ĐỀ ÔN THI THPT QG MÔN TOÁN CÓ ĐÁP ÁN

THI MINH HA K THI THPT QUC GIA NM 2017 Mụn: TON s 063 Thi gian lm bi: 90 phỳt Cõu : A Cõu : A Cõu : A C Cõu : A Cõu : A Cõu : A Cõu : A Cõu : A Cõu : A Cõu 10 : A Cõu 11 : Cho chúp u S.ABCD cú tt c cỏc cnh u bng a Th tớch chúp l a3 a3 a3 a3 B C D 6 2 Kt qu rỳt gn s phc z = (2 + 3i) (2 3i) l: z=-24i B z=12i C z=-12i D z=24i Gi s th hm s y = x + 3mx + 3(m + 6) x + cú hai cc tr Khi ú ng thng qua hai im cc tr cú phng trỡnh l: y = 2(m + m + 6) x + m + 6m + B y = 2( m + m + 6) x m 6m + y = x + m + 6m + D y = x + m + 6m + 1 ) cú nghim l : Bt phng trỡnh log ( x + 4) log ( x +8 x=2 B x C x D x Mt phng qua im A(1;0;0), B(0;-2;0), C(0;0,3) cú phng trỡnh l: x y z x y z + + =6 + + =1 B 6x-3y+2z=6 C D x-2y+3z=1 3 2 Gi z1 , z2 l hai nghim phc ca phng trỡnh z + z + 10 = Giỏ tr ca biu thc | z1 | + | z2 | B 20 C 10 D 40 2 ng thng y=x+m ct ng trũn ( x 1) + ( y 2) = 16 theo dõy cung cú di ln nht bng B C D x y z +1 x7 y z = = = = V trớ tng i gia d1 v Cho hai ng thng: d1 : v d1 : 12 d2 l: Trựng B Song song C Ct D Chộo S phc z tha iz+2-i=0 cú phn thc bng B C D Cú bao nhiờu s t nhiờn l gm ch s khỏc lp t cỏc s 1,2,3,4,5? 18 B 36 C 72 D 144 x Tớch phõn e xdx cú giỏ tr bng A Cõu 12 : A Cõu 13 : A Cõu 14 : A Cõu 15 : A 2e + e +1 2e e B C D 2e 2e Th tớch ca t din OABC cú OA, OB, OC ụi mt vuụng gúc, OA=a, OB=2a, OC=3a l 4a3 B a3 C 3a3 D 2a3 Hm s y = x + x + cú bao nhiờu cc tr B C D Hỡnh chiu vuụng gúc ca A(-2;4;3) trờn mt phng x y + z + 19 = cú ta l: 20 37 37 31 37 31 (1;-1;2) B ( ; ; ) C ( ; ; ) D ( ; ; ) 7 5 5 5 x y z +1 x7 y z = = = = Phng trỡnh mt phng cha hai ng thng d1 : v d1 : cú 12 dng: 3x + y = B x + y + z = C Cõu 16 : x + 19 y + z + = D x + 19 y z = 2x l x Khong cỏch nh nht gia hai im bt k thuc hai nhỏnh ca th hm s y = A B 2 C D Cõu 17 : 2x Vi giỏ tr no ca m thỡ ng thng y=x-m ct th hm s y = ti hai im phõn bit x A < m < B m C m>1 D m 3 Cõu 18 : (1 i 3) Cho s phc z tha z = Mụun ca s phc z + iz bng i A B C 2 D Cõu 19 : Cho hm s f ( x ) = (2 x 3)5 Giỏ tr ca f(3) bng A 1320 B 2320 C 3320 D 4320 Cõu 20 : Cho hm s y = x - 2mx + m - Tỡm m th ca hm s cú im cc tr A, B,C ng thi cỏc im A,B,C to thnh nh ca mt tam giỏc u A m = 3 B m=0 C m=3 D m>0 2017 Cõu 21 : S phc + i + i + + i cú giỏ tr bng A i B 2017i C i 2017 D 22017 Cõu 22 : x2 5x + th hm s y = cú tim cn ng l x2 A x=1 B x=2 C x = D x=-2 Cõu 23 : sin x Nguyờn hm dx bng cos x tan x + C tan x + C A B tan x + C C tan x + C D 3 Cõu 24 : Tỡm giỏ tr ln nht v giỏ tr nh nht