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102 C[1]
[ ]
1.()
2.
3. 25 4 100
4. (A)(B)(C)(D)
2B 5.
6.7.
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102 C[1]
1. logx log0.01234logx log6789x
(A) 6789.1234 (B) 6789 (C) 1234 (D) 1234.6789
[]log0.01234 log(1.234 10 2) 2 log1.234
log6789 log(6.789 103) 3 log6.789
logx 3 log1.234 log103 log1.234 log(103 1.234) log 1234
x1234(C)
2f(x)=x
1 +xf2(x)=f(f(x))f3(x)=f(f2(x))fn(x)=f(fn1(x))
f2013(1)= (A)2013 (B)2014 (C)1
2013(D)
1
2014
[]f(1)= 12
f2(1)=f(f(1))=f( 12 )=
1
2
1 +1
2
= 13
= 11 + 2
f3(1)=f(f2(1))=f(1
3)=
1
3
1 +1
3
=1
4=
1
1 + 3
fn(x)=1
1 + nf2013(1)=
1
2014(D)
3f(x) 12(x2)f(x) q(x) 21
(x1) q(x) (A)3 (B)6 (C)9 (D)33
[]f(x)= f(1) =12f(x)= (x2)q(x)+21 )(2
21)(xq
x
xf
q(x)= (x1)Q(x)+rr= q(1)= 921
2112
21
21)1(
f(C)
4 an 2 9 5 243 logba1 logb a2 logb a10
55 b (A)3
1(B)1 (C)3 (D)27
[] a r
243
9
4
5
2
ara
ara
a 3r 3 an 33n1 3n
a1a2a10 33233310 31210 355
logba1 logba2 logba10 logb(a1a2a10) logb355 55logb3 55
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102 C[1]
logb3 1b 3
1 (A)
5 a =(x,2) b =(1,y)x2
+y2
=5 a b
(A)2 (B)5 (C)25 (D)100
[] a b =(x, 2)(1,y)=x + 2y (x + 2y)2 (x2 +y2 )( 12 + 22 ) 55
5 x+2y5 a b 5(B)
6ABC(a+b+c)(a+bc)=(2 3 )abC=
(A)60 (B)90 (C)120 (D) 150
[](a+b+c)(a+bc)=(2 3 )ab (a+b)2c2=(2 3 )ab a2+b2c2= 3 ab
cosC=
ab
cba
2
222 =
3
2
ab
ab=
3
2 C=150(D)
79
22y
k
x 1 4 k
(A)4 (B)8 (C)16 (D)12
[] 9
22 y
k
x
1 4
4 2 k 4 k 4(A)
8(3, 4)y(a, b) a+ b=
(A)7 (B)10 (C)12 (D)5
[] FL
y 3 3
(6, 4) a+ b=10(D)
9 2334 yx (2, 5) 10
(A)(10, 11) (B)(11, 21) (C) (10, 21) (D) (11, 11)
[]
OP(4,3)
)3,4()5,2( tyx )53,24(),( ttyx
),( yx 2334 yx 10
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102 C[1]
105
|23)53(3)24(4|
tt 25t=50, t=2 )11,10(),( yx
(A)
10 3x 4yk 0k>0 4 k
(A)5
12(B)
5
24(C)
5
48(D)5
[]3x 4yk 0
3
k
x
4
k
y 1 22 )
4()
3(
kk 4
9
2k
16
2k 165k2 9 16 16k>0 5k3 4 4 k
5
48(C)
11 = cos2
3sini
2
3
(A)(1 + 2
)(1 +2
) =6 (B) (2 + 3+ 32
)= 1 (C) ( +1 )( 2
+1 )=2
(D)102
+2013
+7
=1
[]= cos2
3sini
2
33=1(1)(2++1)= 0 12++1= 0
(A)(1 +
2
)(1 +
2
)= (
2
2
)( )= ( 2
2
)( 2) = 4
3
= 4(B)(2 + 3+ 3
2
)= ( 1 +3 + 3+ 32
)= 1
(C)( +1 )( 2
+1 )= 3
+2++1=
3
=1
(D)102 +2013+
7=1
+ 1 + =(B)
12 C(x2)2+(y1)2=25 L3x+4y+5=0AB C O OAB (A)48 (B)36 (C)24
(D)12
