• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Mathematics: Post your doubts here!

Aahliya

Aahliya

Messages
57
Reaction score
3
Points
8
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 ??
 
K

kinal

Messages
11
Reaction score
3
Points
3
Keitak said:
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
Click to expand...



kinal said:
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
Click to expand...

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]
 
Snowberry

Snowberry

Messages
32
Reaction score
1
Points
18
Find the cartesian equations of the plane through (--1,1,2) and (2,0,-1) perpendicular to the plane x+y-2z=1.
 
Mobeen

Mobeen

Messages
369
Reaction score
19
Points
28
leadingguy said:
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 ??
Click to expand...
thanks . i was solving it with another method and getting different signs.. this is easy .. just havnt done maths in weeks!
 
johnsth

johnsth

Messages
62
Reaction score
8
Points
18
Keitak said:
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
Click to expand...
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...
 
K

Keitak

Messages
77
Reaction score
13
Points
18
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?
 
R

reina81

Messages
137
Reaction score
9
Points
18
Aahliya said:
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 ??
Click to expand...
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
 
K

Khoo Hing Yee

Messages
6
Reaction score
0
Points
1
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..
 
K

kinal

Messages
11
Reaction score
3
Points
3

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

Snowberry

Messages
32
Reaction score
1
Points
18
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.​
 
H

hassam

Messages
2,619
Reaction score
293
Points
93
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 ....
 
H

hassam

Messages
2,619
Reaction score
293
Points
93
kinal said:
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?
Click to expand...

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
 
K

kinal

Messages
11
Reaction score
3
Points
3
H

hassam

Messages
2,619
Reaction score
293
Points
93
having found the factorised form fro p(Z)....just replace all the Zs by Z^2 ,,......it shudnt be a problem now
 
S

SUMMERLOV35

Messages
40
Reaction score
20
Points
8
can someone please help with q10 part ii b
please
 

Attachments

  • 9709_w11_qp_33.pdf
    99 KB · Views: 9
  • 9709_w11_ms_33.pdf
    219 KB · Views: 0
user

user

Messages
505
Reaction score
739
Points
103
kinal said:
sorry made a typing mistake... October/November 2010 Variant 3.Question 10, part ii b
Click to expand...

hassam said:
having found the factorised form fro p(Z)....just replace all the Zs by Z^2 ,,......it shudnt be a problem now
Click to expand...
Aoa wr wb!

yeah just as hassam said...like in the a part..u got

z = -2..

instead of z it's z^2 now
so z^2 = -2

similarly do for the other 2 roots...when u take the square root u obviously get..2 answer ;)
 
You must log in or register to reply here.
Top