发布时间 : 星期五 文章2015届福州高三上学期教学质量检查数学理试题更新完毕开始阅读b56ab032be1e650e52ea99de
注意到y1?8?x及y2?则y?8?x?10m（1≤m≤4且m?R）均关于x在[6,8]上单调递减， x?210m关于x在[6,8]上单调递减， ····························································· 10分 x?2610m5m5m故y≥8?8?，由······················································ 11分 ?≥2，得m≥， ·
8?23356所以所求的m的最小值为． ······················································································· 12分
521． 本题主要考查抛物线的标准方程与性质、直线与抛物线的位置关系、归纳推理等基础知识，考查推理论证能力、运算求解能力，考查化归与转化思想、数形结合思想、特殊与一般思想等． 解：（Ⅰ）依题意可设抛物线?的方程为：x2?2py（p?0）． ······························· 1分
p由焦点为F(0,1)可知?1，所以p?2． ······································································ 2分
2所以所求的抛物线方程为x2?4y． ················································································ 3分 （Ⅱ）方法一：
2?x12??x2?1设切点A、B坐标分别为?x1,?,?x2,?，由（Ⅰ）知，y??x．
4??4?2?11则切线PA、PB的斜率分别为k1?y?x?x?x1,k2?y?x?x?x2，
122211121故切线PA、PB的方程分别为y?x12?x1(x?x1)，y?x2············ 4分 ?x2(x?x2)， ·
4242x1?x2?x?,?x?x1?2联立以上两个方程，得?.故P的坐标为(12,x1x2)， ··························· 5分
124?y?xx12?4?1因为点P在抛物线?的准线上，所以x1x2??1，即x1x2??4． ································· 6分
4设直线AB的方程为y?kx?m，代入抛物线方程x2?4y，得x2?4kx?4m?0， 所以x1x2??4m，即?4m??4，所以m?1． ································································· 7分
故AB的方程为y?kx?1，故直线AB恒过定点(0,1)． ················································· 8分
2?x12???x2方法二：设切点A、B坐标分别为?x1,?,?x2,?，设P?m,?1?，
4??4??易知直线PA、PB斜率必存在，可设过点P的切线方程为y?1?k?x?m?．
?y?1?k?x?m?,由?2，消去y并整理得x2?4kx?4?km?1??0． ····················· ①
x?4y,?因为切线与抛物线有且只有一个交点，
所以???4k??16(km?1)?0，整理得k2?mk?1?0， ··································· ②
所以直线PA、PB斜率k1,k2为方程②的两个根，故k1?k2??1， ·································· 4分 另一方面，由??0可得方程①的解为x?2k， 所以x1?2k1,x2?2k2． ··································································································· 5分 假设存在一定点，使得直线AB恒过该定点，则由抛物线对称性可知该定点必在y轴 上，设该定点为C(0,c)， ································································································· 6分
2x12x2则CA?(x1,?c),CB?(x2,?c)．
442所以CA//CB，
2x2x12xx所以x1(?c)?(?c)x2?0，整理得c(x1?x2)?12(x2?x1)
444所以x1?x2，
xx4kk所以c??12??12?1 ······························································································· 7分
44所以直线AB过定点?0,1?． ····························································································· 8分
（Ⅲ）结论一：若点P为直线l:y?t（t?0）上的任意一点，过点P作抛物线?:x2?2py（p?0）的切线PA,PB，切点分别为A,B，则直线AB恒过定点(0,?t)． ······························· 12分 结论二：过点Q?0,m?（m?0）任作一条直线交抛物线?:x2?2py?p?0?于A,B两点，分别以点A,B为切点作该抛物线的切线，两切线交于点P，则点P必在定直线y??m上． ·········· 12分
