湖北省麻城市2020届高三10月联考数学(理)试卷(扫描版)(含答案)

{
³¯³ “◇ l
³
²◇ ³ !÷#y $ƒ'$(ª(*ƒ)
¹·³ [
,fl ]$/$012 41fl —ys 9fl ;fi ,=,$0>$fi ,A )· $T =‡¹,³),’=(²,4),(T T ’=‡¹,4)· ²·A [,fl ]$/$0CD / §;,$0‰*I äA )· !"“flfl ~K '½),%+,fi R ” p ¿fi “1$ '½) +,3flfi R ”· ³·B [
,fl ]$/$0K # ;$M ,$0>$fi ,A )· ¯⁄ ,
$ (
⁄*‡¯¹,◇)T (◇,+8)· ¦⁄¦>◇, 4·B [
,fl ]$/$0}y P fl ˛R SP fl ,$0‰*I äA )· @a 6 ’‰9,:4=¹◇a ;6 ’‰9;@46 ’‰9,:a = 4 $;6 ’‰9·®½ B ·
¹◇
’·D [
,fl ]$/$0÷T K #U ¼XY ®fl äD \,$0>$fi ,A )· ƒ@A ¥=²³f i ¹(4⁄+ ¬)k C KA E F ƒH fifi fi ²L M N fi @A ¥=²³f i ¹(²⁄+ ¬),÷²⁄+ ¬ ’ ’ ’ = ¬+R ¬
(R *²),N ⁄ ³¬ R ¬(R *²)· ²
=
+ 4·³ [
,fl ]$/$0;]y ^$,$0>$fi ,A )· ${
¥=槡⁄, {⁄=◇,{
⁄=¹, ,
¹
( ¯ ²)
RN
1 ½ W fi Y ½ ²
,
,
ª N C ƒ ª Ç _ ` f i
⁄ ◇
⁄ Q
⁄= ¥=⁄ , ¥=◇ ¹
¥=¹ (
² ³ ¹ ³) ¹ ⁄² ¯ ⁄ = · ³ ³ ◇
’·D [
,fl ]$/$0÷T K #fl ä# $fiflc ,$0‰*I äA )· $³f i ¹4◇˚<¹<¹O å³4,¹¹◇·4<◇<’Q ¹²²4˚,³O ³(¯²◇˚)=³O ³²◇˚=³f i ¹’◇˚>³f i ¹4’˚,(fl 9 A ,B ,³,’Q ¹4¹◇˚= ’Q ¹’◇˚>¹>³f i ¹8◇˚> ¹>¹O å²,®½ D · ² ’
8·A [,fl ]$/$0K #U de ,$0‰*I äA )· ت (⁄)
fi '½),fl 9 dfi (¬) ,() ²+¬²
²+¬²
,1½fl 9 , ,®½
L ³·
L ² <◇L ¬
=¯ e ¬ >¯ e
³ >¯¹ B D
A ·
9·B [
,fl ]$/$0÷T ;Y ®,$0>$fi ,A )· ³O ³’²˚+³O ³¹8˚=³f i ¹¹8˚+³O ³¹8˚=槡²³f i ¹(¹8˚+4’˚)=槡²³f i ¹4³˚=槡²³O ³²’˚
, 槡²(³O ³’²˚+³O ³¹8˚)“
²×◇·89¹=¹·’8²,1½槡²(³O ³’²˚+³O ³¹8˚) ½f +fi ¹·’8· ¹◇·³ [
,fl ]$/$0K # äD \fl ¼X \,$0‰*I äA )fl g i A )· d fi L (⁄) hƒA (³,◇)fi Y ,1½L (⁄)+L (4¯⁄)=◇·@L (⁄)=L (²¯⁄),1½L (²¯⁄)+L (4¯⁄) =◇,1½L (⁄)=¯L (⁄+4),:L (⁄)=L (
⁄+8),1½L (¹4◇9)=L (9 +¹◇◇×8)=L (9 )·d fi L (9 )+ ² ²
² ²
L (4¯ 9)=◇,L (9)=¯L (³)=¯(³+¹O å9
