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Mn thi: Ton - Ln th 1
Nm hc 2014 - 2015
N.
co
1
b) Gii phng trnh log 27 x 3 + log 3 ( x + 2) = 1 + log 3 ( 4 3 x ) .
2
e
I =
HV
Cu 4 (1,0 im).
x +1
ln xdx.
x2
1 i
= 5 i . Tm mun ca s phc w = 1 + z + z 2 .
1+ i
b) C hai thng ng to. Thng th nht c c 10 qu (6 qu tt v 4 qu hng). Thng th hai c 8
a) Cho s phc z tha mn iu kin
(2 + i) z +
AT
t nht mt qu tt.
Cu 5 (1,0 im). Trong khng gian vi h ta Oxyz , cho hai im A(1; 1;2), B(3;0; 4) v mt
Cu 6 (1,0 im). Cho hnh chp S . ABCD c y l hnh ch nht, AB = a, AD = 2a . Tam gic SAB
w.
ww
16 13
thng AB c phng trnh x 2 y = 0 . Trng tm ca tam gic BCD l im G ; . Tm ta
3 3
bn nh ca hnh ch nht bit im B c tung ln hn 3.
2 x3 3 + 2 y 2 + 3 y = 2 x y + y
Cu 8 (1,0 im). Gii h phng trnh
2
x y + 3 + y = 0
( x, y ).
T = 3 a + b 3 a 2 + b 2 + 2( a + b) ab .
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Th sinh khng c s dng ti liu. Cn b coi
thi khng gii thch g thm.
H v tn th sinh: ..; S bo danh:
p n thang im
thi th thpt quc gia nm 2015
Mn thi: Ton Ln th 1
--------------- p n c 04 trang --------------
Nm hc 2014 2015
N.
co
Cu
p n
1
a) (1,0 im) Kho st s bin thin v v th ca hm s y = x 4 2 x 2 + 1
(2,0)
Tp xc nh: D = R . lim y = +; lim y = +
x +
HV
-1
0
1
+
y'
0
+
0
0
+
y
+
1
+
0
0
th: (Hs c th ly thm im (2;9); (2;9) )
b) (1,0 im) Tm m th (1) ct trc honh ti bn im phn bit c honh nh hn 2.
Phng trnh honh giao im x 4 + ( m 3) x 2 + 2 m = 0 (1)
(2)
AT
t t = x 2 0 t 2 + ( m 3) t + 2 m = 0
(1) c 4 nghim phn bit th (2) c 2 nghim dng phn bit > 0, S > 0, P > 0
m < 2; m 1 .
iu kin: Phng trnh (2) phi c nghim tha mn iu kin 0 < t1 , t2 < 4
Phng trnh (2) c t1 = 1 (tha mn), t2 = 2 m
iu kin: 2 m < 4 m > 2
p s: 2 < m < 2, m 1 .
2cos x 1 = 0 cos x =
ww
w.
+ k , x =
0,25
0,25
0,25
0,25
0,25
0,25
0,25
+ k , ( k )
x = + k 2 , k
2
3
0,25
0,25
2
a) (0,5 im) Gii phng trnh 3cos 2 x + sin x 1 = cos x + sin 2 x sin 2 x .
(1,0)
Phng trnh cho tng ng vi 2cos 2 x cos x + sin x 2sin x cos x = 0
im
0,25
+ k 2 , k .
1
log 3 ( x + 2) = 1 + log 3 ( 4 3 x )
2
4
. Phng trnh cho tng ng vi
3
log 3 x + log 3 ( x + 2 ) = log 3 3 + log 3 ( 4 3x ) log 3 x ( x + 2 ) = log 3 3 ( 4 3 x )
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0,25
1/4
x = 1(tm)
x ( x + 2 ) = 3 ( 4 3 x ) x 2 + 11x 12 = 0
x = 12( L)
p s: x = 1 .
e
I =
1
x +1
ln xdx.
x2
3
(1,0) Tnh tch phn
e
N.
co
1
1
I = ln xdx + 2 ln xdx = A + B
x
x
1
1
e
0,25
1
A = ln xdx = ln xd (ln x)
x
1
1
e 1
1
A = ln 2 x = .
1 2
2
e
1
1
1
1
ln xdx; t u = ln x u ' = ; v ' = 2 v =
2
x
x
x
x
1
e e 1
e 1e
1
1
B = ln x + 2 dx = ln x
1 1x
1 x1
x
x
HV
B=
AT
e2
1 1
2
B = 1 = + 1 =
e e
e
e
1 e 2 3e 4
x +1
=
I = A+ B = +
. ( I 0, 764) (Hs cng c th tnh ngay u = ln x; v ' = 2 )
2
e
2e
x
4
1 i
= 5 i . Tm mun ca s phc w = 1 + z + z 2 .
(1,0) a) (0,5 im) Cho ( 2 + i ) z +
1+ i
5
Phng trnh cho tng ng vi ( 2 + i ) z = 5 z =
= 2i
2+i
T w = 1 + z + z 2 = 6 5i . Suy ra | w |= 36 + 25 = 61 .
0,25
0,25
0,25
0,25
0,25
0,25
w.
( )
Xc sut ca bin c A l: p A =
12 3
=
80 20
( )
ww
3 17
=
.
Suy ra, xc sut ca bin c A l: p ( A ) = 1 p A = 1
20 20
5
Cho A(1; 1; 2), B (3;0; 4) , ( P ) : x 2 y + 2 z 5 = 0
(1,0)
ng thng AB i qua im A v c vtcp AB = ( 2;1; 6 )
x = 1 + 2t
(t R ) .
