NMHIEUPDP.WORDPRESS.COM ỏp ỏn s 17 THI TH K THI THPT QUC GIA Mụn : TON Thi gian lm bi 180 phỳt Cõu 1a (1,0 im) Tp xỏc nh : D = R S bin thiờn : + Gii hn ti vụ cc : lim y = +; lim y = x + y x + Bng bin thiờn : y = 3x2 ủ 6x = 3x ( x 2); y = x y + x=0 x=2 + U + + y O Hm s ng bin trờn (; 0) v (2; +) Hm s nghch bin trờn (0; 2) Hm s t cc i ti x = 0; yC = Hm s t cc tiu ti x = 2; yCT = th : + Ct Oy ti (0; 4) + Nhn im un U (1; 2) lm tõm i xng Cõu 1b (1,0 im) Ta cú y = 3x2 6mx; y = x = hoc x = 2m th hm s ct trc honh ti ba im phõn bit v ch đ 2m = y(0).y(2m) < đ m=0 m>1 4(8m3 12m3 + 4) < Cỏc giao im cú honh nh hn v ch đ 2m < y (4) > Kt hp ta cú < m < đ 17 m 12 17 12 Cõu 2a (0,5 im) ằ Vỡ < < nờn ta cú cos = sin = ầ2 ó 63 Do ú cos + = cos + = cos cos sin sin = 3 3 x Cõu 2b (0,5 im) Gi z = a + bi, ( a, b R), ta cú : |iz 3| = |z i | |i ( a + bi ) 3| = | a + bi i | |b + | = | a + (b 1)i | ( b + 3)2 + a2 = ( a 2)2 + ( b 1)2 a = 2b + b2 = 5b2 ầ + 4b + = 5b + ồ2 + 2 Du bng xy 5b + = b = z = i 5 5 Khi ú |z| = a2 Vy s phc cú mụun nh nht tha yờu cu bi toỏn l z = i 5 Cõu (0,5 im) Phng trỡnh ó cho tng ng vi : ầ ồ2x ầ ồx = ầ ồx ầ3ồ x = 2 = (vụ nghim) x = Vy phng trỡnh ó cho cú nghim nht x = Cõu (1,0 im) iu kin1 x 1. t u = x2 , v = + x2 (0 Do ú phng trỡnh tr thnh : u 1, v 1), ta cú u2 + v2 = v x2 = v2 (u 2v) + uv + 3v2 = 2u 4v + uv + 2v2 u2 = 2u 4v uv + 2v2 + 2uv u2 = (u 2v) v (u 2v) u (u 2v) = ủ u = 2v (loi) (u 2v) (2 v u) = u+v = Vi u + v = x2 + + x2 = x4 = x = Vy phng trỡnh cú nghim x = Cõu (1,0 im) Ta cú I = 1 d (cos x ) = ln |1 + cos x | + cos x = ln 2 Cõu (1,0 im) Gi O l giao im ca AC v BD; gi H l trng tõm tam giỏc ABD a T din SABD u nờn AO = AC = a 3, BD = a v SH ( ABD ) a2 Do ú din tớch ỏy ABCD l S ABCD = AC.BD = 2 a a Li cú AH = AC = SH = SA2 AH = 3 a3 Vy th tớch chúp S.ABCD l VS.ABCD = SH.S ABCD = S K C B A H O D Gi K l hỡnh chiu ca O trờn SC ta cú BD AC v BD SH nờn BD OK T ú ta cú OK l on vuụng gúc chung ca BD v SC hay d( BD, SC ) = OK a OK OC OC.SH = Ta cú OKC SHC, suy = OK = SH SC SH + HC2 a Vy khong cỏch gia BD v SC l d( BD, SC ) = OK = Cõu (1,0 im) |6 + + 2| = MB = 2d ( M, BD ) = 10 (1) 4+1 im B BD B(t; 2t 2) MB = (tủ+ 3; 2t 4) MB = 5t2 10t + 25 (2) t = (loi) T (1) v (2) ta cú 5t2 10t + 25 = 40 B(3; 4) MB = (6; 2) t=3 ng thng AB qua M(3; 2) v nhn MB = (6; 2) lm mt vect ch phng Do ú AB cú phng trỡnh 2( x + 3) 6(y 2) = x 3y + = ng thng AD qua N (3; 2) v nhn MB = (6; 2) lm mt vect phỏp tuyn Do ú AD cú phng trỡnh 6( x + 3) + 2(y ) = 3x + y = x = x 3y + = A(0; 3) Ta A l nghim h y=3 3x + y = Vy A(0; 3) Ta cú d ( M, BD ) = Cõu (1,0 im) Mt cu (S) tip xỳc vi ( P) nờn cú bỏn kớnh R = d( I, ( P)) = |2 + 9| = 1+4+4 Do ú (S) cú phng trỡnh ( x 2)2 + (y 3)2 + (z + 2)2 = Mt phng ( Q) song song ( P) nờn cú phng trỡnh dng x 2y 2z + D = ( D = 9) ủ |2 + + D | D=9 Li cú ( Q) tip xỳc (S) nờn ta cú d( I, ( Q)) = R =3 D = (loi) 1+4+4 Vy (S) : ( x 2)2 + (y 3)2 + (z + 2)2 = v ( Q) : x 2y 2z + = Cõu (0,5 im) S t nhiờn cú ớt nht ba ch s phõn bit thuc E gm cỏc trng hp: TH1: S cú ba ch s cú A35 = 60 s TH2: S cú bn ch s cú A45 = 120 s TH2: S cú nm ch s cú A55 = 120 s Do ú hp M cú 60 + 120 + 120 = 300 s Phộp th l ly ngu nhiờn mt s t M nờn || = 300 Gi A l bin c "ly c s cú tng cỏc ch s bng 10" Cỏc ca E cú tng cỏc phn t bng 10 bao gm : E1 = {1, 2, 3, 4} , E2 = {2, 3, 5} , E3 = {1, 4, 5} Cỏc s cú tng cỏc ch s bng 10 ln lt l cỏc hoỏn v ca E1 , E2 , E3 Do ú s kt qu thun li cho bin c A l | A | = 4! + 3! + 3! = 36 36 | A | = Vy xỏc sut ca bin c A l P ( A) = = 300 25 || Cõu 10 (1,0 im) T gi thit a + b + c = ta cú ab + c = ab + c( a + b + c) = ( a + c)(b + c) Khi ú ầ b ab a b a = + (1) ab + c a+c b+c a+c b+c Tng t c bc b c b = + b + a c + aó bc + a b + a c + a ca c a a c = + ca + b c+b a+b c+b a+b ầ Cng theo v (1), (2) v (3) ta cú bt ng thc cn chng minh Ht (2) (3)