数学试卷答案与解析东北三省精准教学联盟2025年12月高三联考(12.9-12.10)

7}，故选C.2.B3.D底面圆半径为·、i2，所以圆锥SO的侧面积为故选D.4.C解得m且m≠-2，即m的取值范围为故选C.5.C726或63，故选C.6.A【解析】因为函数f，则f又因为函数f(x)是定义在(-∞,0)(0,+∞)上的奇函数，则f(x)+f(-x)=0，即解得a=1，则f所以f故选A.7.D8.C单调递减，在(ln3,+∞)上单调递增，所以函数g(t)在t=ln3处取得最小值3-3ln3，即函数f(x)在(0，+∞)上的最小值为3-3ln3，所以a≥3-3ln3，即a的最小值为3-3ln3，故选C.03-ex)+a≥0在(0，+∞)上有解，即a≥-lnx3-x(3-exx是增函数.因为ee-3<0，e所以由函数零点存在定理，设f,(x0)=0，则ex0- f(x)在(0，x0)上单调递减，在(x0，+∞)上单调递增，所以f(x)min=0f=-3lnx0-3x0+x0ex0=-3lnx0+xln3+3x0-数f(x)在(0，+∞)上的最小值为3-3ln3，所以a≥3-3ln3，即a的最小值为3-3ln3，故选C.9.AC【解析】对于A选项，因为z4+1>0，x>y，所以成立，所以A选项正确；对于B选项，当z2=0时，xz2=yz2，所以B选项不正确；对于C选项，因为x>y，所以x-y>0，所以x-y，当且仅当x-y=1时，等号成立，所以C选项正确；对于故选AC.10.BCD【解析】因为f(4x-1)为偶函数，则f(4x-1)=f(-4x-1)，则f(x-1)=f(-x-1)，所以函数f(x)的图象关于直线x=-1对称，所以A选项不正确；因为f(x-1)=f(-x-1)，所以f,(x-1)=-f,(-x-1)，即f,(x-1)+f,(-x-1)=0，所以函数f,(x)的图象关于点(-1,0)对称，所以B选项正确；若f(x+4)+f(-x+4)=0，则函数f(x)的图象关于点(4,0)对称，又因为函数f(x)的图象关于直线x=-1对称，所以函数f(x)的一个周期为20，所以f(-5)=f(15)，所以C选项正确；若f,(x+4)+f,(-x+4)=0，则函数f,(x)的图象关于点(4,0)对称，又因为函数f,(x)的图象关于点(-1,0)对称，所以函数f,(x)的一个周期为10，所以f,(-5)=f,(5)，所以D选项正确，故选BCD. 所以A不正确；f cosα+sinα3cosα(π72scosα+sinα3cosα(π72s【解析】因为(-1)2+(-)2<4，所以点P(-1,-)在圆内.设O为坐标原点，由题意得，是以2为首项，1为公差的等差数列，所以2nan=2+(n-1)=n+1，所以an14.-18【解析】在△ABC中，由cos上ACB可得sin上ACB又由正弦定理可得cos上ABC，整理得BC2-4BC-12=0，解得BC=6（负值舍去）.因为O为△ABC所在平面内一点，且OA=OB=OCcos上ABC，整理得BC2-4BC-12=0，解得BC=6（负值舍去）.因为O为△ABC所在平15.（1）证明见解析（5分2【解析】（1）因为AB=AC，上BAC=，D为BC的中点，所以AD丄BC，·······2分【另解】（1）因为AB=AC，上BAC=，D为BC的中点，所以AD丄BC，·······2分因为PA丄底面ABC，AB平面ABC，AC平面ABC，所以PA丄AB，PA丄AC，22+AC2，所以PB=PC.又因为D为BC的中点，所以PD丄BC.···························································4分 因为PA丄底面ABC，上BAC，所以AB，AC，AP两两垂直，故以A为坐标原点，AB，AC，AP所在直线分别为x，y，z轴，建立如图所示的空间直角坐标系，00【另解1】（2）因为PA丄底面ABC，上BAC，所以AB，AC，AP两两垂直，故以A为坐标原点，AB，AC，AP所在直线分别为x，y，z轴，建立如图所示的空间直角坐标系，因为BC=2，AB=AC，上BAC，D为BC的中点，所以AB=AC.00················································································································6分，令x=a，则y=a，z=，所以.··············································7分【另解2】（2）取AB的中点M，连接PM和DM，由（1）知，上PDA即为二面角P-BC-A的平面角，所以上PDA·················7分因为BC=2，AB=AC，上BACD为BC的中点，所以AB=ACAD=1，所因为D为BC的中点，M为AB的中点，所以DM为△ABC的中位线，DM∥AC，DM=AC所以上PDM为异面直线PD与AC所成的角.····································9分在△PMD中，由余弦定理可得cos上PDM=PD2+DM2-PM22PD.DM2=评分细则： 16.（1）开始充电时的电量为27.5%（3分）；充满电需min t-1【解析】（1）由题意知，当t=0时电池开始充电，此时Q(t)=55.210，将t=0代入得-1=27.5，····································································2分当Q(t)=100，即电量达到100%时，电池充满电，已知Q(60)=80，且Q(t)单调递增，故充满电所需的时间必大于60min，·····································································4分故充满电需min.·····················································································7分t2b，因为当t=10时，电量不产生跃变，故在t=10附近，第一段和第二段函数连续.················································································································9分评分细则： 17.（14分25分3）0（6分）所以sinAcosC+sinAsinC=sin(A+C)，整理得sinC(sinAcosA)=0，··································································2分226222，·············································6分6解得AC=2，即AC=2，所以AB=2.·······················································7分△ABC2（3）由正弦定理得 =sin2C-cosC-sinC+cos-cosC-sinC+cosCsinC 442=sin2CcosCπ(5π)π(π)(π5π)cosCπ(5π)π(π)(π5π)).令f则易知fmax=f【另解】（3）由余弦定理得a2=b2+c2-2bccosA，22bπ(5π)π(π)(π5π)bπ(5π)π(π)(π5π),令f=-3t，则易知fmax=f3，AB23，AB2评分细则：导出导出A=给1分.f,(x)的极小值为lna+2，无极大值（7分3）证明见解析（6分） 则曲线y=f(x)在(1,f(1))处的切线方程为y-0=3(x-1)，即3x-y-3=0.··········4分················································································································6分当a>0时，令g,(x)==0，解得x=a，···················································8分(0,a)时，g,(x)<0，所以函数g(x)在(0,a)上单调递减；（3）不等式f(x)-2x>-2，即(x+a)lnx-2x+2>0，即(x+a)lnx-2(x-1)>0，令由（2）可知，a>0时，函数y=lnx+0,a)上单调递减，在(a,+∞)上单调递增，即当a>1时，f(x)-2x>-2在(1,+∞)上恒成立.···············································17分评分细则：f,(x)的极小因为an+1，an+2，an成等差数列，所以2an+2=an+1+an，即an·········2分所以a234a34所以数列{bn}的前4项和为.··················································4分所以cnn，所以cnn.n所以m<p，·······························································································6分*，则→m.