IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: mousa on October 01, 2010, 02:24:32 pm
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Hello ppl,
I need help with those Questions.Please show full working out and reasoning.
June 2009.....Q3,8,9,10
November 2009 ,32,...Q2,6,10
November 2008, Q7,8,9
Thanx in advance, I know, they are alot of questions, but I neeed hellp. :-[
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Hello ppl,
I need help with those Questions.Please show full working out and reasoning.
Thanx in advance, I know, they are alot of questions, but I neeed hellp. :-[
November 2008 Paper 3
Q-7 i)
We have,
2x-y-3z=7 =>r.(2 -1 -3) therefore normal, n1=(2 -1 -3)
x+2y+2z=>r.(1 2 2)=0 therefore normal,n2=(1 2 2)
The angle between two planes is the angle between their respective normals
therefore n1.n2=|n1|.|n2|cosQ
(2 -1 -3).(1 2 2)=(/(2)^2+(-1)^2+(-3)^2)(/(1)^2+(2)^2+(2)^2)cosQ
2-2-6=(/4+1+9)(/1+4+4)cosQ
-6=3/14 cosQ
Q=122.3 degrees
acute angle=180-122.3=57.7 degrees.
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Part ii)
2x-y-3z=7----(i)
x+2y+2z=0----(ii)
elminating two variables one by one from the equation of planes,
eq.(i)
eq (ii) * 2
we get z= -5y-7
------- ---------- (iii)
7
eq (i) *2
eq (ii)
we get z=5x-14
-------- --------(iv)
4
from equations (iii) and (iv), we have
z=-5y-7 5x-14
-----= ------
7 4
z-0 -5(y+7/5) 5(x-14/5)
----=-----------= --------
1 7 4
z-0 y+7/5 x-14/5
----=------=---------
1 -7/5 4/5
Therefore the line of intersection of the two planes is
r=(14/5 i +7/5 i +0k)+ dont know how to draw the sign (4/5 i-7/5 j +k)
1
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PS:This is my teacher's work.
I have just started studying this chapter.
More will come tomorrow
Sorry i have exams
and the working was too much.
Cheers
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Hello ppl,
I need help with those Questions.Please show full working out and reasoning.
June 2009.....Q3,8,9,10
November 2009 ,32,...Q2,6,10
November 2008, Q7,8,9
Thanx in advance, I know, they are alot of questions, but I neeed hellp. :-[
Jun 09
3.
(i) Let pheta be x
Prove cosec 2x +cot 2x = cot x
cosec 2x + cot 2x = 1/(sin 2x) + (cos 2x)/(sin 2x)
Take sin 2x as common denominator to get (1 + cos 2x)/(sin 2x)
Use double angle formula sin 2x = 2sinx cosx and 1 + cos 2x = 2cos2x
You'll get (2cos2x)/(2sinx cosx)
Eliminate cosx from the numerator and denominator to obtain cosx / sinx = cot x
(ii) cosec 2x + cot 2x = 2
This also implies that cot x = 2 ----> 1/tanx = 2
therefore tanx = 1/2
Key angle x = tan-1(1/2) = 26.6o
tan is positive in first and third quadrant.
Hence x = pheta = 26.6o and 206.6o
8.
Let 100/x2(10-x) = A/x + B/x2 + C/(10-x)
Take x2(10-x) as common denominator on the right hand side :
A(x2(10-x)) + B((10-x) + C(x2) = 100
Now to find C : take x=10 in the equation above -----> C=1
Now to find B : take x=0 ----> B 10
Now to find A : take x=1 ----> 9A + 9B + C = 100
Replace values of B and C calculated above to obtain A=1
Hence 100/x2(10-x) = 1/x + 10/x2 + 1/(10-x)
(ii) Give dx/dt = 1/100*x2(10-x)
Send the terms in x to the left side and you'll obtain :
100/x2(10-x) dx = dt
Use the answer obtained in part (i) to replace 100/x2(10-x)
Integrate on both sides to get :
ln x - 10/x + ln(10-x) = t + c
Given when x=1, t=0 ----> replace in the equation to get the value of c
c = -10 - ln 9
Therefore ln x - 10/x - ln(10-x) = t -10 - ln 9
Make t subject of formula and use logarithm rules to obtain answer as :
t = ln (9x/(10-x)) - 10/x + 10
9
Given L lies in the plane ----> Vector 4i + 2j - k should fit in the equation of the plane.
Am sorry I don't know how to type it in column vectors. It's damn easier to explain in column vectors :(
(4i + 2j - k).(2i + bj + ck) = 1
From this dot product you'll obtain equation : 2b - c = -7
When L lies in the plane ---> It also implies that the direction vector of line is perpendicular to the normal vector of the plane. perpendicular implies dot product = 0.
