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第1页共5页2023—2024学年高三质量检测(一)数学参考答案一、选择题本题共8小题,每小题5分,共40分。在每小题给出的四个选项中,只有一项是符合题目要求的。题号12345678答案CADDACBB二、选择题本题共4小题,
每小题5分,共20分。在每小题给出的选项中,有多项符合题目要求。全部选对的得5分,部分选对的得2分,有选错的得0分。题号9101112答案ABACACDACD三、填空题本题共4小题,每小题5分,共20分。13.3√5
514.16015.5316.[43,2]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)设该等差数列{𝑎𝑛}的公差为𝑑,等比数列{𝑏𝑛}的
公比为𝑞,由已知得{𝑎1+𝑑=𝑎1𝑞𝑎1+3𝑑=𝑎1𝑞2,.............................................................................................................
..2分因为数列{𝑎𝑛}为正项数列,{𝑏𝑛}为正项递增数列,所以𝑞=2,𝑑=1,...................................................................................................
........................4分所以𝑎𝑛=1+(𝑛−1)×1=𝑛,𝑏𝑛=1×2𝑛−1=2𝑛−1...............................................................6分(2)由已知得𝑐𝑛={𝑛
,𝑛为奇数2𝑛−1,𝑛为偶数,............................................................................................7分所以数列
{𝑐𝑛}的前2𝑛项和为𝑇2𝑛=(𝑎1+𝑎3+⋯+𝑎2𝑛−1)+(𝑏2+𝑏4+⋯+𝑏2𝑛)=(1+3+⋯+2𝑛−1)+(21+23+⋯+22𝑛−1)..................................................
.........................8分=(1+2𝑛−1)𝑛2+21×(1−4𝑛)1−4=3𝑛2+22𝑛+1−23...............................................
....................................................................................10分18.(12分)证:(1)∵底面𝐴𝐵𝐶𝐷为正方形,∴𝐴
𝐵⊥𝐴𝐷,.........................................................................................................
..............................1分又∵𝐴𝐵⊥𝑃𝐷,𝐴𝐷∩𝑃𝐷=𝐷,𝐴𝐷,𝑃𝐷⊂平面𝑃𝐴𝐷,.................................................................
..3分∴𝐴𝐵⊥平面𝑃𝐴𝐷,.............................................................................................
.................................4分{#{QQABAQgAoggIQBAAABhCQQ1yCgGQkBACCAgOwBAEoAAACQFABAA=}#}第2页共5页∵𝐴𝐵⊂平
面𝐴𝐵𝐶𝐷,...........................................................................................................................5分∴平
面𝑃𝐴𝐷⊥平面𝐴𝐵𝐶𝐷...........................................................................................
.......................6分解:(法一)(2)取𝐴𝐷中点为𝑂,连结𝑃𝑂,∵在△𝑃𝐴𝐷中,𝑃𝐴=𝑃𝐷,∠𝑃𝐷𝐴=60°,∴𝑃𝑂⊥𝐴𝐷,△𝑃𝐴𝐷为等边三角形.∵平面𝑃𝐴𝐷⊥平面𝐴𝐵𝐶𝐷,平面𝑃𝐴𝐷∩平面�
�𝐵𝐶𝐷=𝐴𝐷,𝑃𝑂⊂平面𝑃𝐴𝐷,∴𝑃𝑂⊥平面𝐴𝐵𝐶𝐷,............................................................
...............................................................7分以𝑂为坐标原点,建立如图所示的空间直角坐标系,设底面正方形𝐴𝐵𝐶𝐷的边长为2,∴𝑃(
0,0,√3),𝐴(1,0,0),𝐵(1,2,0),𝐶(−1,2,0),𝐷(−1,0,0),∴𝑃𝐵⃗⃗⃗⃗⃗=(1,2,−√3),𝑃𝐶⃗⃗⃗⃗⃗=(−1,2,−√3),................
............................................................9分设平面𝑃𝐵𝐶的一个法向量𝒎=(𝑥,𝑦,𝑧),则{𝑃𝐵⃗⃗⃗⃗⃗⋅𝒎=0𝑃𝐶⃗⃗⃗⃗⃗⋅𝒎=0,即{𝑥+2�
�−√3𝑧=0−𝑥+2𝑦−√3𝑧=0,令𝑦=3,则𝑥=0,𝑧=2√3,∴𝒎=(0,3,2√3),.........................................................
............................................................10分由(1)可知平面𝑃𝐴𝐷的一个法向量𝒏=(0,1,0),.....................................
