烟台期末-数学答案_2.2025数学总复习_2023年新高考资料_3数学高考模拟题_新高考_山东烟台高三2022-2023学年度第一学期期末学业水平诊断数学

2022～2023 学年度第一学期期末学业水平诊断 高三数学参考答案及评分标准 一、选择题 D B B C A C D A 二、选择题 9.BC 10. ACD 11. ACD 12. ABD 三、填空题 3 1 13. 1 14. 15. 67 16. 2 2 四、解答题 17.解：（1）由正弦定理可得sinAcosC+sinAsinC =sinB， ······················· 1分 因为A+B+C =π，所以sinAcosC+sinAsinC =sin(A+C)， 即sin AcosC+sin AsinC =sinAcosC+cosAsinC， ···························· 2分 整理得：sin AsinC =cosAsinC， 因为0= = ， 31× 12 31 3 93 故平面ABD与平面BCD夹角的余弦值为 . ······································ 12分 31 2 20.解：（1）设该容器的体积为V ，则V =πr2l+ πr3， 3 160 160 2 又V = π，所以l = − r， ·········································································· 2分 3 3r2 3 因为l ≥6r ，所以0 ,所以m−1>0,令r3 − =0，得r = 3 . ··························· 7分 4 m−1 m−1 高三数学答案（第 3 页，共 6 页）40 40 若 3 <2 ，即m>6，当r∈(0,3 ) 时， y′<0 ， y(r) 为减函数，当 m−1 m−1 40 40 r∈(3 ,2)时， y′>0， y(r)为增函数，此时r = 3 为函数 y(r)的极小值点， m−1 m−1 也是最小值点. ······················································································································ 9分 40 9 若3 ≥2，即 6时，建造费用最小时 4 40 r = 3 . ························································································································ 12分 m−1 21.解：（1）设A(−a,0)，B(a,0)，P(x ,y )， 1 1 y −0 y −0 y2 1 则k k = 1 × 1 = 1 = ， ·············································· 1分 AP BP x +a x −a x2 −a2 4 1 1 1 x2 y2 又因为点P(x ,y )在双曲线上，所以 1 − 1 =1. ····································· 2分 1 1 a2 b2 1 a2 b2 于是y2 = x2 − = x2 −b2，对任意x ≠0恒成立， 1 4 1 4 a2 1 1 b2 1 所以 = ，即a2 =4b2. ··································································· 3分 a2 4 又因为c= 5 ，c2 =a2 +b2，可得a2 =4，b2 =1， x2 所以双曲线C的方程为 − y2 =1. ······················································ 5分 4 （2）设直线l的方程为：x=ty+ 5，M(x ,y ),N(x ,y )，由题意可知t ≠ ±2， 3 3 4 4 高三数学答案（第 4 页，共 6 页）x2  − y2 =1 联立 4 ，消x可得，(t2 −4)y2 +2 5ty+1=0，  x=ty+ 5 −2 5t 1 则有y + y = ，y y = ， ··················································· 6分 3 4 t2 −4 3 4 t2 −4 假设存在定点D(m,0)，   则DMDN =(x −m)(x −m)+ y y 3 4 3 4 =(ty + 5−m)(ty + 5−m)+ y y ······················································ 7分 3 4 3 4 =(t2 +1)y y +( 5−m)t(y + y )+( 5−m)2 3 4 3 4 t2 +1 2 5( 5−m)t2 = − +( 5−m)2 t2 −4 t2 −4 (m2 −4)t2 −(4m2 −8 5m+19) = ···························································· 8分 t2 −4 7 5 令4m2 −8 5m+19=4(m2 −4)，解得m= ， ·········································· 10分 8   245 11 此时DMDN =m2 −4= −4=− ， ························································ 11分 64 64   7 5 11 所以存在定点D( ,0)，使得DMDN为定值− . ··································· 12分 8 64 22.解：（1） f(x)= xex −ax2 −2ax，则 f′(x)=(x+1)(ex −2a)， ··················· 1分 当a>0时，方程ex −2a=0的根为x=ln(2a). 1 当ln(2a)>−1，即a> 时，当x∈(−∞,−1)和x∈(ln(2a),+∞)时， f′(x)>0， 2e f(x)单调递增，当x∈(−1,ln(2a))时， f′(x)<0， f(x)单调递减. ············· 2分 1 当ln(2a)<−1，即00， 2e f(x)单调递增，当x∈(ln(2a),−1)时， f′(x)<0， f(x)单调递减. ············· 4分 1 当ln(2a)=−1，即a= 时，y′≥0恒成立，函数在R上单调递增， ·············· 5分 2e 高三数学答案（第 5 页，共 6 页）1 综上所述，当0 时， 2e 2e f(x)在(−∞,−1)，(ln(2a),+∞)上单调递增，在(−1,ln(2a))上单调递减. ········ 6分 （2）存在实数a使得 f′(x)≥b−2a对任意x恒成立，即b≤xex +ex −2ax恒成立. 令g(x)= xex +ex −2ax，则b≤ g(x) . ················································ 7分 min 因为 g′(x)=(x+2)ex −2a ，当 x≤−2 时， g′(x)<0 恒成立；当 x>−2 时， g′′(x)=(x+3)ex >0，函数g′(x)在(−2,+∞)上单调递增， 且g′(−2)=−2a<0，g′(2a)=(2a+2)e2a −2a>0， 所以，存在x ∈(−2,2a)，使得g′(x )=0，且g(x)在(−2,x )上单调递减， 0 0 0 在(x ,+∞)上单调递增，所以g(x) = g(x )=(x +1)ex 0 −2ax . ··············· 9分 0 min 0 0 0 于是，原命题可转化为存在a使得b≤(x +1)ex 0 −2ax 在(−2,+∞)上成立， 0 0 又因为g′(x )=(x +2)ex 0 −2a =0，所以2a =(x +2)ex 0. 0 0 0 所以存在x ∈(−2,+∞)，使得b≤(x +1)ex 0 −(x2 +2x )ex 0 =ex 0(−x2 −x +1)成立. 0 0 0 0 0 0 ···················································· 10分 令h(x)=ex(−x2 −x+1)，x∈(−2,+∞)，则h′(x)=ex(−x2 −3x)，所以当x∈(−2,0) 时，h′(x)>0，h(x)单调递增，当x∈(0,+∞)时，h′(x)<0，h(x)单调递减，所以 h(x) =h(0)=1，所以b≤1. ··························································· 12分 max 高三数学答案（第 6 页，共 6 页）