F.4 A.maths

(| 2x-1 |-3)( | 2x-1 |+2)=0
| 2x-1 |=3 or | 2x-1 |=-2(reject)
2x-1=3 or 2x-1=-3
2x=4 or 2x=-2
x=2 or x=-1

(i)
2x2 <= 3x + 5
2x2-3x-5<=0
(2x-5)(x+1) <=0
-1<=x<=5/2

(ii)
x2> 2x –1
x2-2x+1>0
(x-1)2>0
All real namber, except x=1

2x2<= 3x+5 and X2 > 2x –1
i.e. 1<x<=5/2
k=2

3a.
For n=1,
L.H.S.= 1x3
=3
R.H.S.= 1/6(1)(1+1)(2+7)
=3
=L.H.S.
∴The statement is true for n=1.
Assume1x3+2x4+3x5+......+k(k+2) = 1/6k(k+1)(2k+7),for some positive integer k.
For n=k+1,
L.H.S.
=1x3+2x4+3x5+......+k(k+2)+(k+1)(k+3)
=1/6k(k+1)(2k+7) +(k+1)(k+3)
=1/6(k+1)[k(2k+7)+6(k+3)]
=1/6(k+1)(2k2+13k+18)
=1/6(k+1)(k+2)(2k+9)
=R.H.S.
∴The statement is also true for n=k+1 if it is true for n=k.
By the principle of mathematical induction,the statement is true for all positive integer n.

(b)
1x2+2x3+3x4+......+100(101)= 1x(1+1)+2x(2+1)+3x(3+1)+......+100(100+1)
=(12+22+32+…+1002)+(1+2+3+…+100)
=1/6(100)(100+1)[2(100)+1]+1/2(100)(100+1)
=338350+5050
=343400

4.(a)

A+B=2
AB=-k(k-1)

(A-B)2=(A+B)2-4AB
=22-4[-k(k-1)]
=4+4k2-4k
=4(k2-k+1)
=4(k2-k+1/4)+3
=4(k-1/2)2+3
>0

(b)
|A-B|2=4(k-1/2)2+3
A-B=√(4(k-1/2)2+3)
∴the minimum value of |A-B|=√3

5. ai
k = (x2 - x +1)/(x2 +1)
kx2+k= x2 - x +1
(k-1)x2+x+(k-1)=0____________(1)

ii.
x is all real number.
i.e. (1)△>=0.
12-4(k-1)(k-1)>=0
1-4(k2-2k+1) >=0
1-4k2+8k-4>=0
4k2-8k+3>=0
(2k-3)(2k-1)>=0
k=<1/2 or k>=3/2







2009-05-29 23:45:10 補充：
(2x-1)^2 - | 2x-1 | = 6
| 2x-1 |^2- | 2x-1 | = 6
| 2x-1 |^2- | 2x-1 | - 6=0
(| 2x-1 |-3)( | 2x-1 |+2)=0
| 2x-1 |=3 or | 2x-1 |=-2(reject)
2x-1=3 or 2x-1=-3
2x=4 or 2x=-2
x=2 or x=-1

2009-05-30 14:13:25 補充：
aii.

x is all real number.

i.e. (1)△>=0.

1^2-4(k-1)(k-1)>=0

1-4(k^2-2k+1) >=0

1-4k^2+8k-4>=0

4k^2-8k+3=<0

(2k-3)(2k-1)=<0

1/2=k=<3/2

2009-05-30 14:18:07 補充：
5b.
3/2 = (x^2 - x +1)/(x^2 +1)
3(x^2 +1)=2(x^2 - x +1)
3x^2 +3=2x^2 -2x+2
x^2+2x+1=0
(x+1)^2=0
x+1=0
x=-1(repeated root)