Mathematics: Post your doubts here!

[Maz]

Search the thread before posting a question .. just put 9702_s12_qp_12 in the search bar and you should get posts related to that year...

https://www.xtremepapers.com/commun...st-your-doubts-here.9599/page-384#post-513067

thnx! But i got it upto then. But, wasn't able to solve simultaneously.

 

[Maz]

I've got another doubt in the same paper...
Q10 (iii)...
I got the critical values= 0 and 64/9
but then what would be the inequality... i reckoned that since "a" in negative in -kx^2 + 3kx -16 i.e. -k, the graph would have an upward parabola, but it turned out to be wrong.

 

[Dug]

Wouldn't the graph be something like which starts at 3 and the curve cutting x axis at 2Pi instead of pi?

They asked only for the domain 0 < x < π

 

[Maz]

Wouldn't the graph be something like which starts at 3 and the curve cutting x axis at 2Pi instead of pi?

Actually the curve cannot cut the x- axis at pi, tan is undefined for pi. Nor 2pi since they haven't asked for it.

 

[Dug]

Dug..if u rnt busy, could u help me out with ths pls??
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s08_qp_3.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s08_ms_3.pdf

ques 3 n 4 ii)

pls??

Q3)
i) Perimeter of sector = ½(Perimeter of rectangle)
2r + rx = 4a
Sinx = a/r
a = rSinx
2r + rx = 4(rSinx)
2 + x = 4Sinx
Sinx = ¼(2 + x)

ii) x1 = 0.8
x2 = 0.7754
x3 = 0.7668
x4 = 0.7638
x5 = 0.7628
x6 = 0.7625

x = 0.76 (2 dp)

Q4)
i)

ii) tan²x + (6√3)tanx - 5 = 0
Let y = tanx
y² + (6√3)y - 5 = 0

Now you have to use quadratic formula. I am sure you can do that!

 

[Nurul Huda sabid]

i have problem with question 7(ii).....papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w12_qp_12.pdf

 

[prekshya]

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

 

[A star]

how many of u are giving s1 this year

 

[prekshya]

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

 

[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