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0,25
0,25
0,25
2/4
Gi I = AB ( P ) I AB I (1 + 2t ; 1 + t ; 2 6t )
1
6
4 5
Suy ra ta giao im ca AB v ( P ) l im I ; ;1 .
3 6
Mt phng (Q) qua A v c vtpt nQ = AB, nP , trong nP l vtpt ca ( P )
Ta c nP = (1; 2; 2 )
Suy ra AB, nP = (10;10;5 ) . Chn nQ = ( 2; 2;1)
6
(1,0)
N.
co
0,25
Gi H l trung im ca AB SH AB SH ( ABCD ) ,
0,25
= 450 .
suy ra HC l hnh chiu ca SC ln ( ABCD ) SCH
a 2 a 17
SH = HC = 4a +
=
4
2
2
HV
S ABCD = 2a 2
0,25
0,25
0,25
AT
I
a 3 17
1
1 a 17
2
E
H
VS . ABCD = .SH .S ABCD = .
.
.2a =
B
3
3
3 2
1
1
d ( M ,( SAC ) ) = d ( D,( SAC ) ) = d ( B,( SAC ) ) = d ( H ,( SAC ) )
2
2
K HI AC , HK SI HK AC HK ( SAC ) d ( H ,( SAC ) ) = HK .
K BE AC HI =
0,25
1
1
1
1
1
1
5
2a
a
BE .
=
+
= 2 + 2 = 2 BE =
HI =
2
2
2
2
BE
BA
BC
a
4a
4a
5
5
ww
w.
1
1
1
5
4
89
a 17 a 1513
T suy ra
=
+
= 2+
=
d ( M ,( SAC ) ) =
=
.
2
2
2
2
2
89
17 a
HK
HI
HS
a 17 a
89
7
Trong mt phng ta Oxy , cho hnh ch nht ABCD c din tch bng 15
(1,0)
A
N
10
3 10
BC = .
= 5 AB = 3 5
Ta c d (G, AB) =
2 3 5
3 5
I
ng thng d qua G v vung gc vi AB d : 2 x + y 15 = 0
G
1
Gi N = d AB N ( 6;3) . Suy ra NB = AB = 5
K
D
3
b = 2( L)
Gi B ( 2b; b ) AB NB 2 = 5 b 2 6b + 8 = 0
B ( 8; 4 )
=
b
4
Ta c BA = 3BN A ( 2;1)
3
AC = AG C ( 7;6 ) . CD = BA D (1;3)
2
p s: A ( 2;1) , B ( 8; 4 ) , C ( 7;6 ) , D (1;3) .
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0,25
0,25
0,25
C
0,25
0,25
3/4
8
2 x3 3 + 2 y 2 + 3 y = 2 x y + y (1)
(1,0) Gii h phng trnh
x y + 3 + y = 0
2
( x, y ).
(2)
x 2x + 2x y y = 0 ( x
) ( y)
2 2
2x x y = 0 x 2x + y
2
)( x
y = x 2 : (2) x 2 + 3 = 2 x 2
4 x 4 x 2 3 = 0 x 2 = 1 ( x; y ) = (1; 1),(1; 1) .
y+3 y
= x4
0,25
(2) 3 + ( 2 x x 2 ) = 2 x
2
y = 2 x x 2 : (3)
HV
x 0
x = 1
4
( x 1)( x 3 3 x 2 3 x 3) = 0 3
3
2
x 4x + 3 = 0
x 3x 3x 3 = 0
x = 1 y = 1.
x3 3 x 2 3 x 3 = 0 x 2 ( x 3) 3x 3 = 0 (4)
T (3) suy ra 2 x x 2 0 0 x 2 (4) v nghim.
p s: ( x; y ) = (1; 1), (1; 1).
9
(1,0)
0,25
y =0
N.
co
iu kin: y 0, (1) 2 x3 2 x y + y = y + 3 2 y ( y + 3) + y =
0,25
0,25
a, b 0 : 3 ( a + b ) + 2 ( ab + 1) 5 a 2 + b 2 . Tm max: T = 3 a + b 3 a 2 + b 2 + 2 ( a + b ) ab
Ta c 3( a + b) + 2(ab + 1) 5(a 2 + b 2 ) 2 ( a + b ) + 3 ( a b ) 3 ( a + b ) + 2
2
V 3 ( a b ) 0 a, b 2 ( a + b ) 3 ( a + b ) + 2
2
1
t 2 . V t 0 0 t 2 .
2
AT
t t = a + b 0 2t 2 3t 2 0
2
a+b
Ta c T = ab + 3 a + b 2 ( a + b ) + 1 ( a + b 1)
+ 3 a + b (a + b) +1
2
3
T t 2 + 3 t + 1 = f (t ), t [ 0; 2]
4
3
3
3 t t 1
Ta c f '(t ) = t +
= .
2 2 t
2
t
f '(t ) = 0 t = 1
13
f (0) = 1; f (1) = ; f (2) = 3 2 2
4
13
1
t =1 a = b = .
T : MaxT =
t[ 0;2]
4
2
2
ww
w.
Lu :
0,25
0,25
0,25
0,25
-------------------- Ht --------------------
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4/4
Mn thi: Ton - Ln th 2
Nm hc 2014 - 2015
thng (d ) bng
co
I=
sin 2 x
dx.
HV
N.
Cu 2 (1,0 im).
