2015广东理[名师原创].doc

一.选择题：本大题共8个小题,每小题5分,共40分.在每小题给出的四个选项中，只有一项 是符合题目要求的. 1.因为，，所以，故选A． EMBED Equation.DSMT4 2.因为，所以，故选D． EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 3.令，则，即，，所以既不是奇函数也不是偶函数，而BCD依次是奇函数、偶函数、偶函数，故选A． EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 4.从袋中任取2个球共有种，其中恰好1个白球1个红球共有种，所以恰好1个白球1个红球的概率为，故选B． EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 5.设所求切线方程为，依题有，解得，所以所求切线的直线方程为或，故选D EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 6. 不等式所表示的可行域如下图所示， xy x y O A l 由得，依题当目标函数直线：经过时，取得最小值即，故选C EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 7. 因为所求双曲线的右焦点为且离心率为，所以，，所以所求双曲线方程为，故选B EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 E