Đề thi giữa học kì 1 môn Toán lớp 12 năm 2023-2024 có đáp án - Trường THPT Võ Văn Kiệt

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“Đề thi giữa học kì 1 môn Toán lớp 12 năm 2023-2024 có đáp án - Trường THPT Võ Văn Kiệt” giúp các bạn học sinh có thêm tài liệu ôn tập, luyện tập giải đề nhằm nắm vững được những kiến thức, kĩ năng cơ bản, đồng thời vận dụng kiến thức để giải các bài tập một cách thuận lợi. Chúc các bạn thi tốt!

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x −1 y = x4 − x2 y = x3 − x y= y = x3 + x x+2 y = x3 + x  y = 3x2 + 1  0 x  y = ( x) ( 2; + ) ( 0; + ) ( −;0 ) ( −1; 2 ) f ( x)  0 x  ( 2; + ) ( 2; + ) y = f ( x) ( 0; + ) (1; + ) ( −1;0 ) ( 0;1) y = f ( x) 1
(1;3) ( 3;1) ( −1; −1) (1; − 1) y = f ( x) ( −1; −1) y = f ( x) f ( x ) f ( x) y = f ( x) 3 0 2 1 x=0 y=0 y = ax + bx + c ( a, b, c  4 2 ) 0 −1 −3 2 2
x=0 f ( 0 ) = −1 f ( x ) = x − 3x 3  −3;3 18 −18 −2 2 f  ( x ) = 3x 2 − 3  x = −1  −3;3 f ( x) = 0    x = 1  −3;3 f ( −3) = −18; f ( 3) = 18; f ( −1) = 2; f (1) = −2 min f ( x ) = f ( −3) = −18  −3;3 y = f ( x) M m  −1;3 M −m M = max y = 3 m = min y = −2 [ −1;3] [ −1;3] M −m =5 3x − 1 y= x−2 1 x= x = −2 x=3 x=2 2 3x − 1 lim =  x=2  x →2 x−2 y = f ( x) 3
x = −1 x = −3 x =1 x=3 lim y = − + lim y = + − x →1 x →1 x =1 4x −1 y= x +1 y = −4 y =1 y=4 y = −1 4x −1 lim y = lim =4 y=4 x → x → x +1 y = − x4 + 2x2 y = x 3 − 3x 2 y = − x 3 + 3x 2 + 1 y = x4 − 2 x2 + 1 y = ax4 + bx 2 + c a  0, b  0, c = 0 y = f ( x) f ( x) = 2 1 0 2 3 4
y = f ( x) y=2 f ( x) = 2 6 10 12 11 4;3 4;3 4 3 3;3 4 6 4 4;3 8 12 6 3; 4 6 12 8 5;3 20 30 12 3;5 12 30 20 h B 1 1 1 V = Bh V = Bh V = Bh V = Bh 3 6 2 1 h B V = Bh 3 S. ABCD 4 ABCD 3 7 5 4 12 5
1 1 VS . ABCD = .h.S ABCD = .4.3 = 4 3 3 3a 3 3 27a 3a 9a 3 a3 V = (3a)3 = 27a3 2; 4; 6 16 12 48 8 2.4.6 = 48. y = x − 2x 4 2 ( −;1) ( −1;0 ) ( −; −1) (1; + ) y = 4 x 3 − 4 x x = 0 y = 0  4 x − 4 x = 0   x = 1 3   x = −1  y = x4 − 2 x2 ( −; −1) ( 0;1) y = f ( x) y = f ( x) ( −2;0 ) ( −; −2 ) ( 0;2 ) ( 0;+ ) ( −2;0 ) ( 2; + ) y = ax 4 + bx 2 + c ( a, b, c  ) 6
