Qualification > Math
Pure 3 help.
mousa:
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:
--- Quote from: mousa on October 01, 2010, 02:24:32 pm ---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. :-[
--- End quote ---
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:
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:
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
Deadly_king:
--- Quote from: mousa 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. :-[
--- End quote ---
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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