调递减．（II）由（I）知，a2．xx2因为fx1fx2x1x21al
x1l
x2，所以x1x2fx1fx21l
xl
x2k1a1x1x2x1x2x1x2l
xl
x2又由I知，x1x21．于是k2a1x1x2l
x1l
x2若存在a，使得k2a则1．即l
x1l
x2x1x2．亦即x1x21x22l
x20x21x21再由（I）知，函数htt2l
t在0∞上单调递增，而x21，所以t11x22l
x212l
10这与式矛盾．故不存在a，使得k2ax21
3当a2时0gx0的两根为x1
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