Mathematics: Post your doubts here!

[kaz619]

Okay one more doubt, O/N/11 p32 q6 pt. (ii)

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_32.pdf

thanks

Edit: Changed p31 to p32 , apologies!

 

[angelicsuccubus]

Okay one more doubt, O/N/11 p31 q6 pt. (ii)

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_32.pdf

thanks

you said p31 but you linked p32.. which is it?

 

[kaz619]

you said p31 but you linked p32.. which is it?

Oh god my bad! I meant P32 D:

 

[angelicsuccubus]

Oh god my bad! I meant P32 D:

if you got the first part .. then just substitute the values of R and alpha in Rcos(x-a)=2 and then your x=2theta.. so then get theta

 

[kaz619]

if you got the first part .. then just substitute the values of R and alpha in Rcos(x-a)=2 and then your x=2theta.. so then get theta

Thank you!

 

[Khoo Hing Yee]

Q4(ii).. Why only 2/3 accepted?
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_31.pdf

 

[ALM&B]

Okay one more doubt, O/N/11 p32 q6 pt. (ii)

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_32.pdf

thanks

Edit: Changed p31 to p32 , apologies!

I found Only 1 angle

(2x-71.6) = 50.76
2x = 50.76 + 71.6
2x = 122.36
2x = 122.36 - 90
2x = 32.36
--> (32.36+90) / 2 = 61.2

But I didnt know how they got the second angle :/

 

[angelicsuccubus]

Thank you!

uh I just attempted it and I can't seem to get 10.4... how did they get 10.4?

 

[user]

I found Only 1 angle

(2x-71.6) = 50.76
2x = 50.76 + 71.6
2x = 122.36
2x = 122.36 - 90
2x = 32.36
--> (32.36+90) / 2 = 61.2

But I didnt know how they got the second angle :/

uh I just attempted it and I can't seem to get 10.4... how did they get 10.4?

assalamoalaikum wr wb!

alright...now u know the other angle would be 360 - 50.76 ...right? but that is out of the range given..now remember that in the fourth quadrant angle can either be 360 - x or alternatively be written as "-x "
so we have
(2x-71.6) = 50.76 and (2x-71.6) = - 50.76

that's how u'll get the other angle

 

[GoodRobot]

I found Only 1 angle

(2x-71.6) = 50.76
2x = 50.76 + 71.6
2x = 122.36
2x = 122.36 - 90
2x = 32.36
--> (32.36+90) / 2 = 61.2

But I didnt know how they got the second angle :/

for second angle : 2x-71.57=360-50.768
2x=380.8-360
2x=20.802
x=10.4

 

[sea_princess]

Q4(ii).. Why only 2/3 accepted?
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_31.pdf

because other factors are negative ( -2, - 3/4) and there is no log of negative values

 

[RGBM211]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_31.pdf
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_ms_31.pdf

can someone do no.6 mmm confused


question no.8 how to show the arguement u is PIE/2 i knw its pie by 2 since i got u=-3i but how to show should i draw a graph or i should just say that the cordinates (0,-3i) lies on the y axis thus the angle made with the x axis is -pie/2
Anyone :?


and question no.10

well the above questions are my prev post which no1 has answerd

 

[ALM&B]

assalamoalaikum wr wb!

alright...now u know the other angle would be 360 - 50.76 ...right? but that is out of the range given..now remember that in the fourth quadrant angle can either be 360 - x or alternatively be written as "-x "
so we have
(2x-71.6) = 50.76 and (2x-71.6) = - 50.76

that's how u'll get the other angle

Thank you that totally makes sense!

 

[angelicsuccubus]

for second angle : 2x-71.57=360-50.768
2x=380.8-360
2x=20.802
x=10.4

wth? don't you just divide it by 2 after getting 380.8?

 

[ALM&B]

for second angle : 2x-71.57=360-50.768
2x=380.8-360
2x=20.802
x=10.4

umm well aren't we supposed to take into consideration the ranges they gave us?
I dont think they'll give us full on this
but at least u came to know it!
thanks!!

 

[angelicsuccubus]

assalamoalaikum wr wb!

alright...now u know the other angle would be 360 - 50.76 ...right? but that is out of the range given..now remember that in the fourth quadrant angle can either be 360 - x or alternatively be written as "-x "
so we have
(2x-71.6) = 50.76 and (2x-71.6) = - 50.76

that's how u'll get the other angle

ooh now that makes more sense!

 

[Aarjit]

Okay one more doubt, O/N/11 p32 q6 pt. (ii)

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_32.pdf

thanks

Edit: Changed p31 to p32 , apologies!

sqr(10).cos [2 θ - 71.6'] = 2 ; 0' < θ < 90'
>> 2 θ - 71.6' = cos–1 [2 /sqr(10)].....
......................= 50.8'
>> θ = (50.8' + 71.6'] / 2 = 61.2'
also,
2 θ - 71.6' ~ 360' - 50.8' [since, cosθ = cos(360-θ) = cos(-θ)]
>> 2θ = 380.8' ~ 20.8' [since, 380.8' = 360' (1 complete rotation) + 20.8']
>> θ = 10.4'

θ = 61.2', 10.4'

 

[sea_princess]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_31.pdf
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_ms_31.pdf

can someone do no.6 mmm confused


question no.8 how to show the arguement u is PIE/2 i knw its pie by 2 since i got u=-3i but how to show should i draw a graph or i should just say that the cordinates (0,-3i) lies on the y axis thus the angle made with the x axis is -pie/2
Anyone :?


and question no.10

well the above questions are my prev post which no1 has answerd

question 6 :
first: the area of the minor sector AOB =0.5 X 10^2 theta = 50 theta
the area of the major sector AOB = 0.5 X 10 ^2 X( 2 pi - theta) = 100 pi - 50 theta
the area of triangle AOB = 0.5 X 10^2 sin(theta) = 50 sin (theta)
area of the small segmant AB = area of minor sector AOB - area of triangle AOB
area of the large segmant AB = area of the major sector AOB + area of triangle AOB
now put them all together
50 theta - 50 sin( theta) = 0.25 X ( 11 pi - 50 theta + 50 sin (theta) )
50 theta - 50 sin(theta) = 25 pi - 12.5 theta +12.5 sin( theta)
62.5 theta = 25 pi + 62.5 sin (theta) ( divide by 62.5)
theta = 2/5 pi + sin (theta)
and continue normally

question 10 is answered here http://www.xtremepapers.com/community/threads/mathematics-post-your-doubts-here.9599/page-69
post 1366

 

[Chooi1993]

How to get the initial value for q4 ?​

http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s11_qp_32.pdf
Please help me !

 

[iqraa nadeem]

I'll be so much thankful, if someone helps me with these =/
here's the link for nov11/33 i need help with q10
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_33.pdf

Also with http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_31.pdf
q8 part ii and iii