高三数学答案(1)_2023年11月_0211月合集_2024届山东省青岛市四区统考高三上学期期中_山东省青岛市四区统考2024届高三上学期期中数学

2023-2024 学年度第一学期期中学业水平检测高三数学评分标准 一、单项选择题：本题共8小题，每小题5分，共40分。 1--8：DAAC BCAB 二、多项选择题：本题共4小题，每小题5分，共20分。（选不全得2分） 9．BCD 10．AC 11．BCD 12．ABD 三、填空题：本题共4个小题，每小题5分，共20分。（16题第一个空2分，第二个空3分） 13． y e3x； 14．6； 15．3； 16．（1） 14；（2）2． 四、解答题：本题共6小题，共70分。解答应写出文字说明，证明过程或演算步骤。 17.（10分） 解：（1）因为CD PA，CD  AD ，PA AD A， 所以CD平面PAD····················································································3分 又因为CD平面ABCD，所以平面ABCD平面PAD···································4分 （2）记AD中点为F ，因为PAPD，所以PF  AD ····································5分 又因为平面ABCD平面PAD AD，所以PF 平面ABCD···························6分 故以F 为坐标原点，分别以FM,FD,FP所在射线为x轴,y轴,z轴，建立如图空间直角坐标系， 所以P(0,0,2),M(2,0,0),D(0,2,0),C(2,2,0)，··············································· 7分  设平面PDM 的法向量为n (x ,y ,z )， 1 1 1 1    n PD 0 2y 2z 0 z 则 1  , 1 1 ， n 1 PM 0 2x 1 2z 1 0 P  令z 1 ，可得n (1,1,1)··················8分 1 1  设平面PCD的法向量为n (x ,y ,z ) 2 2 2 2   则    n  2   P  C  0 , 2x 2 2y 2 2z 2 0 ， A F D n 2 PD 0 2y 2 2z 2 0 y  令z 1 ，可得n (0,1,1)················9分 2 2 B C M 设平面PDM 与平面PCD夹角为， x     |n n | 2 6 则cos|cos n ，n | 1 2   ·············································10分 1 2 |n ||n | 6 3 1 2 18.（12分） 解：（1）因为sin2 Asin2 B4sin AsinBcosC 0 ， 由正弦定理知：a2 b2 4abcosC ①···························································2分 又由余弦定理知：c2 a2 b2 2abcosC ②····················································3分 c2 由①②得：cosC  ·············································································4分 6ab c2 3ab 1 又因为c2 3ab，所以cosC    ··········································5分 6ab 6ab 2 2π 因为c(0，π)，所以C  ········································································ 6分 3 π π 5π 2π 5π 14 （2）因为T 2(  ) ，所以  ， ， 2 7 7  7 5 因为N*，所以1或2···································································8分 高三数学答案 第1页（共4页） {#{QQABBQIAogAgABIAARgCEwHiCEKQkBAACKoGAEAIsAAAwAFABCA=}#}1 2π 1 若1，则 f(x)sin(x)，因为 f(C) ，所以sin( )  ， 2 3 2 π 2π π 7π 因为|| ，所以 ( , )，所以无解············································· 10分 2 3 6 6 1 4π 1 若2，则 f(x)sin(2x)，因为 f(C) ，所以sin( )  ， 2 3 2 π 4π 5π 11π 4π 7π π 因为|| ，所以 ( , )，所以  ，解得 ··········11分 2 3 6 6 3 6 6    此时 f(x)sin(2x )在( ， )上不单调，所以无解······························12分 6 7 2 19.（12分） 1 解：（1）由题 f(x)a ，x0································································1分 x 当a0时， f(x)0， f(x)在(0,)上单调递减；·······································3分 1 当a0时，由 f(x)0解得x ································································4分 a 1 1 所以，当x(0, )时， f(x)0；当x( ,)时， f(x)0； a a 1 1 所以， f(x)在(0, )上单调递减，在( ,)上单调递增；·································6分 a a 1 （2）由（1）知：当a0时， f(x)  f( )1lna·····································7分 min a a2 a2 所以，存在a0，使1lna b成立，即存在a0，使1lna b成立··· 8分 2 2 a2 1 1a2 令g(a)1lna ，则g(a) a ··············································9分 2 a a 所以，g(a)在(0,1)上单调递增，在(1,)上单调递减，···································10分 1 所以g(a) g(1) ···················································································11分 2 1 所以b的取值范围为(, ]··········································································12分 2 20.