y2x2同理由椭圆C2:2?2?1(a?b?0)的上、下焦点(0,c),(0,?c)及左、右顶点
ab(?b,0),(b,0)均在圆O:x2?y2?1上可解得：b?c?1,?a?2．得椭圆
y2C2:x??1． ??????????4分
22（2）设直线AB的方程为y?k(x?1),A(x1,y1),B(x2,y2)，则N(0,?k)．
?y2?4x联立方程组?，消去y得：k2x2?(2k2?4)x?k2?0,
?y?k(x?1)?2k2?4?x?x????16k2?16?0,且?12k2 ??????????5分
?xx?1?12????????????????由NA??1AF,NB??2BF得：?1(1?x1)?x1,?2(1?x2)?x2,
整理得：?1?x1x,?2?2 1?x11?x22k2?4?22x1?x2?2x1x2k??1??2????1． ??????????8分 22k?41?(x1?x2)?x1x21??12k（3）设P(xp,yp),Q(xQ,yQ),?S(xp?xQ,yp?yQ)，则P'(xp,0),Q'(xQ,0)
??????????????????yp22由OP?OQ?OP'?OQ'?1?0得2xpxQ?ypyQ??1；① xp??1；②
2xQ?2yQ22?1；③ ??????????11分
(yp?yQ)22由①+②+③得(xp?xQ)?2?1
∴S(xp?xQ,yp?yQ)满足椭圆C2的方程，命题得证． ??????????13分
22．解：（1）解：F(x)??lnx?11?x,则F'(x)?2, xx当x∈（0，1）时F'(x)?0，x∈（1，＋∞）时F'(x)?0，
∴F(x)在（0，1）递增，在（1，＋∞）递减，
F(x)max?F(0)?0 ??????????2分
∴当x?0时f(x)?g(x)恒成立，即x?0时lnx?1?∴f(x2)?f(x1)?ln1恒成立。 xx1x1?1?1??(x2?x1)?f'(x1)(x2?x1) ???????4分 x2x2x2证明：?1f(x1)??2f(x2)?f(?1x1??2x2)，
（2）证明：设?1,?2?0，且?1??2?1，令x3??1x1??2x2,，则x3?0，且
x1?x3??2(x1?x2)，x2?x3??1(x2?x1)，
由（1）可知f(x1)?f(x3)?f'(x3)(x1?x3)??2f'(x3)(x1?x2) ???①
f(x2)?f(x3)?f'(x3)(x2?x3)??1f'(x3)(x2?x1) ???②
①??1+②??2，得
?1f(x1)??2f(x2)?(?1??2)f(x3)??1?2f'(x3)(x1?x2)??1?2f'(x3)(x2?x1)?0
∴?1f(x1)??2f(x2)?(?1??2)f(x3)?f(x3)?f(?1x1??2x2)???????8分 猜想：若?i?0,xi?0(i?1,2,?,n)，且?1??2????n?1时有
?1f(x1)??2f(x2)????nf(xn)?f(?1x1??2x2????nxn) ???????9分
（3）证明：令?i?ai1,xi?(i?1,2,?,n)
a1?a2???anai由猜想结论得
ana1a2lna1?lna2???lnan
a1?a2???ana1?a2???ana1?a2???an??ln(=ln(ana1a2111???????)
a1?a2???ana1a1?a2???ana2a1?a2???anana1?a2???an)
na1?a2???an)，
n∴a1lna1?a2lna2???anlnan?(a1?a2???an)ln(即有a11a22?anaaan?(a1?a2???ana1?a2???an)。 ???????14分
n