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1、2010 年高考大纲全国卷年高考大纲全国卷 II 第第 22 题题设函数设函数 1xf xe ()证明)证明:当:当时,时,;x-1 1xf xx()设当)设当时,时,求,求 a 的取值范围的取值范围0x 1xf xax 解析:解析: (1)采用综合与分析相结合的办法:采用综合与分析相结合的办法:当当时,要证时,要证1xxx xxxf1e1-1 1)(x即证:只需证只需证xxexexx1 111 11即:) 1( 01 1 0 01 xxexeexxxx即证:只需证Q令令) 1( 1)( ) 1( 1)(xexgxxexgxx则令令 令令0 0)( xxg则01 0)( xxg则命题得证时,当
2、上为增函数上为减函数,在在0)0()(1), 0()0 , 1()(gxgxxg(2) 有解恒成立,而时:易知当。)0(01)0(0)(01xaxxxxfa不满足不满足)0(01 )(02.xxexxfax即:时,当0)(1)( )0( 1)( xfexhxxexhxx则令显然成立时,当上为增函数在0)0()(0 ), 0)(hxhxxh上恒成立在时:当。), 001)(03 axxxfa0111)()( axxeaxxxfxtx令(我们的思路是这样的:注意到我们的思路是这样的:注意到t(0)=0 这就意味着这就意味着 如果如果t(x)在区间内为减函数,则一定有上式在在区间内为减函数,则一定有
3、上式在;但是一旦函数;但是一旦函数上恒成立), 0 t(x)在开始的某段区间内位增函数在开始的某段区间内位增函数 则函数则函数t(x)在这一区间内函数值是在这一区间内函数值是大于大于t(0)=0的,则与函数在的,则与函数在小于等于小于等于0恒成立矛盾,而我们恒成立矛盾,而我们上), 0 判断单调性就会采用求导的办法,通过讨论其导函数的正负情况来判断单调性就会采用求导的办法,通过讨论其导函数的正负情况来研究原函数的单调性)研究原函数的单调性)2222) 1(1) 1( ) 1(1 ) 1(1) 1()( axaxe axeaxaxaxextx xx(注意到此时(注意到此时的正负不是确定的,不能直
4、接确定函数的正负不是确定的,不能直接确定函数)( xtt(x)的单调性,但是注意到的单调性,但是注意到这就意味着为了确定这就意味着为了确定的的0)( xt)( xt正负必须知道正负必须知道的单调性。由于的单调性。由于,所以只要,所以只要在区在区)( xt0)( xt)( xt间间为减函数,则为减函数,则,如果,如果在在的某个子区的某个子区上), 0 0)( xt)( xt), 0 间内位增函数,在间内位增函数,在在这个子区间内大于等于在这个子区间内大于等于 0 此为矛盾此为矛盾 )( xt此时不满足题意)此时不满足题意)令0)0( 1) 1()(2raxexrx则)12()1() 1(2) 1
5、()( 2aaxaxeaaxeaxexrxxx01112, 0)( aaaaxxr易知或则令当(如图所示))0(0)( 210 012xxraaa时,即a1aa12 恒成立时当上为减函数,在时,当时,当即:时,当上为减函数在0)0()(0 ), 0)(00)()( 0 0)0()(0 ), 0)(txtxxtxxrxtxrxrxxr(如下图所示)aaxxraaa120 , 0)( 21012则时:令即当a1aa12 0)( 12, 0 )0()( 12, 0xtaaxrxraax时当即:时,当为增函数时,当)( 12, 0xtaax不成立!时,当 0)0()( 12, 0txtaax(由(由我
6、们知道我们知道是满足题意的,由是满足题意的,由我们知道我们知道是不是不210 a21a满足题意的)满足题意的)综上210 a内部资料,请勿外传!#XuyUP2kNXpRWXmA&UE9aQGn8xp$R#͑GxGjqv$UE9wEwZ#QcUE%&qYpEh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89AmYWpazadNu#KN&MuWFA5uxY7JnD6YWRrWwcvR9CpbK!zn%Mz849GxGjqv$UE9wEwZ#QcUE%&qYpEh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89A
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