[]C(x2)2+(y1)2=52 O(2,1) r=5L3x+4y+5=0 d
d=2 2
| 3 2 4 1 5 |
3 4
=3 AB = 2 22 5 3 =8
OAB=1
2AB d=
1
283=12(D)
13ABCA(2 8)B( 6 2)C(6 5)A
BCDD(A) (2 4) (B) (24) (C) (3 4) (D) (3
2 5)
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102 C[1]
[]AB 10AC 5ADA CD
BD
AC
AB
5
10
1
2
D(xy)
412
)5(2)2(1
212
62)6(1
y
x
D(2 4)(A)
142
)153()2( 23
x
xxx0
(A) 1 (B) 2 (C) 3 (D) 4
[]2
)153()2(23
x
xxx0
2
)5()2(4
x
xx0 (x5)( x2)0,x=2 2
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102 C[1]
(1
x+
1
y)(4x +y) =( 22 )
1()
1(
yx )( 22 )()2( yx )
( yy
xx
1
21
)2
=( 2 + 1 )2
=9 4x +y 9(D)
17y = mx2
+ 3(m 4)x 9 x PQ PQ
(A)3 3
2(B)
3
2(C)
3 3
4(D)
3
4
[]y = mx2
+ 3(m 4)x 9x PQ
D = 9(m 4) x2 4m( 9) > 0 m2 4m + 16 > 0 (m 2)2 + 12 >
0
P( 0)Q( 0) mx2
+ 3(m 4)x 9= 0
+= m
m )4(3 =
m
9 PQ =||
PQ 2 =||2 =(+)24= )4
3)
2
14((9)1
416(9
36)4(9 222
2
mmmmm
m
214
m( i.e., m = 8 )PQ 2 9 34 = 274 PQ 3 32
(A)
18 4 10
(A) 410H (B)12
4C (C)13
10C (D)!4!9
13
13
P
[]10x1,x2,,xn
321 xxx + nx = 4 715!9!4
13
1313
4
1410
4
10
4
P
CCH
(D)
19 i x1
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102 C[1]
54
5
63
6
3 C
(D)
20 a b f(x) = a (x 1)2 + bf(4) > 0f(5) < 0
(A) f(0) < 0 (B)f( 1) > 0 (C)f( 2) > 0 (D)f( 3) <
0
[]f(4) = 9a + b > 0b > 9af(5) = 16a + b < 0 b < 16a
9 a < b < 16aa < 0b > 0 f(0) = a + b > 8a > 0f(
1) = 4a + b > 5a > 0f( 2) = 9a + b > 0f( 3) = 16a + b <
0(A)
21f(x)=)8(log
)6)(4)(2(
2
x
xxxxf(0)= (A)2 (B)4 (C)6 (D)8
[]f(0)=
)8(log
)6)(4)(2(lim
0
)0()(lim
200 x
xxx
x
fxf
xx
246
log2 8
=48
3
=16
21
2lim
)1(
)1(1lim
11
1
lim000
xxx
xx
x
x
x
xxx(A)
22 ( )y f x 5
1( )f x
__________
(A)2 (B)4 (C)5 (D)7
[]5
1( )f x
(2 4) 2 1 2 22 2 6 2 4
(B)
23f(x)=3(x1 )(x21 )(x + 9 )sin(
2x) )1(f =
(A)100 (B)20 (C)101 (D)18
[]f(x)=3(x1 )(x21 )(x + 9 )sin(
2x)
=3x31(x21 )(x + 9 )sin(
2x)f(1)=0
)1(f =1
)1()(lim
1
x
fxf
x=
1lim
x
3x31(x1 )(x + 1)(x + 99 )sin(
2x)
x1
=1
limx
3x31(x + 1)(x + 9 )sin(
2x)=12101=20(D)
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102 C[1]
24
1
0
x xlog1
(x2 +1
x)(x>0)= (A)
70
3(B)
1
12(C)
67
3(D)
77
3
[]Firstly , we note thatx xlog1
=x xlog10log
=x10logx
=10
1
0
xxlog
1
(x2 +1
x)=10
1
0
(x2+
1
x)=10( 1 0
3
|23
xxx
)=107
3=
70
3
(A)
25f(x) x2 7x1 3xn+1xn)(
)(
n
n
xf
xf
(A) x348
127(B) xn 1 7 (C)xn< xn 1 (D)
n
n
x
x 1 x2> x3>>xn> xn 1n
n
x
x 1
