së gi¸o dôc vµ ®µo t¹o hµ tÜnh
Tr"êng THPT NguyÔn Trung Thiªn
§Ò THi thö ®¹i Häc LÇN I n¨m 2014
Môn thi: To¸n - KHỐI A, A1, B
Thời gian làm bài: 180 phút
I. PhÇn chung cho tÊt c¶ thÝ sinh(7,0 ®iÓm)
C©u I (2,0 ®iÓm) Cho hµm sè 3
1
x
yx
−
=+ cã ®å thÞ (C)
1. Kh¶o s¸t sù biÕn thiªn vµ vÏ ®å thÞ (C) cña hµm sè.
2. ViÕt phB¬ng tr×nh tiÕp tuyÕn cña (C) biÕt khong c¸ch tõ giao ®iÓm I cña 2 tiÖm cËn cña (C) ®Õn
tiÕp tuyÕn b»ng 22
.
C©u II (2,0 ®iÓm)
1. Gi¶i phB¬ng tr×nh 12sin(2)coscos3
4
x
xx
π
++=+ .
2. TÝnh: I = 2
tanx
1os dx
cx+
∫
C©u III (1,0 ®iÓm) Gi¶i hÖ phB¬ng tr×nh:
22
2
41
2
1
xyxyy
y
xyx
++=−
+=+
+
C©u IV (1,0 ®iÓm) Cho h×nh chãp S.ABCD cã ®¸y ABCD lµ h×nh thang.
§¸y lín AB = 2a ; BC = CD = DA = a; SA vu«ng gãc víi ®¸y, mÆt ph¼ng(SBC) t¹o víi ®¸y mét
gãc 60o. TÝnh thÓ tÝch khèi chãp S.ABCD theo a.
C©u V (1,0 ®iÓm) Cho 3 sè thùc dB¬ng x, y, z. T×m gi¸ trÞ nhá nhÊt cña biÓu thøc :
222
222
()()()
333
xyz
Pxyz
yzxzxy
=+++++.
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 träng t©m G (2;-1).
§Bêng trung trùc cña c¹nh BC cã phB¬ng tr×nh d:3xy40−−=. §Bêng th¼ng AB cã phB¬ng tr×nh
1:10310dxy++=. T×m täa ®é c¸c ®Ønh A, B, C.
C©u VII. a. (1,0 ®iÓm) Trong mÆt ph¼ng víi hÖ täa ®é Oxy cho hai ®iÓm A(2;0), B(6;4). ViÕt
phB¬ng tr×nh ®Bêng trßn (C) tiÕp xóc víi trôc hoµnh t¹i ®iÓm A vµ kho¶ng c¸ch tõ t©m (C) ®Õn B b»ng
5.
C©u VIII. a. (1,0 ®iÓm ) T×m sè h¹ng kh«ng chøa x trong khai triÓn
()
32
n
Pxx
x

=+


(0)x>.
BiÕt r»ng n tháa m·n: 67898
2
332
nnnnn
CCCCC
+
+++= .
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 tam gi¸c ABC vu«ng c©n t¹i A(1;2).
ViÕt phB¬ng tr×nh ®Bêng trßn (T) ngo¹i tiÕp tam gi¸c ABC biÕt ®Bêng th¼ng d:xy10−−= tiÕp xóc víi
(T) t¹i B.
C©u VII. b. (1,0 ®iÓm) Trong mÆt ph¼ng víi hÖ täa ®é Oxy cho hai ®Bêng th¼ng 1:350dxy++=;
2:310dxy++= vµ ®iÓm I(1;-2). ViÕt phB¬ng tr×nh ®Bêng th¼ng ®i qua I c¾t 12
,dd lÇn lBît t¹i A vµ B
sao cho 22AB =.
C©u VIII. b. (1,0 ®iÓm) Gi¶i phB¬ng tr×nh:
3
22
2
loglog2
2
x
x
x
 