.............................11分设平面𝑃𝐴𝐷与平面𝑃𝐵𝐶的夹角为𝜃,则𝑐𝑜𝑠𝜃=|𝒎⋅𝒏||𝒎||𝒏|=3√21×1=√217,∴平面𝑃𝐴𝐷与平面𝑃𝐵𝐶夹角的余弦值为√217...
..............................................................................12分(法二)(2)设平面𝑃𝐴𝐷与平面𝑃𝐵𝐶的交
线为𝑙,∵𝐵𝐶//𝐴𝐷,𝐴𝐷⊂平面𝑃𝐴𝐷,𝐵𝐶⊄平面𝑃𝐴𝐷,∴𝐵𝐶//平面𝑃𝐴𝐷,又∵𝐵𝐶⊂平面𝑃𝐵𝐶,∴𝐵𝐶//𝑙,𝐴𝐷//𝑙,∵平面𝑃𝐴𝐷与平面𝑃𝐵𝐶有一个交点𝑃,∴𝑙为过点𝑃且与𝐵𝐶平行的一条直线,如下图
,..............................................................................7分取𝐴𝐷中点为𝑂,取𝐵𝐶中点为𝑀
,连结𝑃𝑂,𝑃𝑀,𝑂𝑀,∵底面四边形𝐴𝐵𝐶𝐷为正方形,𝑂,𝑀分别为𝐴𝐷,𝐵𝐶的中点,∴𝑂𝑀//𝐴𝐵,又∵𝐴𝐵⊥平面𝑃𝐴𝐷,∴𝑂𝑀⊥平面𝑃𝐴𝐷,...........
..................................................................................................................8分∵
𝑙⊂平面𝑃𝐴𝐷,PABCD(第18题图1)Oyzx{#{QQABAQgAoggIQBAAABhCQQ1yCgGQkBACCAgOwBAEoAAACQFABAA=}#}第3页共5页∴𝑂𝑀⊥𝑙,∵
在△𝑃𝐴𝐷中,𝑃𝐴=𝑃𝐷,𝑂为𝐴𝐷的中点,∴𝑃𝑂⊥𝐴𝐷,𝑃𝑂⊥𝑙,又𝑃𝑂∩𝑂𝑀=𝑂,𝑃𝑂,𝑂𝑀⊂平面𝑃𝐴𝐷,∴𝑙⊥平面𝑃𝑂𝑀,∴𝑙⊥𝑃𝑀,又∵∠𝑂𝑃𝑀为锐角,∴∠𝑂
𝑃𝑀为平面𝑃𝐴𝐷与平面𝑃𝐵𝐶的夹角,..........................................................................
...........10分设底面正方形𝐴𝐵𝐶𝐷的边长为2,在△𝑃𝑂𝑀中,𝑃𝑀=√𝑃𝑂2+𝑂𝑀2=√7,𝑐𝑜𝑠∠𝑃𝑂𝑀=𝑃𝑂𝑃𝑀=√3√7=√217,∴平面𝑃𝐴𝐷与平面𝑃𝐵𝐶夹角的余弦值为√217.............
....................................................................12分19.(12分)解:(1)由正弦定理得𝑠𝑖𝑛𝐶𝑐𝑜𝑠𝐵+3𝑠�
�𝑛𝐵𝑐𝑜𝑠𝐶=𝑠𝑖𝑛𝐴−𝑠𝑖𝑛𝐵,.........................................2分因为𝑠𝑖𝑛𝐴=𝑠𝑖𝑛[𝜋−(𝐵+
𝐶)]=𝑠𝑖𝑛(𝐵+𝐶),所以𝑠𝑖𝑛𝐶𝑐𝑜𝑠𝐵+3𝑠𝑖𝑛𝐵𝑐𝑜𝑠𝐶=𝑠𝑖𝑛(𝐵+𝐶)−𝑠𝑖𝑛𝐵,.............................................
...................3分即𝑠𝑖𝑛𝐶𝑐𝑜𝑠𝐵+3𝑠𝑖𝑛𝐵𝑐𝑜𝑠𝐶=𝑠𝑖𝑛𝐵𝑐𝑜𝑠𝐶+𝑐𝑜𝑠𝐵𝑠𝑖𝑛𝐶−𝑠𝑖𝑛𝐵,2𝑠𝑖𝑛𝐵𝑐𝑜𝑠𝐶=−𝑠𝑖𝑛𝐵,而𝑠𝑖𝑛𝐵≠
0,所以𝑐𝑜𝑠𝐶=−12,............................................................................................................................
...5分又因为𝐶∈(0,𝜋),所以𝐶=2π3....................................................................................
