thi th thpt quc gia nm 2015

Trng thpt lng th vinh

H ni

Mn thi: Ton - Ln th 1

Nm hc 2014 - 2015

Cu 1 (2,0 im). Cho hm s y = x 4 + ( m 3) x 2 + 2 m

Thi gian lm bi: 180 pht, khng k thi gian pht

-------------- Ngy 8.2.2015 --------------

(1), vi m l tham s thc.

a) Kho st s bin thin v v th ca hm s (1) khi m = 1 .

N.
co

b) Tm m th hm s (1) ct trc honh ti bn im phn bit c honh nh hn 2.

Cu 2 (1,0 im).

a) Gii phng trnh 3cos 2 x + sin x 1 = cos x + sin 2 x sin 2 x .

1
b) Gii phng trnh log 27 x 3 + log 3 ( x + 2) = 1 + log 3 ( 4 3 x ) .
2
e

Cu 3 (1,0 im). Tnh tch phn

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 +

qu (5 qu tt v 3 qu hng). Ly ngu nhin mi thng mt qu. Tnh xc sut hai qu ly c c

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

phng ( P ) : x 2 y + 2 z 5 = 0 . Tm ta giao im ca ng thng AB v mt phng ( P ) . Lp

phng trnh mt phng (Q) cha ng thng AB v vung gc vi mt phng ( P ).

Cu 6 (1,0 im). Cho hnh chp S . ABCD c y l hnh ch nht, AB = a, AD = 2a . Tam gic SAB

cn ti S v nm trong mt phng vung gc vi y. Gc gia ng thng SC v mt phng

( ABCD ) bng 450 . Gi M l trung im ca SD . Tnh theo a th tch ca khi chp S . ABCD v

w.

khong cch t im M n mt phng ( SAC ) .

Cu 7 (1,0 im). Trong mt phng ta Oxy , cho hnh ch nht ABCD c din tch bng 15. ng

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 ).

Cu 9 (1,0 im). Cho cc s thc a, b khng m v tha mn: 3 ( a + b ) + 2 ( ab + 1) 5 a 2 + b 2 .

Tm gi tr ln nht ca biu thc

T = 3 a + b 3 a 2 + b 2 + 2( a + b) ab .

---------------- Ht ---------------www.MATHVN.com
- www.DeThiThuDaiHoc.com
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:

Trng thpt lng th vinh

H ni

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 +

o hm: y ' = 4 x 4 x ; y ' = 0 x = 0 hoc x = 1 .

3

Cc khong ng bin: ( 1;0 ) ; (1; + ) . Khong nghch bin: ( ; 1) ; ( 0;1)

HV

Cc tr: Hm s t cc tiu ti x = 1 , yCT = 0 ; t cc i ti x = 0 , yC = 1.

Bng bin thin:
x

-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.

( 2cos x 1)( cos x sin x ) = 0

cos x sin x = 0 tan x = 1 x =

+ k , x =

b) (0,5 im) Gii phng trnh log 27 x 3 +

0,25

0,25

0,25

0,25
0,25
0,25

0,25

+ k , ( k )

x = + k 2 , k
2
3

Vy phng trnh cho c nghim: x =

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 )

iu kin: 0 < x <

www.MATHVN.com - www.DeThiThuDaiHoc.com

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

b) (0,5 im) Tnh xc sut c t nht 1 qu tt

Gi A l bin c C t nht 1 qu tt, suy ra A l bin c: C 2 qu u hng

S bin c ng kh nng: 10.8 = 80
S cch chn 2 qu hng: C41 .C31 = 4.3 = 12

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

Phng trnh tham s ca AB l y = 1 + t

z = 2 6t

(t R ) .

www.MATHVN.com - www.DeThiThuDaiHoc.com

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)

Phng trnh mt phng (Q) : 2( x 1) + 2( y + 1) + 1( z 2) = 0 2 x + 2 y + z 2 = 0 .

Cho hnh chp S . ABCD c y l hnh ch nht, AB = a, AD = 2a ...
I ( P) (1 + 2t ) 2(1 + 6t ) + 2(2 6t ) 5 = 0 t =

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) .

www.MATHVN.com - www.DeThiThuDaiHoc.com

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 --------------------

- Hc sinh lm theo cch khc, nu ng vn cho im ti a.