+=
 

 .
--------------------- HÕt --------------------
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§¸p ¸n K.A gåm cã 6 trang.
L"u ý : Mäi c¸ch gii ®óng ®Òu cho ®iÓm tèi ®a.
C©u §¸p ¸n vµ h"íng dÉn chÊm §iÓm
1 (1,0 ®iÓm)
________________________________________________________________________
+ TËp x¸c ®Þnh:
{ }
\1DR=−
+ Sù biÕn thiªn: 2
4
'0
(1)
yx
=>
+, 1x∀≠− , suy ra hµm sè ®ång biÕn trªn c¸c kho¶ng
(
)
;1−∞− vµ
(
)
1;−+∞ .
________________________________________________________________________
+ Giíi h¹n: lim1
xy
→−∞ =; lim1
xy
→+∞ = => TiÖm cËn ngang: y=1
1
lim
xy
−
→− =+∞; lim
xy
→+∞ =−∞ => TiÖm cËn ®øng: x=-1.
________________________________________________________________________
+ B¶ng biÕn thiªn:
x ∞− -1 ∞+
y’ +
y +∞ 1
1 −∞
________________________________________________________________________
+ §å thÞ :
Giao víi Ox: (3;0), giao víi Oy: (0;-3).
§å thÞ nhËn I(-1;1) lµm t©m ®èi xøng.
x
0
-13
1
-3
0,25
0,25
0,25
0.25
C©u
I.
2,0
®iÓm
2 (1,0 ®iÓm)
________________________________________________________________________
Gi¶ sö
(
)
00
;Mxy thuéc (C), 0
0
0
3
1
x
yx
−
=+, 01x≠− .
Khi ®ã phB¬ng tr×nh tiÕp tuyÕn ∆ t¹i M lµ:
()
()
0
0
20
0
3
4
1
1
x
yxx x
x
−
=−+
+
+
()
(
)
22
000
41630xxyxx⇔−++−−=
________________________________________________________________________
Theo ®Ò :
(
)
,22dI∆=
()
(
)
()
0
2
2
00
4
0
416322
161
xxx
x
−−++−−
⇔=
++
()()
42
00
181160xx⇔+−++=
0.25
0.25
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0
0
1
3
x
x
=
⇔=−
________________________________________________________________________
Víi 01x=, phB¬ng tr×nh :2yx∆=−;
Víi 03x=− , phB¬ng tr×nh :6yx∆=+.
0,5
1 (1,0 ®iÓm)
________________________________________________________________________
PT 1sin2cos22coscos2
xxx⇔++=
2
2cos2sincos2coscos20xxxxx⇔+−=
(
)
(
)
22
2coscossincossin0xxxxx⇔+−−=
(
)
(
)
coscossin1cossin0xxxxx⇔+−+=
________________________________________________________________________
cos0
cossin0
cossin1
x
xx
xx
=
⇔+=
−=
2
tan1
1
cos 42
xk
x
x
π
π
π
=+
⇔=−

+=


________________________________________________________________________
2
4
2
xk
xk
xk
ππ
ππ
π
=+
⇔=−+
=
k∈
0,25
0,5
0,25
C©u
II.
2,0
®iÓm
2 (1,0 ®iÓm)
________________________________________________________________________
Ta cã: 22
tansincos
1coscos(1cos)
xxx
Idxdx
xxx
==
++
∫∫
§Æt 2
costx= 2sincosdtxxdx⇒=−
Suy ra: 1
2(1)
dt
Itt
=− +
∫
________________________________________________________________________
11111
ln
212
t
IdtC
ttt
+

=−=+

+

∫
________________________________________________________________________
KÕt luËn:
2
2
11cos
ln
2cos
x
IC
x

