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A. PHẦN CHUNG CHO TẤT CẢ CÁC THÍ SINH (7.0Đ)
I. TRẮC NGHIỆM: (2 Đ)
II. TỰ LUẬN: (5 Đ)
Câu 1:2k,!"2L/2C2!"6/ 2C2!"m /+[t
j
j
x
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− =
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1!"6/
@ 2=/!m23D(
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&j'
J[k+, /+01%}
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B. PHẦN RIÊNG (3 Đ)
Thí sinh chỉ được làm mt trong hai phần (phần 1 hoặc 2)
1.Theo chương trnh Chuẩn
Câu 4a:12>,D(7EU2>C#7EU 1@=/+YG26(C226HD7$DE$
D$DU$
(JBK +0/23DU(
:/7@=//>2@ADEU(
Z[/2C2+\D*7EU(
2.Theo chương trnh Nâng cao
Câu 4b:12>,D(7EUG2>C#7EU 1++YG26G/>2
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G-s/2
DY$(
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X!"g
A. PHẦN CHUNG CHO TẤT CẢ CÁC THÍ SINH (7.0Đ)
I. TRẮC NGHIỆM: (2 Đ)
II. TỰ LUẬN: (5 Đ)
Câu 1:2k,!"
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x
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1!"6/+H
@ 2=/LN232k,!">(
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B. PHẦN RIÊNG (3 Đ)
Thí sinh chỉ được làm mt trong hai phần (phần 1 hoặc 2)
1.Theo chương trnh Chuẩn
Câu 4a:+:m?Y7E2>Y7GYEGYG=L+@=//>2@ Y7$YE$Y$G +0/
E(
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g12>,D(7EU2>C#7EU 1+/@=/+67@ UG
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A. PHẦN CHUNG CHO TẤT CẢ CÁC THÍ SINH (7.0Đ)
I. TRẮC NGHIỆM: (2 Đ)
II. TỰ LUẬN: (5 Đ)
Câu 1: !"
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B. PHẦN RIÊNG (3 Đ)
Thí sinh chỉ được làm mt trong hai phần (phần 1 hoặc 2)
1.Theo chương trnh Chuẩn
Câu 4a::m?D7E2>7E +/C2@=/2‹E@ 7$G26D7@=//>2@A
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(:/D7E⊥DE
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(JBY +0/237(Z[/2C2+\Y*DE
2.Theo chương trnh Nâng cao
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I.PHẦN CHUNG CHO TẤT CẢ THÍ SINH
Bài 1 (12C2/A6!
(
g
→−∞
− +
(
j
+
→
−
−
Bài 2.
( QR++H+S223 !"!+0HQ23>
d
Z
%
Z
− +
>
=
−
+ ≤
( :/0;/, /+01!2>+k+/?
<− + + =
(
Bài 3 ( !"
#
−
=
+
(
)*+,+++234+5 !"+62> L$&(
)*+,+++234+5 !"*++*,+#*!/!/@Am#$
−
(
Bài 4. 12>,D(7EU2>C#7EU 1@=/26GD7@=//>2@AC#GD7$
(
:/0;/2C2I+H12>, O/+/C2@=/(
KTD7
⊥
DEU(
2/>2/OD@ ,D7E(
m/>2/OI+,M/DEU@ 7EU(
II.PHẦN RIÊNG
Thí sinh chỉ được làm mt trong hai phần (phần 1 hoặc 2)
1.Theo chương trnh Chuẩn
Bài 5a(
x
x
→−
+
+ +
(
Bài 6a(
# d x
= − − −
(J[k+, /+01
8
# <≤
(
2.Theo chương trnh NC
Bài 5b(
→
− −
− +
(
Bài 6b(
#
− +
=
−
(J[k+, /+01
8
# <>
(
X!"j
A. PHẦN CHUNG CHO TẤT CẢ CÁC THÍ SINH
(1
(
)
g − + −
g
( !"%$+G
/
=
−
(
% l<
/ l<
(
(12>,D(7EU2>C#7EU 12O+(E*+D7$G7E$GE$G26H
D7@=//>2@A%7EU(
/>2/O2C2I+,M/DE@ DU@A7EU(
JBY /23-s/2R7@ EU(Z[/2C2+\Y*%DU(
B. PHẦN RIÊNG CHO TỪNG BAN .
I.Ban cơ bản
1/A6
<
2!
