SỞ GD&ĐT QUẢNG NAM ĐỀ THI THỬ QUỐC GIA 2015 Trường THPT Phan Bội Châu Môn: TOÁN - ĐỀ SỐ 1 Thời gian làm bài: 180 phút Câu 1( y f x x x m= = + + m R∈ !"#$% m &' ()*+),- m !"./0#1##*"2345 m )1# · 263 +)' Câu 2(7##89:)*5 ; .) .) x x− − = #<<=> #<&' Câu 34??#8@A& Câu4 8(#BCDE/%FGB=;G z &;G458H# 8H#$B 4*)%*"(#I<&JG;< )/KL9:)DK?M)N###F .OKP.Q#$< Câu 55#18()#E/R23S#1#TK+))1#)U#T VK+)' ' 4?W?#%#18R23S%)##)U9X) Y)2R3 Câu 6 4*)%0)),FTZOxyz#2[J['3[[\ 9X)Y)∆ x y z1 1 2 1 2 + − = = − ]89:)*5V8Y)I^/2/0))1#,9X)Y)∆ 45TZ_*∆#)#∆_23#1LF?#CN Câu 74*)V8Y),FTZOxy#2['3'[9X) Y)∆ ' =+− yx `a889:)*59X)*b^/23#c9X) Y)∆T8/KTKT,/Z)1#\' ο Câu 87N89:)*5 ' ; ;x x x x+ + + ≤ + Câu 9L9:) x y z C x y z+ + &45)*".,N#$/(# x y z P x y z = + + + + + &&&d&&& e x y -2 4 0 1 fgIghfi46ghRj @/ k hZL/) f 'J 'J &'K&< =< =4lfS&m =E/Kn&< =\< Kn&' ⇔ <&'K<&G . . x x y y →+∞ →−∞ = +∞ = −∞ =334 < G ∞ G'= ∞ Kn ='G'= K ;= ∞ G ∞ ' d)"#*)G['!)*)o %) −∞ [G'[ +∞ d0T##T+);%<>##/+)' %<&' =f!" y f x x x m= = + + =Kn&< =\<#1)F<&G<&',-/K *!"./0#1##*" 4-Z##*".2'[3G[;= = · 263 &' ⇔ OA OB OA OB = − uuur uuur ; e 'm m m m m ⇔ − + = + + ⇔ ' m m − + = = f ' ≠< x I4 xx .)pp.) =−⇔ 459q#)F<& =3ME8 #<G <&#< 'J 'J 'J 'J 'J 'J 'J 'J 'J 'J 'J 'J r ; J \ 'J 'J 'J 'J =459q#)F8<& =% [<& = [ fV . Lu x dv x x= = /K* x du dx v x = = SsA& G 4? ( ) L e e x x x e= = − ∫ ^/A& ( ) ; e + 7-8(##H5B&= biaz −=⇒ [ R ∈ =ME9q#\=G;=&eG\ =45&[& 4#1I<& = <= < == < I& = == IG& G = G ==G I&JG; &IG&J=; &r M)##FW.OKP.Q. = = J =& r n PP − = −− O A B C D S K 7-6.@5/0)23S ABCDSO ⊥⇒ 71#R6+)' .)1#)U#TK 4?9q# R6& \ \a aK] R23S & Q6 ⊥ R3 2 ⊥ R3S ⇒ 2 ⊥ 6 #6.%)##)U2R3 4?9q#L2[R3&6& a =459q#]4I4 ( ) [[ −= n #$8I =]9q#88I<GK=B=&' 7- M t t t( 1 2 ;1 ;2 )− + − ∈∆ 'J 'J 'J 'J 'J 'J 'J 'J 'J 'J 'J 'J 'J 'J 'J ' t e r SF?#∆_23. S AM AB t t 2 1 , 18 36 216 2 = = − + uuur uuur & t 2 18( 1) 198 − + u 198 ]aK_R& 198 % t 1 = K_['[ =459q#89X)*/)*#L#$T23K&< =f9X)*b#1@A [ 7-S.H.9q.)#$∆8/KT S#c/TvT,/Z)1#\' ο \' = CJD V#)1# ο ' = CJD )1# \' = CJD 1 · AS &' IAIH = d. 5#/#$A.∆ IAId [ =∆⇔ 459q#@A'['3%?m& I9X)*b< =K & @Aw[3%?mw& J I9X)*b<G =KG &J )1# ο ' = CJD 459q# IAIH = d.5#/ #$fA.