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
⇔ + ≥ +
2 2
x y
x y;(1)
y x
49:)
+ ≥ +
2 2
z y
z y;(2)
y z
+ ≥ +
2 2
x z
x z;(3)
z x
'J
4P
≥ + +P 2(x y z)
&
'J
]aK
=P min 2
%
= = =
1
x y z
3
'J
e