Mathematics: Post your doubts here!

[A star]

daredevil syed1995 @mostafamotani asd Rutzaba part (ii) ASAP

 

[syed1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w13_qp_13.pdf

Q7b please somone explain briefly .. thanks

let me solve it in terms of degrees first and you'll understand better. Sin is Positive is 1st and 2nd Quadrant while negative in 3rd and 4th quadrant then again positive in the 5th and 6th Quadrant then negative in the 7th and 8th Quadrants.

1st Quad = Alpha 2nd = 180-Alpha 3rd = 180+Alpha 4th = 360-Alpha 5th = (1stQuad+360) 6th = (2ndQuad+360)

Sin(2x+60)=1/2
sin^-1(1/2) = 30 Degrees //1st quadrant and (180-30) //2nd quadrant and 360+30 //(5th Quadrant) 360+150 //(6th Quadrant)
(2x+60)= 30 or 2x+60= 150 or 2x+60=390 or 2x+60=510 Degrees
2x=-30 or 2x=90 Degrees or 2x=330 Degrees or 2x=450 Degrees
x=-15 Degrees or x= 45Degrees or x=165 Degrees or x=225 Degrees.
since first and last is outside the range.. their are only two valid answers x=45 Degrees or 0.25Pi and x= 165 Degrees or 11Pi/12

PS: the reason we took 5th and 6th Quadrant was because it said 2x in the question.. meaning the angles we get after sine inverse will be divided by 2.
so if the range was 0<x<180 for x
it will be 0*2<2x<2(180) for 2x which is 0<2x<360

Our question was 2x+60 .. so add 60 to the above range.
for 2x+60 it becomes 0+60<2x+60<360+60
for 2x+60 the range will be 60<2x+60<420 so all the angles in between that range will result in a value of x.

 

[A star]

I'm re-posting this. Someone please help me

you just have to consider the same formulae but take it as Rcos(theeta/2 -alpha)

 

[A star]

let me solve it in terms of degrees first and you'll understand better. Sin is Positive is 1st and 2nd Quadrant while negative in 3rd and 4th quadrant then again positive in the 5th and 6th Quadrant then negative in the 7th and 8th Quadrants.

1st Quad = Alpha 2nd = 180-Alpha 3rd = 180+Alpha 4th = 360-Alpha 5th = (1stQuad+360) 6th = (2ndQuad+360)

Sin(2x+60)=1/2
sin^-1(1/2) = 30 Degrees //1st quadrant and (180-30) //2nd quadrant and 360+30 //(5th Quadrant) 360+150 //(6th Quadrant)
(2x+60)= 30 or 2x+60= 150 or 2x+60=390 or 2x+60=510 Degrees
2x=-30 or 2x=90 Degrees or 2x=330 Degrees or 2x=450 Degrees
x=-15 Degrees or x= 45Degrees or x=165 Degrees or x=225 Degrees.
since first and last is outside the range.. their are only two valid answers x=45 Degrees or 0.25Pi and x= 165 Degrees or 11Pi/12

mine pls

 

[GCE As and a level]

let me solve it in terms of degrees first and you'll understand better. Sin is Positive is 1st and 2nd Quadrant while negative in 3rd and 4th quadrant then again positive in the 5th and 6th Quadrant then negative in the 7th and 8th Quadrants.

1st Quad = Alpha 2nd = 180-Alpha 3rd = 180+Alpha 4th = 360-Alpha 5th = (1stQuad+360) 6th = (2ndQuad+360)

Sin(2x+60)=1/2
sin^-1(1/2) = 30 Degrees //1st quadrant and (180-30) //2nd quadrant and 360+30 //(5th Quadrant) 360+150 //(6th Quadrant)
(2x+60)= 30 or 2x+60= 150 or 2x+60=390 or 2x+60=510 Degrees
2x=-30 or 2x=90 Degrees or 2x=330 Degrees or 2x=450 Degrees
x=-15 Degrees or x= 45Degrees or x=165 Degrees or x=225 Degrees.
since first and last is outside the range.. their are only two valid answers x=45 Degrees or 0.25Pi and x= 165 Degrees or 11Pi/12

mine plz
hey guys
how are u all ?
solve as much as u can, even if only one question !!
i need a help in
1)
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w12_qp_12.pdf