0 ( sin x + 2 )
Cu 4 (1,0 im).
a) Gi z1 , z2 l hai nghim phc ca phng trnh z 2 4 z + 9 = 0 ; M , N ln lt l cc im biu din
z1 , z2 trn mt phng phc. Tnh di on thng MN .
b) Mt t c 7 hc sinh (trong c 3 hc sinh n v 4 hc sinh nam). Xp ngu nhin 7 hc sinh
Cu 5 (1,0 im).
AT
Cu 6 (1,0 im).
Cho hnh lng tr ABC. A ' B ' C ' c y ABC l tam gic vung ti B ;
im ca ( P ) v ( S ) .
AB = a,
ACB = 300 ; M l trung im cnh AC . Gc gia cnh bn v mt y ca lng tr bng 600 .
Hnh chiu vung gc ca nh A ' ln mt phng ( ABC ) l trung im H ca BM . Tnh theo a th tch
w.
khi lng tr ABC. A ' B ' C ' v khong cch t im C ' n mt phng ( BMB ').
Cu 7 (1,0 im). Trong mt phng ta Oxy, cho hnh thang ABCD vung ti A v D ; din tch
hnh thang bng 6; CD = 2 AB , B(0; 4) . Bit im I (3; 1), K (2; 2) ln lt nm trn ng thng AD v
ww
x + x( x 2 3 x + 3) = 3 y + 2 + y + 3 + 1
Cu 8 (1,0 im). Gii h phng trnh
3 x 1 x 2 6 x + 6 = 3 y + 2 + 1
Cu 9 (1,0 im). Cho cc s thc x, y dng v tha mn x y + 1 0 .
Tm gi tr ln nht ca biu thc
T=
x + 3y2
2
2x + y2
5x + 5 y 2
( x, y ).
x +y
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Th sinh khng c s dng
ti liu. Cn b coi
thi khng gii thch g thm.
H v tn th sinh: ..; S bo danh:
p n thang im
thi th thpt quc gia nm 2015
Mn thi: Ton Ln th 2
Nm hc 2014 2015
x +
HV
+
y'
+
0
0
+
y
2
+
im
0,25
N.
co
Cu
p n
1
a) (1,0 im) Kho st s bin thin v v th ca hm s y = x 3 3 x 2 + 2
(2,0)
Tp xc nh: D = R . lim y = ; lim y = +
0,25
0,25
-2
0,25
2.
0,25
0,25
2m m = 1 m = 1(ktm)
p s: m = 1 .
0,25
AT
0,25
w.
2
a) (0,5 im) Gii phng trnh sin x ( 2sin x + 1) = cos x 2 cos x + 3 .
(1,0)
Phng trnh cho tng ng vi
1
3
sin x
cos x = cos 2 x
2
2
0,25
ww
sin x = sin 2 x .
3
5
2
+k
,(k ) .
3 2
18
3
5
x = + 2 x + k 2 x =
+ k 2 , ( k ) .
3 2
6
5
2
5
+k
,x =
+ k 2 , k .
Vy phng trnh cho c nghim: x =
18
3
6
2 x + k 2 x =
0,25
1/4
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t = 9
t = 3(l )
p s: x = 2 .
co
Vi t = 9 3x = 9 x = 2 (tmk).
I =
( sin x + 2 )
dx.
I=
0
sin 2 x
sin 2 x
( sin x + 2 )
dx =
0
2sin x cos x
( sin x + 2 )
dx.
I = 2
0
tdt
(t + 2)
= 2
0
t +22
(t + 2)
x=
t = 1.
HV
N.
3
(1,0)
27
27
. t t = 3x > 0 t 6 =
t 2 6t 27 = 0
x
t
3
3x 6 = 33 x 3x 6 =
1
1 1
I = 2 ln(t + 2) + 4
0
t+2 0
0,25
( I 0.144) .
0,25
4
a) (0,5 im) Cho z 2 4 z + 9 = 0 . M, N biu din z1 , z2 . Tnh di on MN.
(1,0)
Phng trnh cho c ' = 4 9 = 5 = 5i 2 nn c hai nghim z1,2 = 2 i 5 .
p s: MN = 2 5 .
b) (0,5 im) Tnh xc sut c 3 hc sinh n cnh nhau.
Gi A l bin c 3 hc sinh n cnh nhau
+ S bin c ng kh nng: Xp 7 hc sinh ngu nhin, c s hon v l 7!
+ S cch xp c 3 hc sinh n cnh nhau:
Coi 3 hc sinh n l 1 phn t, kt hp vi 4 hc sinh nam suy ra c 5 phn t, c 5! cch sp xp.
Vi mi cch sp xp li c 3! cch hon v 3 hc sinh n. Vy c 5!.3! cch sp xp.
w.
0,25
0,25
AT
3 2
1 1
I = 2(ln 3 ln 2) + 4 = 2 ln .
2 3
3 2
5!.3! 1
= .
7
7!
0,25
0,25
0,25
( p ( A) 0.14) .
ww
5
(1,0)
0,25
dt
dt
.
4
2
t+2
0
0 (t + 2)
dt = 2
+ Xc sut ca bin c A l: p ( A ) =
0,25
| 3 + 12 + 14 11|
=6.
3
Phng trnh mt cu ( S ) : ( x 3)2 + ( y 6) 2 + ( z 7) 2 = 36 .
Mt cu ( S ) tm I c bn knh R = d ( I , ( P )) =
x = 3 + t
0,25
0,25
0,25
(t R) .
