2021届四川省泸州市高三上学期文科数学第一次教学质量诊断性试题答案

2021届四川省泸州市高三上学期文科数学第一次教学质量诊断性试题
答案
13
．3；
14．0；
15.43
-；
16
三、解答题：
17．解：（Ⅰ）因为2()2cos 12
x f x x =-+
cos x x =- ·
················································································· 1分 2sin()6x π
=-， ·
·················································································· 2分
因为()()6f παα=+，所以sin()6π
αα-=， ······························· 3分
1
cos 2
ααα-=，·························································· 4分
即cos αα-=， ·········································································· 5分
所以tan α=··············································································· 6分 （Ⅱ）()f x 图象上所有点横坐标变为原来的1
2
倍得到函数()g x 的图象，
所以函数()g x 的解析式为()2sin(2)6g x x π
=-， ········································· 8分
因为02
x π
≤≤
，所以526
6
6
x πππ
-
-
≤≤
， ··········································· 9分 所以1()2g x -≤≤， ········································································· 11分
故()g x 在[0,]2
π
上的值域为[1,2]-． ·························································· 12分
18．解：（Ⅰ）因为()sin cos f x k x kx x '=+， ··································································· 2分
所以()sin cos 2222
f k k k ππππ
'=+⨯=， ·························································· 3分
又因为()sin 2222
k
f k b b ππππ=⨯+=+， ························································ 4分
点(,())22
f ππ
处的切线方程为230x y --=．
所以2k =, ····························································································· 5分 3b =-；
································································································ 6分 （Ⅱ）()f x 在(0,)2
π
上有且只有一个零点， ························································ 7分
因为()2sin 2cos f x x x x '=+， ··································································· 8分
当(0,)2x π
∈时，()0f x '>， ······································································ 9分
所以()f x 在(0,)2
x π
∈上为单调递增函数且图象连续不断， ···························· 10分
因为(0)30f =-<，()302f π
π=->， ······················································· 11分
所以()f x 在(0,)2
π
上有且只有一个零点． ··················································· 12分
19
························································ 2分
因为sin 0C ≠，所以sin cos 2A
A =， ···························································· 3分
······································································· 4分
cos 02
A
≠，
······················································································ 5分
·
······································································ 6分 （Ⅱ）解法一：设ABD △的AB 边上的高为1h ，ADC △的AC 边上的高为2h ，
因为3,3,1ABD ADC S S c b ===△△， ·
······························································· 7分
所以1211
322c h b h ⋅=⨯⋅， ·
········································································· 8分
所以12h h =，AD 是ABC △角A 的内角平分线，所以30BAD ∠=， ·················· 9分 因为3ABD ADC S S =△△，可知3
4ABD ABC S S =△△， ·
············································· 10分
所以
131
sin 30sin 60242
AB AD AB AC ⨯⨯=⨯⨯⨯， ·
········································· 11分
所以33
AD =
. ···················································································· 12分
解法二：设=(0)3BAD παα∠<<，则=3
DAC πα∠-， ························································ 7分 因为3ABD ADC S S =△△，3,1c b ==，
所以11sin 3sin()223c AD b AD παα⨯⨯=⨯⨯⨯-， ·
· 8分 所以sin sin()3παα=-， ·
······························ 9分 所以31sin cos sin 2ααα=
-，3
tan α∴=
， 因为03
πα<<，所以30BAD ∠=， ························································· 10分 3ABD ADC S S =△△，可知3
4ABD ABC S S =
△△， ·
···················································· 11分
所以
131
sin 30sin 60242
AB AD AB AC ⨯⨯=⨯⨯⨯， 所以33
AD =
. ···················································································· 12分
解法三：设AD x =，=BDA α∠，则=ADC πα∠-，
在ABC △中，由3,1c b ==及余弦定理可得：2222cos a b c bc A =+-，
所以7a =， ·
······················································································· 7分 因为3ABD ADC S S =△△，可知37
3=
BD DC =， ··············································· 8分
在ABD △中2222cos AB BD AD BD AD α=+-⋅⋅， 即26337
9cos 16AD AD α=
+-⋅⋅， ·
···························································· 9分 在ADC △中，277
1cos()16AD AD πα=+-⋅⋅-， ·
······································ 10分 即277
1+cos 16AD AD α=
+⋅⋅， ·
······························································ 11分 所以33
AD =
． ·
