{F4 A.Math} Absolute functions ~

a+b=-k+2
| a | = | b |
a=b or a=-b
a-b=0 or a+b=0
-k+2=0
k=2

|a|=|b|
a=b or a=-b
when a=b
delta=0
(k-2)^2+4(k-1)=0
k^2-4k+4+4k-4=0
k^2=0
k=0
when a=-b
a+b=0
-(k-2)=0
-k+2=0
k=2
Hence, k=0 or 2

a+b = 2-k
ab=1-k

lal=lbl
lal-lbl=0
a^2-2labl+b^2=0
labl=(a^2+b^2)/2
labl=[(a+b)^2-4ab]/2
labl=[k^2-4k+4-4+4k]/2
k=+/-開方2labl

| a | = | b |
a=b or a=-b

when a=b
(k-2)^2+4(k-1)=0
k=0

when a=-b
a+b= 2-k =0
k=2

Case 1 a=b
a+b = -(k-2)
a+b = 2-k.....(1)
2a = 2-k

ab = -(k-1)
ab = 1-k.....(2)
a^2 = 1-k

[(1)/2]^2 - (2)

[(2-k)/2]^2 = 1-k
(4-4k+k^2)/4 = 1-k
4-4k+k^2 = 4-4k
k^2 = 0
k=0

Case 2
-a = b
sum of roots
= a +b = a-a = 0

so -(k-2) = 0
2-k = 0
k = 2

so k = 0 or 2

HKCEE 1992