0,25
2/4
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Gi s M = (d ) ( P) (3 + t ) + (12 + 4t ) + (14 + 4t ) 11 = 0 9t + 18 = 0 t = 2
M (1; 2;3) .
0,25
6
Cho hnh lng tr ABC. A ' B ' C ' c y ABC l tam gic vung ti B ; AB = a,
ACB = 300 ;
(1,0)
A ' H ( ABC ) A ' H l ng cao ca hnh lng tr.
C'
A'
0
0,25
AH l hnh chiu vung gc ca AA ' ln ( ABC ) A ' AH = 60
VABC . A ' BC ' = A ' H .S ABC
B'
1
1
a2 3
.
S ABC = .BA.BC = .a.a 3 =
2
2
2
3a a 2 3 3a 3 3
.
VABC . A ' BC ' = .
=
4
2
2
co
a 3
3a
.
A' H =
2
2
N.
AC = 2a, MA = MB = AB = a AH =
A'
HV
Suy ra
1
1
a2 3
.
BB '.BM = .a 3.a =
2
2
2
1
a 3
.
VA.BMB ' = VB '. ABM = VABC . A ' BC ' =
6
8
Do BM ( AHA ') nn BM AA ' BM BB ' BMB ' vung ti B
S BMB ' =
0,25
3a 3 3 a 2 3 3a
d ( C ',( BMB ') ) =
:
=
.
4
8
2
C'
0,25
B'
0,25
A
M
H
AT
B
a 3
3a
0
(Cch 2: d ( A, ( BMB ')) = AE = AH .sin AHE =
.sin 60 =
).
2
4
7
Trong mt phng ta Oxy , cho hnh thang ABCD vung ti A v D ; din tch hnh
(1,0) thang bng 6; CD = 2 AB , B(0; 4) . I (3; 1), K (2; 2) . Vit phng trnh ng thng AD.
E
0,25
0,25
K
b = 1
| 3 + 5b| |b + 1|
5
2
.
=63
.
= 6 | 5b 3 | . | b + 1|= 2(b + 1) b =
3
b 2 + 1 b2 + 1
1 2 2
b =
7
ww
S ABCD
w.
p s: x + y 2 = 0;3 x 5 y 14 = 0;7 x (1 + 2 2) y 2 2 22 = 0; 7 x (1 2 2) y + 2 2 22 = 0 .
8
x +
(1,0) Gii h phng trnh
x( x 2 3x + 3) =
y + 2 + y + 3 +1
3 x 1 x 6 x + 6 = y + 2 + 1
2
(1)
0,25
0,25
( x, y ).
(2)
3/4
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iu kin: 1 x 3 3; x 3 + 3; y 3
y+2 +
y+2
0,25
+1
Xt hm f (t ) = t + t + 1, t 1 . Ta c f '(t ) = 1 +
3t 2
2 t3 +1
(1) x 1 + ( x 1)3 + 1 =
co
t 1 , suy ra x 1 = 3 y + 2 .
0,25
t t = x 1 +
0,25
t 3
1
5
> 2 t + t2 6 = 3 t2 6 = 3 t 2
t= .
2
2
x 1
t 6 = (3 t )
N.
x 1 +
HV
x 1 = 2
x = 5 y = 62
5
1
5
Vi t = x 1 +
=
.
x 1 = 1
x = 5 y = 127
2
x 1 2
4
64
2
5 127
).
p s ( x; y ) = (5; 62), ( ;
4
64
0,25
9
x + 3 y2
2x + y2
.
x2 + y4
5x + 5 y
x 1 1 1 1 1 1
x
1
Ta c x y 1 0 < 2 2 = . t t = 2 0 < t
y
y y
4 y 2
4
y
4
AT
x
+3
y2
0,25
x
+1
1 y2
t +3
1 2t + 1
1
Ta c T =
vi 0 < t .
.
T = f (t ) =
.
2
2
4
5 x +1
t +1 5 t +1
x
2
+
1
y
y2
1 3t
1
1
.
2
3
( t 2 + 1) 5 ( t + 1)
1
1
1 3t ;
4
4
w.
f '(t ) =
(t
0,25
3
3
17 17 17
+ 1) =
16 16 16
1 3t
(t
+ 1)
4
17
17
16
1
1
1
V .
> . Do f '(t ) >
2
5 (t + 1)
5
ww
4
1
> 0.
17 5
17
16
1
6
1 13
T f (t ) ng bin t (0; ] f (t ) f =
.
4
17 25
4
=
p s: MaxT
1
t(0; ]
4
13
6
1
t = x = 1; y = 2 .
4
17 25
0,25
0,25
-------------------- Ht --------------------
4/4
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Nm hc 2014 - 2015
3x 2
.
x 1
a) Kho st s bin thin v v th ( C ) ca hm s cho.
Cu 1 (2,0 im). Cho hm s y =
co
2+i
b) Cho s phc z tha mn h thc: (i + 3) z +
= (2 i ) z . Tm mun ca s phc w = z i .
i
N.
HV
AT
Cu 8 (1,0 im).
x +1 y 1 z
=
=
. Lp phng trnh mt phng ( P ) cha A v d . Tm ta im B thuc trc Ox
2
1
3
sao cho khong cch t im B n mt phng ( P ) bng 3 .
w.
d:
ww
li mi hc sinh l nh nhau, nh trng quyt nh bc thm ngu nhin 30 hc sinh t 300 hc sinh ni
trn. Tm xc sut trong s 30 hc sinh chn trn c ng 90% s hc sinh t yu cu vo lp 6A.