- Hc sinh trnh by khc, song vn , khng c du hiu lm tt th khng tr im.

www.MATHVN.com - www.DeThiThuDaiHoc.com

4/4

 thi th thpt quc gia nm 2015

Trng thpt lng th vinh

H ni

Mn thi: Ton - Ln th 2

Nm hc 2014 - 2015

Cu 1 (2,0 im). Cho cc hm s y = x 3 3mx 2 + 2 ( Cm ),

Thi gian lm bi: 180 pht, khng k thi gian pht

-------------- Ngy 29.3.2015 -------------y = x + 2 (d ) , vi m l tham s thc.

a) Kho st s bin thin v v th ca hm s ( Cm ) khi m = 1.

thng (d ) bng

co

b) Tm cc gi tr ca m ( Cm ) c hai im cc tr v khong cch t im cc tiu ca ( Cm ) n ng

2.

a) Gii phng trnh sin x ( 2sin x + 1) = cos x 2cos x + 3 .

b) Gii phng trnh log 3 ( 3x 6 ) = 3 x .

I=

sin 2 x

dx.

HV

Cu 3 (1,0 im). Tnh tch phn

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

thnh mt hng ngang. Tm xc sut 3 hc sinh n ng cnh nhau.

Trong khng gian vi h ta Oxyz , cho im I (3;6;7) v mt phng

( P) : x + 2 y + 2 z 11 = 0 . Lp phng trnh mt cu ( S ) tm I v tip xc vi ( P ). Tm ta tip

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

DC . Vit phng trnh ng thng AD bit AD khng song song vi cc trc ta .

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
---------------- HT ---------------www.MATHVN.com
- www.DeThiThuDaiHoc.com
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

Trng thpt lng th vinh

H ni

Mn thi: Ton Ln th 2

--------------- p n c 04 trang --------------

Nm hc 2014 2015

x +

o hm: y ' = 3 x 6 x ; y ' = 0 x = 0 hoc x = 2 .

2

HV

Cc tr: Hm s t cc tiu ti x = 2 , yCT = 2 ;

t cc i ti x = 0 , yC = 2.
Bng bin thin:
x
0
2

+
y'
+
0
0
+
y
2
+

im

0,25

N.

Khong ng bin: ( ;0 ) ; ( 2; + ) . Khong nghch bin: ( 0; 2 )

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

th: (Hs c th ly thm im (1; 2); (1; 0); (3; 2) ).

0,25

b) (1,0 im) Tm cc gi tr ca m ( Cm ) c k/c im cc tiu ca ( Cm ) n (d ) bng

2.
0,25

Ta hai im cc tr: A(0; 2) v B(2m; 2 4m3 ) .

0,25

m < 0 : A l im cc tiu. Khi d ( A, d ) = 0 2 (loi).

m > 0 : B l im cc tiu. Khi :
2m3 m = 1
m = 1(tm)
d ( B, d ) = 2 | 2m3 m |= 1 3

2m m = 1 m = 1(ktm)
p s: m = 1 .

0,25

AT

y ' = 3 x 2 6mx = 3 x( x 2m) . y ' = 0 x = 0; x = 2m

iu kin hm s c hai cc tr l m 0 .

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

sin x 3 cos x = 2 ( cos 2 x sin 2 x ) sin x 3 cos x = 2cos 2 x

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

www.MATHVN.com - www.DeThiThuDaiHoc.com

b) (0,5 im) Gii phng trnh log 3 3x 6 = 3 x

iu kin: x > log 3 6 . Phng trnh cho tng ng vi

t = 9

t = 3(l )
p s: x = 2 .

co

Vi t = 9 3x = 9 x = 2 (tmk).

I =

Tnh tch phn

( 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

t t = sin x dt = cos xdx . x = 0 t = 0;

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 .

T M (2; 5), N (2; 5) MN = 2 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) .

