Đề thi thử Đại học môn Toán khối D năm 2014 - THPT Nguyễn Trung Thiên

së gi¸o dôc vµ ®µo t¹o hµ tÜnh<br /> <br /> Tr­êng THPT NguyÔn Trung Thiªn<br /> <br /> §Ò THi thö ®¹i Häc LÇN I n¨m 2014 Môn thi: To¸n - KHỐI D Thời gian làm bài: 180 phút<br /> <br /> I. PhÇn chung cho tÊt c¶ thÝ sinh (7,0 ®iÓm)<br /> C©u I (2,0 ®iÓm) Cho hµm sè y = − x 4 + 2mx 2 − 4 cã ®å thÞ (Cm ) víi m lµ tham sè thùc 1. Kh¶o s¸t sù biÕn thiªn vµ vÏ ®å thÞ cña hµm sè khi m = 2 2.T×m tÊt c¶ c¸c gi¸ trÞ cña m dÓ c¸c ®iÓm cùc trÞ cña ®å thÞ (Cm ) n»m trªn c¸c trôc täa ®é. C©u II (2,0 ®iÓm) 1. Gi¶i ph­¬ng tr×nh: sin 2 x + cos 2 x = cos x − 3sin x + 2 . 2. Gi¶i ph­¬ng tr×nh : log 2 ( x 2 + 4) = 2log 1 8 x 2 + 32 + 6 4<br /> 2<br /> <br />  xy − 7 y + x + 1 = 0 C©u III (1,0 ®iÓm) Gi¶i hÖ ph­¬ng tr×nh:  2 2 2  x y − 13 y + xy + 1 = 0<br /> <br /> C©u IV (1,0 ®iÓm) Cho h×nh chãp S.ABCD cã ®¸y ABCD lµ h×nh thang vu«ng t¹i A vµ D. BiÕt AB = 2a ; AD = CD = a; SA = 3a vµ SA vu«ng gãc víi ®¸y. TÝnh thÓ tÝch khèi chãp S.BCD vµ kho¶ng c¸ch tõ B ®Õn mp (SCD) theo a. C©u V (1,0 ®iÓm) Cho 3 sè thùc d­¬ng x, y, z. Chøng minh r»ng :<br /> x 2 − xy y 2 − yz z 2 − zx + + ≥ 0. x+ y y+z z+x<br /> <br /> II. PhÇn riªng (3,0 ®iÓm): ThÝ sinh chØ ®­îc lµm mét trong hai phÇn (phÇn A hoÆc phÇn B) A. Theo ch­¬ng tr×nh chuÈn C©u VI. a. (1,0 ®iÓm) Trong mÆt ph¼ng víi hÖ täa ®é Oxy cho tam gi¸c ABC cã trùc t©m H (1;-1). Trung ®iÓm cña c¹nh AC lµ E(-1; 2); c¹nh BC cã ph­¬ng tr×nh d : 2x − y +1 = 0 . X¸c ®Þnh täa ®é c¸c ®Ønh cña ∆ABC . C©u VII. a. (1,0 ®iÓm) Trong mÆt ph¼ng víi hÖ täa ®é Oxy cho ®­êng trßn (C ) : x 2 + y 2 − 2 x − 2 y − 2 = 0 . LËp ph­¬ng tr×nh ®­êng th¼ng d c¸ch gèc täa ®é mét kho¶ng b»ng 2 vµ tiÕp xóc víi (C). C©u VIII. a. (1,0 ®iÓm ) T×m hÖ sè cña x 2 trong khai triÓn nhÞ thøc Niut¬n cña:<br /> 2  P ( x ) =  x 2 +  ( x ≠ 0) . x  3 BiÕt r»ng n tháa m·n: 3 Cn+1 − 3 An2 = 52(n − 1) .<br /> n<br /> <br /> B. Theo ch­¬ng tr×nh n©ng cao C©u VI. b. (1,0 ®iÓm) Trong mÆt ph¼ng víi hÖ täa ®é Oxy cho h×nh b×nh hµnh ABCD t©m I, cã diÖn tÝch b»ng 4. Ph­¬ng tr×nh ®­êng th¼ng BC : x − y = 0 . BiÕt M(2;1) lµ trung ®iÓm cña c¹nh AB. T×m täa ®é ®iÓm I. C©u VII. b. (1,0 ®iÓm) Trong mÆt ph¼ng víi hÖ täa ®é Oxy cho ®­êng th¼ng d : 4 x − 3 y + 2 = 0 vµ ®­êng trßn (C ) : x 2 + y 2 − 2 x + 6 y − 15 = 0 .ViÕt ph­¬ng tr×nh ®­êng th¼ng d1 vu«ng gãc víi d c¾t ®­êng trßn (C) t¹i A vµ B sao cho AB = 6 . C©u VIII. b. (1,0 ®iÓm) Cã 12 häc sinh giái gåm 3 häc sinh khèi 12; 4 häc sinh khèi 11 vµ 5 häc sinh khèi 10. Chän ngÉu nhiªn 6 häc sinh. TÝnh x¸c suÊt sao cho mçi khèi cã Ýt nhÊt 1 häc sinh.