2019高考数学理科(全国通用)：压轴大题突破练1-含解析.pdf

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1、压轴大题突破练1.导数1.(2017安徽“皖南八校”联考)已知函数f(x)exax22ax1.(1)当 a 1 时，求曲线yf(x)在点(1，f(1)处的切线方程；(2)当 x0 时，f(x)0 恒成立，求实数a 的取值范围.解(1)当 a1 时，f(x)exx22x1，f(1)1e，所以切点坐标为1，1e，f(x)ex 2x2，所以 f(1)1e，故曲线 yf(x)在点(1，f(1)处的切线方程为y1e1ex 1，即 y1ex2e.(2)f(x)exax22ax 1 求导得 f(x)ex2ax2a，令 g(x)f(x)ex2ax2a，则 g(x)ex2a(x0).当 2a1，即 a12时，g

2、(x)ex2a12a0，所以 g(x)f(x)ex2ax2a 在(0，)上为增函数，g(x)g(0)12a0，即 g(x)f(x)0，所以 f(x)exax22ax1 在(0，)上为增函数，所以 f(x)f(0)10 010，故 a12时符合题意.当 2a1，即 a12时，令 g(x)ex2a 0，得 xln 2a0，x(0，ln 2a)ln 2a(ln 2a，)g(x)0g(x)减函数极小值增函数当 x(0，ln 2a)时，g(x)g(0)12a0，即 f(x)0，所以 f(x)在(0，ln 2a)上为减函数，所以f(x)f(0)0，与条件矛盾，故舍去.综上，a 的取值范围是，12.2.(2

3、017广东惠州调研)已知函数f(x)x2(a2)x aln x(aR).(1)求函数 yf(x)的单调区间；(2)当 a 1 时，证明：对任意的x 0，f(x)exx2x2.(1)解函数 f(x)的定义域是(0，)，f(x)2x(a2)ax2x2 a2 xaxx1 2xax.当 a0 时，f(x)0 对任意 x(0，)恒成立，所以函数f(x)在区间(0，)上单调递增.当 a0 时，由 f(x)0，得 xa2，由 f(x)0，得 0 xa2，所以函数f(x)在区间a2，上单调递增，在区间0，a2上单调递减.(2)证明当 a1 时，f(x)x2x ln x，要证明f(x)exx2x2，只需证明ex

4、ln x20，设 g(x)ex ln x2，则问题转化为证明对任意的x0，g(x)0，令 g(x)ex1x0，得 ex1x，容易知道该方程有唯一解，不妨设为x0，则 x0满足0ex1x0，当 x 变化时，g(x)和 g(x)的变化情况如下表：x(0，x0)x0(x0，)g(x)0g(x)单调递减单调递增g(x)ming(x0)0 xe ln x021x0 x02，因为 x00，且 x01，所以 g(x)min2120，因此不等式得证.3.(2017荆、荆、襄、宜四地七校联考)已知函数f(x)ln xx.(1)求函数 f(x)的单调区间；(2)假设方程f(x)m(m 2)有两个相异实根x1，x2

5、，且 x1x2，证明：x1 x222.(1)解f(x)ln xx 的定义域为(0，)，f(x)1x 11xx0?x1，当 x(0，1)时，f(x)0，所以 yf(x)在(0，1)上单调递增，当 x(1，)时，f(x)0，所以 yf(x)在(1，)上单调递减.(2)证明由(1)可知，f(x)m 的两个相异实根x1，x2满足 ln xxm0，且 0 x11，x21，ln x1x1mln x2x2m0，文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O

6、1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M

7、10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6

8、E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3

9、I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5

10、K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码

11、:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2

12、I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1由题意可知ln x2x2m 2ln 22，又由(1)可知 f(x)ln xx 在(1，)上单调递减，故 x2 2，所以 0 x11，02x221.令 g(x)ln xxm，则 g(x1)g2x22(ln x1x1)ln 2x222x22(ln x2x2)(ln 2x222x22)x22x223ln x2 ln 2，令 h(t)t2t23ln tln 2(t 2)，则

13、 h(t)14t33tt33t24t3t22t1t3.当 t2 时，h(t)0，h(t)在(2，)上单调递减，所以h(t)h(2)2ln 2320.所以当 x22 时，g(x1)g2x220，即 g(x1)g2x22，因为 0 x11，02x221，g(x)在(0，1)上单调递增，所以 x12x22，故 x1 x222.综上所述，x1 x222.4.(2017 届重庆市一中月考)已知函数f(x)aln xax3(aR).(1)当 a 0 时，求函数f(x)的单调区间；(2)假设函数yf(x)的图象在点(2，f(2)处的切线的倾斜角为45，且函数g(x)12x2 nxmf(x)(m，nR)，当且

