三明市2024年普通高中高三毕业班质量检测数学参考答案及评分细则评分说明：1．本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分标准制定相应的评分细则．2．对计算题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应给分数的一半；如果后继部分的解答有较严重的错误，就不再给分．3．解答右端所注分数，表示考生正确做到这一步应得的累加分数．4．只给整数分数．选择题和填空题不给中间分．一、选择题：本大题考查基础知识和基本运算．每小题5分，满分40分．1．C2．C3．D4．A5．A6．B7．B8．C二、选择题：本大题考查基础知识和基本运算．每小题6分，满分18分．全部选对的得6分，部分选对的得部分分，有选错的得0分．9．BC10．ACD11．BCD三、填空题：本大题考查基础知识和基本运算．每小题5分，满分15分．112．613．,314．6,7,8,9,21（第一空2分，第二空3分）e四、解答题：本大题共5小题，共77分．解答应写出文字说明、证明过程或演算步骤．15.解法一：（1）证明：取BD的中点M，连接PM、MC,·······················1分∵△BPD和△BCD均为等边三角形，∴BDPM，BDCM.··································································2分又PMCMM，∴BD平面CPM，·········································································3分又CP平面CPM，∴BDCP.····················································································4分（2）以M为原点，MB,MC所在直线为x,y轴，过M作平面BCD的垂线所在直线为z轴，如图所示建立空间直角坐标系，···········································5分∵平面ABD平面PBD，平面ABD平面PBDBD，PM平面PBD，PMBD∴PM平面ABD.∵△PBD和△CBD均为等边三角形，∴PMMCPC3，PMC60，第1页共11页{#{QQABDYaEoggoAJJAABhCQQUgCkAQkAEAAKoGwAAIMAAAiBFABCA=}#}33∴P0,,，C0,3,0，B1,0,0，··············································6分223333∴BP1,,，BC1,3,0.MP0,,2222设平面PBC的法向量为m(x,y,z)33mBP0,xyz0,∴即22mBC0x3y0取z1,则m3,3,1，···································································8分33平面ABD的法向量MP0,,，·················································10分22设平面ABD与平面PBC的夹角为，MPn339∴coscosMP,n··································12分MPn3131339∴平面ABD与平面PBC夹角的余弦值为.····································13分13解法二：（1）同解法一······································································4分（2）如图，取MC的中点E为原点，连接PE，过点E作EF//MB,交BC于点F,由（1）知CMBD,EFMC,又由（1）知BD平面CPM，又PE平面CPM，∴BDPE，∵△PBD和△CBD均为等边三角形且棱长为2，∴PMMCPC3，PEMC,BDMCMPE平面CBD以E为原点，EF,EC,EP所在直线为x,y,z轴，建立空间直角坐标系，如图所示··························································5分∵平面ABD平面PBD，平面ABD平面PBDBD，PM平面PBD，PMBD∴PM平面ABD，第2页共11页{#{QQABDYaEoggoAJJAABhCQQUgCkAQkAEAAKoGwAAIMAAAiBFABCA=}#}33平面ABD的法向量MP0,,···················································7分22333∴P0,0,，C0,,0，B1,,0·············································8分22233∴CB1,3,0，CP0,,，22设平面PBC的法向量为mx,y,z，x3y0mCP0∴,即33，取z1,则m3,3,1，·················10分mCB0yz022设平面ABD与平面PBC的夹角为，MPm339∴coscosMP,m，······························12分MPm3131339∴平面ABD与平面PBC夹角的余弦值为.····································13分1316.解法一：13（1）由题意f(x)sinxcos(x)sinxcosxsin(x)6223·····································································································2分π因为fx图象的两条相邻对称轴间的距离为，22πππ所以周期T2，故2，所以fxsin2x，·····················4分23πππ当x0,m时，2x,2m，·················································5分333ππ3π因为fx在区间0,m上有最大值无最小值，所以2m，·········6分232π7ππ7π解得m，所以m的取值范围为,.···································7分12121212第3页共11页{#{QQABDYaEoggoAJJAABhCQQUgCkAQkAEAAKoGwAAIMAAAiBFABCA=}#}（2）将函数fx图象向右平移个单位长度，6得到ysin2(x)sin2x的图象，············································8分63再将图象上所有点的横坐标变为原来的2倍（纵坐标不变），得到g(x)sinx的图象，···································································9分11所以函数h(x)xsinx，所以h(x)cosx，································10分221令h(x)0得cosx，2因为x(2,)，4所以当x(2,)时，h(x)0，h(x)单调递增，····························11分342当x(,)时，h(x)0，h(x)单调递减，································12分3322当x(,)时，h(x)0，h(x)单调递增，··································13分332当x(,)时，h(x)0，h(x)单调递减.·········································14分342所以函数h(x)的极大值点为和.··············································15分33解法二：（1）同解法一.·····································································7分（2）将函数fx图象向右平移个单位长度，6得到ysin2(x)sin2x的图象，············································8分63再将图象上所有点的横坐标变为原来的2倍（纵坐标不变），得到g(x)sinx的图象，···································································9分第4页共11页{#{QQABDYaEoggoAJJAABhCQQUgCkAQkAEAAKoGwAAIMAAAiBFABCA=}#}11所以函数h(x)xsinx，所以h(x)cosx，································10分221令h(x)0得cosx，222当2kx2k时，h(x)0，h(x)单调递增，33因为x(2,)4所以k1时，2x，h(x)单调递增，··································11分322k1时，xh(x)单调递增·················································12分3324当2kx2k时，h(x)0，h(x)单调递减，33因为x(2,)2k0时，x，h(x)单调递减，··············································13分342k1时，x，h(x)单调递减，······································14分3342所以函数h(x)的极大值点为和.··············································15分33解法三：（1）同解法一.·····································································7分（2）将函数fx图象向右平移个单位长度，6得到ysin2(x)sin2x的图象，············································8分63再将图象上所有点的横坐标变为原来的2倍（纵坐标不变），得到g(x)sinx的图象，···································································9分11所以函数h(x)xsinx