(Cch 2: - - - - - - - 7 v tr. Xp 3 n cnh nhau c 5 cch: (123)(567). Mi cch xp li c 3! cch

hon v 3 n. C 4! cch hon v 4 nam. Vy P(A) = 5.3!.4!/7! = 1/7)
Cho ( P ) : x + 2 y + 2 z 11 = 0 , I (3;6;7)

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

ng thng (d ) qua I v vung gc vi ( P) c phng trnh y = 6 + 2t

z = 7 + 2t

0,25

0,25
0,25
(t R) .

0,25

2/4

www.MATHVN.com - www.DeThiThuDaiHoc.com

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 =

d ( C ',( BMB ') ) = d ( C ,( BMB ') ) = d ( A,( BMB ') ) =

3VA. BMB '

.
S BMB '

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.

V AD khng song song cc trc ta nn gi vc t php tuyn ca AD l


A
n = (1; b), b 0; suy ra: Phng trnh AD :1( x 3) + b( y + 1) = 0 .
Phng trnh AB : bx ( y 4) = 0 .
I
AB + CD
3 AB
3
S ABCD =
. AD =
. AD = .d ( B, AD).d ( K , AB )
2
2
2
3 | 3 + 5b| |2b + 2|
.
= .
.
2 b2 + 1 b2 + 1
D

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

www.MATHVN.com - www.DeThiThuDaiHoc.com

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 =

> 0 t > 1 , suy ra f (t ) ng bin

co

t 1 , suy ra x 1 = 3 y + 2 .

0,25

Thay vo (2) ta c 3 x 1 x 2 6 x + 6 = ( x 1) + 1 ( x 1) + 1 + ( x 1)2 4( x 1) + 1 = 3 x 1

x 1 > 0 ta c:

Do x = 1 khng tha mn nn chia c 2 v cho

1
1
+ x 1 4 +
= 3.
x 1
x 1

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
.

(1,0) Cho x, y > 0 : x y + 1 0 . Tm max: T =

2

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.

Nhn xt: 0 < t

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

www.MATHVN.com - www.DeThiThuDaiHoc.com

Trng thpt lng th vinh

H ni

thi th thpt quc gia nm 2015

Mn thi: Ton - Ln th 3
Thi gian lm bi: 180 pht, khng k thi gian pht
-------------- Ngy 16.5.2015 --------------

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

b) Tm cc gi tr ca m ng thng d : y = x + m ct th ( C ) ti hai im phn bit.

Cu 2 (1,0 im).

v tan = 2 . Tnh M = sin 2 + sin + + sin

2 .
2
2

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.

a) Cho gc tha mn: < <

Cu 3 (0,5 im). Gii bt phng trnh: log 2 ( x 2) + log 0,5 x < 1 .

HV

Cu 4 (1,0 im). Gii bt phng trnh: x x 2 > x3 4 x 2 + 5 x x3 3 x 2 + 4 .

Cu 5 (1,0 im). Tnh tch phn: I = x ( x + cos 2 x ) dx.
0

AT

Cu 6 (1,0 im). Cho hnh chp S . ABCD c y l hnh thang vung ti A v B ; AB = BC = a;

AD = 2a ; SA ( ABCD ) . Gc gia mt phng ( SCD) v mt phng ( ABCD) bng 450 . Gi M l trung
im AD . Tnh theo a th tch khi chp S .MCD v khong cch gia hai ng thng SM v BD .
Cu 7 (1,0 im). Trong mt phng ta Oxy, cho tam gic ABC c phng trnh ng phn gic
trong gc A l d : x + y 3 = 0 . Hnh chiu vung gc ca tm ng trn ni tip tam gic ABC ln
ng thng AC l im E (1;4) . ng thng BC c h s gc m v to vi ng thng AC gc

450 . ng thng AB tip xc vi ng trn (C ) : ( x + 2 ) + y 2 = 5 . Tm phng trnh cc cnh ca tam

gic ABC .

Cu 8 (1,0 im).

Trong khng gian vi h ta Oxyz , cho im A (1; 1;0 ) v ng thng

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:

Cu 9 (0,5 im). Trong t xt tuyn vo lp 6A ca mt trng THCS nm 2015 c 300 hc sinh ng

k. Bit rng trong 300 hc sinh c 50 hc sinh t yu cu vo lp 6A. Tuy nhin, m bo quyn

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.

