Mathematics: Post your doubts here!

[user]

Hey first of all thanks for replying and those notes have really helped me but sadly there is no mention of greatest value of argument there.

It does:

Complex No. max/min IzI and arg(z) - P3

 

[user]

Wa iyyakum,
Happens sometimes, dont get dishearted!

haha i'm not
thnx though

 

[2pac]

It does:

Complex No. max/min IzI and arg(z) - P3

oh ok thanks,I thought u were talking about the pdf file.

 

[Keitak]

Can someone please explain how to do this type of question. This is the only problem in maths for me, I could never understand how to do it even though I have tried all the past paper questions. I always get the 2nd part wrong.Please post a detailed explanation.
Thank you

 

[Chooi1993]

Next time post the link to the m.s as well!
View attachment 9589
After a long time, as t becomes very large, lets suppose t = 200
e^(t/2) = e^(200/2) = e^100 = very large value so large that adding 5 to it won't make any difference,
So (5 + e^(t/2)) will become equal to e^(t/2)
The numerator and denominator will cancel out and the answer left will be 1800.
You may verify this using a calculator!

Initial value in which question ?
If its the iteration question, take initial value as pie

Q4 P32 the second link

 

[kinal]

hi...could any1 please help me.........
O/N 2011 V3 number 10, part ii b)
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_33.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_ms_33.pdf
Thanks...​

 

[Aahliya]

6 The equation of a curve is
x ln y = 2x + 1.
(i) Show that dy/dx = − y/x^2

=> I differentiated both the sides and i got : ln y +x/y (dy/dx) = 2

but don't know how to do it further.. can any1 pls help me out ??

 

[kinal]

View attachment 9602 Can someone please explain how to do this type of question. This is the only problem in maths for me, I could never understand how to do it even though I have tried all the past paper questions. I always get the 2nd part wrong.Please post a detailed explanation.
Thank you

Can someone please explain how to do this type of question. This is the only problem in maths for me, I could never understand how to do it even though I have tried all the past paper questions. I always get the 2nd part wrong.Please post a detailed explanation.
Thank you

In the first part u found R as 10^(0.5) and theta as 71.57.

10^0.5 cos(2X-71.58)=2

then just make X the subject..[/quote]

 

[kinal]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w10_qp_33.pdf
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w10_ms_33.pdf

please need help in Q.10 ii) b)

hi, i have a problem in this question too, did u solve it?

 

[reina81]

ON11, P33, Q8 PLEASE HELP

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_33.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_ms_33.pdf

 

[Snowberry]

Find the cartesian equations of the plane through (--1,1,2) and (2,0,-1) perpendicular to the plane x+y-2z=1.

 

[Mobeen]

w is having magnitude 1 and argument 2/3pi

means we can represent w by r (cos(x + sin( x)

now r is magnitude .... 1
cosx and sinx wil be cos2/3pi and sin 2/3pi so w wil be 1(cos(2/3pi) + sin(2/3pi))

u is given as 2i means the real part is zero so u can be represented by 0 + 2i

now u.w wil be ( 0 + 2i ).(cos(2/3pi) + sin(2/3pi))

solve it to get the ans

and u/w wil be ( 0 + 2i )/(cos(2/3pi) + sin(2/3pi)) any cmplex in fractional frm when is to be changed into x+yi is always to be mutiplied and divided by conjugate base.

so u/w wil be ( 0 + 2i )/(cos(2/3pi) + sin(2/3pi)) * ((cos(2/3pi) - sin(2/3pi)) )/((cos(2/3pi) - sin(2/3pi)) )

solve it to get the ans

hope u gt what i said ??

thanks . i was solving it with another method and getting different signs.. this is easy .. just havnt done maths in weeks!

 

[johnsth]

View attachment 9602 Can someone please explain how to do this type of question. This is the only problem in maths for me, I could never understand how to do it even though I have tried all the past paper questions. I always get the 2nd part wrong.Please post a detailed explanation.
Thank you

express Rcos(X-alpha) as cos(A-B)..i.e R(cosX x cos alpha+sinX x sin alpha) then multiply.......
Rcos aplha x cosX + R sin aplha x sin X...................
and then u compare R cos alpha=1 and Rsin alpha=3...........
solve those 2 equations and u get da answer
da 2nd part is simple just express cos 2theta + 3 sin 2theta in the answer u got above and solve it...i guess u know how to solve it...

 

[Keitak]

Thanks for replying!
when you make X the subject does the answer you get have to be within the new range or the one specified in the question?

 

[reina81]

6 The equation of a curve is
x ln y = 2x + 1.
(i) Show that dy/dx = − y/x^2

=> I differentiated both the sides and i got : ln y +x/y (dy/dx) = 2

but don't know how to do it further.. can any1 pls help me out ??

i had the same problem
what i did first was to make lny the subject ---> lny=(2x+1)/x
now differentiate each side ------> lny becomes 1/y (dy/dx)
differentiate (2x+1)/x using quotient rule ------- [2x-(2x+1)]/x^2 = -1/x^2
therefore------- 1/y (dy/dx)= -1/x^2,
take 1/y to the other side becomes ------− - y/x^2

 

[Khoo Hing Yee]

Need help for O/N 09 paper 32 Q8 (ii) ... ><
Is there any mistake for C in part (i) ? Becoz i got -1/3 and i cant get 1/3 in the marking scheme..

 

[kinal]

Thanks for replying!
when you make X the subject does the answer you get have to be within the new range or the one specified in the question?[/quote]

I get (2X-71.57)= 50.77 and 2x-71.57=(360-50.77)

2X=122.3 and 2X=380.8
X=61.2 X=190.4
in 2nd quadrant, only sin is +ve
then, do 360-190.4=169.6
then, 180-169.6=10.4

final answer should be in range specified in the question
Could u please help me in this question:

O/N 2011 V3 number 10, part ii b)
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_33.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_ms_33.pdf
Thanks...

 

[Snowberry]

Can someone help me?

Find the cartesian equations of the plane through (--1,1,2) and (2,0,-1) perpendicular to the plane x+y-2z=1.​

 

[hassam]

use the two points to obtain a direrction vector in the plane....and usee the given plane equation's normal as another vector in the required plane....cross these two vectors and there u go ....

 

[hassam]

Thanks for replying!
when you make X the subject does the answer you get have to be within the new range or the one specified in the question?

I get (2X-71.57)= 50.77 and 2x-71.57=(360-50.77)

2X=122.3 and 2X=380.8
X=61.2 X=190.4
in 2nd quadrant, only sin is +ve
then, do 360-190.4=169.6
then, 180-169.6=10.4

final answer should be in range specified in the question
Could u please help me in this question:

O/N 2011 V3 number 10, part ii b)
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_33.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_ms_33.pdf
Thanks...[/quote]
well u have posted links of 2010.....m cnfused ...cn u tell again clearly which question