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山东省烟台市2021-2022学年高一上学期期末考试数学试题答案

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2021~2022学年度第一学期期末学业水平诊断高一数学参考答案一、选择题ACBCDABB二、选择题9.BD10.ACD11.AC12.ABD三、填空题3113.214.15.≤<a116.1,2<<c222π6四、解答题322117.解:(1)原式=23×−()33························································1分2−×+2211=−+21·····························································································3分21=−····································································································5分2lg32lg2(2)原式=×+×lg(25)·····························································8分lg2lg3=+=213·····························································································10分a418.解:(1)由题意得:cosα==−,··········································2分a2+95解得a=−4.···························································································4分3所以tanα=−.····················································································6分4−+sinαα2cos(2)原式=·····································································8分−+3cosααsin−+tanα2分子分母同除以cosα,则上式=.··············································10分−+3tanα11=−·······················································12分15πππ19.解:(1)令−+22kxππ≤−≤+2k,···········································3分232高一数学参考答案(第1页,共5页)\nππ5解得−+≤≤+kxππk,·································································5分1212ππ5所以,函数fx()单的调递增区间为[−+kππ,+kk](∈Z).·····················6分12121π1(2)不等式fx()≥,即sin(2x−≥).232πππ42ππ令zx=2−,x∈−[,],所以z∈−[,],····································7分3223347ππππ21因为yz=sin在[−−,][,]上满足sinz≥.······························8分3663247ππππππ2由−≤2x−≤−或≤−≤2x,···········································9分336633ππ5ππ解得−≤≤−x或≤≤x.·····························································11分212421π5πππ所以不等式fx()≥的解集为{|xx−≤≤−或≤≤x}.····················12分2212422220.解:(1)由sinxx+=cos1得:2fx()=−+2cosxcosx,········································································1分1令fx()0=,解得cosx=0或cosx=,···················································2分2π当cosx=0时,xk=+π,k∈Z;·······················································3分21π当cosx=时,xk=2π±,k∈Z.·······················································5分23ππ所以函数fx()的零点为+kπ,2kπ±,k∈Z.····································6分232(2)因为fx()=−+2cosxcosx,2令cosxt=,t∈−[1,1],则ft()=−+2tt,···············································7分高一数学参考答案(第2页,共5页)\n2因为fx()的最小值为−1,所以−21tt+≥−,11解得−≤≤t1,即−≤cosx≤1,··························································9分222π21π因为x∈[,]α,且cos=−,··························································10分33222ππ所以α的取值范围为[,)−.······························································12分3321.解:(1)若选用y=log(axmb++),1log(2++=mb)a45则依题意得:log(3a++=mb),··························································2分49log(5++=mb)a411解得a=2,m=−1b=,所以yxx=log(−+1)(≥2).····························4分244若选用函数ycxnd=++模型,1c2++=nd45则依题意得:c3++=nd,·······························································6分49c5++=nd4151151解得c=2,n=−,d=−,所以yx=2−−≥(x2).·················8分8484113(2)对于函数yx=log(−+1),当x=9时,y==3.25万元.················9分244151571对于函数yx=2−−,当x=9时,y=−万元.······················10分8424571因为|−−≈3.3|0.225>|3.253.3|=0.05−,··········································11分24高一数学参考答案(第3页,共5页)\n1所以选用yxx=log(−+1)(≥2)模型更合理.···········································12分2422.解:(1)当x∈(1,+∞)时,logx>0,211所以4logxx+≥24log⋅=4,···········································2分22logxxlog221当且仅当4logx=,即x=2时,等号成立.·····································3分2logx2所以,函数fx()在区间(1,+∞)上的最小值为4.············································4分xx+12xx(2)gxm()=⋅+−=42mm(2)+⋅−22m,x∈[1,2].x2令2=t,则上述函数化为yt()=+−mt2tm,t∈[2,4].·····························5分1因为m<0,所以对称轴t=−>0,m11当−≤2,即m≤−时,函数yt()在[2,4]上单调递减,所以当t=2时,ym=34+;maxm2·········································································································6分11111当24<−<,即−<<−m时,函数gt()在[2,−]上单调递增,在[−,4]上m24mm11单调递减,所以yy=−=−−()m;···············································7分maxmm11当−≥4,即−≤<m0时,函数gt()在[2,4]上单调递增,所以m4yym=(4)15=+8;max111综上,当−≤<m0时,gx()的最大值为15m+8;当−<<−m时,gx()的最大42411值为−−m;当m≤−时,gx()的最大值为34m+..·····························8分m2(3)对∀∈x(1,+∞),∃∈x[1,2],使得fx()()7+>gx成立,等价于1212gx()7()>−fx成立,即gx()>−(7fx())···································9分21max1max由(1)可知,当x∈(1,+∞)时,(7−=fx())7−=fx()3.maxmin高一数学参考答案(第4页,共5页)\n因此,只需要gx()>3.max111所以当−≤<m0时,15m+>83,解得m>−,所以−≤<m0;434111−−35−+35当−<<−m时,−−>m3,解得m<或<<m0,所以,24m22−+351<<−m;2411当m≤−时,343m+>,解得m>−,此时解集为空集;························11分23−+35综上,实数m的取值范围为<<m0.··············································12分2高一数学参考答案(第5页,共5页)

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所属: 高中 - 数学
发布时间:2022-08-10 13:00:05 页数:5
价格:¥5 大小:262.20 KB
文章作者:fenxiang

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