湖北省2026年5月高三下学期壮行考试数学+答案

数学试卷参考答案第1页，共10页高三数学参考答案题号123456789选项DABCABDAABBCDACD12.21.【答案】D2.【答案】A【解析故其对应的点位于第一象限.3.【答案】B【解析】(2x_1)5的展开式的通项为T5_r(_1r令C(_1)r25_r=_40，得r=3，所以该项含x的次数为5_r=2.4.【答案】C【解析】以抛物线的顶点为原点，对称轴为x轴建立平面直角坐标系设抛物线方程为y2=2px(p>0)接收天线的口径为12m，深度为3m，则抛物线上有一点的坐标为(3,6)，代入抛物线方程中，解得p=6:y2=12x，.所以信号处理中心与抛物线顶点的距离为m5.【答案】A6.【答案】B数学试卷参考答案第2页，共10页7.【答案】D【解析】观察函数可知，f(x)为定义在R上的偶函数，在(0,+∞)上递增，8.【答案】A【解析】设正方体棱长为1，以D为原点，以DA,DC,DD1所在直线分别为x,y,z轴，建立空间直角坐标系，可得D(0,0,0),A(1,0,0),B(1,1,0),C(0,1,0),D1(0,0,1)，因为点E为CD中点，可得E，又设=λ1，F(0,0,z)，可得所以(0,0,z)=λ(0,0,1_z)，可得z=λ(1_z)，解得z所以F由三棱锥D1_ADE的外接球即为DA,DE,DD1为棱的长方体的外接球，由DA=1,DEDD1=1，得长方体的对角线长为l所以D1_ADE的外接球的半径为R即球心O为长方体的对角线的中点，所以O，设平面BEF的法向量为所以数学试卷参考答案第3页，共10页因为O∈α,所以向量与平面BEF的法向量为垂直，9.【答案】AB【解析】对于A，若众数为5，则x=5，A正确；对于B，原平均数为6，若平均数不变，则x=6，B正确；对于C，原中位数为6，若中位数不变，则x=6，C错误；对于D，若极差为9，则x=0或1210.【答案】BCD【解析】对于A，(_2)2+(_2)2=8>4，点A在圆O外，A错误；对于B，kBC，则直线BC的方程为y对于C，圆心O到ΔABC的三边的距离均为半径2，则与三边均相切，C正确；,22+(2sinθ_6)2+(2cosθ_4)2+(2sinθ+2)2=80_8sinθ22的最大值为88，D正确.11.【答案】ACD【解析】对于A，小于等于9的正整数中与9不互质的数为3的倍数，即3，6，9，即对于C，若p为质数(p≥2)，则小于等于pn的正整数中与pn不互质的数只有p的倍数，所以互质的数的数目为pnpn_pn_1，数学试卷参考答案第4页，共10页3n)n_1，则Snn_1，C正确；对于D，由φ(xy)=φ(x)φ(y)，可知x，y互质，则满足条件的(x，y)有12,13,14,15,16,23,25,34,35,45,56共11种，故PD正确.,可得c2a13.【答案】0【解析】依题意sincossincos则cos=cosαcosβ+sinαsin14.【答案】[_e2,0]【解析】由题意得g(x)=axex+e与y=x+b的图象最多有一个交点，故关于x的方程axex+e=x+b最多有一个根，所以H(x)=axex+e_x与y=b的图像最多有一个交点，所以H,(x)=a(x+1)ex_1恒大于等于0或恒小于等于0，当x=_1时，H,(x)=_1<0，所以H,(x)≤0恒成立，令n(x)=H,(x)=a(x+1)ex_1，则n,(x)=a(x+2)ex，令n,(x)=0，解得x=_2，当a=0时，n(x)=_1满足题意；当a>0时，n(x)在(_∞,_2)上单调递减，在(_2,+∞)上单调递增，当x0>1_1，且x0>0时a(x0+1)>1，则n(x0)=a(x+1)ex_1>0,a而n(_2)=_ae_2_1<0，故H,(x)=a(x+1)ex_1恒大于等于0或恒小于等于0不成立，不满足题意；当a<0时，n(x)在(_∞,_2)上单调递增，在(_2,+∞)上单调递减，所以n(x)max=n(_2)=_ae__2_1≤0，解得a≥_e2，综上，a的取值范围为[_e2,0].数学试卷参考答案第5页，共10页又因为四边形ABCD是菱形，LABC=60，因为PC=2，所以PC2=PO2+CO2，即PO丄CO，····························································································4分因为ABC平面ABCD，COC平面ABCD，AB∩CO=O，所以PO丄平面ABCD，又POC平面PAB，所以平面PAB丄平面ABCD．··········································6分（2）以O为原点，OC，OB，OP分别为x轴，y轴，z轴建立空间直角坐标系，则B(0,1,0)，D(3,_2,0)．PC=(3,0,_1)，PD=(3,_2,_1)．··························································7分设平面PAD的法向量为=(x1,y1,z1)，设平面PBC的法向量为=(x2,y2,z2)，数学试卷参考答案第6页，共10页所以cos························································12分17由正弦定理可得···············································2分故bccosA+2accosB=3abcosC．························································4分由余弦定理可得···············6分化简得a2+2b2=3c2．···········································································7分（2）由题意可知，C为锐角，2ab2ab2ab3解得a···················································································15分数学试卷参考答案第7页，共10页当k>1时，令f,(x)=ex+2_2k>0得x>ln(2k_2)．······························3分故当x∈(_∞,ln(2k__2))时，f,(x)<0，函数f(x)单调递减，综上，当k≤1时，f,(x)>0恒成立，函数f(x)在R单调递增；当k>1时，函数f(x)在(_∞,ln(2k__2))上单调递减，在(ln(2k_2),+∞)上单调递增．·······························································6分（2）令F(x)=ex+2x_2kx_sinx_1，x≥0，F(0)=0，······························7分故当F,(0)=2_2k≥0，即k≤1时，F,(x)≥F,(0)=0故由F(0)=0可知，x∈(0,x0)时，F(x)<0与F(x综上，实数k的取值范围是(_∞,1]．························································15分18．解（1）设事件M=“小明在前3次射击中得到2分”，数学试卷参考答案第8页，共10页（2）设第n次在场景A下射击为事件Mn，41由全概率公式可得P(Mn+1)=P(Mn)P(Mn+1|Mn)+P(Mn)P(Mn+1|Mn)，···········7分则············································································9分且，可知数列是以首项为，公比为的等比数列，······10分（3）设第k轮得分期望为Ek，则Ekk-1，········14分1-L(3)n1所以前n轮期望总得分为Sn=Ek7||1-L(3)n1所以前n轮期望总得分为Sn=Ek7||1-1-191）设椭圆的半焦距为c，数学试卷参考答案第9页，共10页（2）由已知直线l的斜率不为0，故设直线l的方程为x=my+1，4)y2由已知yy1y·················································4分所以直线OR:3mx+4y=0，其中m·················································6分点A到直线OR的距离d····························7分令u=x1_4∈(_6,_2)，···············································································8分3u_3_3u2_42u_6363422) 133|63422) 133|4___2((22所以_y1=λy2，即，数学试卷参考答案第10页，共10页所以