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扫描全能王创建扫描全能王创建扫描全能王创建扫描全能王创建浙江省Z20联盟2021届高三第三次联考数学参考答案第1页共4页浙江省Z20联盟(名校新高考研究联盟)2021届高三第三次联考数学参考答案一、选择题1-5:DAACD6-10:BBAAC
二、填空题11.1,20−−12.13,6013.1683,314.215.533,10316.2217.+6232,21三、解答题18.解:(1))6sin(3cos23sin23)(+=+=xxxxf...................................
.....4分Z))(k,π(kπ−06对称中心为.........................................6分(2))6sin(3)(++=xxg...............
..........................8分)4,0(,54cos,53sin,43tan),2,0(===...................................10分]2,0[x
,]32,6[6++++x当26=++x时,3)(max=xg,当+=++326x时,103312)32sin(3)(min−=+=xg.−3,103312)(xg......................14分19.解(1)证明:取M
D中点N,连接PN,QN.在MBD中,PN//BD.........................................2分在ACD中,3AQQC=,3ANND=NQ//CD............
.............................4分平面PQN//平面BCDPQ平面PQNPQ//平面BCD.........................................6分(2)取BC中点E,连接DE,AE,则DE⊥BC,BC⊥AD,A
DDE=DBCAED⊥平面且AED为二面角A-BC-D的平面角.........................................8分浙江省Z20联盟2021届高三第三次联考数学参考答案第2页共4页不妨设CD=1,则AD=23,DE=32,由余弦定理可得AE=32...
..................................9分方法一:(定义法)由题意得:面⊥ABC面AED,过点M作AEMF⊥,连接BF则⊥MF面ABC,所以MBF是直线BM与面ABC所成角................................
.12分由题意得:1==ACAB,所以8321,47===AMMFBM,1473sin==MBMFMBF............................15分方法二:(坐标法)以E为原点,建系。B(12−,
0,0),C(12,0,0),D(0,32,0),A(0,34−,34)设平面ABC的法向量()n=,,xyz0,1330.244nBCxnABxyz===−+−=()0,3,1n=............................
13分设所求角则sin=37cos,14nBM=............................15分20.解(1)112111G()()()1nnnnnnaaaaaaaa+−+=−+++−+=−.....
.......................2分12111Tnnnnnnbbbbbbb++−==............................4分nG22nT=−1211−=++nn
ba............................5分nn31S2−=nn-11n13131b3(2)22nnnSSn−−−−=−=−=1b1=1nb3n−=121231nnnab−=−=−.......
.....................7分(2)nSna=n-11Sn2)na−=(n1b2)nnaan−=−(21nnab=−............................9分112)2nnnaaan−+
=−()2(121+=−naannn121,2nnnab−=−=.......................................11分011121223n112231222(21)(21)(21)(21)(21)(21)nnnnnbbbaaaaaa−+++++=+
+−−−−−−121223nn112231n111111112212121212121111-221nnnbbbaaaaaa++++++=−+−+−−−−−−−=−()()()()..............13分n1n11111-(1
)22121n++−−−()n21)(1)n−−(浙江省Z20联盟2021届高三第三次联考数学参考答案第3页共4页n21)(1)n−−(13−........................................
..................15分21.解(1)焦点坐标)1,0(,准线方程1−=y.......................................................4分(2)已知2004xy=,则点A处的切线方程:20024xxyx
=−,...........................6分同(1)得:2000222200412()()4xtxrxxxrtr−=−−−+−=,化简得:224200030216xttxx+−−=...
................8分由0t得:2420002000422(0)2xxxtyyyt−++==−++..........................................10分
设1122(,),(,)ExyFxy则由120kk+=得:1020044xxxx+++=,即0122xxx−=+,................................................................
.............................11分所以021212EFxyykxx−==−−,由8OMNO=得(0,)8tN−所以,直线0:28xtlyx=−−,则2200000200231||2884114xyt
yyydyx++++==++.....................................................................13分0002323181841yyy=+−++关于0y单调递增所以,当01y=时,min232416d+=,....
.........................................................................15分此时,直线l与抛物线相交.22.解:(1)设1)12()12ln()12()(2+−−−=−=xxxfxg,2)1(),12(412
2)(−=−−−=gxxxg,且0)1(=g,切线方程:)1(2−−=xy................................................................4分(3)
1()2fxaxx=−,若0a,则()fx单调,至多一个零点;若0a,则212()axfxx−=,11()(0,),(,)22fxaa+在浙江省Z20联盟2021届高三第三次联考数学参考答案第4页共4页1
11()ln(2)002222efaaa=−+................................................................8分由极值点偏移证得exx221+....................
............................11分12222222122−+−+−xxexxxx只需证222211xxaa++,即证21xa,即证21()()fxfa即证10()fa,即证11ln1
aa−成立..................................................................15分