21【解析】(1)对f(x)求导得

2x?ex?x2?exx(2?x)f?(x)?a??a?(ex)2ex，

设直线

， 21与曲线y?f(x)切于点P(x0,y0)，则 ?1ax0y?xx?0?xe?ee0??1?a?x0(2?x0)?ex0?e解得a?x?1．所以a的值为1．

0(2)记函数

，下面考察函数y?F(x)的符号．对1x21F(x)?f(x)?(x?)?x?x?,x?0xex．当x?2时F?(x)?0恒成立． x(2?x)1F?(x)??1?2,x?0xex函数y?F(x)求导得当0?x?2时，从而

， x?(2?x)2x(2?x)?[]?12． x(2?x)11111F?(x)??1?2?x?1?2?1?1?2??2?0exxexxx∴F?(x)?0在(0,??)上恒成立，故y?F(x)在(0,??)上单调递减． ∵

，∴F(1)?F(2)?0． 143F(1)??0,F(2)?2??0ee2又曲线y?F(x)在[1,2]上连续不间断，所以由函数的零点存在性定理及其单调性知 使F(x)0∴x?(0,x),F(x)?0;x?(x,??),F(x)?0．∴?惟一的x0?(1,2)，

000??1x?，0?x?x01??xg(x)?min{f(x),x?}??x?x2,x?x0??ex1?2x?-cx，0?x?x0??xh(x)?g(x)?cx2??2?x?cx2,x?x0??ex∴

，从而

1?1??2cx，0?x?x0??x2h?(x)???x(2?x)?2cx,x?x0??ex由函数h(x)?g(x)?cx2为增函数，且曲线y?h(x)在

(0,??)上连续不断知h?(x)?0在(0,x0)，(x0,??)上恒成立．

①当x?x时，x(2?x)0ex立． 记

?2cx?0在(x,??)上恒成立，即

02?x在(x0,??)上恒成2c?xe，则，当x变化时，u?(x)，u(x)变化情况如下2?xx?3u(x)?x,x?x0u?(x)?x,x?x0ee表：

x (x0,3) 3 (3,??) ? u?(x) u(x) ∴

? 0 极小值 u(x)min?u(x)极小1． ?u(3)??3e故“

2?x在(x0,??)上恒成立”只需1，即1． 2c?x2c?u(x)min??3c??3ee2e，当c?0时，h?(x)?0在(0,x)上恒成立． 10h?(x)?1?2?2cxx②当0?x?x时，

0综合(1)(2)知，当

1时，函数h(x)?g(x)?cx2为增函数． c??32e故实数c的取值范围是

1．

(??,?3]2e请考生在第(22)，(23)，(24)题中任选一题作答，如果多做，则按所做的第一题计分，作答时请写清题号．

(22)选修4?1：几何证明选讲

(Ⅰ)设?ABE外接圆的圆心为O?， 解：连结BO?并延长交圆O?于G点，连结GE，则?BEG?90?，?BAE??BGE．

······················· 2分 因为AF平分∠BAC，所以BF=FC，所以?FBE??BAE， ·

BDAO'OEFCG所以?FBG??FBE??EBG??BGE??EBG?180???BEG?90?，

····································· 5分 所以O?B?BF，所以BF是?ABE外接圆的切线． ·(Ⅱ)连接DF，则DF?BC，所以DF是圆O的直径， 因为BD2?BF2?DF2，DA2?AF2?DF2，

······························································· 7分 所以BD2?DA2?AF2?BF2． ·

因为AF平分∠BAC，所以?ABF∽?AEC，

所以ABAF，所以AB?AC?AE?AF?(AF?EF)?AF，

?AEAC因为?FBE??BAE，所以?FBE∽?FAB，从而BF2?FE?FA，所以AB?AC?AF2?BF2，

····························································· 10分所以BD2?DA2?AB?AC?6． ·(23)选修4?4；坐标系与参数方程 试题解析：(1)

33

(0,0),(,)22 (2) C:???(??R,??0,0????)

1A(2sin?,?),B(23cos?,?)

AB?|2sin??23cos?|?4|sin(??)|3当

?

5?时，|AB|max?4 ??6考点：极坐标方程化为直角坐标方程， 三角函数性质

(24)选修4?5：不等式选讲 解：(Ⅰ)由|x?3|?2x?1得，

?x??3,???(x?3)?2x?1, ·································································· 2分 或x??3,???x?3?2x?1,解得x?2．

·························································································· 5分 依题意m?2． ·(Ⅱ)因为

11?11?x?t?x?…?x?t???x???t??t?tt?tt?，

当且仅当

···························································· 7分 时取等号， ·1???x?t??x???0t??(t?0)有实数根， 1|x?t|?|x?|?2t因为关于x的方程

所以

······················································································· 8分 ． ·1t??2t， 1t?…2t另一方面，

所以

························································································· 9分 ， 1t??2t10分

所以t?1或t??1．