（12分） 解：（1）因为APBD，PC BD，APPCP，所以BD平面APC········ 1分 所以BD  AC ····························································································2分 因为四边形ABCD是圆柱底面的内接四边形，且AC 为其直径 所以BE ED，AB  AD，BC CD，ABC ADC  90 ·····························4分 又因为AC BCCD，所以AC 2BC， 1 所以在RTABC中，sinBAC  ，所以BAC 30 2 所以BAD 60，BAD是等边三角形························································5分 （2）因为AC 4，由（1）知，在RTABC中，AE 3，EC 1， 所以CE:EA1:3······················································································ 6分 高三数学答案 第2页（共4页） {#{QQABBQIAogAgABIAARgCEwHiCEKQkBAACKoGAEAIsAAAwAFABCA=}#}因为PF :FA1:3，所以PC //EF 又因为PC 平面BFD，EF 平面BFD，所以PC // 平面BFD····················· 7分 所以点P到平面FBD的距离等于点C到平面FBD的距离···································9分 因为CE PC，所以CE EF，又因为CE BD，EFBDE， 所以CE 平面BFD，··············································································· 10分 所以点C到平面FBD的距离为CE 1，点P到平面FBD的距离为1··················12分 21.（12分） CDsinACD 1 解：（1）由题知，在△ACD中，由正弦定理得sinCAD   ····2分 AD 2 π 因为AD CD，所以ACD CAD，所以CAD  ·································3分 6 π 所以D πACDCAD ，所以AD CD········································4分 2 （2）在△ABC中，AC 2··········································································5分 AB2 BC2 AC2 3 由余弦定理知：cosABC   ·······································6分 2ABBC 2 所以AB2 BC2  4 3ABBC ，所以2ABBC 4 3ABBC 4 解得：ABBC  84 3，等号当仅当AB BC 时取··························7分 2 3 1 所以S  ABBCsinB  2 3 ····························································8分 ABC 2 5π AB AC （3）在△ABC中，设BAC (0, )，由正弦定理知：  6 sinACB sinB 5π 5π 所以AB 4sin( )，AE 2sin( ) ····················································9分 6 6 在△ADE中，由余弦定理知：DE2  AD2 AE2 2ADAE cosDAE 5π 5π π 所以DE2 34sin2( )4 3sin( )cos( ) 6 6 6 π π π 34sin2( )4 3sin( )cos( ) 6 6 6 π π π π 52cos( 2)2 3sin( 2)54sin( 2 ) 3 3 3 6 π 54sin( 2)54cos2(1,9]···········································11分 2 π 所以DE2 9，等号当仅当BAC  时取，所以DE 的最大值等于3················12分 2 22.（12分） 解：（1）因为 f(x)ex 2ax1，································································ 1分 设g(x) f(x)，则g(x)ex 2a，·····························································2分 高三数学答案 第3页（共4页） {#{QQABBQIAogAgABIAARgCEwHiCEKQkBAACKoGAEAIsAAAwAFABCA=}#}当a0时，g(x)0，所以 f(x)在(,)上单调递增且 f(0)0； 所以，若x0，则 f(x)0；不合题意·······················································3分 当a0时，令g(x)0解得：xln2a， 所以 f(x)在(,ln2a)上单调递减，在(ln2a,)上单调递增， 所以 f(x) f(ln2a)2a2aln2a1(a0)·············································4分 令h(x) xxlnx1(x 0) ，则h(x)lnx； 所以h(x)在(0,1)上单调递增，在(1,)上单调递减，····································5分 所以 f(x)  f(ln2a)h(1)0， min 1 1 即只有a ，满足 f(x)0，所以，a的值为 ··········································6分 2 2 1 （2）由题知a ，ln2a0，由（1）知： f(x)  f(ln2a)0且 f(0)0 2 min 当x(,0)时， f(x)0，所以 f(x)在(,0)上单调递增； 当x(0,ln2a)时， f(x)0，所以 f(x)在(0,ln2a)上单调递减； 所以x  x 0为 f (x)的极大值点··························································· 7分 1 x x2 由（1）知：ex  x1 x，所以ex (e2)2  4 x2 所以 f(x) 2ax1，即，当x10a时， f(x)5a2 10·················8分 4 所以，存在x (ln2a,10a)使得 f(x )0 2 2 当x(lna,x )时， f(x)0， f (x)在(lna,x )上单调递减； 2 2 当x(x ,)时， f(x)0， f (x)在(x ,)上单调递增； 2 2 所以，x  x 为 f (x)的极小值点，即 f (x)有两个极值点·····························9分 2 ex 2 1 x x 因为2a  ，所以 f(x ) (1 2)ex 2  2 ， x 2 2 2 2 sinx x x sinx 所以，要证 f(x )1 2 2 ，只需证(1 2)ex 2 1 2 2 2 2 2 2sin x 即证2x  2 0 ···········································································10分 2 ex 2 2sin x 2sin x cosx 设n(x ) 2x  2,x 0 ，则n(x )  2 2 1 2 2 ex 2 2 2 ex 2 2(cosx 1) 再设(x )n(x )，则(x ) 2  0··········································· 11分 2 2 2 ex 2 所以n(x )在(0,)上单调递减，所以n(x ) n(0) 0 2 2 所以n(x )在(0,)上单调递减，所以n(x ) n(0)0 2 2 命题得证···································································································12分 高三数学答案 第4页（共4页） {#{QQABBQIAogAgABIAARgCEwHiCEKQkBAACKoGAEAIsAAAwAFABCA=}#}