高考数学（理）二轮专题突破演练（浙江专版）第3部分 专题2 保温训练卷4 Word版含解析

1、保温训练卷(四)一、选择题1a为正实数，i为虚数单位，2，则a()a2 b.c. d1解析：选b由已知2，得|(ai)(i)|1ai|2，2，a0，a.2已知tan()，tan，那么tan()a. b.c. d.解析：选btantan.3已知双曲线1(a0，b0)的一条渐近线方程是yx，则该双曲线的离心率为()a. b.c. d.解析：选c依题意，所以ba，ca.故e.4如图所示的程序框图输出的所有点都在函数()ayx1的图像上 by2x的图像上cy2x的图像上 dy2x1的图像上解析：选d依题意，运行程序框图，输出的点依次为(1,1)，(2,2)，(3,4)，(4,8)，易知这四个点均在y2

2、x1的图像上5把函数ysin的图像向左平移个单位后，所得函数的单调递增区间为()a.(kz)b.(kz)c.(kz)d.(kz)解析：选b依题意，把函数ysin的图像向左平移个单位后，所得函数为ysin，由2k2x2k，得kxk(kz)，所以所得函数的单调递增区间为(kz)6已知实数a、b满足等式2a3b，下列五个关系式：0ba；ab0；0ab；ba0.(1)若曲线yf(x)在点(1，f(1)处的切线与直线yx1垂直，求a的值；(2)若函数f(x)在区间1,2上的最小值为，求a的值解：f(x)(x0)(1)曲线yf(x)在点(1，f(1)处的切线与直线yx1垂直，f(1)2，即1a2，解得a3

3、.(2)当00在1,2上恒成立，这时f(x)在1,2上为增函数，f(x)minf(1)a1，a1，a，与0a1矛盾，舍去；当1a2时，可知当x(1，a)时，f(x)0，f(x)在(a,2上为增函数，f(x)minf(a)ln a，ln a，a，满足题设；当a2时，f(x)0在(1,2)上恒成立，这时f(x)在1,2上为减函数，f(x)minf(2)ln 21，ln 21，a32ln 2，与a2矛盾，舍去；综上得a.高考资源网版权所有！投稿可联系qq：108459180129c91afe4ced1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9

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