Mathematics: Post your doubts here!

[danyaljamil]

some please any help with this q7 part 2
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_13.pdf

 

[A star]

i

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_12.pdf

plz solve q11 part v and see if u get the correct answer...
I get it until the sin^-1(1/3) but after that i think the value in the ms is wrong :O

this ones easy u know that its is defined for values greater then p/2 hence the two values u get one is lesser then p/2 other greater i:e 2.8 will be accepted other rejected

 

[daredevil]

i
this ones easy u know that its is defined for values greater then p/2 hence the two values u get one is lesser then p/2 other greater i:e 2.8 will be accepted other rejected

ohh right.... c another one of those imbecile attacks *silly me* thanks a bunch

 

[A star]

ohh right.... c another one of those imbecile attacks *silly me* thanks a bunch

welcome it happens thats y i sleep well before exams

 

[Jiyad Ahsan]

i
this ones easy u know that its is defined for values greater then p/2 hence the two values u get one is lesser then p/2 other greater i:e 2.8 will be accepted other rejected

bla bla guys you should stop worrying so much, im sure everyone will do fine (though i hope not too fine it'll push up the threshold) anyways
umm after you get the sin^-1(1/3) the basic angle you get is 0.340 radians and since the given range is between 1/2pi and 3/2 pi we get,
x=pi - 0.340 = 2.80 radians

therefore g^-1(3) = 2.80

 

[WayneRooney10]

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

Question 7 part (ii), anyone?

 

[A star]

bla bla guys you should stop worrying so much, im sure everyone will do fine (though i hope not too fine it'll push up the threshold) anyways
umm after you get the sin^-1(1/3) the basic angle you get is 0.340 radians and since the given range is between 1/2pi and 3/2 pi we get,
x=pi - 0.340 = 2.80 radians

therefore g^-1(3) = 2.80

ikr i was telling daredevil the same thing just answering her questions

 

[A star]

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

Question 7 part (ii), anyone?

this question has been posted atleast thrice in this thread -_-

 

[WayneRooney10]

this question has been posted atleast thrice in this thread -_-

Well the thread is 389 pages long, so I can't really go looking for it.

 

[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