答案速览
题号
1
2
3
4
5
6
7
8
9
答案
60 60 60 BCD
D
CD
1 / 6 1/6 1/6 4 4 4 56 56 56 BD
C
题号
10
11
12
13
14
15
16
17
18
答案
BD
AB
5 / 3 5/3 5/3 AC
π / 6 \pi/6 π /6 B
4608 4608 4608 3 3 3 B
题号
19
20
21
22
23
24
25
26
27
答案
{ 0 , 1 } \{0,1\} { 0 , 1 } 918 918 918 9 9 9 B
AC
2 2 2 A
ABC
ABCD
题号
28
29
30
31
32
33
34
35
答案
AB
ACD
D
4 / 27 4/27 4/27 BC
AB
0 0 0 1008 1008 1008
逐题解析
第 1 题
题目
在圆周的十等分点 A 1 , A 2 , … , A 10 A_1,A_2,\ldots,A_{10} A 1 , A 2 , … , A 10
解答
采用恰有一组对边平行的梯形定义，不包含矩形。给十个点编号 0 , … , 9 0,\ldots,9 0 , … , 9 10 10 10
5 ( 5 2 ) + 5 ( 4 2 ) = 80. 5\binom52+5\binom42=80. 5 ( 2 5 ) + 5 ( 2 4 ) = 80. 圆内接平行四边形必为矩形，由五条直径中任取两条组成，共 ( 5 2 ) = 10 \binom52=10 ( 2 5 ) = 10 80 − 2 ⋅ 10 = 60 80-2\cdot10=60 80 − 2 ⋅ 10 = 60
第 2 题
题目
如图，过圆 O O O C C C O O O M , N M,N M , N C C C O O O A , B A,B A , B A B AB A B Q Q Q ∠ A M Q = ∠ C N B \angle AMQ=\angle CNB ∠ A M Q = ∠ C N B
A. △ A M Q \triangle AMQ △ A M Q △ M B C \triangle MBC △ M B C
B. △ A Q M \triangle AQM △ A QM △ N B M \triangle NBM △ N B M
C. △ A M N \triangle AMN △ A M N △ B Q M \triangle BQM △ B QM
D. △ A M N \triangle AMN △ A M N △ B N Q \triangle BNQ △ B N Q
解答
此题选项中的“与……相似”只陈述两个三角形相似；以下写明真正的对应顶点，避免依赖排列顺序。
由圆周角与切线弦定理，
∠ M A Q = ∠ M N B , ∠ A M Q = ∠ C N B = ∠ N M B . \angle MAQ=\angle MNB,\qquad
\angle AMQ=\angle CNB=\angle NMB. ∠ M A Q = ∠ M N B , ∠ A M Q = ∠ C N B = ∠ N M B . 所以 △ A Q M ∼ △ N B M \triangle AQM\sim\triangle NBM △ A QM ∼ △ N B M
A Q N B = A M N M = Q M B M . (1) \frac{AQ}{NB}=\frac{AM}{NM}=\frac{QM}{BM}. \tag{1} N B A Q = N M A M = B M QM . ( 1 ) 过同一外点的两条切线相等。也可直接用割线幂与切线弦定理得到
A M B M = A N B N . (2) \frac{AM}{BM}=\frac{AN}{BN}. \tag{2} B M A M = B N A N . ( 2 ) 为补足此比例的证明，由 △ C A M ∼ △ C M B \triangle CAM\sim\triangle CMB △ C A M ∼ △ C M B A M / B M = C M / C B AM/BM=CM/CB A M / B M = C M / C B △ C A N ∼ △ C N B \triangle CAN\sim\triangle CNB △ C A N ∼ △ C N B A N / B N = C N / C B AN/BN=CN/CB A N / B N = C N / C B C M = C N CM=CN C M = C N
圆内接四边形 A M B N AMBN A M B N A B ⋅ M N = A M ⋅ B N + A N ⋅ B M AB\cdot MN=AM\cdot BN+AN\cdot BM A B ⋅ M N = A M ⋅ B N + A N ⋅ B M A Q = A B / 2 AQ=AB/2 A Q = A B /2 A Q = B Q AQ=BQ A Q = B Q