<br /> <br /> --------------------- HÕt --------------------<br /> <br /> §¸p ¸n K-D gåm cã 5 trang.<br /> L­u ý : Mäi c¸ch gi¶i ®óng ®Òu cho ®iÓm tèi ®a. C©u §¸p ¸n vµ h­íng dÉn chÊm 1 (1,0 ®iÓm) ________________________________________________________________________ Víi m=2 hµm sè trë thµnh y = − x 4 + 4 x 2 − 4 +TËp x¸c ®Þnh: D = R + Giíi h¹n: lim y = −∞ ; lim y = −∞<br /> x →−∞ x →+∞<br /> <br /> §iÓm<br /> <br /> C©u I. 2,0 ®iÓm<br /> <br /> 0,25<br /> <br /> ________________________________________________________________________ x = 0 _ Sù biÕn thiªn: y ' = −4 x 3 + 8 x , y , = 0 ⇔  x = ± 2 + B¶ng biÕn thiªn: x y’ y −∞ + − 2 0 0 0 0 + 2 0 0 +∞ 0,25<br /> <br /> −∞ -4 −∞ -----------------------------------------------------------------------------------------------------------Suy ra: Hµm sè ®ång biÕn trªn mçi kho¶ng −∞; − 2 vµ 0; 2 .<br /> <br /> ) ( ) Hµm sè nghÞch biÕn trªn mçi kho¶ng ( − 2; 0 ) vµ ( 2; +∞ ) . §iÓm cùc ®¹i cña ®å thÞ lµ ( − 2; 0 ) vµ ( 2;0 ) . §iÓm cùc tiÓu cña ®å thÞ lµ ( 0; −4 ) . ( ) ( )<br /> <br /> (<br /> <br /> 0,25<br /> <br /> ________________________________________________________________________ + §å thÞ : §å thÞ c¾t trôc tung t¹i ( 0; −4 ) vµ c¾t trôc hoµnh t¹i ®iÓm − 2; 0 vµ 2;0 . + NhËn xÐt: §å thÞ (C) nhËn trôc tung lµm trôc ®èi xøng. 2 (1,0 ®iÓm) ________________________________________________________________________ x = 0 Ta cã: y ' = −4 x 3 + 4mx = 4 x ( − x 2 + m ) ; y ' = 0 ⇔  2 x = m ________________________________________________________________________ NÕu m ≤ 0 th× ( Cm ) chØ cã mét ®iÓm cùc trÞ vµ ®ã lµ ®iÓm cùc ®¹i n»m trªn trôc tung. ________________________________________________________________________ NÕu m > 0 th× ( Cm ) cã 3 ®iÓm cùc trÞ. Mét cùc tiÓu n»m trªn trôc tung vµ hai cùc ®¹i cã täa ®é − m ; m 2 − 4 ,<br /> <br /> 0.25<br /> <br /> 0.25<br /> <br /> 0,25 0,25<br /> <br /> (<br /> <br /> ) (<br /> <br /> m ; m2 − 4 .<br /> <br /> )<br /> <br /> ________________________________________________________________________ §Ó h¹i ®iÓm nµy n»m trªn trôc hoµnh th× m2 − 4 = 0 ⇔ m = ±2 . V× m > 0 nªn chän m = 2 . VËy m ∈ ( −∞; 0] ∪ {2} lµ nh÷ng gi¸ trÞ cÇn t×m tháa m·n yªu cÇu bµi to¸n. C©u II. 2,0 1 (1,0 ®iÓm) ________________________________________________________________________ PT ⇔ 2sin x cos x + 1 − 2sin 2 x = cos x − sin 3 x + 2<br /> <br /> 0.25<br /> <br /> ®iÓm<br /> <br /> ⇔ (2sin x cos x − cos x) − (2sin 2 x − 3sin x + 1) = 0 ⇔ cosx(2sinx-1) – (2sinx-1)(sinx-1) = 0 ⇔ (2sin x − 1)(cos x − sin x + 1) = 0 ________________________________________________________________________ 