(2i - j - 2k).(2 + bj + ck) = 0
From this you'll obtain : b + 2c = 4
Now you need to solve the two equations simultaneously to obtain b = -2 and c = 3
(ii) For PQ to be perpendicular to L ----> PQ.(2i - j - 2k) = 0
Take vector equation of PQ as (4+2t)i - tj - (5+2t)k
The dot product will indicate that t = -2
When t=-2 ----> point where L meets PQ perpendicularly is (4j + 3k)
Perpendicular distance : Square root of ( 02 +(4-2)2 + (3-4)2)
Answer is square root of 5.
10
(i) Since M is a stationary point ---> dy/dx = 0 at M.
Use the product rule dy/dx = u.dv/dx + v.du/dx
Take u = x2 ---> du/dx = 2x
Hence v = (1-x2)-1/2 ---> dv/dx = -x(1-x2)-1/2
therefore dy/dx = x2(-x(1-x2)-1/2) + ((1-x2)-1/2)(2x)
Solve dy/dx = 0 but x>0 (from range)
You'll b getting x as the positive square root of 2/3.
(ii) I'll take pheta as A
x = sin A ----> dx/dA = cos A
Hence dx = cos A dA
When x=0 ---> A = 0 and pie/2
Area = Integration of x2(1-x2)1/2) dx
Substituting x=sinA and dx = cos A dA, you'll be getting :
Area = Integration of sin2A (1-sin2A)1/2 cos A dA with limits 0 and pie/2
Replacing 1- sin2A by cos2A
Then sin2A cos2A = (sin22A)/4
Area is shown to be 1/4 x integration of sin22A dA with limits 0 and pie/2.
(iii) Replace sin22A by (1-cos2A)/2 using double angle formula.
Then integrate normally
A = -1/8 * integration of (cos 2A -1) dA with limits 0 and pie/2
Solution will be pie/16
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PS:This is my teacher's work.
I have just started studying this chapter.
More will come tomorrow
Sorry i have exams
and the working was too much.
Cheers
Thanks Alott MAAN!!!
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Jun 09
3.
(i) Let pheta be x
Prove cosec 2x +cot 2x = cot x
cosec 2x + cot 2x = 1/(sin 2x) + (cos 2x)/(sin 2x)
Take sin 2x as common denominator to get (1 + cos 2x)/(sin 2x)
Use double angle formula sin 2x = 2sinx cosx and 1 + cos 2x = 2cos2x
You'll get (2cos2x)/(2sinx cosx)
Eliminate cosx from the numerator and denominator to obtain cosx / sinx = cot x
(ii) cosec 2x + cot 2x = 2
This also implies that cot x = 2 ----> 1/tanx = 2
therefore tanx = 1/2
Key angle x = tan-1(1/2) = 26.6o
tan is positive in first and third quadrant.
Hence x = pheta = 26.6o and 206.6o
8.
Let 100/x2(10-x) = A/x + B/x2 + C/(10-x)
Take x2(10-x) as common denominator on the right hand side :
A(x2(10-x)) + B((10-x) + C(x2) = 100
Now to find C : take x=10 in the equation above -----> C=1
Now to find B : take x=0 ----> B 10
Now to find A : take x=1 ----> 9A + 9B + C = 100
Replace values of B and C calculated above to obtain A=1
Hence 100/x2(10-x) = 1/x + 10/x2 + 1/(10-x)
(ii) Give dx/dt = 1/100*x2(10-x)
Send the terms in x to the left side and you'll obtain :
100/x2(10-x) dx = dt
Use the answer obtained in part (i) to replace 100/x2(10-x)
Integrate on both sides to get :
ln x - 10/x + ln(10-x) = t + c
Given when x=1, t=0 ----> replace in the equation to get the value of c
c = -10 - ln 9
Therefore ln x - 10/x - ln(10-x) = t -10 - ln 9
Make t subject of formula and use logarithm rules to obtain answer as :
t = ln (9x/(10-x)) - 10/x + 10
Thnaks amillion!! but can you plz help me in the other questions???! ???
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Thnaks amillion!! but can you plz help me in the other questions???! ???
the questions are quite long. Thus i'll take some time to answer all of them. I've already done all of them but i don't have a scanner. So i'll have to post them 1 by 1.
If you could specify the parts that you don't understand we can work it out faster :)
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the questions are quite long. Thus i'll take some time to answer all of them. I've already done all of them but i don't have a scanner. So i'll have to post them 1 by 1.