.
1+ a 1+ b
2a(1 + a) + 2b(1 + b) + 8
---------------- HT ---------------Th sinh khng c s dng ti liu. Cn b coi thi khng gii thch g thm.
H v tn th sinh: ..;
S bo danh:
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p n thang im
thi th thpt quc gia nm 2015
Mn thi: Ton Ln th 3
Nm hc 2014 2015
co
Cu
p n
1
3x 2
.
(2,0) a) (1,0 im) Kho st s bin thin v v th ca hm s y =
im
x 1
x +
x 1
o hm: y ' =
( x 1)
< 0 x 1 .
y'
y
3
0,25
0,25
HV
N.
lim y = +; lim
y = suy ra tim cn ng ca th hm s l ng thng x = 1 .
x 1+
0,25
AT
0,25
3x 2
= x + m ( x 1)
x 1
f ( x ) = x 2 + (2 m) x + m 2 = 0 (1)
> 0
K: (1) c 2 nghim phn bit khc 1
f (1) 0
0,25
0,25
m 2 4m 12 > 0
m > 6; m < 2 .
0,25
0,25
w.
2
3
ww
2
1
1
1
3
Ta c
= 1 + tan 2 = 1 + 4 = 5 cos 2 = cos =
< x <
.
2
cos
5
2
5
1 1
.
M = sin 2 + cos + cos 2 = sin 2 + cos + 2cos 2 1 = cos 2 + cos =
5
5
b) (0,5 im) Cho (i + 3) z +
0,25
0,25
2+i
= (2 i ) z . Tm mun ca s phc w = z i .
i
1/6
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( a , b R, i
Gi z = a + ib
= 1) . T gi thit ta c:
a = 1
a + 1 = 0
4
(a + 1) + (2a + 5b 2)i = 0
4 z = 1 + i.
5
2a + 5b 2 = 0
b = 5
26
1
.
=
5
25
Gii bt phng trnh: log 2 ( x 2) + log 0,5 x < 1 .
1
5
iu kin: x > 2 .
Bpt log 2 ( x 2 ) log 2 x < 1 log 2
x 2 < 2 x x > 2 .
x2
x2
<1
<2
x
x
N.
3
(0,5)
co
T : | z i |=| 1 i |= 1 +
Bpt x x 2 >
2
x ( x 2 ) + 1
( x 2)
( x + 1)
AT
Xt hm f (t ) = t + 1 + t , t > 0 f '(t ) = 1 +
1
1
.
>
x x2
1+ t2
w.
1
1
1
1
.
Chia 2 v cho x .( x 2) < 0 ta c: (1)
1+ <
1+
2
x x2
x
x
2
(
)
ww
0,25
Xt hm f (t ) = t 1 + t 2 , t R f '(t ) = 1
T (1)
0,25
(1)
0,25
(loi).
2
x > 2 : (1) ( x 2) 1 + x + 1 > x 1 + ( x 2 ) + 1
1
1
1
1
Chia 2 v cho x .( x 2) > 0 ta c: (1)
.
+ 1+ >
+ 1+
2
x x2
x
( x 2)
0,25
0,25
0,25
( x 0) .
0,25
x3 4 x 2 + 5 x x 3 3 x 2 + 4 .
HV
4
(1,0) Gii bt phng trnh: x x 2 >
(i + 3)(a + bi ) + 1 2i = (2 i )(a bi )
t
1+ t2
1+ t2 t
1+ t2
0,25
> 0 t f (t ) ng bin t .
1
1
1
. Trng hp ny v nghim v
<
< 0.
x2
x x2
p s: x > 4 .
2/6
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)(
x +1 >
x2 5x + 4
x 3 4 x 2 + 5 x + x3 3 x 2 + 4
x 1
> 0.
x3 4 x 2 + 5 x + x3 3x 2 + 4
x 1
x +1
f ( x) = ( x 4 )
+
x
+
2
x +1
+ Xt g ( x ) =
+
x +2
x3 4 x 2 + 5 x + x 3 3 x 2 + 4
Nu x 1 th g ( x ) > 0 .
x3 3x 2 + 4 =
x + 1 > 1 . Ta c:
( x + 1)( x 2 )
x +1
x +1 1
>
=
x +2 2 x +2 2
(1)
N.
x 2
co
+ (1)
= x 2 x +1 > x 2 = 2 x
AT
5
(1,0)
HV
x3 4 x 2 + 5 x + x3 3x 2 + 4 > 2 x
1 x
1 x
1 x
1 x
1
<
=
<
=
x3 4 x 2 + 5 x + x3 3x 2 + 4 2 x 2 2 x + x 2 2 x 2
x 1
1
>
(2) . T (1) v (2) suy ra g ( x ) > 0 x > 0 .
2
x3 4 x 2 + 5 x + x3 3x 2 + 4
+ f ( x) > 0 x 4 > 0 x > 4 . Kt hp K suy ra p s: x > 4 .
2
1 3
.
I = x 2 dx + x cos 2 xdx . Ta c A = x 2 dx = x3 2 =
0
3
24
0
0
0
2
1
B = x cos 2 xdx. t u = x u ' = 1. v ' = cos 2 x v = sin 2 x .
2
0
0,25
1
12
x sin 2 x 02 sin 2 xdx .
2
20
w.
B=
0,25
1 1
1
2 1
= 0 cos 2 x = ( 1 1) =
2 2
2
0 4
1
. ( I 0,792) .
