数学福建南平市2026届高三年级第二次适应性练习卷(南平高三二检)(5.7-5.9)

南平市2026届高三年级第二次适应性练习卷数学参考答案及评分标准12．_213．514．_1【详解】（1）∵sinB=cosC+cosAsinC,A+B+C=π,·······································1分∴sinAcosC=cosC，即cosC(sinA__1)=0，·········∴cosC=0或sinA=1，当A时，a边最长，与条件a=b__1<b矛盾，故舍去；··································当C时，则c2=a2+b2，又a=b__1,c=b+1，22（2）显然c>b>a，若ABC为钝角三角形，则C为钝角，······································8分a2由三角形三边关系可得b+b__1>b+1，可得b>2，············································【参考答案】(1)证明见解析；由BC是直径可知AB丄AC，则ABC是等腰直角三角形，故AO丄BC，······················1分由圆柱的特征可知BB1丄平面ABC，又AOC平面ABC，所以BB1丄AO，·····················2分因为BB1BC=B，BB1,BCC平面BCC1B1，则AO丄平面BCC1B1，·······························3分而OEC平面BCC1B1，则AO丄OE，·······································································4分OE2E22丄OE，AO丄OE，AOB1O=O，AO,B1OC平面AB1O，所以OE丄平面AB1O，········································································又OEC平面AOE，故平面AOE丄平面AOB1.···························································7分同理可证OE丄OA，···················································在二面角B1_OA__E中B1O丄OA且OE所以LB1OE为二面角B1_OA__E的平面角，······························································6分由思路一知LB1OE=90o，所以二面角B1_OA__E为直二面角，即平面AOE丄平面AOB1.··7分（2）由题意及（1）易知AA1,AB,AC两两垂直，如图所示建立空间直角坐标系，·············8分则B1则有取z=_2，可得3分设平面AB1E与平面B1OE夹角为θ,所以cosθ=cos则平面AB1E与平面B1OE夹角的大小为π.·······························································15分4设A(_x0,y0),B(x0,y0)，其中0<x0<2，0<y0<3，因为A,B在C上，所以，由基本不等式，x0y0|，········································································4分故|x0y0|（2）设P(_4,y0)，Q(0,m)，则P,F1,Q三点共线，(_3,y0)=λ(1,m)，解得m··············································10分则，··························0)20所以································································【参考答案】（1）当b≥0时，f(x)无极值；当b0时，f(x)有极小值为ln无极大值2）当b=1时，f(x)在B处不“等比偏移”3）证明见详解．当b≥0时，f,(x)>0,f(x)在(0,+∞)上当b<0时，f,(x)在(0,+∞)上单调递增，令f,=0→x易知f(x)在上单调递减，在上单调递增，·················································4分:f极小值=flnf(x)无极大值．·······································5分（2）结论：当b=1时，f(x)在B处不“等比偏移”，理由如下：····················6分当b=1时，f=lnx+x2.f,x，f(x)在B处的切线斜率为f7分取xAxClnln8分所以当b=1时，f(x)在B处不“等比偏移”.·············································10分一般的，取A,B,C三点的横坐标成等比数列，设xAxCq>0,q≠1),xB直线AC的斜率为kAC取q=2就是上面的特例;若取q=4，则有所以当b=1时，J(x)在B处不“等比偏移”.要证J(x)是其定义域上的“等比偏移”函数，只要证JkAC．*）······································12分设t2，t>1，则只需证：当t>1时，2lnt_t+<0恒成立．上单调递减，从而g(t)<g(1)=0，命题获证．·······················································17分（3）思路二：以上证法同思路一．········································若设t，t>1，则只需证：当t>1时，lnt恒成立．令g=lnt因为g,，所以g(t)在(1,+∞)上单调递减g(t)<g(1)=0，命题获证．················································································17分x2+x1，·····························································12分x_x22⇋b|(2_lnx1__x22故只要证b路一．·····································································1917分）【详解】（1）依题意可得，X的所有可能取值为0，1，2，P························································································1分则X的分布列为X012P 272719所以E5分（2i）由质点每次等可能地随机沿棱移动1个单位可知，若质点移动了n次，n次后质若质点移动了n+1次，由B，C，D三点等可能地向点A移动，