B Q A N = B M M N = Q M A M , \frac{BQ}{AN}=\frac{BM}{MN}=\frac{QM}{AM}, A N B Q = M N B M = A M QM , 得到 △ B Q M ∼ △ N A M \triangle BQM\sim\triangle NAM △ B QM ∼ △ N A M
B Q B N = A M M N , ∠ Q B N = ∠ A B N = ∠ A M N , \frac{BQ}{BN}=\frac{AM}{MN},\qquad
\angle QBN=\angle ABN=\angle AMN, B N B Q = M N A M , ∠ QB N = ∠ A B N = ∠ A M N , 由两边及夹角判定 △ Q B N ∼ △ A M N \triangle QBN\sim\triangle AMN △ QB N ∼ △ A M N
A 不恒成立。例如取单位圆、C = ( 1 , 2 ) , M = ( 1 , 0 ) , N = ( − 3 / 5 , 4 / 5 ) , A = ( − 1 , 0 ) , B = ( 0 , 1 ) C=(1,2),M=(1,0),N=(-3/5,4/5),A=(-1,0),B=(0,1) C = ( 1 , 2 ) , M = ( 1 , 0 ) , N = ( − 3/5 , 4/5 ) , A = ( − 1 , 0 ) , B = ( 0 , 1 ) Q = ( − 1 / 2 , 1 / 2 ) Q=(-1/2,1/2) Q = ( − 1/2 , 1/2 ) △ M B C \triangle MBC △ M B C 2 , 2 , 2 \sqrt2,\sqrt2,2 2 , 2 , 2 △ A M Q \triangle AMQ △ A M Q 2 , 2 / 2 , 10 / 2 2,\sqrt2/2,\sqrt{10}/2 2 , 2 /2 , 10 /2
第 3 题
题目
已知方程 k x = sin x kx=\sin x k x = sin x ( − 3 π , 3 π ) (-3\pi,3\pi) ( − 3 π , 3 π ) x 1 < x 2 < x 3 < x 4 < x 5 x_1<x_2<x_3<x_4<x_5 x 1 < x 2 < x 3 < x 4 < x 5
A. x 5 = tan x 5 x_5=\tan x_5 x 5 = tan x 5
B. 29 12 π < x 5 < 5 2 π \dfrac{29}{12}\pi<x_5<\dfrac52\pi 12 29 π < x 5 < 2 5 π
C. x 2 , x 4 , x 5 x_2,x_4,x_5 x 2 , x 4 , x 5
D. x 1 + x 2 + x 3 + x 4 + x 5 = 0 x_1+x_2+x_3+x_4+x_5=0 x 1 + x 2 + x 3 + x 4 + x 5 = 0
解答
x = 0 x=0 x = 0 x 1 = − x 5 , x 2 = − x 4 , x 3 = 0 x_1=-x_5,x_2=-x_4,x_3=0 x 1 = − x 5 , x 2 = − x 4 , x 3 = 0
题面没有规定 k > 0 k>0 k > 0 k = 0 k=0 k = 0 − 2 π , − π , 0 , π , 2 π -2\pi,-\pi,0,\pi,2\pi − 2 π , − π , 0 , π , 2 π tan 2 π = 0 ≠ 2 π \tan2\pi=0\ne2\pi tan 2 π = 0 = 2 π − π , π , 2 π -\pi,\pi,2\pi − π , π , 2 π
第 4 题
题目
已知函数 f ( x ) = { x , x ⩾ a , 4 x 3 − 3 x , x < a , f(x)=\begin{cases}x,&x\geqslant a,\\4x^3-3x,&x<a,\end{cases} f ( x ) = { x , 4 x 3 − 3 x , x ⩾ a , x < a ,
A. 若 f ( x ) f(x) f ( x ) a = 0 a=0 a = 0 1 2 < a < 1 \dfrac12<a<1 2 1 < a < 1
B. 若 f ( x ) f(x) f ( x ) a > 1 2 a>\dfrac12 a > 2 1
C. 若 f ( x ) f(x) f ( x ) a > − 1 2 a>-\dfrac12 a > − 2 1
D. 使 f ( x ) f(x) f ( x ) a a a 3 3 3
解答
极值按邻域内大小比较定义，不预先要求该点连续。三次部分导数为 12 x 2 − 3 12x^2-3 12 x 2 − 3 − 1 / 2 -1/2 − 1/2 a > − 1 / 2 a>-1/2 a > − 1/2 1 / 2 1/2 1/2 a > 1 / 2 a>1/2 a > 1/2 x = a x=a x = a
比较左极限与函数值：