24 2
ww
I = A+ B =
6
(1,0)
0,25
0,25
0,25
3/6
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1
1
VS .MCD = .SA.S MCD . SA = AC = a 2; S MCD = a 2 .
3
2
3
a 2
1
1
Suy ra VS .MCD = .a 2. a 2 =
.
6
3
2
Gi N l trung im AB BD //( SMN ) .
0,25
Suy ra:
0,25
HV
N.
0,25
co
7
Tam gic ABC c phn gic trong gc A l d : x + y 3 = 0 . Hnh chiu ca tm ng trn ni
(1,0)
tip tam gic ABC ln AC l E (1; 4) . BC c h s gc m v to vi ng thng AC gc 450 .
ng thng AB tip xc vi (C ) : ( x + 2 ) + y 2 = 5 . Tm phng trnh cc cnh.
2
AT
1 10
E , AC : x 2 y + 7 = 0 d = J ; .
3 3
2
Gi vtpt ca ng thng BC l n = ( a; b), a + b 2 0 . Ta c:
| 2a + b |
cos 450 =
5. a 2 + b 2
2 ( 2a + b ) = 5 ( a + b
2
) 3a
w.
0,25
E
H
+ 8ab 3b = 0
2
a = 0 : suy ra b = 0 (loi)
a 0 : chn a = 1 b = 3 (tha mn h s gc m),
1
b = (loi).
3
Suy ra phng trnh BC : x + 3 y + C = 0 .
0,25
F
J
C
D
ww
0,25
4/6
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x +1 y 1 z
=
=
. Lp ( P) cha A v d . Tm B Ox : d ( B, Ox ) = 3 .
2
1
3
ng thng d qua M ( 1;1;0 ) v c vtcp u = (2;1; 3) . Ta c MA = (2; 2;0) .
( P) qua A (1; 1;0 ) v c vtpt n = MA, u = ( 6;6;6 ) . Chn n = (1;1;1) .
Phng trnh tng qut ca ( P) l: 1( x 1) + 1( y + 1) + 1( z 0) = 0 x + y + z = 0.
|b|
Gi B(b;0;0) Ox; d ( B, ( P )) = 3
= 3.
3
| b |= 3 b = 3 B(3;0;0) .
p s: ( P ) : x + y + z = 0 ; B(3;0;0) .
A (1; 1;0 ) , d :
HV
N.
8
(1,0)
0,25
co
2 10
1
| + 6 | | + 10 + C |
29 + 10 2
29 10 2
3 3
Suy ra
(tha mn); C =
(loi v khi
= 3
C =
3
3
5
10
29 + 10 2
A, J nm 2 pha BC ). T : BC : x + 3 y
= 0.
3
29 + 10 2
= 0.
p s: AB : x + 2 y 3 = 0 ; AC : 2 x + y 6 = 0 ; BC : x + 3 y
3
0,25
0,25
0,25
0,25
AT
0,25
27
3
S cch chn hc sinh t yu cu l: C50
. C250
.
3
C5027 .C250
1,6.1021 .
30
C300
Xc sut ca bin c A l P ( A) =
10
(1,0) Cho a, b > 0 : ab 1 . Tm GTNN ca T =
1
1
2
+
,
1 + a 1 + b 1 + ab
w.
Ta c:
1
1
32
.
+
1+ a 1+ b
2a (1 + a ) + 2b(1 + b) + 8
( ab 1) .
1 + ab
ww
Li c:
0,25
a b
)(
2
ab 1 0 (ng).
0,25
1
1
4
2
2
4
. Suy ra:
.
+
=
1 + a 1 + b ab + 3
1 + ab.1 1 + ab + 1 ab + 3
2
5/6
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Ta c: a (1 + a ) + b(1 + b) = a 2 + b 2 2 + ( a + b + 2 ) ( 2ab 2 ) + 2 ab + 2 2 ab + 2 .
Suy ra: 2a (1 + a ) + 2b(1 + b) + 8 4 ab + 12 .
16
1
1
32
32
=
2a(1 + a) + 2b(1 + b) + 8
2a(1 + a) + 2b(1 + b) + 8 2 ab + 3
4 ab + 12
4
16
T
.
ab + 3
ab + 3
N.
.
0,25
co
4
16
= f (t ).
t +3
t +3
8t
8
(t 2 + 3) 2 t (t + 3) t + 3
f '(t ) = 2
+
= 8. 2
.
(t + 3) 2 (t + 3) t + 3
(t + 3) 2 (t + 3) t + 3
t t = ab 1 T
ab + 3
0,25
0,25
t 1
HV
Xt M = (t 2 + 3) 2 t (t + 3) t + 3 > (t + 3) t 2 + 3 t t + 3 > 0
1
1
4
4
+
=
1+ a 1+ b 1+ a +1+ b u + 2
a(1 + a) + b(1 + b) = a + b + a 2 + b 2 a + b + 2 a 2b 2 a + b + 2 = u + 2
1
1
Suy ra: 2a (1 + a ) + 2b(1 + b) + 8 2u + 12
2a(1 + a) + 2b(1 + b) + 8
2u + 12
4
32
T
u2
AT
ww
w.
-------------------- Ht --------------------
6/6
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Mn thi: Ton - Ln th 4
Thi gian lm bi: 180 pht, khng k thi gian pht
Nm hc 2014 - 2015
co
I = x e3 x +
dx.
3x + 1
0
HV
N.
Cu 2 (1,0 im).
a) Gii phng trnh cos x + cos3 x = 2cos 2 x .
b) Tm s phc z sao cho | z 4 | = | z | v ( z + 4)( z + 2i ) l s thc.