14、仅当在x1 处取得极值，其中f(x)为 f(x)的导函数，求m 的取值范围.解(1)f(x)a 1xx(x0)，当 a0 时，令 f(x)0，得 0 x1，令 f(x)0，得 x1，故函数 f(x)的单调递增区间为(0，1)，单调递减区间为(1，).(2)因为函数yf(x)的图象在点(2，f(2)处的切线的倾斜角为45，则 f(2)1，即 a 2，所以 g(x)12x2nxm 22x，文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M

15、10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6

16、E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3

17、I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5

18、K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码

19、:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2

20、I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O

21、1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1所以 g(x)xn2mx2x3nx22mx2，因为 g(x)在 x 1 处有极值，故g(1)0，从而可得n 12m，则 g(x)x3nx22mx2x1 x22mx2mx2，又因为 g(x)仅在 x1 处有极值，所以 x22mx2m0 在(0，)上恒成立，当 m0 时，2m0，易知?x0(0，)，使得 x202mx02m0，所以 m 0不成立，故m0，当 m0 且 x(0，)时，

22、x22mx2m0 恒成立，所以 m 0.综上，m 的取值范围是(，0.5.(2017湖北沙市联考)已知函数f(x)ex(ln x2k)(k 为常数，e 2.718 28是自然对数的底数)，曲线 yf(x)在点(1，f(1)处的切线与y 轴垂直.(1)求 f(x)的单调区间；(2)设 g(x)1 x ln x1ex，对任意x0，证明：(x1)g(x)exex2.(1)解因为 f(x)1xln x2kex(x0)，由已知得f(1)1 2ke0，所以 k12.所以 f(x)1xln x1ex，设 k(x)1xln x1，则 k(x)1x21x0 在(0，)上恒成立，即 k(x)在(0，)上单调递减，

23、由k(1)0 知，当 0 x1 时，k(x)0，从而 f(x)0，当 x1 时，k(x)0，从而 f(x)0.综上可知，f(x)的单调递增区间是(0，1)，单调递减区间是(1，).(2)证明因为 x0，要证原式成立即证g xex1e2x1成立.当 x1 时，由(1)知 g(x)01 e2成立；当 0 x1 时，ex1，且由(1)知，g(x)0，所以 g(x)1xln xxex1xln xx，设 F(x)1xln x x，x(0，1)，则 F(x)(ln x2)，当 x(0，e2)时，F(x)0，当 x(e2，1)时，F(x)0，所以当 xe2时，F(x)取得最大值F(e2)1e2，所以 g(x

24、)F(x)1e2，文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z

25、5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编

26、码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R

27、2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1

28、O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6

29、M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y

30、6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1即当 0 x1 时，g(x)1e2.综上所述，对任意x0，g(x)1 e2恒成立.令 G(x)exx1(x0)，则 G(x)ex1 0 恒成立，所以G

31、(x)在(0，)上单调递增，G(x)G(0)0 恒成立，即exx10，即 01ex1x1.当 x1 时，有g xex01e2x1；当 0 x1 时，由 式，g xex1e2x1.综上所述，当x0 时，g xex1e2x1成立，故原不等式成立.6.(2017西安模拟)已知函数f(x)k4kln x4x2x，其中常数k0.(1)讨论 f(x)在(0，2)上的单调性；(2)当 k4，)时，假设曲线yf(x)上总存在相异的两点M(x1，y1)，N(x2，y2)，使曲线 yf(x)在 M，N 两点处的切线互相平行，试求x1x2的取值范围.解(1)由已知得，f(x)的定义域为(0，)，且 f(x)k4kx

32、x24x2x2 k4kx 4x2xkx4kx2(k 0).当 0k2 时，4kk 0，且4k2，所以 x(0，k)时，f(x)0；x(k，2)时，f(x)0.所以函数f(x)在(0，k)上是减函数，在(k，2)上是增函数；当 k2 时，4kk2，f(x)0 在区间(0，2)内恒成立，所以 f(x)在(0，2)上是减函数；当 k2 时，04k2，k4k，所以当 x 0，4k时，f(x)0；x4k，2 时，f(x)0，所以函数在0，4k上是减函数，在4k，2 上是增函数.(2)由题意，可得f(x1)f(x2)，x1x20 且 x1x2，即k4kx14x21 1k4kx24x221，化简得，4(x1

33、x2)k4kx1x2.由 x1x2x1x222，文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10

34、G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2

35、 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6

36、G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1

37、文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:C

38、L9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I1

39、0M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1得 4(x1x2)k4kx1x222，即(x1 x2)16k4k对 k4，)恒成立，令 g(k)k4k，则 g(k)14

40、k2k24k20 对 k4，)恒成立.所以 g(k)在 4，)上是增函数，则g(k)g(4)5，所以16k4k165，所以(x1x2)165，故 x1 x2的取值范围为165，.文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文

41、档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL

42、9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10

43、M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 H

44、F6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G

45、6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2

46、ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1文档编码:CL9R2I10M1O1 HF6M10G6Y6E2 ZD3I6G1Z5K1