ca hm s y = sin x - cos x + A y = xẻ D y = xẻ D y = xẻ D t = 1; t = - 1; t = 1; max y = xẻ D t =Cõu 25 : A Cõu 26 : A Cõu 27 : A Cõu 28 : A Cõu 29 : 25 B max y = xẻ D t =- 25 C max y = xẻ D t =- 25 y = xẻ D t = 1; max y = D x ẻ D t =4 x2 + 1 cú giỏ tr bng x x2 + x -2 B C -1 D Hm s y = x + 3( m 1) x + 6( m 2) x tng trờn Ă m B m0 hoc m>4 B m f(3)=1320 f '''( x) = 160(2 x 3)3 Cõu 20 : Cho hm s y = x - 2mx + m - Tỡm m th ca hm s cú im cc trA, B,C ng thi cỏc im A,B,C to thnh nh ca mt tam giỏc u A m = 3 B m=0 C m=3 D m>0 TX: D = Ă Ta cú: y = 4x (x - m ) Cho y ' = x = 0; x = m Hm s cú cc tr phng trỡnh y ' = cú nghim phõn bit m > To im cc tr l A (0; m - 1) , B (- m ; - m + m - 1), C ( m ; - m + m - 1) Ta luụn cú AB=AC nờn tam giỏc ABC u khi: A B = BC m + m = 4m m = (vỡ m > ) Cõu 21 : S phc + i + i + + i 2017 cú giỏ tr bng A i B 2017i C i 2017 D Cõu 22 : x 5x + th hm s y = cú tim cn ng l x2 A x=1 B x=2 C x = D x 5x + Ta cú lim = vy x = l tim cn ng x ( 2) x2 Cõu 23 : sin x Nguyờn hm dx bng cos x A B tan x + C C tan x + C D tan x + C sin x tan x 2 = dx tan x dx = tan xd (tan x ) = = +C cos4 x cos x Cõu 24 : Tỡm giỏ tr ln nht v giỏ tr nh nht ca hm s y = sin x - cos x + A y = B y = C y = D xẻ D xẻ D xẻ D t = 1; t = - 1; t = 1; max y = xẻ D 25 t =4 max y = xẻ D 25 t =4 max y = xẻ D t =4 25 22017 x = tan x + C y = xẻ D t = 1; max y = xẻ D t =- Tp xỏc nh: D = Ă Ta cú y = sin x - cos x + = 2(1 - cos2 x ) - cos x + = - cos2 x - cos x + ự , hm s tr thnh: y = - 2t - t + t t = cos x vi t ẻ ộ ờ- 1;1ỷ ỳ ự Ta cú: y ' = - 4t - ; y ' = t = - ẻ ộ ờ- 1;1ỷ ỳ ổ 1ử 25 ữ = ỗ- ữ Do y ( - 1) = 2; y ( 1) = 0; y ỗ ữ ữ ỗ ố 4ứ y = t = ; max y = 25 t = - Vy xẻ D xẻ D Cõu 25 : x2 + 1 cú giỏ tr bng x x2 + x 1 x( + ) x +1 x x =0 lim = lim x x x +x x (1 + ) x A -2 B Gii hn lim C -1 D Cõu 26 : A Cõu 27 : A Cõu 28 : A Cõu 29 : A C Cõu 30 : Hm s y = x + 3( m 1) x + 6( m 2) x tng trờn R m B m P(A)= 105 Cõu 36 : Hiu s gia giỏ tr cc i v giỏ tr cc tiu ca hm s y = x x + l A B C D Ta cú ycd = 1; yct = => ycd yct = Cõu 37 : x2 Tỡm trờn th y = ca hm s cỏc im cú tng khong cỏch n hai tim cn l x+2 nh nht A M (0, 1) ; M (4,3) B M (0,1) ; M (4,3) C M (0, 1) ; M (4, 3) D M (0, 1) ; M ( 4,3) 10 Cõu 38 : Tớnh o hm ca hm s y = log 2017 ( x + 1) A B 1 2 ( x 1) ln 2017 ( x + 1) ln 2017 ( x + 1) ln 2017 Cõu 39 : dx bng Nguyờn hm 1+ x A x ln | x + 1| +C B ln | x + 1| +C Ta cú y ' = C 2017 ( x + 1) ln 2017 D x +C D 2017 ( x + 1) ln 2017 2 C x ln | x + | +C d ( x )] = x ln | x + 1| +C x 1+ x Cõu 40 : Vi giỏ tr no ca m thỡ hm s y = x + (m 1) x 2m + t cc i ti x=2? A B C D y ' = x + 2(m 1) x 1+ dx = [1 y '' = x + 2(m 1) HS t C ti x=2 m=2 Cõu 41 : Vi giỏ tr no ca m thỡ th hm s y = x + 3(m 1) x + 6(m 2) x cú cc i, cc tiu tha mó |xC+xCT|=2 A m=1 B m=2 C m=-1 D m=-2 Ta cú y ' = x + 6(m 1) x + 6( m 2) (m 1) 4(m 2) > x1 + x2 = m m = HS cú C, Ct tha x1 x2 = m | xCD + xct |= Cõu 42 : Hm s y = x 3x + gim trờn khong no ? A (-1;1) B (-2;0) C (; 1) (1; +) D (0;2) y ' = 3x HS NB y ' = 3x < x (1;1) Cõu 43 : Mt cu tõm I(-1;2;0) ng kớnh bng 10 cú phng trỡnh l A ( x + 1) + ( y 2) + z = 25 B ( x + 1) + ( y 2) + z = 100 C ( x 1) + ( y + 2) + z = 25 D ( x 1) + ( y + 2) + z = 100 R=5, tõm I(-1;2;0) ADCT ta cú ( x + 1) + ( y 2) + z = 25 Cõu 44 : Vi giỏ tr no ca m thỡ th hm s y = x 2m2 x + cú ba cc tr to thnh tam giỏc vuụng cõn A m=1 B m = C m=0 D m = y ' = x 4m x HS cú cc tr m>0 Khi ú x3 4m x = x = 0, x = m HS cú cc tr to thnh tam giỏc vuụng cõn m = Cõu 45 : 4 Qua im A( ; ) k c my tip tuyn n th hm s y = x x + x 3 11 A Cõu 46 : B C 4 PT ng thng qua A cú dng d: y = k ( x ) + d k c cỏc tip tuyn n th l nghim ca phng trỡnh 4 k ( x ) + = x x + 3x 3 cú nghim => C k = x 4x + Bt phng trỡnh log x + log x < cú nghim l : D 2 A (0; + ) B (2;3) C (0;3) D (0;2) K x>0 BPT 0dx=2costdt i cn x=0=> t=0 ; x=2=> t= Vy 0 I = 4sin t cos tdt = cos tdt Cõu 49 : Mt phng i qua A(-2;4;3), song song vi mt phng x y + z + 19 = cú phng trỡnh dng A x + y + z + 19 = B x y + z = C x y + z + = D x y + z = r Vỡ MP qua A // x y + z + 19 = => VT PT n = (2; 3;6) Vy MP x y + z = Cõu 50 : xy + x = m( y 1) Vi giỏ tr no ca m thỡ h phng trỡnh cú nghim nht xy + y = m( x 1) = A m=4 B m=2 C m=0 D m=8 Nhn xột H phng trỡnh l h i xng, ú h cú nghim (x;y) thỡ cng cú nghim (y;x) H phng trinh cú nghim nht x=y 12 Thay vo pt h ta cú x my + m = 0` cú nghim nht v ch m = = m 8m = m = Kim tra m=0 thỡ h cú nghim=> loi; m=8 (t/m) 13 ... log 2017 ( x + 1) A B 1 2 ( x 1) ln 2017 ( x + 1) ln 2017 ( x + 1) ln 2017 Cõu 39 : dx bng Nguyờn hm 1+ x A x ln | x + 1| +C B ln | x + 1| +C Ta cú y ' = C 2017 ( x + 1) ln 2017 D x +C D 2017. .. s cú im cc tr A, B,C ng thi cỏc im A,B,C to thnh nh ca mt tam giỏc u A m = 3 B m=0 C m=3 D m>0 2017 Cõu 21 : S phc + i + i + + i cú giỏ tr bng A i B 2017i C i 2017 D 22017 Cõu 22 : x2 5x +... ABC u khi: A B = BC m + m = 4m m = (vỡ m > ) Cõu 21 : S phc + i + i + + i 2017 cú giỏ tr bng A i B 2017i C i 2017 D Cõu 22 : x 5x + th hm s y = cú tim cn ng l x2 A x=1 B x=2 C x = D x