浙江省杭州市蕭山區(qū)2022年高考數(shù)學(xué)模擬試卷10理新人教版

浙江省杭州市蕭山區(qū)2022年高考模擬試卷10數(shù)學(xué)卷（理科）考生須知：1．本卷滿分150分，考試時間120分鐘。2．答題前，在答題卷密封區(qū)內(nèi)填寫學(xué)校、班級和姓名。3．所有答案必須寫在答題卷上，寫在試題卷上無效。4．考試結(jié)束，只需上交答題卷。參照公式：如果事件A、B互斥，那么球的體積公式S4R2VShV1ShB1DC131開始第6題圖kknkPn(k)P(1P)A1Cn=1V1h(S1S1S2S2)2BCS=03ES4R2V球4R3UR正視圖側(cè)視圖AS≤2022否311是俯視圖第12題輸出S=S2-1A{x|x21,xR}B{x|2x1,xR}ACUB=2結(jié)束1題圖A.[1,1]B.[0,1]C.(0,1]D.[第711,0)a1,a516A.1B.2C.9D.321534ix(a,b),y(3,4)16abiA.x//yB.xyC.|x||y|D.x與y夾角為3A(3,0),B(0,4),C(cos,sin),RABCA.7917D.21R2B.C.、222p:設(shè)q:sincossincos1CCA1E//B1DCADB1DCB1DC1ACC1A1B1DCA.65B.63C.33D.32A.1B.1C.1D.11F1、F2F1PF21200A.1e13e21362765444B.3e121e221C.311D.131yba0,b0444e124e224e124e22xa12xy20xyA.2B.3C.4D.12(2x)68xy40z0,b0)x2(aabx0,y0|OA|2,|OB|1|OC|13OCOAOB2OCOA________f(x)x22ax1,x02011(m,n)|xa|a,x0amn11第17題圖23456789第16題圖101112131415161718192021222324252627282930313233343536.................................圖甲12457910121416f(x)23sinxcosx2cos2x1ABC1719212325262830323436.................................圖乙f(A)3bc4a1,a2,a4S530Tn(n2nbnSmn)3nN,mNTn12ABCDAB2,CD4,ADCBCD600ADEABCEABDCADEABEABCEDDECDECECA圖1BA圖2BA圖3BC1:x2y21(ab0)OAOBa(3,1)k1k3OA4a2b22OMOB是否在橢圓上問55題的探索，從能力上來說考察學(xué)生探究能力；從知識點上來說，考察橢圓上點坐標擁有的特點。（5）對原題作改編，特別是增加第2問的設(shè)置、更改第3小題的問題的設(shè)置，適合減少難度。22（本小題滿分15分）（2022山東卷理科第22題）設(shè)函數(shù)f(x)x2bln(x1)，其中．1（Ⅰ）當b時，判斷函數(shù)在定義域上的單一性；2（Ⅱ）求函數(shù)的極值點；（Ⅲ）證明對隨意的正整數(shù)，不等式ln1111都成立．nn2n3（1）考察函數(shù)的單一性、導(dǎo)數(shù)的應(yīng)用、不等式的證明方法。（2）滲透分類議論思想；（3）考察利用結(jié)構(gòu)函數(shù)證明不等式問題。2022年高考模擬試卷數(shù)學(xué)參照答案（理科）一、（10×5=50）題1234567891號0答CABCADADCB案二、填空題（7×4=28）491213_______141534___（45,38）__三、解答題（共72分）1814分解：f(x)3sin2xcos2x2sin(2x).........分6f(x)最小正周期為T2分2.................................................................................4令2k2x2k得kxk，kZ..........分26263f(x)單一遞增區(qū)間為[k,k],kZ分63.................................................6由知f(A)2sin(2A)3sin(2A)3...........7分32123又0A2A或2A2A或A...9分333332........................................當A3時，由余弦定理可知，a2b2c22bccosAb2c2bc(bc)23bc(bc)23(bc)242當且僅當bc時等號成立此時的最小值為2..........分2a.........................................11當A2時，由勾股定理可知，a2b2c2(bc)22bc(bc)22(bc)282當且僅當bc時等號成立此時的最小值為22..........分2a...........................................13當A時，a的最小值為2；當A時，的最小值為22。..........14分32a..............................19（14分）a2aa4221(a1d)a1(a13d)a1d54解：由條件可知，S55a130S55a110d305a1分19.(1)2d10d30...4d0d0d0解得a1d2.................................................................................................................6分由條件可知，a1d且d0Snna1n(n1)dn(n1)分(2)22d..........................7bn前項和Tn(n2n)3n當時，b1T1,此時b11;..............8分nn1T1....................當時，bnTnTn1bnTnTn11Tn1222分n2TnTnTn33(n1)3....................10由上剖析可知，對nN,bn2分Tn3..........................................................................11要使nN,mN，有bnSm成立，只需22d...........13分Tn3(Sm)min，即..........3等差數(shù)列{an}公差d2分（，0]....................................................................................1432014分DDECDECECD’FooA圖1BA圖2BAB圖3解：（1）在圖3中取AE中點O，成立直角坐標系。平面DAE平面ABCE,DOAE,DO平面DAE,平面DAE平面ABCEAEDO平面ABCE易知BABEBOAC可取O為原點成立直角坐標系分..............................................2得A(1,0,0),B(0,3,0),C(2,3,0),D(0,0,3),E(1,0,0)AB(1,3,0),DB(0,3,3),CB(2,0,0)............................................................3分設(shè)平面ABD法向量為n(x,y,z),平面CBD法向量為m(x,y,z)則ABnx3y0x3y可取平面ABD法向量n(3,1,1).........................4分DBn3y3z0yzCBm2x0x0可取平面ABD法向量m(0,1,1)......5分DBm3y3z0yz...................cosn,mnm10分|n||m|5.............................................................................................72在圖2中作DFD'B,交D'B于F點。易知DOAE,D'OAE,AE平面DD'BAEDF又DFD'BDF平面ABCE.在折動過程中，直線DC與平面所成的角為DCF分ABCE..............................9設(shè)D'Ft[0,23],則FB23t,FCFB2BC2t243t16.FO|t3|,DFDO2FO223tt22DCFDF223tt2分tanFC2t243t16.......................................................................11t223(t43)(t43)設(shè)g(t)23t,t[0,23],則g(t)3t243t16(t243t16)2t(0,4時,g(t)；t(43,23),g(t)033)03t4時,g(t)max1分332,..........................................................................................13即t4時,tanDCF最大值為2分332.........................................................................14備注：（）關(guān)于tan2DCF23tt2也可用基本不等式求最值。1t243t16（）本小題也能夠用向量法解決。2x2y