.....................................................6分(2)因为𝑐𝑜𝑠𝐵=1314,𝐵∈(0,𝜋),所以𝑠𝑖𝑛𝐵=√1−𝑐𝑜𝑠2𝐵=3√314,.................................
.....................................................................7分𝑠𝑖𝑛𝐴=𝑠𝑖𝑛(𝐵+𝐶)=𝑠𝑖𝑛𝐵𝑐𝑜𝑠2𝜋3+𝑐𝑜𝑠𝐵𝑠𝑖𝑛2𝜋3=5√31
4,..............................................................8分由正弦定理𝑎𝑠𝑖𝑛𝐴=𝑏𝑠𝑖𝑛𝐵=𝑐𝑠𝑖𝑛𝐶,得55√314=𝑏3√
314=𝑐√32,PABCD(第18题图2)OMl{#{QQABAQgAoggIQBAAABhCQQ1yCgGQkBACCAgOwBAEoAAACQFABAA=}#}第4页共5页解得𝑏=3,𝑐=7,..........................
................................................................................................10分则𝐴𝐷=𝑐−𝐵�
�=2,所以𝑆△𝐴𝐶𝐷=12×𝐴𝐷×𝑏×𝑠𝑖𝑛𝐴=12×2×3×5√314=15√314........................................................12分20.(12分)解:(1)记“质检员甲认定一箱产
品合格”为事件𝐴,“该箱产品不含次品”为事件𝐵,则𝑃(𝐴)=0.8×1+0.1×𝐶93𝐶103+0.1×𝐶83𝐶103=1112,......................................................
.........................3分𝑃(𝐴𝐵)=0.8=45,...............................................................
...............................................................4分由条件概率公式得𝑃(𝐵|𝐴)=𝑃(𝐴𝐵)𝑃(𝐴)=451112=4855,所以在质检员甲认定一箱产
品合格的条件下,该箱产品不含次品的概率为4855.........................6分(2)由题意可得𝑋可以取0,1,2,....................................
............................................................7分则𝑃(𝑋=0)=𝑃(𝐴)=1112,...........................................................
......................................................8分𝑃(𝑋=1)=0.1×𝐶11⋅𝐶92𝐶103+0.1×𝐶21⋅𝐶82𝐶103=23300,...............................
.....................................................9分𝑃(𝑋=2)=0.1×𝐶22𝐶81𝐶103=1150,......................................
...................................................................10分所以随机变量𝑋的分布列为𝑋012𝑃1112233001150....
..................................................................................................................................
........................11分所以𝐸(𝑋)=0×325+1×23300+2×1150=9100...............................................................................12
分21.(12分)解:(1)𝑓′(𝑥)=𝑒𝑥−𝑚,...............................................................................................................1分当𝑚
⩽0时,由𝑓′(𝑥)>0,𝑓(𝑥)在𝑹上单调递增,.......................................................................2分当𝑚>0时
,由𝑓′(𝑥)=0,可得𝑥=𝑙𝑛𝑚,∴𝑥∈(−∞,𝑙𝑛𝑚)时,𝑓′(𝑥)<0,𝑓(𝑥)单调递减;...................................................................
....3分𝑥∈(𝑙𝑛𝑚,+∞)时,𝑓′(𝑥)>0,𝑓(𝑥)单调递增.....................................................................4分
∴当𝑚⩽0时,𝑓(𝑥)在𝑹上单调递增;当𝑚>0时,𝑓(𝑥)在区间(−∞,𝑙𝑛𝑚)上单调递减,在区间(𝑙𝑛𝑚,+∞)上单调递增.(2)设𝑔(𝑥)=𝑒𝑥−𝑚𝑥+𝑙�
�(𝑥+1)−1(𝑥⩾0),则𝑔′(𝑥)=𝑒𝑥+1𝑥+1−𝑚,.........................5分(i)当𝑚⩽1时,𝑔′(𝑥)=𝑒𝑥+1𝑥+1−𝑚⩾1−𝑚⩾0,.......................
..........................6分∴𝑔(𝑥)在区间[0,+∞)上单调递增,则𝑔(𝑥)⩾𝑔(0)=0恒成立,.............................................7
分(ii)当𝑚>1时,令ℎ(𝑥)=𝑒𝑥+1𝑥+1−𝑚,则ℎ′(𝑥)=𝑒𝑥−1(𝑥+1)2,...........................................8分{#{QQ
ABAQgAoggIQBAAABhCQQ1yCgGQkBACCAgOwBAEoAAACQFABAA=}#}第5页共5页令𝑘(𝑥)=𝑒𝑥−1(𝑥+1)2,则𝑘′(𝑥)=𝑒𝑥+2(𝑥+1)3>0,∴𝑘(𝑥)在区间[0,+∞)上单调
递增,则𝑘(𝑥)⩾𝑘(0)=0,∴ℎ(𝑥)在区间[0,+∞)上单调递增,则ℎ(𝑥)⩾ℎ(0)=2−𝑚,....................................................9分①若1<𝑚⩽2,则𝑔′
(𝑥)⩾0恒成立,则𝑔(𝑥)在区间[0,+∞)上单调递增,∴𝑔(𝑥)⩾𝑔(0)=0,........................................................
