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 xcosec 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 = 2cos
2x
You'll get (2cos
2x)/(2sinx cosx)
Eliminate cosx from the numerator and denominator to obtain
cosx / sinx = cot x(ii)
cosec 2x + cot 2x = 2This also implies that cot x = 2 ----> 1/tanx = 2
therefore tanx = 1/2
Key angle x = tan
-1(1/2) = 26.6
otan is positive in first and third quadrant.
Hence x = pheta = 26.6
o and 206.6
o8.Let 100/x
2(10-x) = A/x + B/x
2 + C/(10-x)
Take x
2(10-x) as common denominator on the right hand side :
A(x
2(10-x)) + B((10-x) + C(x
2) = 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/x
2(10-x) = 1/x + 10/x
2 + 1/(10-x)
(ii) Give dx/dt = 1/100*x
2(10-x)
Send the terms in x to the left side and you'll obtain :
100/x
2(10-x) dx = dt
Use the answer obtained in part (i) to replace 100/x
2(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
9Given 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 : 2
b -
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 + 2
c = 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 ( 0
2 +(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 = x
2 ---> du/dx = 2x
Hence v = (1-x
2)
-1/2 ---> dv/dx = -x(1-x
2)
-1/2therefore dy/dx = x2(-x(1-x
2)
-1/2) + ((1-x
2)
-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 x
2(1-x
2)
1/2) dx
Substituting x=sinA and dx = cos A dA, you'll be getting :
Area = Integration of sin
2A (1-sin
2A)
1/2 cos A dA with limits 0 and pie/2
Replacing 1- sin
2A by cos
2A
Then sin
2A cos
2A = (sin
22A)/4
Area is shown to be 1/4 x integration of sin
22A dA with limits 0 and pie/2.
(iii) Replace sin
22A 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
