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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.
3850.4950.1580++++=..........................................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()22100201535309.0916.63555455050K−=,.........................................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121cos,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()1233221()4333PAPAAA===.........................................3分(2)X的可能取值为0,1,2,3,5...............
..........................4分则()1231111(0)43336PXPAAA====()()112312321211(1)4339PXPAAAPAAAC==+==()1233111(2)43312PXPAAA====()()()1123123
1232122321114(3)433433939PXPAAAPAAAPAAAC==++=+=+=()1233221(5)4333PXPAAA====...............................
..........9分X的分布列为X01235P136191124913则59()18EX=....................................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)xxxhxxx−++=+.........
.........................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002l
n0xx+=..................................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''0,2
2xxxxyy===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2222201020000(1)()()(1)44(1)4PA
PBkxxxxkxkxmkxy=+−−=+−−=+−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220000023222122222222xyymyy+−−−====++−.......
...........................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111222
xx−+−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