)=¯’,1½L (¹4◇9)=¯’· ² ² ² ² ¹¹·A [
,fl ]$/$0÷T ;Y ®fl ÷T K # j M ,$0‰*I äA )· ²³f i ¹(4⁄+ ¬) $L (⁄)= ³ =¯²³f i ¹(²⁄+ ¬),³O ³(²⁄+ ¬)>
◇, ¯²³O ³(²⁄+ ¬) 4 4
4 %fi %m fi ³O ³(²⁄+ ¬)=◇ ,¦³f i ¹(²⁄+ ¬)¦=¹,(L (⁄) +,fi (¯²,²)· 4 4
¹²·B [
,fl ]$/$0¾# l 2^$,$0!fiflfl ! #y pq§>$fi ,A )· ÷L
'(⁄)=²⁄(³⁄¯a ),N ⁄ =◇,⁄ =a (a <◇), ¹ ²
³ fi a <⁄<◇ ,L '(⁄)<◇;fi ⁄<a o ⁄>◇ ,L '(⁄)>
◇· 槡 ²
³³
[›!÷#$$ƒ•'$(¤ø¹½(¤4½)-.K]
=
⁄
(⁄>å
T=¯¯=
flf(⁄)fl⁄
a r$N t}+
(a)a
³
L =³ L ³=¯ ·
²’
$(⁄)a
³
,N(⁄a)²(⁄a),RN⁄a o⁄a
L =¯
²’¯³ ²+³ =◇=³ =¯4 ·
$(⁄)fl(a,a+4)k$v}+,(a a+4¯a,(a“¯4·
L ²³³< ³“4
¹³·¹4[,fl]$/$0y¼K#fi j,$0>$fi,A)·
L(¯L(¹◇))=L(¯²)=4²=¹4·
¹4·³[,fl]$/$0÷T K#U§K#flfi¼,$0#v w2#yfi=·
(
³¬)
¹,
$³O³¯4=²<¯²
($A²+¹=◇fl@A=³O³⁄fl(³¬,³¬)k$³KçA·
¥¥¯4 ²
¹’·¹◇;¹8·’[,fl]$/$0#yflŁz$ƒ}^$,$0#y%•#y??$·zØ~|fi~•Çª?$عƒþ,;?@fi¹²◇+’◇¯⁄=¹8◇ý,:⁄=¹◇·
?zØ~|fi~Ç$õ?fi Tý,fi◇<T<¹’◇,fl$C¿fi?N fi½??3fl¼ƒ???õ?
‡fl·fi T“¹’◇,◇·8(T¯⁄)“◇·’T,%T“8⁄fi T“¹’◇?fi?,:8⁄“¹’◇,⁄“¹8·’’,@²⁄*²
,1½⁄v}+fi¹8·’·
¹4·(◇,+8)[,fl]$/$0¾#^$,$0K#fi?=^$fl‰*IäA)·
?½)å(⁄)⁄L
(⁄)◇),:'(⁄)=
+¹
‡L(⁄)+⁄L'(⁄)‡(⁄+¹)¯⁄L(⁄)
(⁄+¹)²·
dfi(⁄)+⁄'(⁄)
⁄L(⁄)1½‡(⁄)+⁄'(⁄)‡(⁄+¹)¯⁄(⁄)◇,'(⁄)◇,
L L<⁄+¹,L L L < å<
®å(⁄)fl(◇,+8)k???·1½(⁄+¹)(⁄+¹)<⁄+²¤(⁄+¹)L(⁄+¹)¹{⁄+¹>◇,
L
:⁄+¹>¹,%⁄>◇·
⁄+²<¤å(⁄+¹)<¹=å(¹),¹’·R:(¹)d fi L(⁄)=a²⁄¯a⁄+²a(a>◇%a>¹)?‰A A(¹,4),
1½L(¹)=a²+a=4·………………………………………………………………………………………¹y
d fi a>◇%a>¹,1½a=²,………………………………………………………………………………³y
1½L(⁄)R flfifi L(⁄)=4⁄¯²⁄+4(o L(⁄)=²²⁄¯²⁄+4)·………………………………………4y (²)(⁄)=(²⁄¯¹)²
¹’
,…………………………………………………………………………………’y L ²+ 4
fi²⁄=¹,%⁄=¯¹,(⁄)$N vfi+
¹’
