烟台期末-数学答案

2022～2023学年度第一学期期末学业水平诊断高三数学参考答案及评分标准一、选择题DBBCACDA二、选择题9.BC10.ACD11.ACD12.ABD三、填空题3113.114.15.6716.22四、解答题17.解：（1）由正弦定理可得sinAcosC+sinACsin=sinB，·······················1分因为ABC++=π，所以sinAcosC+sinAsinC=sin(AC+)，即sinAcosC+sinACsin=sinAcosC+cosACsin，····························2分整理得：sinACsin=cosACsin，因为0<==，31×1231393故平面ABD与平面BCD夹角的余弦值为.······································12分31220.解：（1）设该容器的体积为V，则V=ππrl23+r，31601602又V=π，所以lr=−，··········································································2分333r2因为lr≥6，所以02<≤r.························································································4分916029所以建造费用y=2πrl×+3ππrm22=2(r−r)×+3πrm，4334r2240π因此y=3π(mr−1)2+,0<≤r2.···································································5分r240ππ6(m−1)40（2）由（1）得y′=6π(mr−1)−=(r3−),0<≤r2.··········6分rr22m−1934040由于mm>,所以−>10,令r−=0，得r=3.···························7分4m−1m−1高三数学答案（第3页，共6页）4040若3<2，即m>6，当r∈(0,3)时，y′<0，yr()为减函数，当m−1m−14040r∈(3,2)时，y′>0，yr()为增函数，此时r=3为函数yr()的极小值点，m−1m−1也是最小值点.······················································································································9分409若3≥2，即<≤m6，当r∈(0,2]时，y′<0，yr()为减函数，此时r=2m−14是yr()的最小值点.···········································································································11分9综上所述，当<≤m6时，建造费用最小时r=2；当m>6时，建造费用最小时440r=3.························································································································12分m−121.解：（1）设，，，Aa(−,0)Ba(,0)Px(,11y)yy−−00y21则=×=111=，1分kkAPBP22··············································x11+−axax1−a4xy22又因为点Px(,y)在双曲线上，所以11−=1.·····································2分11ab221ab22于是y2=x2−=xb22−，对任意x≠0恒成立，1144a211b21所以=，即ab22=4.···································································3分a24又因为c=5，c2=ab22+，可得a2=4，b2=1，x2所以双曲线C的方程为−=y21.······················································5分4（2）设直线的方程为：，，由题意可知，lx=ty+5Mx(33,y),Nx(44,y)t≠±2高三数学答案（第4页，共6页）x2−=y21联立4，消x可得，(t22−4)y+25ty+=10，x=ty+5−25t1则有yy+=，yy=，···················································6分34t2−434t2−4假设存在定点Dm(,0)，则DMDN=(x3−m)(x4−+m)y34y7分=(ty3+−5)(5)mty4+−+my34y······················································22=+(t1)yy34+−(5mt)(y3++−y4)(5m)t22+−125(5mt)=−+−(5m)2tt22−−44(mtm22−−4)(42−85m+19)=····························································8分t2−475令4mm22−85+=194(m−4)，解得m=，··········································10分824511此时DMDN=m2−=44−=−，························································11分64647511所以存在定点D(,0)，使得DMDN为定值−.···································12分86422.解：（1）f()x=−−xexax22ax，则fx′()=+−(