Mathematics: Post your doubts here!

[nepal]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_32.pdf
Q. 8 (ii) from the paper. For some reason I'm getting ln25 + ln8. Please help!

 

[sea_princess]

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 !

if u mean for part ii , then u just assume any value, I assumed it was 2

 

[OrlandoBloom<3]

can anyone please explain about how to find arguments of a complex number :/ Please explain the method it would mean alot ! thanks !

 

[RGBM211]

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

how to do part ii of question 6

 

[2pac]

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

what should we do if Rsin(x-theeta) is the condition.How will we find the second answer in the range?

 

[Manesh]

can any one help me out in June 2011/31 Q.10

 

[redapple20]

How Wud u integrate 1/(1+X^2) ?

 

[ALM&B]

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 !

U need to find OT first!
imagine the non-right angled triangle as a right angle, and thus you can have your new "x" opposing the 90 degree angle and thus you can have your
OT as --> r/cosx
Then substitute OT in the equations!
need more explanation?

 

[sea_princess]

how to do part ii of question 6

u write the " 2/5 pi + sin (theta) " on your calculator , replacing theta with the value 2.1 and press = , write the answer correct to 3 decimal places , then replace the 2.1 value on your calculator , using the " ans" button , and get theta 3 , keep pressing = and writing down the values till they give u the same value correct to 2 decimal places , in this question it will be 2.11
now find AB using the cosin rule
AB^2 = OB^2 + OA ^2 - ( 2X OA X OB X cos (2.11) )
you'll get AB = 17.4

 

[ALM&B]

How Wud u integrate 1/(1+X^2) ?

Integrating this gives you ---> ln (1+x^2)/2

 

[redapple20]

No it doesnt differential of x square is 2x not 2. There's a formula for it.

 

[ALM&B]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_32.pdf
Q. 8 (ii) from the paper. For some reason I'm getting ln25 + ln8. Please help!

you must have this integrated equation

(-3 ln (2-x) + 2 ln (4+x^2))
got this?

 

[ALM&B]

No it doesnt differential of x square is 2x not 2. There's a formula for it.

oh ya right!!

 

[nepal]

you must have this integrated equation

(-3 ln (2-x) + 2 ln (4+x^2))
got this?

Yes exactly this. So the equation is -3 ln (1) +2 ln (4+1) + 3 ln(2-0) - 0 = ln25 + ln8 . What am I doing wrong?

 

[smzimran]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_31.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_ms_31.pdf

question 7 part ii)
can any one give a rough idea of the sketch????

 

[miss irfan]

View attachment 9676

i was just thinking of messaging you! could you please solve that integration question now? November 2011 paper 33 Q10

 

[redapple20]

Got it for the integral of 1/(a square+x square) the integral Wud b 1/a tan inverse of (x/a) +c

 

[smzimran]

i was just thinking of messaging you! could you please solve that integration question now? November 2011 paper 33 Q10

Still very difficult to show the solution here !
I have done the question, it is tricky but very easy.

 

[miss irfan]

Still very difficult to show the solution here !
I have done the question, it is tricky but very easy.

is there any way at all that you could help me?

 

[Manesh]

can any one help june 2011/31 Q.10