Mathematics: Post your doubts here!

[nafis shahRiyar]

thankyou!

 

[daredevil]

aww man this is a pretty weird question.. took me so much time to solve it
well umm you found that 30 and 150 degrees in the first part right, so they say that" theta" here in [n(theta) = ?] is 10 so you plug in theta=10 and the smallest angle was 30
so umm 10n = 30 since 30 was the smallest angle therefore n=3
then umm n(theta) = 720 + 150 (since they ask for the largest solution) put n =3
3(theta)= 870
theta =290

i didn't get it :/
I mean can u walk me through ur thinking process tooo.... i mean the solution of this equation means the value of Sin(theta) right??
urgghhh i'm not grasping this one.... can u plz explain :O _-_

 

[Jiyad Ahsan]

i didn't get it :/
I mean can u walk me through ur thinking process tooo.... i mean the solution of this equation means the value of Sin(theta) right??
urgghhh i'm not grasping this one.... can u plz explain :O _-_

ok well put it this way they say that the smallest possible value of theta in n(theta) is 10 and since the smallest angle was 30, n(theta)=30,
n(10)=30
n=3

and umm the largest angle would be in the range,
since 0<theta<360,
we have 0<3theta<1080
the largest angle would be in the third cycle, right? the third cycle starts from 720 degrees and above
the largest value of theta in one cycle was 150, add 2 cycles(720) to that and you have 870 which is in the third cycle so 3(theta)=870
theta = 290

 

[A star]

i didn't get it :/
I mean can u walk me through ur thinking process tooo.... i mean the solution of this equation means the value of Sin(theta) right??
urgghhh i'm not grasping this one.... can u plz explain :O _-_

well this is the thinking on which value is sin curve the minimum it is minimum at 30 degrees acording to first part hence
ntheta=30 so 10n=30 n=3 now that you know n=3. find the maximum value if the maximum range was three times it comes 870 hence 3theta =870 so theta =290

 

[Sarah22]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w10_qp_12.pdf
Question- 6(ii), 7(iii), 11(ii) please..
Thank you..

 

[daredevil]

ohh okay i get it now.. thanks Jiyad Ahsan and A star
but i sure hope this doesn't raid my paper -__-

 

[Jiyad Ahsan]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_12.pdf
Question- 6(ii), 7(iii), 11(ii) please..
Thank you..

Q6(ii)
from part one, kx^2 - kx+1,
the line is a tangent therefore, b^2-4ac=0
k=0, or k=4,
since k is non zero, k=4
put k=4 in the equation,
(4)x^2-(4)x+1=0
x=1/2
putting it in the equation y=kx,
y=(4)(1/2)
y=2


for 7(iii)
f(x)=(x-2)^2+3
f=hg
h(g(x))=(x-2)^2+3
h(g(x))=(g(x))^2+3, take g(x)=x
h(x)=x^2+3

for 11(ii)
V=pi (integral).... wait i'll snap a photo

 

[daredevil]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_12.pdf
Question- 6(ii), 7(iii), 11(ii) please..
Thank you..

for 6ii take the determinent = 0

b^2 - 4ac = 0

P.S. first rearrange the equation and make it in the form of ax^2 + bx+c

 

[daredevil]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_12.pdf
Question- 6(ii), 7(iii), 11(ii) please..
Thank you..

for 11ii
take the area under line graph.... it will give u all the area under graph shaded + unshaded
take area under curve by integrating the curve equation (gives unshaded are)
subtract the two

 

[Tabi Sheikh]

guys any one know that if there will be a deduction of mark if we directly solve quadratic equation from the calculator and just write down the answer without showing our steps

 

[A star]

11ii) any one http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_12.pdf
any one ???????????????????????????????????????????????????????????????????????????????????

 

[Jiyad Ahsan]

for 11ii
take the area under line graph.... it will give u all the area under graph shaded + unshaded
take area under curve by integrating the curve equation (gives unshaded are)
subtract the two

daredevil the question is volume not area..
sorry for sounding like a know-it-all

 

[A star]

any one answer this question

 

[Jiyad Ahsan]

11ii) any one http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_12.pdf

sorry man stupid webcam isnt taking the stupid photo, wait a sec

 

[Jiyad Ahsan]

oki i'll just type the formula here,
since V=pi ((integral)y^2) dx
Volume here is V=pi ((integral) (y2^2-y1^2))dx

y2 is the line, y1 is the curve
you square y2 then subtract square of y1,
integrate the whole thing with definite boundaries 5 and 0
multiply with pi at the end

 

[A star]

no i know the formulae i know the limits
but my answer is comming wrong idk y ?

 

[A star]

its comming 10pie/3 - 2pie

 

[Jiyad Ahsan]

no i know the formulae i know the limits
but my answer is comming wrong idk y ?

the answer is (11/12)pi units^3
the step before it i have
pi ((-10/3(1/8) - 2/3(4)) - (-10/3 - 2/3))

 

[~`Heba`~ :)]

For the first part you HAVE to sketch .. The. State the number of solutions .. Let's say 2
So then you check the other question it says upto 20pie but u have till 0-2pie .. So u consider o-2pie as one revolution and then count number of revolutions u need .. Till 20 pie it will be around 10 maybe .. And u have for one resolution 2 solutions so for 10 resolutions .. Just multiply

yeah but in (ii) It's asking about a different equation , not the one i already sketched. and it's for one mark , do i have to sketch that as well?

 

[~`Heba`~ :)]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_11.pdf
#10 (iii) I got the g-1(x)=(X+25/4)^0.5 +2 , in the ms it's different , can you please explain.