + pie/6) =1
cos-1(1)=x/2 +pie/6
0=x\2 +pie/6
pie/3 =x
6ii f(g)= 1 hence sub
cosx/2 +pie/6 =1
cosx/2 =1-pie/6
cosx= 0.4764*2
cosx=0. 9528
x=0.31 and for the negative value we have to take x= -0.31
x=+-0.31