AT
Cu 6 (1,0 im). Trong khng gian vi h ta Oxyz , cho bn im A ( 3;2;3) , B (1;0;2), C ( 2;3;4),
D (4; 3;3) . Lp phng trnh mt phng ( BCD) . Tm phng trnh hnh chiu vung gc ca ng
thng AB ln mt phng ( BCD) .
Cu 7 (1,0 im). Cho hnh lng tr ABC. A ' B ' C ' c y ABC l tam gic u cnh a , nh A ' cch
u A, B, C . Gc gia cnh bn v mt y ca lng tr bng 600 . Tnh theo a th tch khi lng tr
ABC. A ' B ' C ' . Xc nh tm v tnh theo a bn knh mt cu ngoi tip hnh chp A '. ABC .
w.
Cu 8 (1,0 im). Trong mt phng ta Oxy, cho tam gic ABC ni tip ng trn tm I (2;1) ,
bn knh R = 5 . Chn ng cao h t B, C , A ca tam gic ABC ln lt l D(4; 2), E (1; 2) v F .
Tm ta tm ng trn ni tip ca tam gic DEF , bit rng im A c tung dng.
ww
T=
y + 3x( x + 1)
x2 z
16
( y + 1)
10 3
y
x3 + 2
---------------- HT ---------------Th sinh khng c s dng ti liu. Cn b coi thi khng gii thch g thm.
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p n thang im
thi th thpt quc gia nm 2015
Mn thi: Ton Ln th 4
Nm hc 2014 2015
x +
HV
+
y'
+
0
0
+
y
4
+
N.
co
Cu
p n
1
a) (1,0 im) Kho st s bin thin v v th ca hm s y = x 3 3 x 2 + 4
(2,0)
Tp xc nh: D = R . lim y = ; lim y = +
0,25
0,25
0,25
AT
x3 3 x 2 + 4 = 5 x + 7 x 3 3 x 2 + 5 x 3 = 0 ( x 1)( x 2 2 x + 3) = 0
x = 1 y = 2 giao im l M (1;2) .
Phng trnh tip tuyn vi (C) ti ( x0 ; y0 ) : y = y '( x0 )( x x0 ) + y0
x0 = 1; y0 = 2
im
0,25
0,25
0,25
0,25
0,25
2
a) (0,5 im) Tm cc nghim ca phng trnh cos x + cos3 x = 2cos x .
(1,0)
Phng trnh cho tng ng vi: 2cos 2 x.cos x = 2cos 2 x 2cos x ( cos 2 x cos x ) = 0
2
ww
w.
cos x = 0
cos 2 x = cos x
x = 2 + k
( k ) .
x = k 2
3
b) (0,5 im) Tm s phc z sao cho | z 4 | = | z | v ( z + 4)( z + 2i ) l s thc.
Gi z = a + bi
( a, b R, i
= 1) . T gi thit ta c:
| z 4 |=| z | ( a 4) 2 + b 2 = a 2 + b 2 a = 2
0,25
0,25
0,25
1/5
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T : z = 2 + bi; z = 2 bi
0,25
3
(0,5)
9
3
TX: D = R . Chia 2 v ca phng trnh cho 4 > 0 ta c: 2. 5. + 3 = 0 .
4
2
x
3
t t = > 0 ta c: 2t 2 5t + 3 = 0
2
3
t = 1; t = .
2
x
x
3 3
3
3
t =1 =1 x = 0.
t = = x = 1.
2 2
2
2
Tp nghim ca phng trnh cho l S = {0; 1} .
4
1
(1,0) Tnh tch phn: I = x e3 x +
Tnh A = xe dx : t u = x u ' = 1;
1
1
1
1
v ' = e v = e3 x A = x.e3 x e3 x dx .
3
3
30
0
1 3 1 3x
1 3 1 3
2e3 + 1
= e ( e 1) =
.
A= e e
3
9
3
9
9
0
2
x = 0 t = 1. Suy ra: B = ( t 2 1) dt
91
x = 1 t = 2;
2
0,25
0,25
t2 1
2
x
dx : t t = 3x + 1 t 2 = 3 x + 1 x =
dx = t.dt
3
3
3x + 1
Tnh B =
0,25
3x
AT
3x
0,25
1
dx.
3x + 1
x
dx .
3x + 1
I = xe3 x dx +
0
HV
N.
co
w.
2
11
2e 3 + 1 8
21
8
.
B = t 3 t = . T : I = A + B =
+
I = e3 +
9
27
93
9
27
1 27
0,25
0,25
5
C 10 i bng (trong c Vit Nam v Thi Lan). Bc thm ngu nhin chia thnh 2
(0,5)
ww
0,25
0,25
4
n( M ) 112
=
= .
n() 252 9
2/5
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6
(1,0) Cho A ( 3;2;3) , B(1;0;2), C (2;3; 4), D(4; 3;3) . Lp phng trnh mt phng ( BCD) . Tm
0,25
co
0,25
HV
N.
0,25
0,25
x 1 y 0 z 2
th khng cho 0,25 im phn cui ny).
=
=
0
0
1
7
Lng tr ABC. A ' B ' C ' c y ABC l tam gic u cnh a , nh A ' cch u A, B, C .
(1,0)
AT
B'
w.
1 a 3 a2 3
ABC u cnh a nn S ABC = .a.
=
.
2
2
4
2 a 3
A ' H = AH .tan 600 = .
. 3 = a.