…………………………………………………………… 8y
²L 4·
d fi²⁄>◇,……………………………………………………………………………………………………9y
1½L(⁄)+,fi‡
¹’
,+8)·……………………………………………………………………………¹◇y
4
¹8·R:(¹)$ªª³’¬(¬)³¬,…………………………………………………………………¹y 4¹²³4
®T=¬,:
²¬
………………………………………………………………………………………³y m=T=²·
@ (⁄)‰A(
’¬
,³),:(
’¬
)=³,N³f i¹(
’¬
)=¹·……………………………………………4y
L¹²L¹²
4 +P
f¦
P¦
<¬,1½P=¯¬·…………………………………………………………………………………4y
²³
(²)$(¹)ª,L(⁄)=³³f i¹(²⁄¯¬),:L(a)=³³f i¹(a¯¬)=9,
³²³’
:³f i¹(a¯¬)=³·…………………………………………………………………………………………’y ³’
d fi a*(¬,
’¬
),1½a¯¬*(◇,¬),1½³O³(a¯¬)=4,…………………………………………9y ³4 ³²³’
1½³f i¹a=³f i¹‡(a¯¬)+¬‡=³f i¹(a¯¬)³O³¬+³O³(a¯¬)³f i¹¬………………………………¹◇y
³³³³³³[›!÷#$$ƒ•'$(¤ø²½(¤4½)-.K]
²
P = ¹× ³+ 4
…………………………………………………………………………… ¹²y ² ’ ’
¹9·R :(¹)L
'(⁄)=¹¯a ²e a ⁄, …………………………………………………………………………………… ¹y 1½L '(◇)=¹¯a ²
· ………………………………………………………………………………………… ²y
@L (◇)=¯a ,1½@A ¥=L (⁄)fl A (◇,L (
◇))r fl A fi ¼fi ¥+a =(¹¯a ²
)⁄, %¥=(¹¯a ²
)⁄¯a · ………………………………………………………………………………………… ’y (²)d fi a ◇ ,1½a ² ◇
·÷ '(⁄)=◇,N ⁄ ²¹¹a ; ………………………………………………… 4y > > L =¯ a ÷ '(⁄)>◇,N ⁄ ²¹¹a ;………………………………………………………………………………… ’y
L <¯ a
÷ '(⁄) ◇
,N ⁄ ²¹¹a ………………………………………………………………………………… 8y L < >¯ a · 1½ (⁄) = ( ²¹¹a ) ²¹¹a +¹
………………………………………………………………… ¹◇y
L ¹Q ~
L ¯ a =¯ a ·
() , ²¹¹a +¹ ,
, ¯¹ , d fi L ⁄ <◇?fi ? 1½¯
a
<◇
d fi a >◇ 1½a >
e ² ®a $+?þfi (e ¯¹
,+8)·…………………………………………………………………………… ¹²y ²◇·R :(¹)$å(⁄)=4³f i ¹⁄
⁄ ¹ ⁄)=槡³³f i ¹²⁄¯(¹¯³O ³²⁄)
³O ³ ² ¯ ³f i ¹ ²
=²³f i ¹(²⁄+ ¬)¯¹
,………………………………………………………………………………………… ²y 4
(L (
