ĐỀ THI VÀ ĐÁP ÁN MÔN TOÁN KHỐI A A1 2013

 BS: NLong  ( Đề thi có 1 trang )  Môn thi: TOÁN   Thời gian làm bài: 180 phút, không kể thời gian giao đề    32 y x 3x 3mx 1 (1)      a)    )   1 tanx 2 2sin x 4          4 4 22 1 1 2 2 ( 1) 6 1 0                  x x y y x x y y y (x, y  R).  Tính tích phân 2 2 2 1 1 ln    x I x dx x   0 ABC 30 , SBC là tam giác     2 (a c)(b c) 4c     3 3 2 2 33 32a 32b a b P (b 3c) (a 3c) c      Thí sinh chỉ được làm một trong hai phần (phần A hoặc phần B) A.     2x y 5 0   và A( 4;8) g -4).   x 6 y 1 z 2 : 3 2 1        và      sao cho AM = 2 30 .         :x y 0    có bán kính R = 10    42     (P):2x 3y z 11 0    và  2 2 2 (S):x y z 2x 4y 2z 8 0           z 1 3i    5 w (1 i)z .    Câu 1: a) -x 3 + 3x 2 - -3x 2  x = 0 hay x = 2; y(0) = -1; y(2) = 3 lim x y    và lim x y    x  0 2 +   0 + 0  y + 3 -1   2) - y" = -   : b. -3x 2  m= 2 2xx =g(x)     0, 0;x     m 2 2xx   0;x        2 0 min 2 , 0; x m x x x          11mg   Câu 2 : 1+tanx=2(sinx+cosx)    sinx+cosx=0 hay cosx = 1 2  tanx=-1 hay cosx = 1 2  2, 43 x k hay x k k          Câu 3 :  1x   22 2 1 6 1 0     x y x y y   2 1 4 0    x y y     2 4 1 *   y x y  0y 4 4 1 1 2     x x y y        44 4 4 1 1 1 1 1 1 **        x x y y  4 11tt   ) Nên (**)  f(x) = f(y 4 + 1)  x = y 4 + 1 (y 4 + y) 2 = y 8 + 2y 5 + y 2  74 01 24 yx y y y           0 1 y y      (vì g(y) = y 7 + 2y 4 )  Cách khác  y  0 Xét 4 10xy     Xét 4 10xy    4 4 ( 1 2) ( 1 ) 0x y x y       y x 2 -1 3 0  44 2 4 4 11 0 ( 1 )( 1 ) 12 x y x y x y x y xy              4 2 4 4 11 ( 1) 0 ( 1 )( 1 ) 12 xy x y x y xy                  x = y 4 + 1 (do y > 0) Câu 4 : 2 2 2 1 1 ln x I xdx x       , , (1) 0, 2 ln2 t dx dt x e t t x        ln2 0 tt I t e e dt       , tt du dt dv e e       tt v e e    I = ln2 ln2 0 0 ( ) ( ) t t t t t e e e e dt        = 5ln2 3 2  Cách khác :  u lnx dx du x  dv = 2 22 x 1 1 dx (1 )dx xx   1 vx x    2 2 1 1 1 1 dx I x lnx (x ) x x x          2 1 51 ln2 (1 )dx 2x      2 1 51 ln2 (x ) 2x    51 ln2 (2 ) 22    53 ln2 22  Câu 5.  (ABC) và SH = 3 2 a  BC=a, 3 , 22  aa AC AB 3 1 1 3 3 3 2 2 2 2 16 a a a a V      HI=a/4, 3 2  a SH  SI thì HK  (SAB), ta có 22 2 1 1 1 3 52 3 4 2 a HK HK a a               = 2HK = 2 3 3 52 13  aa Cách khác : Ta có SI 2 = 2 13 16 a .  SAB = 2 39 16 a  d(C, SAB)= 33 () 13 Va dt SAB   Câu 6. Gi  1 1 4 ab cc              a c ; y = b c thì (x + 1)(y + 1) = 4  S + P = 3 ; P = 3  S P = 3 3 22 32 33 xy xy yx                3 22 8 33 xy xy yx        = 3 2 32 8 39 2 S S P S SP       S A B C H I = 3 2 3 2(3 ) 8 3 (3 ) 9 2 S S S S SS           = 3 3 2 5 6 1 88 2 12 2 22 S S S S S S              = 3 ( 1) , 2 2 S SS   3 (S  1) 2  1 2 > 0, S  2  P min = P (2) = 1  2  Câu 7a. C(t;-2t-5)  4 2 3 ; 22        tt I Ta có: IN 2 = IA 2 , suy ra t =1 -7) -4;-7) Câu 8a. Ptmp (P)   -3; -2; 1). -3(x  1)  2(y  7) + z  3 = 0  3x + 2y  z  14 = 0   M (6 -3t; -1  2t; -2 + t) YCBT  (5  3t) 2 + (-8  2t) 2 + (-5 + t) 2 = 120  14t 2  8t  6 = 0  t = 1 hay t = 3 7  -3; -1) hay ( 51 7 ; 1 7  ; 17 7  ). Câu 9a. trong S là : 3.6.5=90    Câu 7b. Cos(AIH) = 1 5 IH IA   IH = 2  IH = 4 2  M  Oy MI  AB  MI : x + y + c = 0 ; M (0;-c) MH = d (M; ) = 2 c = 4 2  c = 8 hay c =-8  -t + 8) d (I; ) = (8 ) 2 2 tt IH    t = 3 hay t = 5 t = 3  I (3; 5); t = 5  I (5; 3)  -8 : I (t; -t - 8) d (I; ) = 2  t = -3 hay t = -5 t = -3  I (-3; -5); t = -5  I (-5; -3)  -5; -3)   5) 2 + (y  3) 2 = 10 hay (x + 5) 2 + (y + 3) 2 = 10. Câu 8b. (S) có tâm là I (1; -2; 1) và R 2 = 14.  2(1) 3( 2) 1 11 14     = 14 = R  Pt (d) qua I và   : 1 2 1 2 3 1 x y z    , T  (d)  T (1 + 2t; 3t  2; 1 + t) T  (P)   M A B I H Câu 9b. r = 13 = 2; tg = 3  = 3    2(cos sin ) 33 i    z 5 = 5 5 1 3 32(cos sin ) 32( ) 3 3 2 2 ii      w = 32(1 + i) 13 () 22 i = 1 3 1 3 32( ) 32 ( ) 2 2 2 2 i    13 32( ) 22   13 32( ) 22  . .  (ABC) và SH = 3 2 a  BC =a, 3 , 22  aa AC AB 3 1 1 3 3 3 2 2 2 2 16 a a a a V      HI =a/ 4, 3 2  a SH . NLong  ( Đề thi có 1 trang )  Môn thi: TOÁN   Thời gian làm bài: 180 phút, không kể thời gian giao đề . HK  (SAB), ta có 22 2 1 1 1 3 52 3 4 2 a HK HK a a               = 2HK = 2 3 3 52 13  aa Cách khác : Ta có SI 2 = 2 13 16 a .  SAB = 2 39 16 a  d(C,