Author Topic: Pure 3 help.  (Read 3705 times)

Offline mousa

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Pure 3 help.
« on: October 01, 2010, 02:24:32 pm »
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. :-[

Freaked12

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Re: Pure 3 help.
« Reply #1 on: October 02, 2010, 01:12:15 am »
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.

Freaked12

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Re: Pure 3 help.
« Reply #2 on: October 02, 2010, 01:24:43 am »
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

Freaked12

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Re: Pure 3 help.
« Reply #3 on: October 02, 2010, 01:25:42 am »
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

Offline Deadly_king

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Re: Pure 3 help.
« Reply #4 on: October 02, 2010, 04:55:50 am »
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
« Last Edit: October 02, 2010, 07:04:03 am by Deadly_king »

Offline mousa

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Re: Pure 3 help.
« Reply #5 on: October 02, 2010, 05:47:54 am »
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!!!

Offline mousa

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Re: Pure 3 help.
« Reply #6 on: October 02, 2010, 05:50:25 am »
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???! ???

Offline Deadly_king

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Re: Pure 3 help.
« Reply #7 on: October 02, 2010, 05:57:56 am »

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 :)

Offline mousa

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Re: Pure 3 help.
« Reply #8 on: October 02, 2010, 06:00:51 am »
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

Offline Deadly_king

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Re: Pure 3 help.
« Reply #9 on: October 02, 2010, 06:05:09 am »
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??
« Last Edit: October 02, 2010, 07:38:07 am by Deadly_king »

Offline mousa

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Re: Pure 3 help.
« Reply #10 on: October 02, 2010, 06:14:10 am »
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

Offline mousa

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Re: Pure 3 help.
« Reply #11 on: October 02, 2010, 06:19:11 am »
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??

Offline Deadly_king

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Re: Pure 3 help.
« Reply #12 on: October 02, 2010, 06:38:35 am »
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 :)

elemis

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Re: Pure 3 help.
« Reply #13 on: October 02, 2010, 06:49:14 am »
@deadly_king  + rep for all the support you've been giving ;)

Offline Deadly_king

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Re: Pure 3 help.
« Reply #14 on: October 02, 2010, 07:06:42 am »
@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  ;)