山东省烟台市2022-2023学年高三上学期期末学业水平诊断数学参考答案

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
13. 1 14. 32 15. 67 16. 12
四、解答题
17.解：（1）由正弦定理可得sin cos +sin sin sin A C A C B =， ······················· 1分
因为πA B C ++=，所以sin cos +sin sin
sin()A C A C A C =+， 即sin cos +sin sin sin cos cos sin A C A C A C A C =+， ···························· 2分 整理得：sin sin cos sin A C A C =，
因为0C π<<，所以sin 0C ≠，所以tan 1A =， 因为0A π<<，所以4
A π
=
. ································································ 4分
（2）在ABD ∆中，由余弦定理得：2222cos BD AB AD AB AD A =+−⋅， ······ 5分
即229(2AB AD AD AB AD =+−⋅≥⋅，
································ 6分
整理得AB AD ⋅≤AB AD =时，等号成立.
所以1sin 24ABD S AB AD AB AD π=
⋅=⋅≤
△， ························ 8分
因为2AD DC =
，所以32ABC
ABD S S =≤
△△
所以ABC △. ··············································· 10分 18.解：（1）因为12n n n a a S +=*
()n ∈N ，所以
()1122n n n a a S n −−=≥， 两式相减得()()1122n n n n a a a a n +−−≥. ····································································· 1分
又因为0n a ≠，所以()1122n n a a n +−−=≥， ······························································ 2分 所以数列{}21n a −和{}2n a 都是以2为公差的等差数列. 因为11a =，所以在12n n n a a S +=中，令1n =，得22a =，
所以()2112121,n a n n −=
+−=−()22122,n a n n =+−×= ············································· 3分 所以n a n =，
··················································································································· 4分 对于数列{}n b ，因为112n n n b b b b +⋅==，且0n b ≠，所以1
*2()n n
n b
b +=∈N ， ··········· 6分
所以数列{}n b 是以2为首项，2为公比的等比数列，所以2n n b =. ························ 7分 （2）因为2
3
=122232...2n
n T n ×+×+×++×
所以()2
3
4
12=122232122
n
n n T n n +×+×+×++−×+× (8)
分
两式相减得，21
2222
n
n n T n +−=+++−× ··························································· 9分
1
122212
n n n ++−=−×−
12(1)2n n +=−−−× ························································································· 11分
所以()1
122n n T n +=−×+. ····························································································· 12分 19. 解：（1）证明：取BC 中点O ，连接,OA OD ，
因为ABC ∆是以BC 为斜边的等腰直角三角形，所以OA BC ⊥. ························· 1分 因为BCD ∆是等边三角形，所以OD BC ⊥. ··························································· 2分
OA OD O = ，OA ⊂平面AOD ，OD ⊂平面AOD ， ······································ 3分
所以BC ⊥平面AOD . ································································································· 4分 因为AD ⊂平面AOD ，故BC AD ⊥.
（2）在AOD ∆中，1AO =，OD =
AD =cos AOD ∠，故150AOD ∠= . ·············· 6分 如图，以,OA OB
及过O 点垂直于平面ABC 的方向为,,x y z 轴
的正方向建立空间直角坐标系O xyz −，
·············· 7分 可得3(2D −，所以3(,2BD =−
− ，(0,2,0)CB = ，(1,1,0)AB −，
设111(,,)x y z =n 为平面ABD 的一个法向量，则
111110302x y x y z −+=
−−+
=
，令x =
=n ， ······················ 9分 设222(,,)x y z =m 为平面BCD 的一个法向量，则
222220302
y x y z =
−−+=
，令2x =
，可得=m ， ···················· 11分
所以cos
,<>=n m
故平面ABD 与平面BCD
······································ 12分 20.解：（1）设该容器的体积为V ，则
232
3
V r l r ππ=+，
又160
3
V π=
，所以
2160233l r r =−， ·········································································· 2分 因为6l r ≥，所以02r <≤. ························································································ 4分
所以建造费用222916029232()34334
y rl r m r r r m r ππππ=×+=−×+， 因此
2
2403(1),0 2.y m r r r
π
π=−+<≤ ··································································· 5分 （2）由（1）得3222406(1)406(1)(),0 2.1