1  1  sin x = 2 sin x = ⇔ ⇔ 2  π 2   sin x − cos x = 1 sin  x − 4  = 2    ________________________________________________________________________ π   x = 6 + k 2π   x = 5π + k 2π ⇔ k ∈ 6   x = π + k 2π  2  x = π + k 2π  2 (1,0 ®iÓm) ________________________________________________________________________ Ta cã 2 log 1 8 x 2 + 32 = −2log 2 8 x 2 + 32<br /> 2<br /> <br /> 0,5<br /> <br /> 0,5<br /> <br /> 0,25<br /> <br /> = −2log 4 (8 x 2 + 32 )<br /> <br /> Ph­¬ng tr×nh ⇔ log 2 ( x 2 + 4 ) + 2 log 4 ( x 2 + 4 ) − 3 = 0 4<br /> <br />   = −2  log 4 ( x 2 + 4 ) + log 4 8 = −2 log 4 ( x 2 + 4 ) − 3 .<br /> <br /> 0,5<br /> <br /> ________________________________________________________________________ §Æt log 4 ( x 2 + 4 ) = t , ph­¬ng tr×nh trë thµnh t 2 + 2t − 3 = 0 ⇔ t = 1 ; t = −3 . ________________________________________________________________________ Víi t = 1, ta cã log 4 ( x 2 + 4 ) = 1 ⇔ x = 0 . Víi t =-3, ta cã log 4 ( x 2 + 4 ) = −3 (v« nghiÖm). VËy ph­¬ng tr×nh cã nghiÖm x = 0.<br /> <br /> 0,25<br /> <br /> 0,25<br /> <br /> C©u III. 1,0 ®iÓm<br /> <br /> NhËn xÐt y=0 kh«ng tháa m·n hÖ ph­¬ng tr×nh ®· cho. 1 x  x + y + y = 7  HÖ ph­¬ng tr×nh ®· cho ⇔   x 2 + 1 + x = 13  y2 y  ________________________________________________________________________ 1 x 1 x 1 §Æt u = x + ; v = . Suy ra: u 2 = x 2 + 2 + 2 ⇒ x 2 + 2 = u 2 − 2v . y y y y y  u = 4  u + v = 7 u + v = 7 v = 3 Khi ®ã, ta ®­îc:  2 ⇔ ⇔ 2  u = −5 u − v = 13 u + u − 20 = 0   v = 12  ________________________________________________________________________<br /> <br /> 0,25<br /> <br /> 0,25<br /> <br /> 0,25<br /> <br />  y = 1 1   x = 3y u = x + y = 4 x = 3 x = 3y   ⇔  Víi  ⇔ ⇔ 2 1 1 3y + = 4 x 3y − 4 y +1 = 0  y =  v = = 3  y 3    y  x = 1  ________________________________________________________________________ 1  u = x + = −5  x = 12 y  y  x = 12 y   Víi  (v« nghiÖm). ⇔ ⇔ 1 2 12 y + 5 y + 1 = 0 v = x = 12 12 y + y = −5   y  x = 1 x = 3  KÕt luËn: HÖ ph­¬ng tr×nh ®· cho cã hai nghiÖm  vµ  1. y =1 y = 3 <br /> <br /> 0,25<br /> <br /> C©u IV. 1,0 ®iÓm<br /> <br /> 1 3a 2 DiÖn tÝch h×nh thang ABCD lµ S = ( 2a + a ) a = . 2 2 1 DiÖn tÝch tam gi¸c ABD lµ S ∆ABD = AB. AD = a 2 . 2 a2 DiÖn tÝch tam gi¸c BCD lµ S ∆BCD = S − S ∆ABD = . 2 ________________________________________________________________________ 1 1 a2 a3 ThÓ tÝch khèi chãp S.BCD lµ VS .BCD = SA.S ∆BCD = .3a. = . 3 3 2 2 ________________________________________________________________________ Ta cã: SD = 9a 2 + a 2 = a 10 . V× SA ⊥ ( ABCD ) ⇒ SA ⊥ CD . MÆt kh¸c AD ⊥ CD . Suy ra CD ⊥ SD . 