Cu 10 (1,0 im). Cho cc s thc a, b dng v tha mn ab 1 .

1
1
32
Tm gi tr nh nht ca biu thc
T=
+

.
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:
www.MATHVN.com - www.DeThiThuDaiHoc.com

Trng thpt lng th vinh

H ni

p n thang im
thi th thpt quc gia nm 2015
Mn thi: Ton Ln th 3

--------------- p n c 06 trang --------------

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

Tp xc nh: D = R \ {1} . lim y = 3; lim y = 3 suy ra tim cn ngang y = 3 .

x

x +

x 1

o hm: y ' =

( x 1)

< 0 x 1 .

Bng bin thin:

x

y'
y
3

0,25

0,25

HV

Hm s lun nghch bin trn khong ( ;1) v (1; + ) .

Hm s khng c cc tr.

N.

lim y = +; lim
y = suy ra tim cn ng ca th hm s l ng thng x = 1 .

x 1+

0,25

th: (Hs c th ly im (2; 4); (0; 2) ).

b) (1,0 im) Tm cc gi tr ca m d : y = x + m ct th ( C ) ti hai im phn bit.

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

Phng trnh tng giao:

m 2 4m 12 > 0
m > 6; m < 2 .

0,25
0,25

w.

2
3

. Tnh M = sin 2 + sin + + sin

2 .
(1,0) a) (0,5 im) Cho tan = 2 . < <

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

www.MATHVN.com - www.DeThiThuDaiHoc.com

( 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 +

Kt hp iu kin ta c nghim ca bpt l x > 2 .

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.

x 2 > x x 2 5 x + 4 > 0 x > 4; x < 1 .

Kt hp x > 2 x > 4 .
0 < x < 2:
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
(
)

ww

0,25

Xt hm f (t ) = t 1 + t 2 , t R f '(t ) = 1

T (1)

0,25

> 0 t > 0 f (t ) ng bin t > 0

(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) .

( x 2)+ | x 2 | x + 1 > x 1 + ( x 2 ) + 1 . (1)

x = 2 : (1) 0 > 2 2 (loi). x = 0 : (1) 2 > 2

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

www.MATHVN.com - www.DeThiThuDaiHoc.com

Cch 2: K x 0 (mi du + ng vi im)

x = 0 khng l nghim. Xt x > 0 :

)(

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.

+ Nu 0 < x < 1: x + 1 > 1

x 2

co

+ (1)

= x 2 x +1 > x 2 = 2 x

Tnh tch phn: I = x ( x + cos 2 x ) dx.

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)

S . ABCD y l hnh thang vung ti A v B ; AB = BC = a; AD = 2a ; SA ( ABCD) . Gc gia

( SCD) v ( ABCD) bng 450 . M l trung im AD . Tnh th tch S .MCD , d ( SM , BD )

= 450.
Ta c ( SCD) ( ABCD) = CD. CD SA, AC CD ( SAC ) SC CD SCA

0,25

0,25

0,25

3/6

www.MATHVN.com - www.DeThiThuDaiHoc.com

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

AH ( SMN ) d ( A,( SMN )) = AH

1
1
1
Tam gic vung SAP c
=
+
2
2
AH
AS
AP 2
1
1
1
1
1
1
11
=
+
+
= 2+ 2 + 2 = 2
2
2
2
a
AS
AN
AM
2a
a
2a
4
a 22
a 22
Suy ra AH =
.
d ( SM , BD ) =
11
11

N.

0,25

co

d ( SM , BD) = d ( BD,( SMN )) = d ( D,( SMN )) = d ( A, ( SMN )) .