................................................................10分②若𝑚>2,则𝑔′(0)<0,𝑔′(𝑙𝑛𝑚+1)=(𝑒−1)𝑚+12
+𝑙𝑛𝑚>0,∴∃𝑥0∈(0,𝑙𝑛𝑚+1),使得𝑔′(𝑥0)=0,∴𝑔(𝑥)在区间[0,𝑥0)上单调递减,则𝑔(𝑥0)<𝑔(0)=0,与条件矛盾,.................................11分综上所述,实数𝑚的取值范围为(−∞,2]..
..............................................................................12分22.(12分)解:(1)由双曲线定义可知||𝑀𝐹1|−|𝑀𝐹2||=2�
�=2,∴𝑎=1,...........................................1分又由|𝐹1𝐹2|=4,∴𝑐=2,....................................................
...2分∵𝑎2+𝑏2=𝑐2,∴𝑏=√3,......................................................3分∴双曲线𝐶的方程为𝑥2−𝑦23=1..........
.........................................4分(2)(i)设𝑀(𝑥0,𝑦0),𝑃(𝑥1,𝑦1),𝑄(𝑥2,𝑦2),则𝑦1=√3𝑥1①,𝑦2=−√3𝑥2②,将①+②可得𝑦1+𝑦2=√3(𝑥1−𝑥2),将①−②可得𝑦
1−𝑦2=√3(𝑥1+𝑥2),............................5分∴𝑦1+𝑦2√3(𝑥1+𝑥2)=√3(𝑥1−𝑥2)𝑦1−𝑦2,即𝑦1+𝑦2𝑥1+𝑥2=3(�
�1−𝑥2)𝑦1−𝑦2,...................................................................................6分由题可知|𝑀𝑃|=|𝑀𝑄|,∴𝑥1+𝑥2=2𝑥0,𝑦1+𝑦2=2
𝑦0,∴𝑦0𝑥0=3(𝑥1−𝑥2)𝑦1−𝑦2,即𝑘𝑃𝑄=3𝑥0𝑦0,..............................................................
............................................7分∴直线𝑃𝑄的方程为𝑦−𝑦0=3𝑥0𝑦0(𝑥−𝑥0),即3𝑥0𝑥−𝑦0𝑦=3𝑥02−𝑦02,又∵点𝑀在𝐶上,∴3𝑥02−𝑦
02=3,则3𝑥0𝑥−𝑦0𝑦=3,................................................................8分将方程联立{𝑥2−𝑦23=1,3𝑥0𝑥−𝑦0𝑦=3,得(𝑦0
2−3𝑥02)𝑥2+6𝑥0𝑥−3−𝑦02=0,∴−3𝑥2+6𝑥0𝑥−3𝑥02=0,由𝛥=0可知方程有且仅有一个解,∴𝑙与𝐶有且仅有一个交点........................
................................9分(ii)由(2)(i)联立{𝑦=√3𝑥,3𝑥0𝑥−𝑦0𝑦=3,可得𝑥1=√3√3𝑥0−𝑦0,同理可得𝑥2=√3√3𝑥0+𝑦0,.........10分∴|𝑂𝑃|⋅|𝑂𝑄
|=√𝑥12+𝑦12⋅√𝑥22+𝑦22=4|𝑥1𝑥2|=4×33𝑥02−𝑦02=4,............................................11分∴1|𝑂𝑃|+2|𝑂𝑄|=1
|𝑂𝑃|+|𝑂𝑃|2⩾2√1|𝑂𝑃|×|𝑂𝑃|2=√2,当且仅当1|𝑂𝑃|=|𝑂𝑃|2即|𝑂𝑃|=√2时取等号.又∵|𝑂𝑃|∈(0,+∞),∴1|𝑂𝑃|+2|𝑂𝑄|的取值范围为[√2,+∞)....
..........................................12分{#{QQABAQgAoggIQBAAABhCQQ1yCgGQkBACCAgOwBAEoAAACQFABAA=}#}获得更
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