2解：設(shè)F(c,0),則直線方程為yxc,代入橢圓方程：C1:a2b21.l

2

1，得(a2b2)x22ca2xa2c2a2b20易知恒成立設(shè)A(x1,y1),B(x2,y2)0.x1x22a2c,x1x2a2c2a2b2,....................................................2分a2b2a2b2y1y2x1x22c分.....................................................................3OAOB(x1x2,y1y2),a，且(OAOB)//a(3,1)(x1x2)3(y1y2)0,即4(x1x2）6c0..............................................4分42a2c得a23b2...............................5分a2b26c........................................a23(a22),得c6橢圓離心率為6分ca33..............................................6由可知a23b2,c22b2,x1x2a2c2a2b23b2，...........7分(2)(1)a2b24....................b2y1y2(x1c)(x2c)....9分4......................................................................k1k2y1y21x1x23直線OA斜率k1與直線OB斜率k2乘積為定值1.....10分3........................................(3)設(shè)點M為則3(x1,y1)455x03x14x255分3y4y.........................................................................................11y05152且x12y121,x22y22，即223b223y23b2........................12分3b2b23b2b21x13y1,x22且由(2)知x1x23y1y2022x2y222(3x4x)23(3y4y)2x0y000x03y051525152a2b23b2b23b23b29x216x224xx39y216y224yy251252251225125225123b29(x23y2)16(x2316y2)24(xx3yy)25112522522512123b293b2163b202525分3b21....................................................................................14點為切合橢圓方程，點M在橢圓上.........................................................15分M(x0,y0)2215If(x)x2bln(x1)1,f'(x)2xb12x22xb1xx1g(x)2x22xb1,1,1222g(x)ming(1)1bb1g(x)min1b032222g(x)2x22xb01,f'(x)0,4b1,1,521II1Ib6212(x1)212b2x1,f'(x)0,f'(x)x1