Q4 all
Q6 ii)
Q7 ii)
Q8 all
Q10 ii)
Q11 all


2)
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s12_qp_12.pdf

Q7 a)

3)
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s09_qp_1.pdf

Q2 all
Q10 iii)


4)
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_S08_qp_01.pdf

Q5 all
Q6 ii) Domain __ explain it plz


5)
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w08_qp_1.pdf

Q5 all
Q6 all
Q9 ii) and iii)
Q10 ii) and iii) and iv)


6)
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s07_qp_1.pdf

Q4 all
Q5 all
Q8 ii)
Q10 ii)
Q11 ii)

7)
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w07_qp_1.pdf

Q11 v)

8)
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w06_qp_1.pdf

Q10 v)


9)
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s12_qp_13.pdf

Q1 all
Q4 all
Q11 iii)

I really need your help
waiting your help
thanks in advance
wish u all good luck

 

[syed1995]

daredevil syed1995 @mostafamotani asd Rutzaba part (ii) ASAP

yaar vectors bilkul nai aatay btw which year is this question from?

 

[A star]

2013 june

 

[Rutzaba]

2013 june

Gimme sum tym

 

[Serenia]

Hello!

I have a doubt in question 8(ii) on this P1 exam.
I could find the value of r when A has a stationary value, but I can't show that there are no straight sections in the track. I looked at the mark scheme but I don't understand that part. The second part (determine the nature of the stationary point) I can do.

Attached I am sending the paper and the mark scheme.

Thanks in advance.

 

[Faaiz Haque]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_11.pdf

Question 5, 9 full please, thanks!

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_12.pdf

Question 9, I keep getting 9.333, answer is 10.666, can anyone do and show working?

 

[A star]

Gimme sum tym

a responce would be VERY nice please

 

[Namehere]

I´ll post this again... please someone help!

I´ve got a question and would appreciate some help.

For Q9(i), if you are trying to find the normal to the plane using vector product, why if you do AC x AB or CA x BA you get a wrong answer, but if you use any other two vectors, you get the correct normal to the plane? The only difference for the normal to the plane I get using AC x AB or CA x BA is a minus sign for the k coordinate, instead of a positive one.

I would also like if someone could clarify me how to use vector product correctly, ie. what requisites should the vectors you are going to use for vector product have?

Also, how do you do Q9(ii).

Thank you in advance.

 

[daredevil]

a responce would be VERY nice please

Oh hey... i just saw ur post.... wait... i will work on it and see if i can work something out...

 

[ZaqZainab]

2013 june

what paper and what variant

 

[Namehere]

Can someone answer my above post please? ^^

 

[daredevil]

what paper and what variant

it's Paper 3 FOR SURE *shudders*

not sure about the vairant though... i think it's 31 but i'm not sure... just check

 

[ZaqZainab]

it's Paper 3 FOR SURE *shudders*

not sure about the vairant though... i think it's 31 but i'm not sure... just check

*doesn't take paper3*

 

[daredevil]

I´ll post this again... please someone help!

I´ve got a question and would appreciate some help.

For Q9(i), if you are trying to find the normal to the plane using vector product, why if you do AC x AB or CA x BA you get a wrong answer, but if you use any other two vectors, you get the correct normal to the plane? The only difference for the normal to the plane I get using AC x AB or CA x BA is a minus sign for the k coordinate, instead of a positive one.

I would also like if someone could clarify me how to use vector product correctly, ie. what requisites should the vectors you are going to use for vector product have?

Also, how do you do Q9(ii).

Thank you in advance.

Use AB x BC and see if u get the correct answr..
i'm getting i+2j+k for that... [talking about part i ]

i think u have to use the vectors where the second alphabet of the first vect and the first alphabet of the second vector are common:
AB x BC

 

[daredevil]

*doesn't take paper3*

count your blessings my friend count your blessings :')

 

[Namehere]

Use AB x BC and see if u get the correct answr..
i'm getting i+2j+k for that... [talking about part i ]

i think u have to use the vectors where the second alphabet of the first vect and the first alphabet of the second vector are common:
AB x BC

Well, I also tried CA x CB and get the correct answer, I just don´t get why using AC x AB or CA x BA I get the wrong answer.