x =1 x = −1 x = −2 x=0 f ( x) f ( x) 4 1 2 3 f ( x) x = −1 x =1 f ( x) x = −1 x =1 M y = x − 2x + 3 4 2 0; 3    M =9 M =8 3 M =1 M =6 y = 4 x3 − 4 x = 4 x ( x 2 − 1)  x=0 (  ) y = 0  4 x x − 1 = 0   x = 1 2  x = −1(l )  y ( 0) = 3 y (1) = 2 y ( 3) = 6 y = x4 − 2x2 + 3 0; 3    M=y ( 3) = 6 M,m f ( x ) = x4 − 2x2 + 3 0; 2 M +m 11 14 5 13 D= f  ( x ) = 4x − 4x 3 7
 x = 0   0; 2  f  ( x ) = 0  4 x − 4 x = 0   x = −1 0; 2 3   x = 1   0; 2 f ( 0 ) = 3; f (1) = 2; f ( 2 ) = 11  M = 11   M + m = 13 m = 2 y = f (x) 4 1 3 2 lim y = 2  y = 2 x →− lim y = 5  y = 5 x →+ lim y = +  x = 1 x →1− 3 x +9 −3 y= x2 + x 3 2 0 1 D =  −9; + ) \ 0; −1 x+9 −3 x +9 −3 lim + y = lim + = + lim − y = lim − = − x →( −1) x →( −1) x2 + x x →( −1) x →( −1) x2 + x  x = −1 x +9 −3 x 1 1 lim+ y = lim = lim 2 = lim = x →0 + x →0 x +x 2 x →0 + ( + ) ( ( x + x ) x + 9 + 3 x→0 ( x + 1) x + 9 + 3 6 ) x +9 −3 x 1 1 lim− y = lim = lim 2 = lim = x →0 − x →0 x +x 2 x →0 − ( − ) ( ( x + x ) x + 9 + 3 x→0 ( x + 1) x + 9 + 3 6 )  x=0 1 8
x −3 y = x 4 − 3x 2 + 2 y= y = x2 − 4x + 1 y = x 3 − 3x − 5 x −1 −x + 5 y = x −1 y= x−2 x1 , x2 x1 + x2 2 3 −1 1 −x + 5  x = −1 = x −1  x2 − 2x − 3 = 0   x−2 x = 3 x1 + x2 = 2 y = ax3 + 3x + d ( a, d  ) a  0, d  0 a  0, d  0 a  0; d  0 a  0; d  0 + a0 d 0 3;5 35 30 15 20 9
3;5 20 20.3 = 30 2 3 9 4 6 6 A A A D C D C D C H H H B B B A A A D D D C C C B B B V ABCD. ABCD AC = a 3 3 3 6a 1 V = a3 V= V = 3 3a3 V = a3 4 3 x; ( x  0 ) A' B 'C ' B' A ' C ' = A ' B ' + B ' C ' = x + x = 2x2  A ' C ' = x 2 2 2 2 2 2 A ' AC ' A' 10
AC '2 = A ' A2 + A ' C '2  3a 2 = x 2 + 2 x 2  x = a ABCD. ABCD V = a3 S. ABC A AB = 2 SA SA = 3 12 2 6. 4. 1 1 1 1 1 1 V = B.h = SABC .SA = . AB. AC.SA = . .2.2.3 = 2 3 3 3 2 3 2 f ( x) f ( x) y = f (5 − 2x ) ( 2;3) ( 0;2 ) ( 3;5) ( 5; +  ) y = f (5 − 2x ) y =  f ( 5 − 2 x )  = −2 f  ( 5 − 2 x )    −3  5 − 2 x  −1 3  x  4 y  0  f  ( 5 − 2 x )  0    5 − 2 x  1 x  2 y = f (5 − 2x ) ( −;2) ( 3;4) ( 0;2)  ( −;2) 1 m f ( x ) = x3 + mx 2 + 4 x + 3 3 5 4 3 2 D= f  ( x ) = x 2 + 2mx + 4 11