3 2
Suy ra: VABC . A ' B ' C '
a3 3
a2 3
.
= a.
=
4
4
P
I
0,25
A
H
M
C
N
ww
B
Xc nh tm mt cu:
Gi P l trung im AA. K ng trung trc d ca AA trong (AAH). d ct AH ti I.
I d IA ' = IA. I A ' H IA = IB = IC I l tm mt cu cn tm.
0,25
A'
2a
A' P
2 1
1
2 a 3
.
=
. AA ' =
.2. AH =
.
=
0
3
cos 30
3 2
3
3 3
0,25
0,25
3/5
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8
Cho tam gic ABC ni tip ng trn tm I (2;1) , bn knh R = 5 . Chn ng cao h t
(1,0)
co
= BCD
.
T gic BCDE ni tip ng trn nn AED
= EAt
K tip tuyn At ca ( I ; R ) ta c: BCD
= EAt
At / / DE AI DE.
AED
Tm ta im A:
Phng trnh AI qua I, vung gc vi DE : 3 x + 4 y 10 = 0 .
0,25
D
N.
C
F
t = 6 A(6; 2) (L)B
10 3t
2
2
AI
AI
=
25
4
t
12
=
0
t = 2 A(2; 4) (TM)
4
Chng minh trc tm H ca tam gic ABC l tm ng trn ni tip tam gic DEF :
= DBC
= HEF
EC l phn gic trong ca DEF
.
DEC
H = BD CE l tm ng trn ni tip DEF .
Tng t: DB l phn gic trong ca EFD
Tm ta im H :
Phng trnh CE qua E v vung gc vi AE : x 2 y 5 = 0
Phng trnh BD qua D v vung gc vi AD : 3 x y 10 = 0
T H = BD CE H ( 3; 1) .
HV
A t;
0,25
0,25
0,25
10
(1) 4 ( 2 x 2 + x 1) +
4 ( 2 x + x 1)
AT
9
Gii phng trnh 8 x 2 + 10 x + 11 + 14 x + 18 = 11 . (1)
(1,0)
11
K: x .
) (
10 x + 11 2 x 3 +
2 ( 2 x 2 + x 1)
14 x + 18 2 x 4 = 0
2 ( 2 x 2 + x 1)
=0
10 x + 11 + 2 x + 3
14 x + 18 + 2 x + 4
1
2 x 2 + x 1 = 0 x = 1; (tmk).
2
1
1
f ( x) = 2
=0
10 x + 11 + 2 x + 3
14 x + 18 + 2 x + 4
11
11
Ta c f '( x ) > 0 x f ( x) ng bin trn [ ; +) .
10
10
1
11
T f ( x) f > 0 nn trng hp ny v nghim. p s: S = 1; .
2
10
0,25
0,25
w.
0,25
ww
11
4
9
11
11
x 10 x + 11 + 2 x + 3 2 + 3 = ; 14 x + 18 + 2 x + 4 2 + 4 =
10
5
5
10
10
5 5
f ( x) > 2 > 0 .
4 9
C th nhm nghim v tch thnh tch: (1) ( x + 1)(2 x 1)h( x) = 0 ri CM h( x) v nghim.
0,25
4/5
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10
2
(1,0) Cho cc s thc x, y , z dng v tha mn 4 x 2 x + 1 16 x 2 yz 3 x ( y + z ) .
T=
y + 3 x( x + 1)
2
x z
16
( y + 1)
10 3
T gi thit ta c:
x +2
Ta c: yz 1
x 2 yz
y 2 + 3xy
x2
16
y
y
= +3
x
x
3
16
16
+
3y +
= ( y + 1) + ( y + 1) + ( y + 1) +
3 4.2 3 = 5
3
3
z ( y + 1)
( y + 1)
( y + 1)3
10 3.
y
x3 + 2
10 3.
HV
y 2 + 3xy
x3 + 1 + 1
10 3.
0,25
N.
co
1
2
2
4 x 2 x + 1 16 x yz 3 x ( y + z ) 4 x + 1 16 yz 3 ( y + z ) 16 yz 3.4 yz
x
1
1
4 yz 3 yz x + 1 1, t = yz > 0 3t 2 4t + 1 0 t 1 yz 1 y.
x
z
2
y + 3x( x + 1)
16
y
y + 3xy 3
16
y
T=
+
10 3 3
=
+ +
10 3 3
.
2
3
2
3
z ( y + 1)
x z
x +2
x yz
x +2
( y + 1)
0,25
y
y
= 10
3x
x
AT
y
y
y
T : T + 3 + 5 10 .
x
x
x
y
t t =
> 0 T f (t ) = t 4 + 3t 2 10t + 5 .
x
Ta c: f '(t ) = 4t 3 + 6t 10 = 2(t 1)(2t 2 + 2t + 5) .
f '(t ) = 0 t = 1 .
BBT:
0
1
0
t
f '(t )
f (t )
0,25
-1
w.
Suy ra T 1 MinT = 1 t = 1 x = y = z = 1.
t >0
Cch 2:
y
=
x +2
3
ww
Ta c:
y
y
1
=
x +1+1
3x
3
3
y
+1
y
1 x
.
.1
.
x
3 2
y2
y2
y
y2
y
2 + 1 2 2 = 2 2 2 1 .
x
x
x
x
x
y
+1
1 x
y
y
Suy ra: T 2 1 + 3 + 5 10 3.
.
T 1 MinT = 1 x = y = z = 1.
x
x
3 2
0,25
-------------------- Ht --------------------
5/5
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