眉山市2026届高三第二次模拟测试数学答案_2024-2026高三（6-6月题库）_2026年04月高三试卷_260417四川省眉山市2026届高三第二次模拟测试（全科）

2023 级高三第二次模拟测试 数学试题参考答案及评分意见 一、选择题：（每小题5分，共40分） 1.D； 2.A； 3.B； 4.A； 5.B； 6.C； 7.C； 8.B. 二、选择题：（每小题6分，共18分） 9.AD； 10.BCD； 11.ACD； 三、填空题：（每小题5分，共15分） 1 12.2； 13. ； 2  1 1 14. f(x)sin2 x 或 f(x) cosx 或 f(x)|x2k|,x[2k1,2k1],kZ（答 2 2 2 案不唯一）； 四、解答题：（共77分） 15.解：⑴因为ABC ,所以sin(BC)sin(A)sinA·························1分 因为sin2Asin(BC)0，所以2sinAcosAsinA0， 所以sinA(2cosA1)0·······································································3分 因为0 A，所以sinA0 ，···························································4分 1  所以cosA ，所以A ··································································6分 2 3  1   （2）因为 AD 为 BC 边上的中线，有AD (AB AC)，两边同时平方可得： 2  A  D  2  1 (  A  B  2   A  C  2 2  A  B    A  C  ) ，即 7 1 (c2 b2 bc) ，························8分 4 4 28 由重要不等式： c2 b2≥2bc （当且仅当 bc 时取得等号)得bc≤ ．·········10分 3 由三角形面积公式 S  1 bcsinA≤ 1  28  3  7 3（当且仅当 bc 2 21 时取等）， 2 2 3 2 3 3 7 3 故ABC 存在最大值，最大值为 ．··················································13分 3 16.解：（1）取PA中点F ，连接EF,DF . 1 又因为1时，E 是PB中点，所以EF∥AB且EF  AB.·················· 1分 2 1 又因为CD∥AB且CD AB，所以CD∥EF且CDEF ，····················2分 2 所以四边形CDFE是平行四边形， 所以CE∥DF .················································································3分 又因为CE平面PAD，DF 平面PAD，·········································4分 所以CE∥平面PAD.········································································5分 S AB 2 （2）解：因为 ABC   ,三棱锥E ABC与四棱锥PABCD的高之 S ABCD 3 梯形ABCD 第1页共6页BE 1 比等于  ，所以三棱锥 E ABC 与四棱锥 PABCD 的体积之比等于 BP 1 2 1 2   ， 3 1 9 解得2······················································································7分 取AD中点O，则在等腰直角三角形PAD中，PO AD. 又因为平面PAD平面ABCD，平面PAD平面ABCD AD，PO平面PAD， 所以PO 平面ABCD. 取BC中点G，则在梯形ABCD中，OG∥AB. 又因为AB AD，所以OG AD.    以点O 为坐标原点，分别以OD,OG,OP的方向为x轴，y轴，z轴的正方向，建立 如图所示的空间直角坐标系Oxyz.···················································8分 则A(1,0,0)，B(1,2,0)，C(1,1,0)，P(0,0,1).   2 4 1 因为PE2EB，所以E( , , ).····················································9分 3 3 3  易知平面PAD的一个法向量m(0,1,0) ·············································· 10分  设平面ACE的法向量为n(x,y,z)，   1 4 1 因为AC (2,1,0)，AE( , , )， 3 3 3   2x y0  nAC 0  由  得1 4 1 ，令x1，则y2,z7 nAE 0  x y z0 3 3 3  所以n(1,2,7).············································································12分     |mn| 6 所以|cosm,n|    ······················································ 14分 |m||n| 9 6 所以平面ACE与平面PAB的夹角的余弦值为 ··································15分 9 17.解：（1）当a1时， f(x)lnxx23x，·················································1分 1 则 f(x) 2x3，········································································2分 x 所以切线斜率k  f(1)0，·······························································3分 又 f(1)2，切点为(1，2)···························································4分 所求切线方程为y2···································································5分 第2页共6页1 2ax2(2a1)x1 (2ax1)(x1) (2) f(x) 2ax(2a1)  ··················· 6分 x x x (x1) 当a0时， f(x) ，满足题意················································· 7分 x 当a0时，2ax10，x(0，1)时 f(x)0，x(1，)时 f(x)0满足题意 ····································································································8分 1 当a0时， f(x)0的根为 ，1 2a 1 1 若 1，即a 时， f(x)0恒成立， f(x)在(0，)单调递增，与题意矛盾舍掉. 2a 2 1 1 1 1 若 1，即0a 时， f(x)在(0，1)递增，(1， )递减，( ，)递增，与题意矛 2a 2 2a 2a 盾舍掉. 1 1 1 1 若 1，即a 时，f(x)在(0， )递增，( ，1)递减，(1，)递增，与题意矛盾舍 2a 2 2a 2a 掉·································································································10分 综上所述，a的取值范围为(,0]····················································11分 由题 f(x)在区间(0,1)上单调递增，在区间(1，)单调递减，x0时， f(x)， x时， f(x)，又f(x)  f(1)a1·································12分 max a10，即a1时，函数 f(x)的仅有一个零点，·····························13分 a10，即a1时，函数 f(x)的有两个零点，···································14分 a10，即1a≤0时，函数 f(x)的无零点·······································15分 18.