∆ IAId [ =∆⇔ tt +−=⇔ I4]h fE/%F< 'x =<&'.)F8 =<y'#8# 9q# = = fV& = &y<= & G; 3N89:)*5 G 'J 'J 'J 'J 'J 'J 'J 'J 'J 'J 'J ], #1 = < ;[' q8,E/%Fx)F<&'9q#a8 )F8.R&z'[{ z;[= { =4#1 x y z P x y z = + + + + + x y z = − + − + − + + + &G x y z + + + + + _ x y z + + + + + x y z ≥ + + + 0G =] x y z x y z + + + + + ≥ + + + ⇔ ; x y z x y z ≥ = + + + + + + + + 4PR/K* r ;x y z + + ≥ + + + R/K* r ; ; P ≤ − = ]aK _<I ; = T9q#%#% 'J 'J SỞ GD&ĐT QUẢNG NAM ĐỀ THI THỬ QUỐC GIA 2015 Trường THPT Phan Bội Châu Môn: TOÁN - ĐỀ SỐ 2 Thời gian làm bài: 180 phút Câu 1 (2,0 điểm). ( ) − = + x 2 y 1 x 1 #1!" !"#$ ()9X)Y)L + + =2x y m 0 ./0#c!"T8@F 23f" m %)##23|N Câu 2 (1,0 điểm).789:)*5 + = 3 4 sin x cos x 1 Câu 3 (1,0 điểm).4??#8@ + = + + ∫ 4 0 2x 1 I dx 1 2x 1 Câu 4 (1,0 điểm). 458(# z ( ) + − =1 i z 1 0 _ZZ})F#1J)9X)!'JUdC#1/##.a8Z1 !)#)!e)9X*+)*)118#1?N)9XU Câu 5 (1,5 điểm). 4*) %0) ) , F - Z 6<KB # V 8Y) − + + =(P) : x 2y 2z 1 0 V#H/ + + − + + + = 2 2 2 (S) : x y z 4x 6y 6z 17 0 ]89:)*59X)Y)L^/@#$R/0))1#,I 7-.9X)*b)/K#$RI`a889:)*5V#H/Rw#( 9X)*b#1@/Z#V8Y) + + + =(Q) : x y z 3 0 Câu 6 (1,0 điểm). 5#18R23S#1K.5,)1# · 32S +) 0 120 = BD a V8Y)R23R2S#~)/0))1#,VK)1#)UV8Y) R3VK+) 0 60 4?W?#%#18R23S%)##PS 8R3 x y 1 1 -1 Câu 7 (0,5 điểm). 4*) V 8Y) , F - Z 6<K # )# ABC #1 A(2;3),B(2;1),C(6;3) 7- D .#@9X)8@)#*))1# A #$)# ABC 45 M *9X)*b ( ) ( ) ( ) − + − = 2 2 C : x 3 y 1 25 #LF?#)# MCD )N8 0LF?#)# ABD. Câu 8 (1,0 điểm).7F89:)*5 ( ) + + = + + + + + − = 4 2 2 2 2 32 y 2y 4x 4x y 2x xy 2x y 1 2x 1 1 Câu 9 (1,0 điểm). <KB.###L9:)C + + =x y z 1 4?)*"CN #$/(# ( ) ( ) ( ) + + + = + + 2 2 2 x y z y x z z y x P yz xz yx Hết ĐÁP ÁN ĐỀ THI THỬ SỐ 2 - PBC @/ f8 f ' =4a8<#" { } = −D R \ 1 =RE/ ( ) = > ∀ ≠ − + ' ' 2 3 y ,y 0 x 1 x 1 d!)*P)%) ( ) ( ) −∞ − − +∞; 1 ; 1; 'J =7,TF#a →−∞ →+∞ = = x x limy 1;limy 1 ⇒ =y 1 .F#a)) − + →− →− = +∞ = −∞ x 1 x 1 limy ;limy ⇒ = − x 1 .F#a() 'J =3) <G ∞ G= ∞ ' y == = ∞ KG ∞ 'J =f!" 'J ; ' =I9:)*5Z)#$L − = − − + x 2 2x m x 1 'J ( ) ⇔ + + + − = ≠ − 2 2x m 3 x m 2 0,x 1 x <&G%0).)F#$89:)*5x_V%#x#1 ( ) ∆ = − + = − + > ∀ 2 2 m 2m 25 m 1 24 0, m ]aKx./0#1 )F8@F%#G/K*L./0#cT8@F 'J = ( ) ( ) − + = − + − = 2 2 2 2 B A B A m 2m 25 AB x x y y 5 4 'J 23 ⇔ − + 2 m 2m 25 ⇔ =m 1 'J ' I9:)*5D#9:)9:), ( ) ( ) + = + ⇔ − = − 3 4 2 2 2 2 2 sin x cos