1 2 a 10 . 2 ________________________________________________________________________ Gäi d lµ kho¶ng c¸ch tõ B ®Õn (SCD). Ta cã: 1 a3 3a 3 3a 10 VS .BCD = d .S∆SCD = ⇔d = 2 = . 3 2 10 a 10 x 2 − xy x ( x + y ) − 2 xy 2 xy Ta cã = = x− . x+ y x+ y x+ y 2 xy x + y ≤ Do x,y d­¬ng nªn . x+ y 2 x 2 − xy x+ y x− y Suy ra ≥ x− = . x+ y 2 2 ________________________________________________________________________ y 2 − yz y − z z 2 − zx z − x T­¬ng tù ta còng cã ≥ ; ≥ . y+z 2 z+x 2 x 2 − xy y 2 − yz z 2 − zx x − y y − z z − x Tõ ®ã suy ra + + ≥ + + = 0 (®pcm). x+ y y+z z+x 2 2 2 DiÖn tÝch tam gi¸c SCD lµ S ∆sCD =<br /> <br /> 0,25<br /> <br /> 0,25<br /> <br /> 0,25<br /> <br /> 0,25<br /> <br /> C©u V. 1,0 ®iÓm<br /> <br /> 0,5<br /> <br /> 0,5<br /> <br /> A. Theo ch­¬ng tr×nh chuÈn C©u Gi¶ sö C ( m; 2m + 1) .  VI. a. V× E ( −1; 2 ) lµ trung ®iÓm AC nªn A ( −2 − m;3 − 2m ) ; AH = ( 3 + m; −4 + 2m ) . 1,0 ®iÓm ________________________________________________________________________  <br /> V× vect¬ chØ ph­¬ng cña BC lµ uBC = (1; 2 ) .    V× AH ⊥ BC nªn AH .uBC = 2 + m + 2 ( −4 + 2m ) = 0 ⇔ m = 1 . VËy A(−3;1) vµ C (1;3) . Gi¶ sö B ( n; 2n + 1) .    Cã BH = (1 − n; −2 − 2n ) ; uBC = ( 4; 2 ) .    V× BH ⊥ AC nªn AH .uBC = 4(1 − n) + 2(−2 − 2n) = 0 ⇔ n = 1 . VËy B(0;1) .<br /> <br /> 0,25<br /> <br /> 0,5<br /> <br /> ______________________________________________________________<br /> 0,25<br /> <br /> C©u VII. a. 1,0 ®iÓm<br /> <br /> Gäi ph­¬ng tr×nh ®­êng th¼ng d lµ ax + by + c = 0 ( a 2 + b2 ≠ 0 ) , a +b ________________________________________________________________________ §­êng trßn cã t©m I(1;1) b¸n kÝnh R =2. a +b+c V× d tiÕp xóc víi (C) nªn d (d ; O) = 2 ⇔ =2 a2 + b2 ________________________________________________________________________ b = − a Suy ra a + b + c = c ⇔  c = − a + b  2<br /> 2 2<br /> <br /> d ( d ; O) = 2 ⇔<br /> <br /> c<br /> <br /> = 2.<br /> <br /> 0,25<br /> <br /> 0,25<br /> <br /> 0,25<br /> <br /> ______________________________________________________________<br /> Víi b = − a , chän a = 1 ⇒ b = −1; c = ±2 2 ta ®­îc ph­¬ng tr×nh x − y ± 2 2 = 0 . a+b Víi c = − , ta cã 15a 2 − 2ab + 15b 2 = 0 ⇔ a = b = 0 (kh«ng tháa m·n). 2 Tõ ®Ò bµi, t×m ®­îc n =13. ________________________________________________________________________ 0,25 0,5 0,25<br /> <br /> C©u VIII. a. 1,0 ®iÓm<br /> <br /> 13  2 2 k 2(13− k ) k − k 2 x .  x +  = ∑ C13 x x  k =0 Gi¶i t×m ®­îc k =8. ________________________________________________________________________ 8 HÖ sè cÇn t×m lµ 28.C13 .<br /> <br /> 13<br /> <br /> 0,25<br /> <br />