四川省2024届绵阳三诊数学理数答案

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绵阳市高中2021级第三次诊断性考试
理科数学参考答案及评分意见
一、选择题:本大题共12小题,每小题5分,共60分.
CBACB
DDCBC
DA
二、填空题:本大题共4小题,每小题5分,共20分.
13.10
1415.45π16.2
三、解答题:本大题共6小题,共70分.17.解:
(1)由列联表可计算()2
2
10024124816 4.762 3.84140607228
K ⨯⨯-⨯=≈>⨯⨯⨯,········4分
∴有95%的把握认为参数调试能够改变产品合格率.···························5分(2)根据题意,设备更新后的合格概率为0.8,淘汰品概率为0.2.········6分可以认为从生产线中抽出的6件产品是否合格是相互独立的,···············8分
设X 表示这6件产品中淘汰品的件数,则(60.2)X B ,
~,·······················9分可得:60166015(1)C 0.80.2C 0.80.2p P X =≤=⨯⨯+⨯⨯·································10分
50.8(0.8 1.2)0.65536=⨯+=.······················································12分
18.解:
(1)设{}n a 的公差为d ,则1,1+d ,2+2d 成等比数列,·················1分∴2(1)1(22)d d +=⨯+,解得:d =1或d =−1,·····································3分而d =−1,不满足1a ,2a ,31a +成等比数列,
∴d =1,·······················································································4分∴数列{}n a 的通项公式n a n =.······················································5分(2)令121112 31n n n n n n a b a b a b a b D --=++++=- ,·······························6分∴1112112131 31n n n n n n n a b a b a b a b a D b ++-++++++==+- ,·······················7分两式相减有:11111(2) 3n n n n n n a b b b b D D +-+-=++++=⋅ ,····················8分∴数列{}n b 的前n +1项和为23n ⋅,即123n n T +=⋅,·································9分又1112D a b ==,所以12b =,··························································10分∴112123n n n b b b b --++++=⋅ ,·····················································11分∴123n n T -=⋅.·············································································12分
19.解:(1)过C 作CH ⊥1BB 交1BB 于H ,·············································1分
∵C 在平面11ABB A 内的射影落在棱1BB 上,
∴CH ⊥平面11ABB A ,又AB ⊂平面11ABB A ,···································2分∴CH AB ⊥,··············································································3分
又1AB B C ⊥,且1B C CH C = ,····················································4分∴AB ⊥平面11BCC B ;···································································5分(2)∵111111
2
ABC A B C ABB A V S CH -=
⋅,则3
2213
CH ⨯==,·························6分
过C 作1CQ AA ⊥交1AA 于Q ,连结HQ ,
∵AA 1与CC 1
则CQ =,
又∵CH ⊥平面11ABB A ,则CH HQ ⊥,···········································7分在Rt △CHQ 中:222211HQ CQ CH =-=-=,则1HQ =,又1AA CH ⊥且1AA CQ ⊥,
∴1AA ⊥平面CHQ ∴1AA HQ
⊥又由(1)知:AB ⊥平面11BCC B ,∴AB ⊥1BB ,∴1AB AA ⊥,则四边形ABHQ 为矩形,∴1AB HQ ==,
又四边形ABB 1A 1的面积为3,则BB 1=3,···········································8分分别以HB HQ HC ,,为x 轴、y 轴、z 轴建立如图所示空间直角坐标系,设(0)BH x x =>,
∴(00),,B x ,(10)A x ,,,1(301)C -,,,
∵1AC =∴22221(3)1127AC x =+++=,
解得2x =,············································9分
∴B (2,0,0),1(110)A -,,,(001)C ,,
,∴1(310)A B =- ,,,(201)BC =- ,,,
设平面1A BC 的法向量为n 1()x y z =,
,,∴11130
20
A B n x y BC n x z ⎧⋅=-=⎪⎨⋅=-+=⎪⎩
,令1x =,则n 1(132)=,,,·········易知平面11ABB A 的法向量n 2(001)=,
,,·······································11分
∴12cos 7
n n <>=
,,∴平面1A BC 与平面11ABB A 所成锐二面角的余弦值为
14
7
. (12)