x sin x cos x sin x sinx 1 cos x 1 cos x 'J ( ) ( ) ⇔ − − = ⇔ + − = 2 2 2 2 sin x sinx 1 cos x 0 sin x sin x sinx 2 0 'J ⇔ π π π = = ⇔ = = + ∈ x k sinx 0 sinx 1 x k2 ;k z 2 'J ' fV = + ⇒ = + ⇒ = 2 t 2x 1 t 2x 1 dx tdt fM#a = ⇒ = = ⇒ =x 0 t 1;x 4 t 3 'J + = = + + + ∫ ∫ 4 3 2 0 1 2x 1 t I dx dt 1 t 1 2x 1 'J = − + = − + + ÷ ÷ + ∫ 3 3 2 1 1 1 t t 1 dt t ln t 1 t 1 2 'J = + 2 ln2 'J ; 'J ( ) + − =1 i z 1 0 − ⇔ = ⇔ = + 1 1 i z z 1 i 2 'J = − ⇔ = + 1 1 1 1 z i z i 2 2 2 2 'J 'J -U=J#1 = 3 5 5 10 C C 2520 ## -;U=;#1 = 4 4 5 10 C C 1050 ## 'J J -JU=#1 = 5 3 5 10 C C 120 ## 4W^/c##Z)#1\r'##.a81!)# 'J J J R#14@ − −I(2; 3; 3) %? =R 5 'J =I#1]4I4 [ [n = − r =R/K*L#1]4I. [ [n = − r =I9:)*59X)Y)L = + = − − = − + x 2 t y 3 2t z 3 2t 'J 'J 'J 'J #1@ − − = ⇒ ÷ I 5 7 11 H (d) (P) H ; ; 3 3 3 %?*& 7- ( ) S' #1@•%? ' r 1 = ⇒ − − =I ' E (d) (Q) E(3; 5; 1),r 2 5 'J h ( ) ( ) ( ) ( ) − + + + + = ' 2 2 2 S : x 3 y 5 z 1 20 'J \ =h/9q# ⊥SA (ABCD) 7-A.*/)#$3)1#)U R3VK.)1#RA2[/K* = a AI 2 'J = = 0 a 3 SA AI.tan60 2 ⇒ = = = 3 S.ABCD ABCD 1 1 1 a a 3 a V S .SA . .a. . 3 3 2 2 12 3 'J 4#1 ⇒ =AD / / (SBC) d(D;(SBC) d(A;(SBC)) ⊥ ⇒ ⊥BC (SAI) (SAI) (SBC) 7-d.5#//0))1##$2.#TRA#1 ⊥AH (SBC) =d(A;(SBC)) AH 'J 4*))#/0)R2A#1 = = SA.AI a 3 AH SI 4 'J \ ]aK = a 3 d(D;(SBC)) 4 L t =7-S<[K.#@9X)8@)#*)#$)1#24W? #N9X)8@)#*) = = ⇔ = − ⇒ ÷ uuur uuur DB AB 2 1 10 5 BD CD D ; DC AC 4 2 3 3 23 = ⇒ = ⇒ = = ABD 4 1 4 x 2 d(D;AB) S AB.d(D;AB) 3 2 3 I9:)*5#TS − =x 2y 0 'J 7-_<[K/Z# ( ) ( ) − + − = 2 2 x 3 y 1 25;(1) 4#1 ( ) − = ⇔ = ⇔ − = 2 MCD ABD 2 x 2y 4 S 2S 2. x 2y 16;(2) 3 3 7F)!459q# ( ) ( ) − − − + − + − ÷ ÷ ÷ ÷ 1 2 3 4 18 4 29 1 2 29 18 4 29 1 2 29 M 2;1 ;M 6;5 ;M ; ;M ; 5 5 5 5 'J e fE/%F + ≥ ⇔ ≥ 2 xy 2x 0 x 0 I9:)*5 ( ) ( ) ( ) ( ) ( ) ( ) − + = + + − ⇔ + − − + + = 2 2 2 2 2 2 2 y 2x y 2x y 2x xy 2x 2 y 2x y 2x xy 2x 2 0 'J l| + = 2 y 2x 0 F0)F l| − − + + = 2 2 y 2x xy 2x 2 0 + + + ⇔ − − = ⇔ = 2 2 2 y 2 y 2 y 2 2 0 2 x x x ⇔ + = − 2 y 1 4x 1 h − + − − = 3 4x 1 2x 1 1 0 l| = − + − − 3 f(x) 4x 1 2x 1 1 ⇒ = + > − − ' 2 3 2 2 f (x) 0 4x 1 3. (2x 1) !) > 1 x 4 'J 'J _ = ⇒ = = 1 1 f( ) 0 x ;y 0 2 2 .)FL/KN#$89:)*5 'J r 4#1 ( ) − ≥ 2 x y 0 + − ≥ ⇔ + ≥ + 2 2 3 3 x y xy xy x y xy(x y) 'J t