Author Topic: Pure 3 help.  (Read 3706 times)

Offline Deadly_king

  • <<Th3 BO$$>>
  • Global Moderator
  • SF Farseer
  • *****
  • Posts: 3391
  • Reputation: 65078
  • Gender: Male
  • Hard work ALWAYS pays off.........just be patient!
    • @pump_upp - best crypto pumps on telegram !
Re: Pure 3 help.
« Reply #15 on: October 02, 2010, 07:36:54 am »
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 :)

Offline Deadly_king

  • <<Th3 BO$$>>
  • Global Moderator
  • SF Farseer
  • *****
  • Posts: 3391
  • Reputation: 65078
  • Gender: Male
  • Hard work ALWAYS pays off.........just be patient!
    • @pump_upp - best crypto pumps on telegram !
Re: Pure 3 help.
« Reply #16 on: October 02, 2010, 08:27:10 am »
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 :)

Offline S.M.A.T

  • SF Master
  • ******
  • Posts: 1595
  • Reputation: 18522
  • Gender: Male
Re: Pure 3 help.
« Reply #17 on: October 02, 2010, 09:13:19 am »
Nov 09 No 2, 6

« Last Edit: October 02, 2010, 09:35:19 am by asiftasfiq93 »


"A man's life is interesting primarily when he has failed - I well know. For it's a sign that he tried to surpass himself." Clemenceau, Georges

Offline S.M.A.T

  • SF Master
  • ******
  • Posts: 1595
  • Reputation: 18522
  • Gender: Male
Re: Pure 3 help.
« Reply #18 on: October 02, 2010, 09:18:00 am »
DK!! ...that was really helpful ;D

+rep


"A man's life is interesting primarily when he has failed - I well know. For it's a sign that he tried to surpass himself." Clemenceau, Georges

Offline mousa

  • SF Citizen
  • ***
  • Posts: 233
  • Reputation: 447
Re: Pure 3 help.
« Reply #19 on: October 02, 2010, 11:58:11 am »
God bless you all.  8) thanks for all reponses!!

Offline Deadly_king

  • <<Th3 BO$$>>
  • Global Moderator
  • SF Farseer
  • *****
  • Posts: 3391
  • Reputation: 65078
  • Gender: Male
  • Hard work ALWAYS pays off.........just be patient!
    • @pump_upp - best crypto pumps on telegram !
Re: Pure 3 help.
« Reply #20 on: October 02, 2010, 03:26:35 pm »
DK!! ...that was really helpful ;D

+rep
That's alright br0 :)

Thanks for completing the other numbers  ;)

You're welcome mousa :)

Offline mousa

  • SF Citizen
  • ***
  • Posts: 233
  • Reputation: 447
Re: Pure 3 help.
« Reply #21 on: October 02, 2010, 04:10:07 pm »
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)

Offline Deadly_king

  • <<Th3 BO$$>>
  • Global Moderator
  • SF Farseer
  • *****
  • Posts: 3391
  • Reputation: 65078
  • Gender: Male
  • Hard work ALWAYS pays off.........just be patient!
    • @pump_upp - best crypto pumps on telegram !
Re: Pure 3 help.
« Reply #22 on: October 02, 2010, 04:18:19 pm »
WOW, You NAILED it MAAN!! Thnxxxxxxxxxx 8)
Hehe.....anytime pal :)

I just hope you understand everything  ;)

Freaked12

  • Guest
Re: Pure 3 help.
« Reply #23 on: October 02, 2010, 10:44:28 pm »
November 08
Question8
Part i)


note:in the second image k/4=0.05
« Last Edit: October 02, 2010, 11:04:19 pm by Requiem »

Freaked12

  • Guest
Re: Pure 3 help.
« Reply #24 on: October 02, 2010, 10:48:13 pm »
part ii)

Freaked12

  • Guest
Re: Pure 3 help.
« Reply #25 on: October 02, 2010, 10:59:15 pm »
part iii)


Freaked12

  • Guest
Re: Pure 3 help.
« Reply #26 on: October 02, 2010, 11:02:18 pm »
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

Offline Deadly_king

  • <<Th3 BO$$>>
  • Global Moderator
  • SF Farseer
  • *****
  • Posts: 3391
  • Reputation: 65078
  • Gender: Male
  • Hard work ALWAYS pays off.........just be patient!
    • @pump_upp - best crypto pumps on telegram !
Re: Pure 3 help.
« Reply #27 on: October 03, 2010, 05:03:49 am »
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 :)
« Last Edit: October 03, 2010, 09:02:37 am by Deadly_king »

Offline astarmathsandphysics

  • SF Overlord
  • *********
  • Posts: 11271
  • Reputation: 65534
  • Gender: Male
  • Free the exam papers!
Re: Pure 3 help.
« Reply #28 on: October 03, 2010, 12:49:06 pm »
Those images are all black

elemis

  • Guest
Re: Pure 3 help.
« Reply #29 on: October 03, 2010, 12:52:26 pm »
Those images are all black

They were taken in poor light. Open them up and increase their brightness in MS Word.