解：（1）由题意得|NM ||NB|，|NA||NM ||AM |6， 所以|NA||NB|6|AB|4，···························································3分 所以C是以A(2,0)，B(2,0)为焦点，6为长轴长的椭圆，·······················4分 x2 y2 所以C的方程为  1.·······························································5分 9 5 9 13 （2）当t 时，直线QR恒过定点( ,0)············································6分 2 4 x2 y2   由（1）可知点B为椭圆  1的右焦点，而PBBQ,0所以P,Q为过右 9 5 焦点B的直线与椭圆的交点 因为C 与直线l 均关于x轴对称，所以若直线QR过定点，该点必在x轴上. 2 ····································································································7分 第3页共6页当直线PQ的斜率不为零时，设直线PQ的方程为 xmy2，P(x,y ),Q(x ,y ), 1 1 2 2 R(t,y )··························································································8分 1 5x2 9y2 45 由 得(5m2 9)y2 20my250  xmy2 20m 25 所以 y  y  ,y y  ··············································10分 1 2 5m2 9 1 2 5m2 9 y y 直线QR的方程为y 1 2 (xt) y ···············································11分 tx 1 2 y (tx ) ty y 2y ty 令y0，则x 1 2 t  1 2 1 2 ····································12分 y  y y  y 1 2 1 2 y  y 4 5 因为 1 2  t，所以ty y  (y  y )··············································14分 y y 5 1 2 4 1 2 1 2 5 13 5 (y  y )2y ty y ( t)y 所以x 4 1 2 1 2  4 1 4 2 y y y y 1 2 1 2 13 5 t 当 4  4 ，即t  9 时，x 13 ····················································· 15分 1 1 2 4 13 当直线PQ的斜率为零时，显然直线QR过点( ,0)································16分 4 9 13 所以当t 时，直线QR恒过定点( ,0)··············································17分 2 4 t 方法二：当PQ的斜率存在且不为零时，若直线QR恒过定点H( 1,0)， 2   则HQ∥HR，·················································································10分  t  t 而HQ(x  1,y ),HR( 1,y )， 2 2 2 2 1 t t 所以只需(x  1)y ( 1)y 0， 2 2 1 2 2 t t 只需(x  1)(kx 2k)( 1)(kx 2k)0， 2 2 1 2 2 t 只需kxx ( 1)k(x x )2kt0···················································13分 1 2 2 1 2 t 只需xx ( 1)(x x )2t0 1 2 2 1 2 36k2 36k2 45 因为x x  ，x x  1 2 9k2 5 1 2 9k2 5 36k2 45 t 36k2 所以只需 ( 1) 2t 0 9k2 5 2 9k2 5 只需36k2 4518(t2)k2 2t(9k2 5)0 第4页共6页只需2t90 9 所以t ·······················································································16分 2 13 综上，直线QR恒过定点H( ,0)·······················································17分 4 19.解：为方便表述，设事件B ：第n天该学生晨读打卡为无效打卡 n （1）根据全概率公式，第2天为有效打卡的概率，由第1天有效、无效两种情况拆 分：P(A ) P(A)P(A |A)P(B )P(A |B )， 2 1 2 1 1 2 1 2 1 3 1 （2）代入已知条件：PA  ,PB  ,PA |A  ,PA |B  1 3 1 3 2 1 4 2 1 2 2 3 1 1 1 1 2 计算得：PA        ·············································2分 2 3 4 3 2 2 6 3 下面推导PA 与PA 的递推式 n n-1 对任意n≥2，第n天有效打卡仅与第n1天结果相关，由全概率公式 PA PA PA |A PB PA |B ，且PB 1PA  n n1 n n1 n1 n n1 n1 n1 3 1 3 1 1 代入条件概率得PA n  4 PA n1  2  1PA n-1    4 PA n1  2  2 P A n1  ····································································································4分 化简得PA  1 PA  1 n≥2,nN* ············································5分 n 4 n1 2 （3）（ⅰ）由题知X 0，1，2，·····················································6分 2 1 1 1 PX 0PBB PB PB |B    2 1 2 1 2 1 3 2 6 2 3 1 PX 2PAA PA PA |A    2 1 2 1 2 1 3 4 2 1 1 1 PX 11PX 0PX 21   2 2 2 6 2 3 故X 的分布列为： 2 X 0 1 2 2 1 1 1 P 6 3 2 ····································································································9分 1 1 1 4 所以数学期望EX 0 1 2  ············································10分 2 6 3 2 3 1 2 2 2 （ⅱ）因为PX 0 ，PX 1 ，则EX  ， 即E(X ) 0 ， 1 3 1 3 1 3 1 3 第5页共6页4 2 由（ⅰ）知EX  ，即E(X ) 20 ， 2 3 2 3 2 猜测E(X ) n ，············································································ 12分 n 3 设 为第 n 天有效打卡次数， 1表示有效， 0表示无效 n n n 则X  X ，因此EX EX EEX PA ···············13分 n n1 n n n1 n n1 n 1 1 结合（1）中递推式PA  PA  ，且PA EX EX n≥3 n 4 n1 2 n1 n1 n2 1 1 联立得EX n EX n1  4  EX n1 EX n2    2 ···································14分 5 1 1 整理得：EX  EX  EX  ,n≥3（*） n 4 n1 4 n2 2 1 1 1 即EX  EX EX  EX  ,n≥3， n 4 n1 n1 4 n2 2 1 7 1 7 1 1 1 由EX  EX  得，EX  EX  n2  n ， 2 4 1 6 n 4 n1 6 2 2 6 2 1 2 于是EX  n [EX  n1] ， n 3 4 n1 3 2 2 结合E(X ) ，递推得E(X ) n ，··················································16分 1 3 n 3 2 E(X ) 2 实际意义：长期坚持晨读打卡后，学生有效打卡率趋于稳定值 （ n  ），打卡 3 n 3 状态不再随天数波动，形成稳定习惯.·····················································17分 说明：学生用数学归纳法证明，同样给分 第6页共6页