结论三：已知点P为直线l:y?kx?b上的一点，若过点P可以作两条直线与抛物线?:x2?2py（p?0）相切，切点分别为A,B，则直线AB恒过定点?pk,?b?． ·························· 12分 说明：①以上两结论只要给出其中一个即可或给出更一般性的结论； ②以上两结论中的抛物线开口方向均可改变； ③该小题评分可对照以下表格分等级给分： 得分 0分 1分 2分 3分 4分 答题情况 写出与命题（ⅰ）无关的结论． 所给命题的条件与结论均存在问题． 将准线或抛物线改为其它特殊情况，结论正确. 将准线或抛物线其中一个一般化，但结论中的定点（或定直线）有误． 写出命题的逆命题，结论正确.( 其它分点逆命题相应给分) 将准线和抛物线都推广成一般情况，但结论中的定点（或定直线）有误． 将准线和抛物线其中一个推广成一般情况，结论正确． 将准线和抛物线都推广成一般情况，但m?0,p?0中漏写一个或两个． 将准线和抛物线都推广成一般情况，结论正确． 22．本题主要考查函数的零点、函数的导数、导数的应用、不等式的恒成立等基础知识，考查推理论证能力、运算求解能力等，考查函数与方程思想、化归与转化思想、数形结合思想等．
π解：（Ⅰ）函数y?f?x?在(0,)上的零点的个数为1． ·············································· 1分
2理由如下：
因为f?x??exsinx?cosx，所以f??x??exsinx?excosx?sinx． ······························· 2分
π，所以f?(x)?0， 2π所以函数f(x)在(0,)上是单调递增函数． ··································································· 3分
2ππ因为f(0)??1?0，f()?e2?0，
2根据函数零点存在性定理得
π函数y?f?x?在(0,)上的零点的个数为1． ································································· 4分
2（Ⅱ）因为不等式f(x1)?g(x2)≥m等价于f(x1)≥m?g(x2)，
ππ所以 ?x1?[0,],?x2?[0,]，使得不等式f(x1)?g(x2)≥m成立，等价于
22················································· 6分 f(x1)min≥?m?g(x2)?min，即f(x1)min≥m?g(x2)max． ·因为0?x?
ππ当x?[0,]时，f??x??exsinx?excosx?sinx?0，故f(x)在区间[0,]上单调递增，所以x?0时，
22····································································································· 7分 f?x?取得最小值?1．·
又g??x??cosx?xsinx?2ex，由于0≤cosx≤1,xsinx≥0,2ex≥2，
π所以g??x??0，故g?x?在区间[0,]上单调递减，
2因此，x?0时，g?x?取得最大值?2．······································································· 8分 ?2?1． 所以?1≥m??2，所以m≤-??所以实数m的取值范围是??,?1?2?···································································· 9分 ?． ·
（Ⅲ）当x??1时，要证f?x??g?x??0，只要证f?x??g?x?， 只要证exsinx?cosx?xcosx?2ex， 只要证exsinx?2??x?1?cosx，
excosx?由于sinx?2?0,x?1?0，只要证． ············································· 10分 x?1sinx?2excosx?下面证明x??1时，不等式成立． x?1sinx?2???ex?x?1??exxexex令h?x??， ??x??1?，则h??x??22x?1?x?1??x?1?当x???1,0?时，h??x??0，h?x?单调递减； 当x??0,???时，h??x??0，h?x?单调递增．
所以当且仅当x?0时，h?x?取得极小值也就是最小值为1． 令k?cosxsinx?2，其可看作点A?sinx,cosx?与点B?2,0连线的斜率，
??所以直线AB的方程为：y?kx?2，
由于点A在圆x2?y2?1上，所以直线AB与圆x2?y2?1相交或相切， 当直线AB与圆x2?y2?1相切且切点在第二象限时，
直线AB取得斜率k的最大值为1． ··············································································· 12分
2?1?h?0?；x?0时，h?x??1≥k． ·故x?0时，k?·········································· 13分 2综上所述，当x??1时，f?x??g?x??0成立． ························································ 14分
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