→
− +
−
( !"
g
#
= + − −
2>4+5()*+, /+01+*,+#*234+5
*+0;/+*,+#*>!/!/@A-s/+M/m#$g'(
II .Ban khoa hc tự nhiên
(2k,!"
( )
+ut
x
d
d<
− + =
+ =
1!"6/+H
@ 2=/LN232k,!">(
( !"
2! 2!
+
%
!
− π
<
= + ≤ ≤
− + +
2
nÕu 0 < x
6
nÕu -1 x 0
x nÕu x < -1
+0/>G +!"(
1G%H+S2+62C2$&@ $<(
X!"x
A. PHẦN CHUNG CHO TẤT CẢ CÁC THÍ SINH (7.0Đ)
I. TRẮC NGHIỆM: (2 Đ)
II. TỰ LUẬN: (5 Đ)
Câu 1:E"!"GG2Gm+6+ D2>+w/;/<<G+2;/&d(1g!">
Câu 2:12C2/A6!
(
→−∞
− +
−
(
<
→
+ − + +
(
Câu 3::m?D7E2>+/C27E@=/+6EG7E$GE$GD7
⊥
7EGD7$(JBK
+0/237E(
(/>2/OI+,M/DE@ 7E
(-s/27|23+/C27K
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g(Z[/2C2+\7*DK
B. PHẦN RIÊNG (3 Đ)
Thí sinh chỉ được làm mt trong hai phần (phần 1 hoặc 2)
1.Theo chương trnh Chuẩn
Câu 4a: !"%$!y2!y
g
+
(J[, /+01%}$<(
Câu 5a: !"#$
&'
)*+, /+01+*,+#*@A4+5 !"+5$
2.Theo chương trnh Nâng cao
g( !"#$
y
'
()*+, /+01+R,+#*23Z•+\7<9
(1+0H-s/+M/#$2C2+\>2>+Z•-h2+*,+#*@=//>2@A(
X!"f
A. PHẦN CHUNG CHO TẤT CẢ CÁC THÍ SINH (7.0Đ)
I. TRẮC NGHIỆM: (2 Đ)
II. TỰ LUẬN: (5 Đ)
Câu 1:D`+[
g
d<
x<
+ =
+ =
(1
d g
GD
Câu 2: !"% =
Z
Z
+ −
<
−
+ + ≥
1+k+2[2C2/C+0523 !"H+S2+0HQ(
Câu 3:12>,D(7E(C# +/C27E2>7E$G7$xG
·
<
E7 d<=
(
6HD7@=//>2@AC#(E*+D7$E(
mEGD7(
12C2XgDG7GEG(
2 JBKG`+]+:+ 12*237+0HDEGD(/>2/OI+,M/7K`@
7E(
B. PHẦN RIÊNG (3 Đ)
Thí sinh chỉ được làm mt trong hai phần (phần 1 hoặc 2)
1.Theo chương trnh Chuẩn
Câu 4a !"#$2!