················································································ 12分 20.解：（Ⅰ）第一步：在平面ABCD 内作GH ‖BC 交CD 于点H ； ······································· 2分
第二步：在平面SCD 内作HP ‖SC 交SD 于P ；
········································· 4分
第三步：连接GP ，点P 、GP 即为所求． ······················································· 5分 （Ⅱ）因为P 是SD 的中点，//HP SC ，
所以H 是CD 的中点，而//GH BC ，
所以G 是AB 的中点， ················································································ 6分
所以1sin1202GBC S BC GB ︒=
⋅⋅=
△， 连接AC ，GD 交于O ，连SO ，设S 在底面ABCD 的射影为M ， 因为SA SB SD ==，
所以MA MB MD ==， ································ 7分 即M 为ABD △的外心，
所以M 与O 重合， ····································· 8分
因为OD 2SD =，
所以SO ， ········································ 9分
所以13S GBC GBC V S SO -=⋅⋅△ ·
················ 10分 因为GP //平面SBC ， ·
································ 11分
所以S PBC P SBC G SBC S GBC V V V V ----====
． ··················································· 12分 21.解：（Ⅰ）当5
2m =
时，152
)ln 2(5x x x f x =---， ·
························································· 1分 所以()222
15252
122x x f x x x x -+'=+-=
， ························································ 2分 因为0x >,
由()0f x '>得22520x x -+>， ·································································· 3分 所以1
02
x <<
，或2x >， 所以()f x 在[1,2)上单减，(2,e]上单增， ·
····················································· 4分 所以函数()f x 在[1,e]上的最小值为5
1ln 22
--； ·
············································ 5分 （Ⅱ）原不等式()1ln ln m x x x x n
k x
+-++⇔≤
. ····················································· 6分
因[]1,e m ∈,[],e 1x ∈,所以
()1ln ln 1ln ln m x x x x n
x x x x n
x
x
+-+++-++≥
,
令()1ln ln x x x x n
g x x +-++=
, ···································································· 7分
即()2
ln x x n g x x -+-'=
，令()ln p x x x n =-+-，即()
1
1p x x '=-+， 所以()p x 在[],e 1x ∈上递增； ····································································· 8分 ①当()10p ≥即1n ≤时, 因为[]1,e n ∈,所以1n =，
当[],e 1x ∈,()0p x ≥,即()0g x '≥，所以()g x 在[]1,e 上递增， 所以()()min 1c g x g n ===，
故22n c n +==， ·
·············································································· 9分 ①当()e 0p ≤即[]e 1,e n ∈-时， 因为[],e 1x ∈,()0p x ≤,即()0g x '≤，
所以()g x 在[]1,e 上递减，所以()()min 2e e
n c g x g +===， 故21
2e ,e 1e e
e n n c n +⎡⎤+=
+∈+++⎢⎥⎣⎦ ······················································· 10分 ①当()()1e 0p p <即()1,1n e ∈-时, 又()ln p x x x n =-+-在[]1,e 上递增，
所以存在唯一实数()01,e x ∈,使得()00p x =,即00ln n x x =-，
则当()01,x x ∈时()0p x <，即()0g x '<，当()0,e x x ∈时()0p x >即()0g x '>， 故()g x 在()01,x x ∈上减，()0,e x x ∈上增， 所以()()00000mi 000
n 1ln ln 1
ln x x x x n x x x c g x g x +-++=+===.························· 11分
所以000000
11
ln ln n c x x x x x x +=+
+-=+， 设()001x x x u =+（()01,e x ∈），则()2'
0200
1110x u x x x -=-=
>， 所以()u x 在[]1,e 上递增,所以12,e e n c ⎛
⎫+∈+ ⎪⎝
⎭.
综上所述22,e 1e n c ⎡
⎤+∈++⎢⎥⎣
⎦. ·
································································· 12分 22.解： (Ⅰ) 解法一：设曲线1C 与过极点且垂直于极轴的直线相交于异于极点的点E ，且曲线1C 上
任意点F (,)ρθ，边接OF ，EF ，则OF ⊥EF ， ·············································· 2分
在△OEF 中，4cos()4sin 2
π
ρθθ=-=， ······················································ 4分
解法二：曲线1C 的直角坐标方程为22(2)4x y +-=， ······································ 2分
即2240x y y +-=， 所以曲线1C 的极坐标方程为4sin ρθ=； ·························· 4分
（Ⅱ）因曲线2C
的参数方程为)4
x t
y t π⎧=⎪
⎨=-⎪⎩与两坐标轴相交，
所以点(2,0),(0,2)A B ， ············································································ 6分 所以线段AB 极坐标方程为cos sin 20(0)2
π
ρθρθθ+-=≤≤
， ·························· 7分
12
||sin cos OP ρθθ
==
+，2||4sin OM ρθ==,
sin cos 4sin 2
OM
OP θθ
θ+=⨯ 22sin 2sin cos θθθ=+
······················ 8分 1cos2sin2θθ=-
+)14
π
θ=-+， ·
····················································· 9分 当38
π
θ=
1． ··························································· 10分 23.解：(Ⅰ)
由3222,ab a b =++≥ ······································································· 2分
220-≥，
， ··························································· 4分 当且仅当1,2a b ==时取得“=，
即k 的最小值为2． ···················································································· 5分
（Ⅱ）由2k =，2()(2)2x m x x m x m -+-≥---=-， ·········································· 7分
因0,x R ∃∈使不等式22x m x -+-≤成立， 所以22,m -≤
即222m -≤-≤， ····················································································· 9分 即m 的取值范围是[0,4] ············································································ 10分。