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哈三中2020-2021学年度高三年级线上学习阶段性考试理科数学答案一、选择题123456789101112BBAADBDCABAC二、填空题13.14.3415.3516.4三、解答题17.(1)由已知得()0.0150.03100.85b++=,解得0.04b=
,.........................................2分又()0.0051010.85a+=−,解得0.01a=,.........................................2分所以评分的平均值为550.05650.1750.3
850.4950.1580++++=..........................................6分(2)由题意可得,22列联表如下表:态度性别满意不满意合计男生203050女生351550合计5545100因此()2210020153530
9.0916.63555455050K−=,.........................................11分∴能有99%的把握认为对“线上教学是否满意与性别有关”.........................
................12分18.(1)直棱柱111CBAABC−1CCABC⊥平面BCABC平面1BCCC⊥1111,ACBCACCCCACCCAACC⊥=且、平面11BCAACC⊥平面111DCAACC平面1BCDC⊥由勾股定理可得1DCDC⊥
由因为,DCCBCDCCBBCD=、平面⊥1DC平面BCD;.........................................5分..............................8分..............................2分.......
.......................2分(2)以C为坐标原点建立空间直角坐标系,则1(0,1,2)BC=−,11(1,0,)2DC=−.................................6分易求
平面1BCD的一个法向量为(1,4,2)m=,.........................................8分同理平面1BCC的一个法向量为(1,0,0)n=.........................................9分12
1cos,2121mnmnmn===........................................11分二面角CBCD−−1的余弦值为2121......................
...................12分19.设小李同志第i次套圈命中为事件iA(1)恰好命中三次为事件A.........................................1分()123
3221()4333PAPAAA===.........................................3分(2)X的可能取值为0,1,2,3,5.........................................4分则()1231111(0)43336PXPA
AA====()()112312321211(1)4339PXPAAAPAAAC==+==()1233111(2)43312PXPAAA====()()()11231231232122321114(3)433433939PXPAAA
PAAAPAAAC==++=+=+=()1233221(5)4333PXPAAA====.........................................9分X的分布列为X01235P136191124913则59()18
EX=....................................12分..............................10分20.(1)'1()fxmx=−,则'(1)10fm=−=1m
=........................................2分()lnfxxx=−,'11()1xfxxx−=−=令'()0fx,得01x;令'()0fx,得1xx(0,1)1
(1,)+'()fx+0−()fx单调递增极大值单调递减......................4分()fx的增区间为(0,1),减区间为(1,)+,且()fx的极大值为(1)1f=−.................
.................6分(2)2()1112fxxmx++对于任意的0x恒成立,即21ln122xxmxx+++设2ln1()2xxhxxx++=+,则'22(1)(2ln)()(2)xxx
hxxx−++=+..................................8分令'()0hx,即2ln0xx+,设()2lngxxx=+,()2lngxxx=+单调增,且1()02g,(1)0g01(,1)2x使0()0gx=,即'0()
0hx=,所以002ln0xx+=..................................10分()hx在0(0,)x单调递增,在0(,)x+单调递减000max022000001ln1112(
)()(,1)2222xxxhxhxxxxxx+++====++02()(1,2)mhxm的最小整数值为2..................................12分21.(1)设抛物线2:4Cxy=上切
点为200(,)4xPx24xy=0''0,22xxxxyy===上,切线为20001()42xyxxx−=−..................................1分整理得200240xxyx−−=,圆22:(1)1Exy++=的圆心(1,0)−到切线的
距离为202041416xdx−==+..................................2分化简得4200120xx−=,解得00x=或23,即0k=或3切线为0y=或33yx=−或33yx=−−..........................
........4分(2)设切线为ykxm=+,则圆22:(1)1Exy++=的圆心(1,0)−到切线的距离为2111mdk+==+22(1)1mk+=+..........................
........6分联立24ykxmxy=+=,得2440xkxm−−=,得121244xxkxxm+==−..................................8分2222201
020000(1)()()(1)44(1)4PAPBkxxxxkxkxmkxy=+−−=+−−=+−22(1)MEm=+,222200(1)4(1)1kxymk+−=+=+20041xy−=P在214yx=下部20040xy−20041xy−=−联立200220041(1)1xyxy
−=−++=,即200610yy++=,0223y=−或0223y=−−(舍)..........................10分221PMME+=,即22200()1(1)xymm+−+=+化简得2200000232
22122222222xyymyy+−−−====++−..................................12分22.(1)①当12x−时,1201xxx−−+−,所以1x−②当102x−时,12103xxx++−,所以103x−③当0x时,101
xx+−,所以0x综上,不等式的解集为(1,1,3−−−+..................................5分(2)原式即121222axxaxx+−+−由绝对值三角不等式,1122xx+−,即111222xx−+−
122a−,即1a−.................................10分23.(1)曲线1222:212xtCyt=−=+(t为参数),消去参数t,得30xy+−=,其极坐标方程为()cossin3+=,即32sin42+=
.曲线2C:4cos=,即24cos=,即2240xyx+−=,所以曲线2C的直角坐标方程为()2224xy−+=...................................5分(2)2(2,0)C到1:3Cxy+=的距离223222dCN−===又21CM=
,得222222MNCMCN=−=..................................10分