AP MN ( P MN ) , AH SP ( H SP )

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

Gi F l im i xng vi E qua d F (1;2) . Nhn xt: (C ) c tm I (2;0), bn knh R = 5

v F (C ) .
T AB qua F v vung gc vi IF nn c phng trnh AB : x + 2 y 3 = 0 .
AB d = A(3;0) AC : 2 x + y 6 = 0 .
Gi J l tm ng trn ni tip ABC . ng thng qua

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

www.MATHVN.com - www.DeThiThuDaiHoc.com

Do J l tm ng trn ni tip ABC nn d ( J , AC ) = d ( J , BC )

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

C 300 hc sinh ng k. C 50 hc sinh t yu cu vo lp 6A. Bc thm ngu nhin 30 hc sinh t

9
(0,5) 300 hc sinh ni trn. Tm xc sut c ng 90% s hc sinh t yu cu.
Gi A l bin c: Chn c 90% hc sinh t yu cu.
30
Chn ngu nhin 30 hc sinh t 300 hc sinh c C300
cch chn.
27
Chn c 90% hc sinh t yu cu, tc l chn c 27 em. Chn 27 hc sinh t 50 hc sinh c C50
cch.
3
Chn nt 3 em t 250 em cn li c C250
cch.

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) .

Tht vy: Quy ng, chuyn v, bt trn tng ng vi

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

www.MATHVN.com - www.DeThiThuDaiHoc.com

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

t 2 + 3 > t t + 3 t 4 + 6t 2 + 9 > t 3 + 3t 2 (t 4 t 3 ) + 3t 2 + 9 > 0 (ng t 1 ).

Suy ra f '(t ) > 0 t 1 f (t ) ng bin t 1 .
T : MinT = f (1) = 7 t = 1 a = b = 1.

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

Cch 2: C th dn bin v u = a + b 2 ab 2 nh sau:

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

= f (u ), u 2. Chng minh f '(u ) > 0 u 2 tng t cch 1.

u+2
2u + 12
Kt lun: MinT = f (2) = 7 u = 2 a = b = 1.

u2

AT

ww

w.

-------------------- Ht --------------------

6/6

www.MATHVN.com - www.DeThiThuDaiHoc.com

 thi th thpt quc gia nm 2015

Trng thpt lng th vinh

H ni

Mn thi: Ton - Ln th 4
Thi gian lm bi: 180 pht, khng k thi gian pht

Nm hc 2014 - 2015

-------------- Ngy 13.6.2015 --------------

Cu 1 (2,0 im). Cho hm s y = x 3 3x 2 + 4 .

a) Kho st s bin thin v v th (C ) ca hm s cho.

co

b) Vit phng trnh tip tuyn ca th (C ) ti giao im ca (C ) vi ng thng (d ) : y = 5 x + 7 .

Cu 3 (0,5 im). Gii phng trnh

2.9 x + 3.4 x = 5.6 x .

1

I = x e3 x +
dx.
3x + 1
0

HV

Cu 4 (1,0 im). Tnh tch phn

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.

Cu 5 (0,5 im). Ti mt k SEA Games, mn bng nam c 10 i bng tham d (trong c i

Vit Nam v i Thi Lan). Ban t chc bc thm ngu nhin chia 10 i bng ni trn thnh 2 bng
A v B, mi bng 5 i. Tnh xc sut i Vit Nam v i Thi Lan cng mt bng.

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

Cu 9 (1,0 im). Gii phng trnh 8 x 2 + 10 x + 11 + 14 x + 18 = 11 .

Cu 10 (1,0 im). Cho cc s thc x, y, z dng v tha mn 4 x 2 x + 1 16 x 2 yz 3x ( y + z ) .

Tm gi tr nh nht ca biu thc

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.
www.MATHVN.com - www.DeThiThuDaiHoc.com

H v tn th sinh: ..; S bo danh:

p n thang im
thi th thpt quc gia nm 2015

Trng thpt lng th vinh

H ni

Mn thi: Ton Ln th 4

--------------- p n c 05 trang --------------

Nm hc 2014 2015

x +

o hm: y ' = 3 x 6 x ; y ' = 0 x = 0 hoc x = 2 .

2

HV

Cc tr: Hm s t cc tiu ti x = 2 , yCT = 0 ;

t cc i ti x = 0 , yC = 4.
Bng bin thin:
x
0
2

+
y'
+
0
0
+
y
4
+

N.