⁄)=²³f i ¹(²⁄+ ¬+² )¯¹·…………………………………………………………………………… ³y 4
@L (⁄)fi '½),: ¬+²P = ¬+R ¬(R *²),$◇<P “ ¬,(P
= ¬, ………………………………… 4y
() ( 4 ² ¬) ² ²(³O ³²⁄¯³f i ¹²⁄) 4 ²(¹¯’Q ¹²⁄)
…………………… y (L ⁄ =²³f i ¹²⁄+² ¯¹=²³O ³²⁄¯¹= ³O ³²⁄+³f i ¹²⁄ ¯¹= ¹+’Q ¹²
⁄ ¯¹· ’ $’Q ¹ ²,( ()= 4 ¯³ 4 ¹¹,…………………………………………………… 4y a > L a ¹+’Q ¹²
< ²¯³=¯ ’
a ¹+² @ (a
)= 4 ¯³ ¯³,( () $+?þfi (
, ¹¹)·………………………………………… ’y L
¹+’Q ¹²
a > L a ¯³ ¯’ (²)$⁄*(¬,’¬),(²⁄+ ¬+²P *(²¬+ ¬+²P ,²¬+ ¬+²P )
·………………………………………… 8y 4 4 4 ²
$◇<P “ ¬,( ¬+²P *(¬,’¬‡,¬+²P *(¬,³¬‡
· ………………………………………………… 9y ² 4 4 4 ² ² ²
‘¬+²P “ ¬, $L (⁄)fl (¬,’¬)k %??½),( 4 ² ……………………………………………………… ¹◇y 4 < ¬,
L
◇<P “² (P *‡¬,¬‡·……………………………………………………………………………………………… ¹²y 4 ²
²¹·(¹)R :÷L (
¬⁄)=¬⁄(¹¯³f i ¹¬⁄)=◇,N ⁄=◇o ³f i ¹¬⁄=¹, …………………………………………… ¹y %⁄=◇o ¬⁄= ¬+²R ¬
(R *²),%⁄=◇o ⁄= ¹+²R (R *²),………………………………………… ³y ² ²
1½L (¬
⁄)fl (¯²◇,²◇)k 3 A $ ª fi ¯³9¯³’¯³¹¯ … ¯ ³ +◇+ ¹ + ’ + … +³’=
( ³9 ³’) ² ² ² ² ² ² ² ¯²+²
ײ◇ ²
=¯¹◇·…………………………………………………………………………………… ’y
(²)½?:?å(⁄)=L '(⁄),å
'(⁄)=⁄³f i ¹⁄¯²³O ³⁄, ¹(⁄)=å
'(⁄),¹'(⁄)=⁄³O ³⁄+³³f i ¹⁄, …………………………………………………………………… 4y
fi ⁄*(◇,¬) ,¹'(⁄)>◇,:¹(⁄)=å'(⁄)fi ?½)·…………………………………………………… ’y ²
[›!÷#$$ƒ•'$(¤ ø ³ ½(¤4½)-.K ]
⁄ ² d fi å'(◇)=¯²<◇,å'(¬)= ¬>◇,1½E m *(◇,¬),å'(m )=◇,…………………………………… 8y ² ² ²
1½fi ⁄*(◇,m ) ,å'(⁄)<◇;fi ⁄*(m ,¬) ,å'(⁄)>◇, …………………………………………… 9y ² fl f å(⁄)fl (◇,m )k ?? ?,fl (m ,¬)k ?? ?· ² @å(◇)=¹>◇,å(¬)=◇,1½;flfl ?~ ⁄◇*(◇,¬), N å(⁄◇)=◇, ………………………… ¹◇y ² ²
fi ⁄*(◇,⁄◇) ,å(⁄)>◇;fi ⁄*(⁄◇,¬) ,å(⁄)<◇·………………………………………………… ¹¹y ² ®L (⁄)fl (◇,¬)k ¼$¹K t +A ⁄◇· ………………………………………………………………… ¹²y ² ()R :'(⁄) a ⁄ ²a ²
