x－4x＋3k1 (x－1)(x－3)＝k1( )＝ ， ········································ 7分
x x
又∵当1≤x≤3时， (x－1) (x－3) ≤0，
(x－1)(x－3)
) ≥0.
x
∴ PE－NE≥0. ······································································· 8分 ∴ k1( ∴
PNPE
＝－1 NENE
14
＝－x2＋x－1. ··························································· 9分
33
PN1
∴ 当x＝2时，的最大值是. ··············································10分
NE31
由题意，此时PN＝，
2
3
∴ NE＝. ············································································· 11分
23
∴ 点N(2，) . ∴ k2＝3.
2
3∴ y＝. ··················································································12分
x
k
解3：∵ 点A(1，c)、B(3，d)是直线y＝k1x＋b与双曲线y＝2（k2＞0）的交点，
x
∴ c＝k2,，3d＝k2，c＝k1＋b，d＝3k1＋b. ································ 5分 k2＝3d， k1＝－d，b＝4d. ∴ 直线y＝－dx＋4d，双曲线y＝
3d. x
2
∵ A(1，c)和点B (3，d )都在第一象限，∴ 点P在第一象限. 3d
∴ PN＝PE－NE＝－dx＋4d－
x
x2－4x＋3d (x－1)(x－3)
＝－d( )＝－ ，··································· 6分
x x 又∵当1≤x≤3时，(x－1) (x－3) ≤0，
d (x－1)(x－3)
≥0.
x
∴ PN＝PE－NE≥0. ······························································· 7分 ∴－
PN
＝NE
－dx＋4d－
3dx
3dx
∴ ··························································· 8分
124
＝－x＋x－1. ··························································· 9分
33
PN1
∴ 当x＝2时，的最大值是. ··············································10分
NE3
- 16 -
1
由题意，此时PN＝，
2
3
∴ NE＝. ··············································································· 11分
23
∴ 点N(2，) .
2∴ k2＝3.
3∴ y＝. ··················································································12分
x
- 17 -