If you could specify the parts that you don't understand we can work it out faster :)
Sorry man, but I dont know all of the parts, they are quite annoying.I have been studying the subject alone this summer and my exams are this NOV session!!
Please do you have any good resources to revise from or tips??
Thanks
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Sorry man, but I dont know all of the parts, they are quite annoying.I have been studying the subject alone this summer and my exams are this NOV session!!
Please do you have any good resources to revise from or tips??
Thanks
Well the CIE textbook can be quite helpful. But alone it is quite difficult to understand all that. A teacher would have been really helpful!
Anyway i'll be modifying my posts now and then until I answer all the questions.
However if you do not understand something.......do ask me and i'll try to elaborate more....ok??
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Well the CIE textbook can be quite helpful. But alone it is quite difficult to understand all that. A teacher would have been really helpful!
Anyway i'll be modifying my posts now and the until I answer all the questions.
However if you do not understand something.......do ask me and i'll try to elaborate more....ok??
thanks alot, I will make sure to do that :P
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Part ii)
2x-y-3z=7----(i)
x+2y+2z=0----(ii)
elminating two variables one by one from the equation of planes,
eq.(i)
eq (ii) * 2
we get z= -5y-7
------- ---------- (iii)
7
eq (i) *2
eq (ii)
we get z=5x-14
-------- --------(iv)
4
from equations (iii) and (iv), we have
z=-5y-7 5x-14
-----= ------
7 4
z-0 -5(y+7/5) 5(x-14/5)
----=-----------= --------
1 7 4
z-0 y+7/5 x-14/5
----=------=---------
1 -7/5 4/5
Therefore the line of intersection of the two planes is
r=(14/5 i +7/5 i +0k)+ dont know how to draw the sign (4/5 i-7/5 j +k)
1
I didnt understand this part,can you help me with it?? That is ,How to solve these sort of questions in general??
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I didnt understand this part,can you help me with it?? That is ,How to solve these sort of questions in general??
Ok......i'll be finishing Jun 09 and i'll try to explain that :)
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@deadly_king + rep for all the support you've been giving ;)
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@deadly_king + rep for all the support you've been giving ;)
That's alright br0 :)
We all form part of the huge SF family. So we gotta support each other ;)
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I am back now!
So you need to use the two equations. By elimination method, you eliminate one of the variables.
Eqn 1 : 2x - y - 3z = 7
Eqn 2 : x + 2y + 2z = 0
Step 1
Multiply Eqn by 2 and you'll be getting 4x - 2y -6z =14 while Eqn 2 is still x +2y +2z = 0.
Add both equations and you'll note that -2y cancels 2y. You'll thus be getting 5x - 4z = 14
Make z subject of formula to get z = (5x - 14)/4
Now you need to find z in terms of y. So you should eliminate the values of x.
Step 2
Multiply Eqn 2 by -2 and you'll be getting -2x - 4y - 4z = 0 while Eqn 1 remains 2x - y -3z = 7
Add both equations and this time 2x will cancel -2x. Hence you'll obtain -5y - 7z = 7
Again make z subject of formula to get z = (-5y - 7)/7
Now you equate all the equations of z you obtained just like Requiem previously did.
Then you should make the coefficients of both x and y = 1
I guess by now, you'll be able to understand the workings of Requiem :)
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Nov 09 No 2, 6, 10
2.
3(x+2) = 3x + 32
Divide everything by 3x and you'll get : ( dividing implies powers are subtracted)
32 -1 = 3(2-x)
Therefore 3(2-x) = 9 - 1
Apply ln on both sides
ln 3(2-x) = ln 8
(2-x)ln 3 = ln 8
Hence x = 2 - (ln 8/ln3) = 0.107
Sorry but I need to go to tuition now. Will complete the other numbers once am back :)
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Nov 09 No 2, 6
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DK!! ...that was really helpful ;D
+rep
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God bless you all. 8) thanks for all reponses!!
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DK!! ...that was really helpful ;D
+rep
That's alright br0 :)
Thanks for completing the other numbers ;)
You're welcome mousa :)
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Jun 09
3.
(i) Let pheta be x
Prove cosec 2x +cot 2x = cot x
cosec 2x + cot 2x = 1/(sin 2x) + (cos 2x)/(sin 2x)
Take sin 2x as common denominator to get (1 + cos 2x)/(sin 2x)
Use double angle formula sin 2x = 2sinx cosx and 1 + cos 2x = 2cos2x
You'll get (2cos2x)/(2sinx cosx)
Eliminate cosx from the numerator and denominator to obtain cosx / sinx = cot x
(ii) cosec 2x + cot 2x = 2
This also implies that cot x = 2 ----> 1/tanx = 2
therefore tanx = 1/2
Key angle x = tan-1(1/2) = 26.6o
tan is positive in first and third quadrant.
