江苏省盐城市2024-2025学年高二下学期期末考试数学试卷参考答案

e56，··············································································································2分=9，所以公差d=2，···························································································4分16.解1）:正三棱柱ABC一A1B1C1，点E，G分别为棱AA1，CC1的中点，又:EG面EB1G,AC丈面EB1G,:AC//面EB1G.又:AC∩CF=C,且AC面ACF，CF面ACF,:面AFC//面EB1G.··································································································7分（2）法1：取AC的中点H，连接BH，B1H，AB1，CB1，:正三棱柱ABC一A1B1C1的体积为4，且AB=2，:H为AC的中点，正三角形ABC，:B1H丄AC,BH丄AC，:上BHB1为二面角B一AC一B1的平面角.··········································································12分法2：取BC的中点O，连接AO，过点O在平面BB1C1C中作OM//CC1,:正三棱柱ABC一A1B1C1，:CC1丄面ABC，:AO面ABC，BC面ABC，:CC1丄AO,CC1丄BC，:OM丄AO,OM丄BC，:O为BC的中点，正三角形ABC，:AO丄BC，设面ACB1的法向量n1=(x,y,z)，:n1.AC=0且n1.AB1=0,:CC1丄面ABC，:面ABC的法向量n2=CC1=(0,4,0)，分17.解1）记盲盒的外层包装A型为事件A，盲盒的外层包装B型为事件B，盲盒中含限量版商品为事件C，则P(C)=P(C|A).P(A)+P(C|B).P(B)····························································2分（2）小王抽中含限量版商品的盲盒数量为随机变量X，X~B(5,)，··································6分则随机变量X的概率分布为：(2,(2,(2,(2,(2,(2,X012345P1 1（3）若单个盲盒含限量版商品，该盲盒外层包装为A型的概率为条件概率··········································································12分2518.解1）:椭圆的离心率为e==，又经过点(1，)，得又:a2=b2+c2，解得a=2,b=:椭圆方程为.···································4分4又点A，B在椭圆上，:{()24l4443此时直线l的斜率为±.·····································································8分②设A(x1,y1)，B(x2,y2)，当直线l斜率不为0时，设直线l的方程：x=my+4，与椭圆E：联立，72m72m3(x21)所以f所以f(x)在(0,)上单调递减；在(,+∞)上单调递增.····················································4分22所以由与（1）同理可得f(x)在(0,)上单调递减；在(,+∞)上单调递增，所以···············································7分令只需证g(a)≥0即可.于是，类似可得φ(t)在(0,1)上单调递减；在(1,+∞)上单调递增，（3）不等式f(x)≥sinx恒成立，即ax2—lnx≥sinx恒成立，类似可得m(x)≥m(1)=0，所以x2—lnx≥x；······························································14分又令h(x)=xsinx(x>0)，所以h(x)>h(0)=0，所以x