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.112.613.,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,∴BDPM,BDCM.··································································2分又PMCMM,∴BD平面CPM,·········································································3分又CP平面CPM,∴BDCP.····················································································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PBDBD,PM平面PBD,PMBD∴PM平面ABD.∵△PBD和△CBD均为等边三角形,∴PMMCPC3,PMC60,第1页共11页{#{QQABDYaEoggoAJJAABhCQQUgCkAQkAEAAKoGwAAIMAAAiBFABCA=}#}33∴P0,,,C0,3,0,B1,0,0,··············································6分223333∴BP1,,,BC1,3,0.MP0,,2222设平面PBC的法向量为m(x,y,z)33mBP0,xyz0,∴即22mBC0x3y0取z1,则m3,3,1,···································································8分33平面ABD的法向量MP0,,,·················································10分22设平面ABD与平面PBC的夹角为,MPn339∴coscosMP,n··································12分MPn3131339∴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CMBD,EFMC,又由(1)知BD平面CPM,又PE平面CPM,∴BDPE,∵△PBD和△CBD均为等边三角形且棱长为2,∴PMMCPC3,PEMC,BDMCMPE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PBDBD,PM平面PBD,PMBD∴PM平面ABD,第2页共11页{#{QQABDYaEoggoAJJAABhCQQUgCkAQkAEAAKoGwAAIMAAAiBFABCA=}#}33平面ABD的法向量MP0,,···················································7分22333∴P0,0,,C0,,0,B1,,0·············································8分22233∴CB1,3,0,CP0,,,22设平面PBC的法向量为mx,y,z,x3y0mCP0∴,即33,取z1,则m3,3,1,·················10分mCB0yz022设平面ABD与平面PBC的夹角为,MPm339∴coscosMP,m,······························12分MPm3131339∴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)sinxcos(x)sinxcosxsin(x)6223·····································································································2分π因为fx图象的两条相邻对称轴间的距离为,22πππ所以周期T2,故2,所以fxsin2x,·····················4分23πππ当x0,m时,2x,2m,·················································5分333ππ3π因为fx在区间0,m上有最大值无最小值,所以2m,·········6分232π7ππ7π解得m,所以m的取值范围为,.···································7分12121212第3页共11页{#{QQABDYaEoggoAJJAABhCQQUgCkAQkAEAAKoGwAAIMAAAiBFABCA=}#}(2)将函数fx图象向右平移个单位长度,6得到ysin2(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)xsinx,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342当x(,)时,h(x)0,h(x)单调递减,································12分3322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342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fx图象向右平移个单位长度,6得到ysin2(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)xsinx,所以h(x)cosx,································10分221令h(x)0得cosx,222当2kx2k时,h(x)0,h(x)单调递增,33因为x(2,)4所以k1时,2x,h(x)单调递增,··································11分322k1时,xh(x)单调递增·················································12分3324当2kx2k时,h(x)0,h(x)单调递减,33因为x(2,)2k0时,x,h(x)单调递减,··············································13分342k1时,x,h(x)单调递减,······································14分3342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fx图象向右平移个单位长度,6得到ysin2(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)xsinx
2024届福建省三明市普通高中高三毕业班5月质量检测数学参考答案
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