Khong ng bin: ( ;0 ) ; ( 2; + ) . Khong nghch bin: ( 0; 2 )

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

th: (Hs c th ly thm im (1; 0); (1; 2); (3; 4) ).

b) (1,0 im) ) Vit phng trnh tip tuyn ti giao im ca (C ) vi (d ) : y = 5 x + 7 .
Phng trnh honh giao im:

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

y ' = 3x 2 6 x y '( x0 ) = y '(1) = 3

Phng trnh tip tuyn cn tm: y = 3( x 1) + 2 y = 3 x + 5

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

www.MATHVN.com - www.DeThiThuDaiHoc.com

T : z = 2 + bi; z = 2 bi
0,25

3
(0,5)

( z + 4)( z + 2i ) = (6 + bi ) [ 2 + (2 b)i ] = 12 b(2 b) + (12 4b)i

Suy ra: 12 4b = 0 b = 3 .
p s: z = 2 + 3i .
Gii phng trnh 2.9 x + 3.4 x = 5.6 x .

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)

bng A v B, mi bng 5 i. Tm xc sut Vit Nam v Thi Lan cng mt bng.

ww

Gi M l bin c: Vit Nam v Thi Lan cng mt bng.

S bin c ng kh nng: S cch chia 10 i bng thnh 2 bng u nhau n() = C105 .C55 = 252 .
Xt s cch chia m Vit Nam v Thi Lan cng mt bng:
Chn bng (A hoc B): c 2 cch
Chn nt 3 i cn li: c C83 cch. Chn 5 i ca bng kia: c C55 cch.

n( M ) = 2.C83 .C55 = 112. Suy ra: xc sut ca bin c M: p ( M ) =

0,25

0,25

4
n( M ) 112
=
= .
n() 252 9
2/5

www.MATHVN.com - www.DeThiThuDaiHoc.com

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

phng trnh hnh chiu vung gc ca ng thng AB ln mt phng ( BCD) .



Ta c BC = (3;3;2), BD = (3; 3;1) .

 

Mp ( BCD) i qua B (1;0; 2) v c vtpt n = BC , BD = ( 9;9;0 ) . Chn n = (1;1;0) .
Phng trnh ( BCD) : 1( x 1) + 1( y 0) + 0( z 2) = 0 x + y 1 = 0 .
ng thng AB ct ( BCD) ti B (1;0; 2) . Ta i tm hnh chiu A ' ca im A ln ( BCD) .
x = 3 + t

ng thng i qua A v vung gc vi ( BCD) c phng trnh y = 2 + t (t R).

z = 3

A ' = ( BCD) (3 + t ) + (2 + t ) 1 = 0 t = 2 A '(1;0;3) .

 
Hnh chiu vung gc ca AB i qua B, A ' nn c vtcp u = BA ' = (0;0;1) .
x = 1

Phng trnh BA ' : y = 0

(t R).
z = 2 + t

0,25

co

0,25

(Lu : Hc sinh vit

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

Gc gia cnh bn v mt y ca lng tr bng 600 . Tnh 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 .
Xc nh gc 600 :

Gi H l hnh chiu ca A ln (ABC) HA = HB = HC =

trn ngoi tip tam gic ABC.

AA '2 A ' H 2 suy ra H l tm ng

C'

AH l hnh chiu ca AA ln (ABC), suy ra 

A ' AH = 600.
Tnh th tch lng tr: VABC . A ' B ' C ' = A ' H .S ABC

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.

Tnh bn knh R: R = IA ' =

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

www.MATHVN.com - www.DeThiThuDaiHoc.com

8
Cho tam gic ABC ni tip ng trn tm I (2;1) , bn knh R = 5 . Chn ng cao h t
(1,0)

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.
Chng minh AI DE :

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

Lu : + Hc sinh ch tm c 1 nghim, cho im

+ Hc sinh tm c 2 nghim m khng CM c phn cn li v nghim, cho im
C th CM f ( x ) > 0 nh sau:

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

www.MATHVN.com - www.DeThiThuDaiHoc.com

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=

Tm gi tr nh nht ca biu thc

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

www.MATHVN.com - www.DeThiThuDaiHoc.com