m y m r r r r r m πππ−′=
−−=−<≤− ·········· 6分 由于9,10,4
m m >
−>所以令3
4001r m −=−
，得r =··························· 7分
若
2<，即6m >
，当r ∈时，0y ′<，()y r
为减函数，当2)r ∈时，0y ′>，()y r
为增函数，此时r =为函数()y r 的极小值点，也是最小值点. ······················································································································ 9分
2≥，即964m <≤， 当(0,2]r ∈时，0y ′<，()y r 为减函数，此时2r =是()y r 的最小值点. ··········································································································· 11分
综上所述，当
9
64
m <≤时，建造费用最小时2r =；当6m >
时，建造费用最小时r =
························································································································ 12分 21.解：（1）设(,0)A a −，(,0)B a ，11(,)P x y ， 则211122111001
4
AP BP
y y y k k x a x a x a −−=×==+−−， ·············································· 1分 又因为点11(,)P x y 在双曲线上，所以22
11221x y a b −=. ····································· 2分
于是222
22
21
112144a b y x x b a
=−=−，对任意10x ≠恒成立，
所以221
4b a
=，即224a b =. ··································································· 3分
又因为c =
，222c a b =+，可得24a =，21b =，
所以双曲线C 的方程为2
214
x y −=
. ······················································ 5分 （2）设直线l
的方程为：x ty =+，3344(,),(,)M x y N x y ，由题意可知2t ≠±，
联立22
14x y x
ty −=
= ，消x
可得，22(4)10t y −++=，
则有34y y +，3421
4
y y t =−， ··················································· 6分 假设存在定点(,0)D m ，
则3434()()DM DN x m x m y y =−−+
3434()()ty m ty m y y =+−+ ······················································ 7分
22
3434(1))())t y y m t y y m =
++−++−
22
21)4t m t +=+−−
=
8分
令224194(4)m m −+=−
，解得m =
·········································· 10分 此时224511446464
DM DN m =−=−=− ， ························································ 11分
所以存在定点D ，使得DM DN 为定值11
64−. ··································· 12分
22.解：（1）2
()e 2x
f x x ax ax =−−，则()(1)(e 2)x
f x x a ′=
+−， ··················· 1分 当0a >时，方程e 20x a −=的根为ln(2)x a =. 当ln(2)1a >−，即1
2e
a >
时，当(,1)x ∈−∞−和(ln(2),)x a ∈+∞时，()0f x ′>， ()f x 单调递增，当(1,ln(2))x a ∈−时，()0f x ′<，()f x 单调递减. ············· 2分
当ln(2)1a <−，即1
02e
a <<，当(,ln(2))x a ∈−∞和(1,)x ∈−+∞时，()0f x ′>，
()f x 单调递增，当(ln(2),1)x a ∈−时，()0f x ′<，()f x 单调递减. ············· 4分
当ln(2)1a =−，即1
2e
a =时，0y ′≥恒成立，函数在R 上单调递增， ·············· 5分
综上所述，当1
02e
a <<
时，()f x 在(,ln(2))a −∞，(1,)−+∞上单调递增， 在(ln(2),1)a −上单调递减；当12e a =时，()f x 在R 上单调递增，当12e
a >时， ()f x 在(,1)−∞−，(ln(2),)a +∞上单调递增，在(1,ln(2))a −上单调递减. ········ 6分
（2）存在实数a 使得()2f x b a ′−≥对任意x 恒成立，即e e 2x x b x ax +−≤恒成立. 令()e e 2x
x
g x x ax +−，则min ()b g x ≤. ················································ 7分
因为()(2)e 2x
g x x a ′=
+−，当2x −≤时，()0g x ′<恒成立；当2x >−时，()(3)e 0x g x x ′′=+>，函数()g x ′在(2,)−+∞上单调递增，
且(2)20g a ′−=−<，2(2)(22)e
20a
g a a a ′=+−>，
所以，存在0(2,2)x a ∈−，使得0()0g x ′=，且()g x 在0(2,)x −上单调递减， 在0(,)x +∞上单调递增，所以0min 000()()(1)e 2x
g x g x x ax ==+−. ··············· 9分 于是，原命题可转化为存在a 使得000(1)e 2x
b x ax +−≤在(2,)−+∞上成立， 又因为000()(2)e 20x
g x x a ′=+−=，所以002(2)e x
a x =+.
所以存在0(2,)x ∈−+∞，使得00022
00000(1)e (2)e e (1)x
x
x
b x x x x x +−+−−+≤成立.
···················································· 10分
令2
()e (1)x
h x x x =−−+，(2,)x ∈−+∞，
则2
()e (3)x
h x x x ′=−−，所以当(2,0)x ∈−时，()0h x ′>，()h x 单调递增，当(0,)x ∈+∞时，()0h x ′<，()h x 单调递减，所以
max
()(0)1h x h ==，所以1b ≤. ··························································· 12分。