20.解:(1
)离心率3
e 2=,则12
b a =,①································1分
当x =1
,y =±
||=2AB ,②····························3分联立①②得:21,a b ==,···························································4分
故椭圆C 方程为:2
214
x y +=;·······················································5分
(2)设过F ,A ,B 三点的圆的圆心为Q (0,n ),1122()()A x y B x y ,,,,
又(0)F ,
则22||=||QA QF
,即222211(0)()(0(0)x y n n -+-=++-,·················6分
又11(,)A x y 在椭圆2214x y +=上,故221114
x y +=,
带入上式化简得到:2113210y ny +-=,③··········································7分
同理,根据2
2
=QB QF 可以得到:2
223210y ny +-=,④···················8分
由③④可得:12,y y 是方程23210y ny +-=的两个根,则121
3
y y =-,·····9分
设直线AB :1x ty =+,联立方程:22
1
4
1x y x ty ⎧+=⎪⎨⎪=+⎩
,整理得:22(4)230t y ty ++-=,⑤················································10分故1223143
y y t -=
=-
+,解得:2
5t =,
∴t =,···············································································11分∴直线AB
的方程为:10x ±-=.············································12分
21.解:(1)当1=a 时,x x x x x x f --+=224
1
ln )21()(,
∴x x x f ln )1()(+=',则切线斜率1e +=k ,······································2分∴曲线)(x f 在(e ,(e)f )处的切线方程:)e )(1e (e 4
12
-+=-x y ,···········4分即:0e e 4
3)1e (2
=--
-+y x ,························································5分(2)证明方法一:因为)ln )(ln ()(a x a x x f -+=',·····························6分由()0f x '>得到a x >;由()0f x '<得到a x <<0.
∴()f x 在)0(a ,单调递减,在)(∞+,
a 单调递增.∴2min 4
5
)()(a a f x f -==,·····························································7分
要证105()(2ln )e 8a f x a -<-+,即证:2155
(2ln )e 48
a a a --<-+,
只需证:2
12ln 20e
a a a --->(12)a <<(*)·········································8分
设)21(2ln e
2)(12
<<--=-x x x x g x ,
则2231
1142142e ()e e x x x x x x x g x x x ------'=-=
,···········································9分设23
1
42()1(12)e x x x h x x --=-<<,
则321121082(1)(4)
()e e x x x x x x x x h x ---+--'==
,易知:()h x 在(1,2)上单调递减,而(1)10h =>,(2)10h =-<,
故必存在唯一)21(0,∈x ,使得0()0h x =,·······································10分∴当)1(0x x ,∈时,()0h x >,即()0g x '>;当)2(0,x x ∈时,()0h x <,即()0g x '<,
∴()g x 在0(1)x ,上单调递增,在)2(0,x 上单调递减.························11分而0)1(=g ,022ln e
8
)2(>--=
g ,∴()0g x >在(1,2)上恒成立,即(*)式成立,原命题得证.··············12分
由()0f x '>得到a x >;由()0f x '<得到a x <<0.
∴()f x 在)0(a ,单调递减,在)(∞+,
a 单调递增.∴2min 45
)()(a a f x f -==,·····························································7分
要证105
()(2ln )e 8a f x a -<-+,即证:2155(2ln )e 48
a a a --<-+,
只需证:122ln 0(12)e a a a a a -+-><<,················································8分设122ln ()(12)e x x x
g x x x -+=-<<,即证()0g x >在x ∈(1,2)恒成立.则12
221ln ()(12)e x x x
g x x x --+'=
+<<,令()()h x g x '=,则13
2(2)12ln ()(12)e x x x
h x x x --+'=-<<,························9分又∵12x <<,∴
13
2(2)12ln 00e x x x
x --+<-<,,∴13
2(2)12ln ()0e x x x
h x x --+'=
-<在(1,2)上恒成立.······························10分∴()g x '在(1,2)单调递减,又8e(1+ln 2)
(1)10(2)04e
g g -+''=>=
<,,
∴存在0(12)x ∈,,使得()g x 在0(1)x ,单调递增,在0(2)x ,
单调递减.又8e(2ln 2)
(1)0(2)02e
g g -+==
>,,·············································11分
∴()0g x >在x ∈(1,2)恒成立,得证.··········································12分