f  ( x )  0; x    = m2 − 4  0  −2  m  2 m  m  −2; −1;0;1; 2 y = − x3 + 3x 2 + 5 A B S OAB O 10 S =9 S= S =5 S = 10 3 x = 0 y ' = −3x 2 + 6 x y ' = 0  −3x 2 + 6 x = 0   x = 2 A(0;5), B(2;9)  AB = (2;4)  AB = 22 + 42 = 20 AB y = 2x + 5 OAB S =5 1 3 m y= x − mx 2 + (m 2 − 4) x + 3 x=3 3 m =1 m = −1 m=5 m = −7 y = x2 − 2mx + (m2 − 4) y = 2 x − 2m 1 3  y(3) = 0 y= x − mx 2 + (m 2 − 4) x + 3 x=3  3  y(3)  0   m = 1( L) 9 − 6m + m 2 − 4 = 0 m 2 − 6m + 5 = 0        m = 5(TM ) 6 − 2m  0) m  3 m  3  1 y = x3 + m 2 x − 2m 2 + 2m − 9, m S 3 m 0;3 3 m S = ( −; −3  1; + ) S = ( −3;1) S = ( −; −3)  (1; + ) S =  −3;1 y ' = x2 + m2 , x  y '  0, x   max y = y (3) = m 2 + 2m 0;3 m + 2m  3  m   −3;1 2 m  −2021; 2021 x +1 y= x − mx + 4 2 4033 4034 2017 2016 12
lim y = 0 y=0 x →   x − mx + 4 = 0 2 −1   m  −4 m 2 − 16  0    m  4  1 + m + 4  0  m  −5 m   −2021; 2021 m 4033 f ( x ) = ax3 + bx 2 + cx + d ( a, b, c, d  ) a, b, c, d lim f ( x ) = +  a  0 x→+ x=0 y = d =1 0  x = −2 f  ( x ) = 3ax 2 + 2bx + c f ( x) = 0   x = 0 −2b f  ( 0) = 0  c = 0 = −2  b = 3a  0 3a a, b, d y = f ( x) f ( f ( x )) = 0 12 10 8 4 13
f ( x ) = a ( a  −1) (1)  f ( x ) = b ( −1  b  0 ) ( 2 ) f ( f ( x )) = 0   f ( x ) = c ( 0  c  1) ( 3) f ( x ) = d ( d  1) ( 4)  (1) ( 2) 4 ( 3) 4 ( 4) f ( f ( x )) = 0 S. ABCD ABCD AB = a AD = a 3 SA ( SBC ) 60 V S. ABCD 3 a a3 3 V= V= V = a3 V = 3a3 3 3 S A a 60 B a 3 D C S ABC = 3a 2 (𝑆𝐵𝐶) ∩ (𝐴𝐵𝐶𝐷) = 𝐵𝐶 { 𝐵𝐶 ⊥ 𝑆𝐵 ⊂ (𝑆𝐵𝐶) ̂ ̂ ⇒ ((𝑆𝐵𝐶), (𝐴𝐵𝐶𝐷)) = (𝑆𝐵; 𝐴𝐵) = ̂ 𝑆𝐵𝐴 𝐵𝐶 ⊥ 𝐴𝐵 ⊂ (𝐴𝐵𝐶𝐷) ̂ = 60 𝑜 𝑆𝐵𝐴 𝑆𝐴 𝑆𝐴𝐵 tan60 𝑜 = 𝐴𝐵 ⇒ 𝑆𝐴 = 𝐴𝐵. tan60 𝑜 = 𝑎√3 1 1 𝑉 𝑆.𝐴𝐵𝐶𝐷 = 3 𝑆 𝐴𝐵𝐶𝐷 . 𝑆𝐴 = 3 𝑎2 √3. 𝑎√3 = 𝑎3 ABCD. A ' B ' C ' D ' BD = 4a ( A ' BD ) ( ABCD ) 30o 14
