四川省绵阳市2026年高考适应性考试（绵阳三诊）数学+答案

2026年高考适应性考试数学参考答案和评分标准一、选择题：本题共8小题，每小题5分，共40分．二、选择题：本大题共3小题，每小题6分，共18分．全部选对的得6分，选对但不全的得部分分，有选错的得0分．三、填空题：本题共3个小题，每小题5分，共15分．994四、解答题：本题共5小题，第15题13分，第16、17小题15分，第18、19小题17分，共77分．解答应写出文字说明、证明过程或演算步骤．由正弦定理得：sinCsinB=sinBcos(C_π)，又sinB≠0，·····························1分6∴sinC=cos············································································2分∴sinC=cosCcossinCsin·····························································4分∴tanC=·3，又C∈(0，π)，则（2）方法一：在△BCD中，∵BD=CD，可得LBCD=B，LADC=2B，又∵C·······················································································7分∴LACDB，sinA=sin=sin····································8分又在△ACD中，由正弦定理得：且CD=2AD，·····································································9分数学答案第1页（共7页）数学答案第2页（共7页）数学答案在△ABC中，由余弦定理得：9x2=a2+b2_2abcosπ,322_ab①平方得：4xab························································10分 16．解1）∵l过F1(−c，0)时，△ABF2的周长为8，则4a=8，a=2，············2分设LAMO=θ,则cos2θ=1_2sin···············································4分∴sin····················································································5分又sin则c=2，·······················································6分22+2)x1··········································12分x1x2x1x2x22)································································13分x1x2=2k_4k=_2k，····································································14分数学答案第3页（共7页）数学答案_ 3 5分∴EF丄BC，EF丄AB，····································································7分又∵ABBC=B，（2）∵EF⊥平面ABC，以E为原点，建立如图所示的空间直角坐标系Exyz，(x0，y0，z0)，由可得设平面A1EC1的法向量为n=(x,y,z)，由A1A=1，AB=AC=2，∠BAA1=∠CAA1=60O，由余弦定理，∴∠AA1B=∠AA1C=90O，且A1B=A1C=3，··············································2分由E为BC中点，则BC⊥A1E，延长A1F交B1C1于点G，则A1G⊥B1C1，则A1G⊥BC，A1E∩A1G=A1，数学答案第4页（共7页）数学答案∴BC⊥平面A1GE，EF≤平面A1GE，在Rt△A1BE中，可得A1E=·2，···························································5分在△A1EG中，EG=1，A1G，则A1G2=EG2+A1E2，1E⊥EG，······················································································6分又F为A1G上靠近点G的一个三等分点，FG，A1F又A1G∩B1C1=G，则EF⊥平面A1B1C1，（2）由（1）知GE=AA1=1，A1E=2，A1G=3，又由（1）知BC⊥A1E，BC∩GE=E，BC≤平面GEC1，GE≤平面GEC1，1E⊥平面GEC1，∴C1E⊥A1E，····················································································13分在Rt△EGC1中，cos∠GEC········································15分P······································································2分P·······································································4分故X的数学分布列为：X012X012P∴n2(n_6)+(n_6)(3n_20)=0，···························································7分∴(n_6)(n2+3n_20)=0，···································································8分∴n=6；·····························································································9分数学答案第5页（共7页）数学答案（3）Iiii中，则XIi，k由于两次抽取相互独立，且每个球被抽到的概率均为，nk可得：E，其中E又E(IiIj)=P(i，j∈M).P(i，∵P(i，····································································12分∵IiIj共有C项，代入得：E·····························14分②当n为奇数时，k或kD(X)最大，D··17分19．解：∵f-flnx3+a+x-2ln令glnx2+x+1-2ln2，glnx2-3x+1，易知g¢(0)=0，数学答案第6页（共7页）数学答案2∴当a≥-3时，xÎ(-1，0)，f(x)<f(1)；·············································4分2（2）∵flnx2+2ax，且f¢(0)=0，（i）∵0为f(x)的极小值点，由于f¢(0)=f(0)=0，2∴必有fⅱ(0)=2a+3>0，即a>-3，··················································6分2由于f∴存在-1<m<0<x0<n，使得在(-1，m)与(n，+∞)上满足fⅱ(x)<0，f¢(x)单调递减；在(m，n)上fⅱ(x)>0，f¢(x)单调递增．··········································8分∴存在-1<s<0<t，使得在(-1，s)与(0，t)上有f¢(x)>0，f(x)单调递增；在(s，0)与(t，+∞)上有f¢(x)<0，f(x)单调递减．···········································9分∴f(x)的极大值点为：x1=s，x2=t，32由（1）得：f(x1)<f(1)，∴f(x1)<f(1)<f(x2)，则f(x1)<f(x2)； （ii）∵x1=0为f(x)的一个极大值点，f¢(0)=0，且f(0)=0，由于f(x1)=f(x2)，所以f(x2)=f¢(x2)=0，2ï(4x2+1)ln(x2+1)+2x2-x2+2ax2=0 2(2x(