(
(#ŽG#Ž}(
(/C+0523+:27$#}}}'d#}'d#yx(
2.Theo chương trnh Nâng cao
g
(2C2/A6!(
→
−
− −
(+w/D$
g <<f
<<f <<f <<f <<f <<f
( ( ( g( ((( (
−
+ + + + +
X!"<
A. PHẦN CHUNG CHO TẤT CẢ CÁC THÍ SINH (7.0Đ)
I. TRẮC NGHIỆM: (2 Đ)
II. TỰ LUẬN: (5 Đ)
d
Câu 1:D
( )
2>
g f
g
dd
+ =
+ =
(+w/gd!"6/+H23D
Câu 2: !"
g
Z
%
d Z
−
≠
=
+ −
≠
(KT !"H+S2+6$(
Câu 3:12>,D(7EU2>C#7EU 12O+GE$G7E$G
D7$DE$D$
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JB +0/236+M/D`(:/K
⊥
DU(
2 JB• +0/236+M/D(:/I+,M/K•
⊥
DK`(
B. PHẦN RIÊNG (3 Đ)
Thí sinh chỉ được làm mt trong hai phần (phần 1 hoặc 2)
1.Theo chương trnh Chuẩn
Câu 4a !"%$
− +
+
()*+, /+01+*,+#*234+5 !"*++*,
+#*>!/!/@A-s/+M/#$−−
Câu 5a: 2C2/A6!
→±∞
− +
− + −
2.Theo chương trnh Nâng cao
g !"#$
y
'
1!2%
}
z<(
0H4+5 !"#$%Gt#+1+6>+*,+#*234+5 !"2>?!"/>2;/
(
Đ 31
Bài 1.
1/?+L26
•<9 ‘π
23, /+01
! (2! 2! + =
(
J[, /+01
! 2! ! g2! − + + = −
(
Bài 2.
1?!"23
g
+0/Z+023+:2
g d j
= + + + + + + +
(
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u(JB7 *2"’z#ŽGE *2"’'#$jŽ(^u*2"7@ E2>L2,#Z=/v
Bài 3.0/I+,M/+BLY#G2-s/+M/m2>, /+01y#'g$<()*+, /
+01-s/+M/m} [23m
“,R,+5+*+]@]2+.
@ 9= −
r
(
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<
(
Bài 4. +:m?7EU(JBJ +0B/++/C27EGK +L226U!2
K$KU(
1/23-s/+M/E@ I+,M/7JK(
:/KJ!/!/@AI+,M/7EU(
21/+#*23I+,M/7JK@ 7EU(
Đ 32 :
j
Bài 1.J[, /+01
2! (2!(2!g(2!x 2!
x
=
(
Bài 2.>H!"+H2>j2O!"ZC2G+0/>2>g2O!"2M2O!"•(
Bài 3.KL+/-s2>21ZC-/2‹21F-h2ZC(`/-s>F+\/21L+2C2
/”H(C2!k++:F-h2ZC*
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Bài 4.1+/27EU2>C#7EGU@A7g9&Ef9&(1+,h,ZU
mL/+0H-s/+0—2>, /+01
# g# <+ − + + =
(
Bài 5.12>,D(7EU2>C#7EU 1+/2>C#A7UGK+L2267E(KI+
,M/
( )
α
NK@ !/!/7UGDE(
QC25+*+m?+6F
( )
α
@A12>,D(7EU 1/1v
:/D88
( )
α
(
Đ 33 :
Bài 1(
!$
@
π
< < π
(2!92!(
1+,C2523 !"!
#$
!
2! −
#$
! −
2#$+
1/C+05Ak+Guk+232C2 !"!
#$&2!&
d
π
#$!'2!'
Bài 2. 0/I+,M/+BLY#G279g@ -s/+M/m'#'$<(1+BL
[237@ 23-s/+M/23mN,R,
Y
9
Y#
9
@
r
@A
( )
@ 9=
r
(
Bài 3.
8 J[, /+01
! g 2! g= +
'!$!'!
22!
y!
$< m
g g
2! ! !
− =
8+/C27E@=/+6+u7E$7'
E(/>2E@ (
Bài 4.
8 wL+2>d6@ j6OG+w2>x6@ g6O(,L+46A,
+0-F/2B/–H˜+w6(C2!k+46/4+4I2+4O(
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8
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2JBJ +0B/++/C2D7E(QC25+*+m?2312>,@AI+,M/^|JG+*+
m?> 1/1v
m1+™!"