16 3 3 16 3 3 a 48 3a3 a 16 3a3 9 3 O BD A ' AB = A ' AD A' B = A' D A ' BD ( A ' BD )  ( ABCD ) = BD   A ' O ⊥ BD  AO ⊥ BD  (( ABD) , ( ABCD)) = AOA = 30. 30o  AA AA AA AA 2a 3 AOA A tan 30o = = = =  A ' A = 2a tan 30 = AO AC BD 2a 3 2 2 ABCD BD = AB 2  AB = 2a 2. 2a 3 ( ) 16 3 3 2 V = A ' A. AB2 . 2a 2 a 3 3 4x + m m f ( x) = 2x + m + 3 ( 0;1) 1 5 4 3  m + 3 2m + 12 D= \ −  f ( x) =  2  ( 2 x + m + 3) 2 ( 0;1)  y '  0, x  ( 0;1)  2m + 12  0 m  −6 − m + 3  1     2    m  −5    m  −5  m  ( −6; −5   −3; + )  m + 3   m  −3   m  −3 − 0     2 m −5; −3; −2; −1 y = f ( x) y = f ( x) x4 g ( x ) = f ( x2 − 2x ) + − 2 x3 + 2 x 2 + 2022 2 15
3 4 5 6 g  ( x ) = ( 2 x − 2 ) f  ( x 2 − 2 x ) + 2 x3 − 6 x 2 + 4 x = 2 ( x − 1) f  ( x 2 − 2 x ) + 2 ( x − 1) ( x 2 − 2 x ) = 2 ( x − 1)  f  ( x 2 − 2 x ) + ( x 2 − 2 x )    t = x2 − 2 x f  (t ) y = −t t = −1 t = 0 t = 1 t = 2 x = 1 x = 1  2   x − 2 x = −1  x = 1 g ( x ) = 0   x2 − 2x = 0   x = 0  x = 2    x2 − 2 x = 1 x = 1 2  2   x − 2x = 2 x = 1 3   x2 − 2x  2 x  1− 3  x  1+ 3   f  ( x 2 − 2 x )  − ( x 2 − 2 x )  0  x 2 − 2 x  1  1 − 2  x  0  2  x  1 + 2  x 2 − 2 x  −1 VN    16
g ( x) y = f ( x) g ( x ) = f ( 4 x − x 2 ) + x 3 − 3x 2 + 8 x + 1;3 1 1 3 3 25 19 3 3 g  ( x ) = ( 4 − 2 x ) f  ( 4 x − x 2 ) + x 2 − 6 x + 8 = ( 2 − x ) 2 f  ( 4 x − x2 ) + 4 − x    x  1;3 4 − x  0 3  4x − x  4 2 f  ( 4x − x2 )  0 2 f  ( 4 x − x 2 ) + 4 − x  0 x  1;3 max g ( x ) = g ( 2 ) = f ( 4 ) + 7 = 12 1;3 ax + 4 f ( x) = (a, b, c  ) bx + c a, b c 2 3 1 0 17
x = −1 y =3  c − b  0 (1)  a  0 (2)  b ac − 4b  0 (3)   −a.c  0  ac  0 (3)  −4b  0  b  0. c  0, a  0, b  0. S. ABC ABC 1 A 6 15 ( SBC ) B ( SCA) C 4 10 30 ( SAB ) S ABC 20 VS . ABC . 1 1 1 1 36 48 12 24 M , N, P H AC , BC , AB 1 3 h 3 SH = h  VS . ABC = .h. = 3 4 12 2S 6VS . ABC h 3 30 AP = SAB = 2S SAB = = : = h 10 AB d ( C; ( SAB ) ) 2 20 HM = 2h, HN = h  PH = SP 2 − SH 2 = 3h 1 3 3 S ABC = S HAB + S HAC + S HBC = ( HP + HM + HN )  3h =  h = 2 4 12 3 3 1 VS . ABC = . = 12 12 48 18