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)a45则依题意得：log(3a++=mb)，··························································2分49log(5++=mb)a411解得a=2,m=−1b=，所以yxx=log(−+1)(≥2).····························4分244若选用函数ycxnd=++模型，1c2++=nd45则依题意得：c3++=nd，·······························································6分49c5++=nd4151151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)
