Mathematics: Post your doubts here!

[Jiyad Ahsan]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_13.pdf

Question 7 part (ii), anyone?

oh umm you have to get umm two expressions for f^-1 and then umm you give the domains for those expressions (the ranges for the f(x) expressions will be the domain for these)

 

[A star]

u dont just lok into three or four pages atleast

 

[WayneRooney10]

oh umm you have to get umm two expressions for f^-1 and then umm you give the domains for those expressions (the ranges for the f(x) expressions will be the domain for these)

Ahan! Are we supposed to give a value to the 'p', or should we express the domain in terms of 'p'?

 

[Magenta]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_11.pdf

Q7, part 2 please!

Edit: please, please, please?

 

[Jiyad Ahsan]

Ahan! Are we supposed to give a value to the 'p', or should we express the domain in terms of 'p'?

you have to give the value of p

didnt you have to find the value of p in umm the first part?

 

[WayneRooney10]

:

you have to give the value of p

didnt you have to find the value of p in umm the first part?

Haha yup! Found it in the first part. Thanks mate!

 

[Jiyad Ahsan]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_11.pdf

Q7, part 2 please!

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

 

[Magenta]

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

Thank you so much! But wait, where did 720 come from?

 

[PANDA-]

Since we're having 3θ ... Then we have to multiply the range by 3, making it 0 < θ < 1080.
So we're going 3 cycles around the graph, now our minimum is within the first cycle, and our maximum is within the last cycle... So we start from the beginning of the second cycle, and add 150, our answer from the previous question... That's 720+150... Now you might ask why we don't start from the third cycle, as in 1080+150... Well that's because we'd be going into a fourth cycle, but we're limited to only the third cycle.... So 720+150 = 870. So maximum of 3θ = 870... So the maximum of θ is 870/3 = 290.

Magenta

 

[Jiyad Ahsan]

Magenta

yup exactly what panda said

 

[daredevil]

thanks kewwwl dude Jiyad Ahsan glad u cud be that chill I am CHILLED to the bones ryt now but definitely not chill
yeah A star has been telling all of us to CHILL too maybe i will ur advice but not at the moment no have to make up for a whole year of lazines...
thanks for the answers both of u #respect

 

[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