a ⁄²¯⁄+²a ²,⁄ (, ) ……………………………………………… y
²²·¹ L = ¯¹+ ⁄ = ⁄ * ◇ +8 · ¹ ?4(⁄)=a ⁄²¯⁄+²a ²(⁄>◇),A =
¹¯8a ³, fi a “ ¹ ² ,A “◇,4(⁄)“◇,:L '(⁄)“◇,L (⁄)fl (◇,+8)k ???· ……………………………… ²y fi ¹ , ,() 3Afi ¹¯ 槡¹¯8a ³ , ¹+ 槡¹¯8a ³ ,% , ◇ <a < ² A >◇4 ⁄ ⁄¹= ²a ⁄²= ²a ◇ <⁄¹<⁄² ÷ '(⁄) ,N ⁄ ⁄ o ⁄ ⁄ ,1½ (⁄)fl (,¹¯ 槡¹¯8a ³ ),(¹+ 槡¹¯8a ³ , )k ?? ?;…
L >◇ ◇ < < ¹ > ² L ◇ ²a ²a +8 …………………………………………………………………………………………………………… ³y ÷ '(⁄) ,N ⁄ ⁄ ⁄ ,1½ (⁄)fl (¹¯ 槡¹¯8a ³ ,¹+ 槡¹¯8a ³ )k ?? ? ………………… y L
<◇ ¹< < ² L ²a ²a · 4 fi a ,
, (⁄) 3A f i ¹¯ 槡¹¯8a ³ ,(⁄)fl (,¹¯ 槡¹¯8a ³ )k ?? ?,
<◇ A >◇4
²a L ◇ ²a fl (¹¯ 槡¹¯8a ³ , )k ?? ? ………………………………………………………………………
y ²a +8 · ’ (²)½?:$(¹)ª,fi ◇<a < ¹ ² ,L (⁄)flfl Q K t +A · ……………………………………………… 4y fl ¾??◇<⁄ <⁄ ,:⁄ +⁄ = ¹· ¹ ² ¹ ² a ?½L (⁄¹)¯L (⁄²)
¹+ ¹ ,¼;½ (⁄ )¯ (⁄ ) (⁄¹¯⁄²)(⁄¹+⁄²) ⁄¹ ⁄²,………………… ’y ⁄ ¯⁄ < ⁄ L ¹ L ² > ⁄⁄ = ¯ ¹ ² ⁄¹ ² ¹ ² ⁄² ⁄¹ ¼;½ ¹(⁄ ¯⁄ )‡a (⁄ +⁄ )¯²‡+²a ² ⁄¹=¯ ¹(⁄ ¯⁄ )+²a ² ⁄¹ ⁄¹ ⁄²,………………… 8y ² ¹ ² ¹ ² ¹¹ ¹ ² ⁄² ¹¹ > ¯ ⁄² ⁄² ⁄¹ %½²a ² ⁄¹ ⁄¹ ⁄²> ¹(⁄ ¯⁄ )·…………………………………………………………………… 9y ¹¹ ¯ + ¹ ² ⁄² ⁄² ⁄¹ ² ?’ ⁄¹( ’ ),?½) (’) a ² ’ ’ ¹,'(’) ’²¯²a ²’+¹, = ◇< <¹ ⁄² å =² ¹¹ ¯ +’ å =¯ ’² d fi A '=4a 4¯4<◇,1½’²¯²a ²’+¹>◇,å'(’)<◇,…………………………………………………… ¹◇y 1½å(’)fl (◇,¹)k ???,:å(’)>å(¹)=◇· ……………………………………………………… ¹¹y @ ¹(⁄ ¯⁄ )<◇,:å(’)>◇> ¹(⁄ ¯⁄ ),:²a ² ⁄¹ ⁄¹ ⁄²> ¹(⁄ ¯⁄ ), ² ¹ ² ² ¹ ² ¹¹ ¯ + ¹ ² ⁄² ⁄² ⁄¹ ² fl f L (⁄¹)¯L (⁄²) ¹+ ¹· …………………………………………………………………………… ¹²y ⁄¹¯⁄²
< ¹ ⁄²
[›!÷#$$ƒ•'$(¤ø4½(¤4½)-.K]。