高三数学试题答案（第1页，共7页）2025~2026学年度第一学期期中学业水平诊断高三数学参考答案一、选择题：1.C2.B3.D4.C5.A6.D7.B8.B二、选择题9.BC10.ABD11.AD三、填空题12.113.6(0,)π14.5,21四、解答题15.解：（1）由题知，=+++fxxx63()sin()2cos()ππ，所以=++−fxxxxx2222()sincos2(cossin)3113，即=−fxxx22()cossin33，所以=+fxx6()3cos()π.····························································2分因为fx()图象关于点3(,0)π对称，所以Z+=+kk362,ππππ，··························································4分所以=+k13，············································································5分又因为02，所以=1.························································6分（2）由（1）知，=+fxx63cos()()π.将函数fx()图象上各点横坐标缩短为原来的21倍（纵坐标不变），得到=+yx63cos(2)π，···································································8分再将得到的图象向左平移12π个单位，故得到函数=+gxx3()3cos(2)π.······10分当x2[0,]π时，+x3332[,]π4ππ，故当+x33]π2[,ππ，即x3[0,]π时，函数gx()单调递减，当+x33,]π2[π4π，即x32[,]ππ时，函数gx()单调递增.··················································11分所以−gx2()[3,]3.·································································13分高三数学试题答案（第2页，共7页）16.解（1）当08x时，()()20054WRxx=−+329720054xaxx=−−−，整理得3243200Wxax=−−；·····························································2分因为5x=时，年利润为890万元，所以2435125200890a−−=，···········3分所以1a=，·······················································································4分所以3243200xxW−−=，··································································5分当8x时，()()20054WRxx=−+5400253920054xx=−−−5400233954xx=−−.·················································6分综上：3243200,085400233954,8xxxWxxx−−=−−.···············································7分（2）当08x时，3243200Wxx=−−，22433Wx=−.所以当08x时，0W恒成立，所以W在()0,8上单调递增，·······························································9分所以当8x=时，max1232W=.························································10分当8x时，5400233954Wxx=−−.因为54005400542541080xxxx+=，当且仅当540054xx=，即10x=时取“=”.·········································································12分所以当10x=时，max1259W=.·······...