Hence x = pheta = 26.6o and 206.6o
8.
Let 100/x2(10-x) = A/x + B/x2 + C/(10-x)
Take x2(10-x) as common denominator on the right hand side :
A(x2(10-x)) + B((10-x) + C(x2) = 100
Now to find C : take x=10 in the equation above -----> C=1
Now to find B : take x=0 ----> B 10
Now to find A : take x=1 ----> 9A + 9B + C = 100
Replace values of B and C calculated above to obtain A=1
Hence 100/x2(10-x) = 1/x + 10/x2 + 1/(10-x)
(ii) Give dx/dt = 1/100*x2(10-x)
Send the terms in x to the left side and you'll obtain :
100/x2(10-x) dx = dt
Use the answer obtained in part (i) to replace 100/x2(10-x)
Integrate on both sides to get :
ln x - 10/x + ln(10-x) = t + c
Given when x=1, t=0 ----> replace in the equation to get the value of c
c = -10 - ln 9
Therefore ln x - 10/x - ln(10-x) = t -10 - ln 9
Make t subject of formula and use logarithm rules to obtain answer as :
t = ln (9x/(10-x)) - 10/x + 10
9
Given L lies in the plane ----> Vector 4i + 2j - k should fit in the equation of the plane.
Am sorry I don't know how to type it in column vectors. It's damn easier to explain in column vectors :(
(4i + 2j - k).(2i + bj + ck) = 1
From this dot product you'll obtain equation : 2b - c = -7
When L lies in the plane ---> It also implies that the direction vector of line is perpendicular to the normal vector of the plane. perpendicular implies dot product = 0.
(2i - j - 2k).(2 + bj + ck) = 0
From this you'll obtain : b + 2c = 4
Now you need to solve the two equations simultaneously to obtain b = -2 and c = 3
(ii) For PQ to be perpendicular to L ----> PQ.(2i - j - 2k) = 0
Take vector equation of PQ as (4+2t)i - tj - (5+2t)k
The dot product will indicate that t = -2
When t=-2 ----> point where L meets PQ perpendicularly is (4j + 3k)
Perpendicular distance : Square root of ( 02 +(4-2)2 + (3-4)2)
Answer is square root of 5.
10
(i) Since M is a stationary point ---> dy/dx = 0 at M.
Use the product rule dy/dx = u.dv/dx + v.du/dx
Take u = x2 ---> du/dx = 2x
Hence v = (1-x2)-1/2 ---> dv/dx = -x(1-x2)-1/2
therefore dy/dx = x2(-x(1-x2)-1/2) + ((1-x2)-1/2)(2x)
Solve dy/dx = 0 but x>0 (from range)
You'll b getting x as the positive square root of 2/3.
(ii) I'll take pheta as A
x = sin A ----> dx/dA = cos A
Hence dx = cos A dA
When x=0 ---> A = 0 and pie/2
Area = Integration of x2(1-x2)1/2) dx
Substituting x=sinA and dx = cos A dA, you'll be getting :
Area = Integration of sin2A (1-sin2A)1/2 cos A dA with limits 0 and pie/2
Replacing 1- sin2A by cos2A
Then sin2A cos2A = (sin22A)/4
Area is shown to be 1/4 x integration of sin22A dA with limits 0 and pie/2.
(iii) Replace sin22A by (1-cos2A)/2 using double angle formula.
Then integrate normally
A = -1/8 * integration of (cos 2A -1) dA with limits 0 and pie/2
Solution will be pie/16
WOW, You NAILED it MAAN!! Thnxxxxxxxxxx 8)
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WOW, You NAILED it MAAN!! Thnxxxxxxxxxx 8)
Hehe.....anytime pal :)
I just hope you understand everything ;)
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November 08
Question8
Part i)
note:in the second image k/4=0.05
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part ii)
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part iii)
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I would like to add again
This is my tutor's work
I will in a month's time be able to master A2 maths.
And my tutor is the guy that got me a A in as maths so worry not
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I would like to add again
This is my tutor's work
I will in a month's time be able to master A2 maths.
And my tutor is the guy that got me a A in as maths so worry not
Hey that's alright dude. We trust you ;)
Anyway thanks for your help :)
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Those images are all black
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Those images are all black
They were taken in poor light. Open them up and increase their brightness in MS Word.
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Hello guys, i will be needing your help today :P, Whe never i face problems, I am gona post the questions here. So be ready :P :P