由()0f x '>得到a x >;由()0f x '<得到a x <<0.
∴()f x 在)0(a ,单调递减,在)(∞+,
a 单调递增.∴2min 45
)()(a a f x f -==,·····························································7分
要证1
05()(2ln )e 8a f x a -<-+,即证:2155(2ln )e 48
a a a --<-+,
只需证:1
2(2ln )e 1(12)2a a a a -+<<<,,··············································8分令1
2
(2ln )()e (12)2a a g a a a
-+=<<,则1
3
(2ln )32ln ()(12)2a a a a g a a a -+--'=
<<,
设()(2ln )32ln (12)h a a a a a =+--<<,··········································9分∴'2
()3ln (12)h a a a a
=+-
<<,易知()h a '在(1,2)单调递增.∴''()(1)10h a h >=>,·································································10分∴()h a 在(1,2)单调递增,又(1)10(2)10h h =-<=>,,
∴存在唯一0(12)a ∈,,使得当0(1)()0a a h a ∈<,,,()0()g a g a '<,单调递减,
当0(2)()0a a h a ∈>,
,,()0()g a g a '>,单调递增,···························11分又(22ln 2)e
(1)1(2)18
g g +==
<,,∴()1g a <在(12)a ∈,
恒成立,原不等式得证.·································12分
22.(1)方法一:
令0=x ,即0sin 3cos =+αα,解得3
3
tan -=α,···························1分∴ππαk 26
5+=或Z k k ∈+=,ππ
α2611,···········································2分当ππ
αk 26
5+=时,4)23(3212=-
⨯-+=y ;···································3分当ππ
αk 26
11+=
时,0233212=⨯
--=y ,······································4分∴曲线C 1与y 轴的交点坐标为(0,4),(0,0).·······························5分方法二:消参:由C 1的参数方程得:
431)cos 3(sin )sin 3(cos )2(2222=+=-++=-+ααααy x ,··············1分
即曲线C 1的普通方程为:4)2(22=-+y x ,······································2分令0=x ,得0=y 或4,·································································4分∴曲线C 1与y 轴的交点坐标为(0,4),(0,0).·······························5分(2)方法一:将曲线C 1:4)2(22=-+y x 化为极坐标方程,
得:θρsin 4=,···········································································6分
联立C 1,C 2的极坐标方程⎪⎩⎪⎨⎧==+θ
ρπθρsin 42)3
sin(,得θsin 42)3sin(=+⋅πθ,从而12sin 2
3
22cos 11)cos 3(sin sin =+-⇒
=+θθθθθ,····················7分整理得:21)62sin(=-πθ,所以6
5662π
ππθ或=-,····························8分
即2

πθ或=
,············································································9分∴∠AOB 3
62π
ππ=-=.······························································10分
方法二:将C 2的极坐标方程2)3
sin(=+π
θρ,
化为直角坐标方程:043=-+y x ,················································6分∴C 2是过点(0,4)且倾斜角为3

的直线,······································7分不妨设B (0,4),则∠OBA 6π=,因为BO 为直径,所以∠BAO 2
π
=,····9分∴∠AOB 3
62π
ππ=-=
.································································10分
23.(1)由b a b a 33+=
+得3)(3=⇒+=+ab ab
b a b a ,①·····························1分又由2)())(=-=---≥-+-=a b b x a x b x a x x f (,························3分且0>>b a ,所以2=-b a ,②······················································4分由①②得:1,3==b a ;··································································5分(2)t t t t bt at +-=+-=+-13333,································6分令2
0sin π
θθ≤
≤=,t ,则θcos 1=-t ,········································7分∴3
sin(2sin cos 313π
θθθ+=+=+-t t ,··································9分∴当6πθ=
时,即4
1
=t 时,bt at +-3的最大值为2.·····················10分。

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