Mathematics: Post your doubts here!

[Banglarbagh]

Hello can anyone solve this maths for me.... June 12/p32 Question no. 2 (ii) .... Please help me......

 

[Maz]

Hello can anyone solve this maths for me.... June 12/p32 Question no. 2 (ii) .... Please help me......

Take f(x) = ø - tan^-1(3ø)

Search for a sign change.
f(0) = 0
f(pi/4) = -0.38
f(pi/2) = 0.21

Therefore, the root is between pi/4 and pi/2...
take x1 = pi/4...
I guess now it's easy...

 

[Banglarbagh]

Take f(x) = ø - tan^-1(3ø)

Search for a sign change.
f(0) = 0
f(pi/4) = -0.38
f(pi/2) = 0.21

Therefore, the root is between pi/4 and pi/2...
take x1 = pi/4...
I guess now it's easy...

Brother i still did not get u..... how do i do it.... at first i take f(O) = O .... den what ? :/

 

[kaushar]

am having a prob with paper 13 may/june 2013 i cant understand where it got ½.3²π have it in the marking scheme plzz help number 2

 

[$~SauD~$]

Brother i still did not get u..... how do i do it.... at first i take f(O) = O .... den what ? :/

dude dude.. It's simple..

Equation is θn+1 = tan^−1 (3θn) between 0< θ <.5π
now take any value between that range... let's take .25π

θ1 = .25π
θ2 = tan^−1 (3(.25π)) = 1.1694
θ3 = tan^-1 (3(1.1694)) = 1.2931 << is the 'ans' right?
θ4 = tan^-1 (3(ans)) = 1.3185
θ5 = tan^-1 (3(ans)) = 1.3231
θ6 = tan^-1 (3(ans)) = 1.3240 do it until we get all 4 decimal point same
θ7 = tan^-1 (3(ans)) = 1.3241
θ8 = tan^-1 (3(ans)) = 1.3241 <<<< this and above have same.. so the answer should be written in 2 decimal as said by question

so the final answer is 1.32

 

[Banglarbagh]

dude dude.. It's simple..

Equation is θn+1 = tan^−1 (3θn) between 0< θ <.5π
now take any value between that range... let's take .25π

θ1 = .25π
θ2 = tan^−1 (3(.25π)) = 1.1694
θ3 = tan^-1 (3(1.1694)) = 1.2931 << is the 'ans' right?
θ4 = tan^-1 (3(ans)) = 1.3185
θ5 = tan^-1 (3(ans)) = 1.3231
θ6 = tan^-1 (3(ans)) = 1.3240 do it until we get all 4 decimal point same
θ7 = tan^-1 (3(ans)) = 1.3241
θ8 = tan^-1 (3(ans)) = 1.3241 <<<< this and above have same.. so the answer should be written in 2 decimal as said by question

so the final answer is 1.32

Thanks man i got it fully!

 

[Banglarbagh]

Can some one help me with june 12 p32/ No. 4..... please.....

 

[kaushar]

its it 2013

 

[Banglarbagh]

its it 2013

Nope 12

 

[kaushar]

i did the 2013 not 2012 sorry

 

[$~SauD~$]

am having a prob with paper 13 may/june 2013 i cant understand where it got ½.3²π have it in the marking scheme plzz help number 2

Link and question?

Can some one help me with june 12 p32/ No. 4..... please.....

link please?

 

[kaushar]

http://maxpapers.com/wp-content/uploads/2012/11/9709_s13_qp_13.pdf

 

[kaushar]

question number 2

 

[Kumkum]

Can some one help me with june 12 p32/ No. 4..... please.....

cosec(2x) = secx + cotx
1/sin(2x) = 1/cosx + cosx/sinx [using the identity; sin(2x) = 2sinxcosx on the right hand side and taking the common denominator on the left hand side]
===> 1/2sinxcosx = (sinx + (cos^2)x)/cosxsinx [sinxcosx cancels out]
1 = 2sinx + 2(cos^2)x [using identity (sin^2)x + (cos^2)x = 1 on left hand side]
1 = 2sinx + 2(1 - sin^2 x)
1 = 2sinx + 2 - 2sin^2 x
2sin^2 x - 2sinx - 1 =0 (i'm using the quadratic formula since there are no factors)

sinx = -(-2) +/- sq.root [(-2)^2 - ( 4*2*-1)] / 2*2
sinx = [2+/-sq.rt 12] / 4

so, sinx = [2+ sq.rt 12] / 4
sinx = 1.366 ( this is out)

or
sinx = [2 - sq.rt 12] / 4
sinx = -0.366 (i often say let sin(a) = 0.366, so that i can work with positive angles)
sin(a) = 0.366
y = 21.47 <----use this angle to find the other angles where 'sin' is negative
u'll get x as 201.5 and 338.5

hope u got it

 

[$~SauD~$]

http://maxpapers.com/wp-content/uploads/2012/11/9709_s13_qp_13.pdf

question number 2

(i) Area of circle = (pie) x r^2
Area of shaded = 1/2 x 9^2 x Q - 1/2 x 3^2 x Q

area of circle = area of shaded region

(pie) x 3^2 = .5 x81 x Q - .5 x 9 x Q

now do the equation and get Q. and Q is Theeta
btw, Paper 1? :s

 

[Banglarbagh]

cosec(2x) = secx + cotx
1/sin(2x) = 1/cosx + cosx/sinx [using the identity; sin(2x) = 2sinxcosx on the right hand side and taking the common denominator on the left hand side]
===> 1/2sinxcosx = (sinx + (cos^2)x)/cosxsinx [sinxcosx cancels out]
1 = 2sinx + 2(cos^2)x [using identity (sin^2)x + (cos^2)x = 1 on left hand side]
1 = 2sinx + 2(1 - sin^2 x)
1 = 2sinx + 2 - 2sin^2 x
2sin^2 x - 2sinx - 1 =0 (i'm using the quadratic formula since there are no factors)

sinx = -(-2) +/- sq.root [(-2)^2 - ( 4*2*-1)] / 2*2
sinx = [2+/-sq.rt 12] / 4

so, sinx = [2+ sq.rt 12] / 4
sinx = 1.366 ( this is out)

or
sinx = [2 - sq.rt 12] / 4
sinx = -0.366 (i often say let sin(a) = 0.366, so that i can work with positive angles)
sin(a) = 0.366
y = 21.47 <----use this angle to find the other angles where 'sin' is negative
u'll get x as 201.5 and 338.5

hope u got it

Thanks Sis got it...i just goofed up in the part where it shuld have been sinx + (cos^2)x ....thats why i culdnt solve it...newys thanks!

 

[kaushar]

(i) Area of circle = (pie) x r^2
Area of shaded = 1/2 x 9^2 x Q - 1/2 x 3^2 x Q

area of circle = area of shaded region

(pie) x 3^2 = .5 x81 x Q - .5 x 9 x Q

now do the equation and get Q. and Q is Theeta
btw, Paper 1? :s

ohh thnks i just mess it up just because of stress and exams

 

[Maz]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_32.pdf

Question 10. part 3.

Hey, did you get the answer? I got the first possibility of P...
Tell me if you have found the second one...

 

[$~SauD~$]

Hey, did you get the answer? I got the first possibility of P...
Tell me if you have found the second one...

here's the solution to 10 part 3 https://www.xtremepapers.com/community/threads/a-maths-doubt.24710/

 

[Maz]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w10_qp_33.pdf

Q) 3 ii) What's the area to be shaded??? Please