7E
U
+*+m? 11 (
x
Đ 34 :
Bài 1.J[, /+01
! 2! ! ! !
<
! (2!
! !
+ − + − +
= =
−
+
Bài 2.1Jp`@ J``*2>23 !"
# ! 2!= −
(
Bài 3.
1?!"23
#
j
+0/Z+0y#
(
J[, /+01
<<
− −
+ + =
(
Bài 4. KL+A,B2/4d<B2!G+0/>g<B2!+2B2=+CG<B2!+2B2
=@„@ <B2!+2B22[=(2B/”HL+B2!(C2!k+232C2*
2"!
7’B2!+2B2=+C’(
E’B2!+2B2=@„’(
2’B2!+2B22[=’(
mU’B2!Z=/+2B22[=’(
Bài 5. 0/I+,M/+BLY#G27&9@ Ed9&(
1[23-s/+M/7EN,R,+5+*+]
@ 9=
r
(
-s/+0—C
g # d g− + − =
(1[23-s/+0—CN,R,4/m6/Z
+2?,R,H+*, ,R,":/+0S2Y#@ ,R,@5++7+‹!"Z$&(
Bài 6. 12>, D(7EU2>C# 11 ( JB J@ ^ -h+ +0B/+
D7 G D7E∆ ∆
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1//O-s/+M/7J@A,DEU(
1//O-s/+M/D7@A,JE(
2:/^J88D7U(
Đ 35 :
Bài 1.
81+,C2523 !"
2!
#
+
−
=
+
(
81Jp`@ J``23 !"#$!
g
'2!
g
8)š4+5 !"#$!+0H6•&π9π‘(D#04+5 !"#$!
+0H6•&π9π‘@ ,
[/*+H23 !"#$!
+0H6•&π9π‘(
Bài 2. J[, /+01
8
! ! <− + + =
8
2! ! g! g
− − = −
8
! 2! + = −
g8
2! ! ! g
+ =
8
2! (2!(2! g(2!x
d(!
=
Bài 3.
8:/'
'
'
'Š'y
&
$
(
81!"6/Z=/2:+0/Z+0
x
−
(
8>„*/1--h2/+\*(pk#/–H*/1@ *,+]+:++\
+0C!/,[(C2!k+23*2"
7’D"+6+ !"•Ž(\>G!#0C2!k+23*2"’!"+6+ !"2›Ž(
E’D"+6+ 2>+w/2C22O!" !"•Ž(
Bài 4.
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f
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Bài 5.12>,D7EU2>C#7EU 11 (JBJ +0B/++/C2D7E@
+0/7E(pk#K+0/67U!27U$7K(
1/+#*23I+,M/D7U@ DE(
:/KJ88DU(
2JBα I+,M/2:KJ@ !/!/D7(QC25+*+m?23I+,M/α@ 1
2>,D7EU(
Đ 36 :
Bài 1(1/C+05Ak+Guk+23 !"#$yg!
2!
(
Bài 2(J[2C2, /+0!
g2!'2!'2!g$& !- 2!$
+
, ∈<9π
+/ +/
2!
+ − =
Bài 3.
J[2C2, /+01!
7
− − −
− =
|*+N[$f
g
j
d
+ +
− =
|*+N[$)$x
>L,/L,/+0€/GjuG9L,/<+0€/GduGf
(k#/–H+\˜L,@H(1C2!k+@Hk#02V/ (
|*+N[$<j8d
Bài 4(+:m?7EU(JBGœ-h+ +0/237GE(KL+I+,M/
α
Nœ2€+
67EG7U-h++6^G|(
:/C2•^| 1/1v
QC25/+#*232C2I+,M/Uœ@ 7EU9^@ 7EU(
2J[!•Y$|
∩
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