GCE O Level Pure Maths

Note that logab = ln b/ ln a (ln = natural log)
So,
the first given equation becomes
ln q / ln p + 4 ln p / ln q = 5
(ln q / ln p)^2 -5(ln q / ln p)+4=0
[(ln q / ln p) - 4][(ln q / ln p)-1]=0
(ln q / ln p) - 4 = 0 since ln q =/=ln p
lnq = 4 ln p................(1)

The second equation:
pq=32
ln pq=ln 32
ln p + ln q = ln 32...............(2)

Putting (1) into (2),
ln p + 4 ln p = ln 32
5ln p = ln 32 = = ln (2^5) = 5 ln2
p = 2
Put it into (1): ln q = 4 ln 2 = ln(2^4) = ln 16
So, p=2, q=16

(d) (i) 6x lnx + 4lnx -2 - 3x = 2lnx(3x+2)-(3x+2) = (3x+2)(2lnx-1)
(ii) 6x lnx + 4lnx -2 - 3x = 0
(3x+2)(2lnx-1)=0
3x+2 =0 or 2lnx-1=0
x = -2/3 or lnx = 1/2
x=-2/3 or x =e^(1/2)7.
Solve y=x^2-4 and y=5 to get intersection points
==>5 = x^2-4 x^2 = 9 ==> x=3 or -3
So, the coordinates of upper right corner point of the shaded region = (3,5)Put y=0 into the curve, x^2-4=0 ==> x=2 or -2
The curve cuts the positive x-axis at (2,0)Draw a straight line from (3,0) to (3,5)
This line together with the positive y and x axis form a new close region and let call it A.
This line together with the positive x-axis and the curve form another close region and let call it B.The required solid is (the solid generated by rotating A) - (the solid generated by rotating B)
The solid generated by rotating A has a volume
= (pi)(5)^2(3)= 75 pi cubic unitsThe solid generated by rotating B has a volume
=(pi) Integral [2--->3] (x^2-4)^2 dx
= (pi) Integral [2--->3] x^4-8x^2+16 dx
= (pi) [x^5 /5-8x^3 /3 +16x]x=3 - (pi) [x^5 /5-8x^3 /3 +16x]x=2
=113/15 pi cubic unitsThe required volume = (75 -113/15) pi = 1012/15 pi = 211.95...=212 cubic units

2011-07-01 17:23:47 補充：
ln q / ln p + 4 ln p / ln q = 5
multiply both sides by ln q / ln p,
(ln q / ln p)^2 + 4 = 5( ln q / ln p)
(ln q / ln p)^2 -5(ln q / ln p)+4=0

2011-07-01 17:29:18 補充：
The solid generated by rotating A is a cylinder which has base radius 5 and height 3.
The volume of this solid is thus
= (pi) (radius)^2 x height = (pi)(5)^2 (3) =75 pi

2011-07-01 17:48:23 補充：
As mentioned by wanszeto, x=-2/3 in d(ii) should be rejected as ln x is undefined
Thx wanszet for pointing out the error.

2011-07-01 17:48:29 補充：
To wanszeto,
In (c), p=/=q

In 7.,
It should be ∫2→3 π[5^2- (x² - 4)^2] dx = 262π/15 instead of ∫2→3 π[5- (x² - 4)] dx=8π/3.
[The later term is just (pi)*(area under between two curves) !!!]
The answer required becomes 262π/15 +50π=1012π/15 instead of 158π/3.

(c)
logp q + 4logq p = 5 ...... (1)
pq = 32 ...... (2)

From (2):
log(pq) = log32
log p + log q = log32 ...... (3)

From (1):
(log q / log p) + (4log p / log q) = 5 ...... (4)

Let (log q / log p) = u
(4) becomes:
u + (4/u) = 5
[u + (4/u)]u = 5u
u² + 4 = 5u
u² - 5u + 4 = 0
(u - 1)(u - 4) = 0
u = 1 or u = 4

When u = 1:
log q / log p = 1
log q = log p
q = p ...... (5)
Put (5) into (3):
log p + log p = log 32
log p² = log 32
p² = 32
p = 4√2 or p = -4√2 (rejected)
Put p = 4 into (6):
q = 4√2

When u = 4:
log q / log p = 4
log q = 4log p
log q = log p⁴
q = p⁴ ...... (6)
Put (6) into (3):
log p⁴ + log p = log 32
4log p + log p = log 32
log p⁵ = log 2⁵
p = 2
Put p = 2 into (6):
q = 2⁴ = 16

Ans: p = 4√2, q = 4√2 or p = 2, q = 16


(d)
(i)
Let y = ln x
Then, 6x ln x + 4 ln x - 2 - 3x
= 6xy + 4y - 2 - 3x
= (6xy + 4y) - (2 + 3x)
= 2y(3x + 2) - (3x + 2)
= (2y - 1)(3x + 2)
= [(2ln x) - 1](3x + 2)

(ii)
6x ln x + 4 ln x - 2 - 3x = 0
[(2ln x) - 1](3x + 2) = 0
2ln x = 1 or 3x = -2
ln x = 1/2 or 3x = -2
x = √e or x = -2/3 (rejected)


7.
The curve: y = x² - 4
x-axis: y = 0
Solve the equations, the curve cuts the x-axis at (2, 0) and (-2, 0)

The curve: y = x² - 4
The line: y = 5
Solve the equations, the curve cuts the line at (3, 0) and (-3, 0)

The required volume
= ∫0→2 π(5)²dx+ ∫2→3 π[5- (x² - 4)]dx
= ∫0→2 25πdx+ ∫2→3 π(9- x²)dx
= [25πx]0→2 + [9πx -(1/3)πx³]2→3
= [25π(2)] - [25π(0)] + [9π(3) - (1/3)π(3)³] - [9π(2) - (1/3)π(2)³]
= 50π + 27π - 9π - 18π + (8/3)π
= 158π/3
= 165