+
=+


.
0,25
0,5
0,25
C©u
III.
1,0
®iÓm
NhËn xÐt y=0 kh«ng tháa m·n hÖ phB¬ng tr×nh.
0,25
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HÖ tB¬ng ®B¬ng víi
2
2
14
2
1
xxy
y
y
xyx
+
++=
+=+
+
________________________________________________________________________
§Æt
21x
uy
+
=, v = x + y. HÖ trë thµnh:
4
12
uv
vu
+=
=+
Gi¶i hÖ ta cã: u =1
v = 3
________________________________________________________________________
Víi
21
12
11
32
35
x
xy
uy
vx
xy y
=
+=
==

⇒⇔

==−
+= =
0,25
0,25
0,25
C©u
IV.
1,0
®iÓm
Gäi N lµ trung ®iÓm AB.
AB
C
D
N
600
Ta cã: AN // DC
AN = DC = a
nªn ADCN lµ h×nh b×nh hµnh.
Suy ra: NC = AD = a
=> NA = NB = NC =a hay ACB∆vu«ng t¹i C suy ra ACBC⊥.
Do ()SAABCD⊥ nªn SABC⊥.
¸p dông ®Þnh lý ba ®Bêng vu«ng gãc ta suy ra SCBC⊥.
Suy ra: Gãc gi÷a (SBC) vµ (ABCD) lµ SCA∠ => 60SCA∠=°
________________________________________________________________________
MÆt kh¸c: NBC∆ ®Òu nªn 60NBC∠=°
33
2
ACABa==
.tan603.33SAACaa=°==
________________________________________________________________________
2
33
4
ABCD
a
S=
________________________________________________________________________
TÝnh ®Bîc thÓ tÝch chãp S.ABCD b»ng
3
33
4
a.
0,25
0,25
0,25
0,25
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C©u
V.
1,0
®iÓm
Ta cã :
333222
2
3
x
yzxyz
Pxyz

++++
=+


¸p dông bÊt ®¼ng thøc 22
2,,ababab+≥∀
222
x
yzxyyzzx⇒++≥++.
(§¼ng thøc x¶y ra khi x=y=z)
3332
3
x
yzxyyzzx
Pxyz

++++
⇒≥+


333
222
333
xyz
P
x
yz

⇒≥+++++


________________________________________________________________________
XÐt hµm sè
32
() 3
t
ft t
=+ víi t > 0 ;
2
2
2
'()fttt
=− ; 4
'()02ftt=⇔= .
________________________________________________________________________
B¶ng biÕn thiªn:
t 0 42 ∞+
y’ - 0 +
y +∞ +∞
4
8
32
________________________________________________________________________
VËy 4
48P≥. §¼ng thøc x¶y ra khi 42xyz=== hay 4
48P=.
0,25
0,25
0,25
0,25
A. Theo ch"¬ng tr×nh chuÈn
C©u
VI. a.
1,0
®iÓm
Gäi M lµ trung ®iÓm BC, v× Md∈ nªn M (m; 3m-4).
Mµ 2GAGM=−

nªn A (6-2m; 5-6m).
________________________________________________________________________
AAB∈ 2m⇒=
(
)
2;2M⇒,
(
)
2;7A−.
________________________________________________________________________
BC qua M vµ vu«ng gãc víi d nªn cã phB¬ng tr×nh x + 3y – 8 = 0.
BABBC=∩ nªn
(
)
1;3B−.
________________________________________________________________________
M lµ trung ®iÓm BC nªn
(
)
5;1C.
0,25
0,25
0,25
0,25
C©u
VII.
a.
1,0
®iÓm
Gäi 00
(;)Ixy lµ t©m cña ®Bêng trßn (C).
Khi ®ã, do (C) tiÕp xóc víi Ox t¹i A nªn víi (0;1)i=
lµ vect¬ ®¬n vÞ trªn trôc Ox, ta cã:
IAi⊥

(
)
(
)
00
1.10.00xy⇔−+−= 02x⇔=.
________________________________________________________________________
Theo gi¶ thiÕt, ta cã:
R = IB – 5 2
;25IB =
()()
22
0
26425y⇔−+−=
043y⇔−=± 0
0
7
1
y
y
=
⇔=
________________________________________________________________________
0,25
0,25
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