lim x → a − f ( x ) − f ( a ) = 4 a ( a 2 − 1 ) . \lim_{x\to a^-}f(x)-f(a)=4a(a^2-1). x → a − lim f ( x ) − f ( a ) = 4 a ( a 2 − 1 ) . 当 − 1 < a < 0 -1<a<0 − 1 < a < 0 a > 1 a>1 a > 1 a = − 1 , 0 , 1 a=-1,0,1 a = − 1 , 0 , 1 a = 0 a=0 a = 0 f ( 0 ) f(0) f ( 0 ) 0 0 0 ( − 1 / 2 , 0 ] ∪ ( 1 / 2 , 1 ] (-1/2,0]\cup(1/2,1] ( − 1/2 , 0 ] ∪ ( 1/2 , 1 ] a = 0 a=0 a = 0 − 1 / 2 -1/2 − 1/2
连续要求 4 a 3 − 3 a = a 4a^3-3a=a 4 a 3 − 3 a = a a = − 1 , 0 , 1 a=-1,0,1 a = − 1 , 0 , 1
第 5 题
题目
空间直角坐标系 O − x y z O-xyz O − x y z 0 ⩽ x ⩽ y ⩽ z ⩽ 1 0\leqslant x\leqslant y\leqslant z\leqslant1 0 ⩽ x ⩽ y ⩽ z ⩽ 1 ( x , y , z ) (x,y,z) ( x , y , z )
解答
单位立方体内除去坐标相等的平面（体积为零），六种严格大小顺序由交换坐标相互对应，体积相等。所给区域是其中一种的闭包，体积为 1 / 6 1/6 1/6
第 6 题
题目
圆 O O O 3 3 3 A B = 4 AB=4 A B = 4 P P P O O O A B → ⋅ B P → \overrightarrow{AB}\cdot\overrightarrow{BP} A B ⋅ B P
解答
取 A = ( − 2 , 5 ) , B = ( 2 , 5 ) A=(-2,\sqrt5),B=(2,\sqrt5) A = ( − 2 , 5 ) , B = ( 2 , 5 ) x 2 + y 2 = 9 x^2+y^2=9 x 2 + y 2 = 9 P = ( u , v ) P=(u,v) P = ( u , v )
A B → ⋅ B P → = 4 ( u − 2 ) ≤ 4 ( 3 − 2 ) = 4. \overrightarrow{AB}\cdot\overrightarrow{BP}=4(u-2)\le4(3-2)=4. A B ⋅ B P = 4 ( u − 2 ) ≤ 4 ( 3 − 2 ) = 4. P = ( 3 , 0 ) P=(3,0) P = ( 3 , 0 ) 4 4 4
第 7 题
题目
集合 A = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 } A=\{1,2,3,4,5,6,7,8,9,10\} A = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 } A A A
解答
设所选元素为 i < j < k i<j<k i < j < k j ≥ i + 2 , k ≥ j + 2 j\ge i+2,k\ge j+2 j ≥ i + 2 , k ≥ j + 2 i < j − 1 < k − 2 i<j-1<k-2 i < j − 1 < k − 2 1 , … , 8 1,\ldots,8 1 , … , 8 ( 8 3 ) = 56 \binom83=56 ( 3 8 ) = 56
第 8 题
题目
已知 z z z
A. 2 ∣ Re ( z 2 − z ) 2\mid\operatorname{Re}(z^2-z) 2 ∣ Re ( z 2 − z )
B. 3 ∣ Re ( z 3 − z ) 3\mid\operatorname{Re}(z^3-z) 3 ∣ Re ( z 3 − z )
C. 4 ∣ Re ( z 4 − z ) 4\mid\operatorname{Re}(z^4-z) 4 ∣ Re ( z 4 − z )
D. 5 ∣ Re ( z 5 − z ) 5\mid\operatorname{Re}(z^5-z) 5 ∣ Re ( z 5 − z )
解答
令 z = a + b i z=a+b\mathrm i z = a + b i a , b a,b a , b
Re ( z 3 − z ) = a 3 − a − 3 a b 2 ≡ 0 ( m o d 3 ) , \operatorname{Re}(z^3-z)=a^3-a-3ab^2\equiv0\pmod3, Re ( z 3 − z ) = a 3 − a − 3 a b 2 ≡ 0 ( mod 3 ) , Re ( z 5 − z ) = a 5 − a − 10 a 3 b 2 + 5 a b 4 ≡ 0 ( m o d 5 ) , \operatorname{Re}(z^5-z)=a^5-a-10a^3b^2+5ab^4\equiv0\pmod5, Re ( z 5 − z ) = a 5 − a − 10 a 3 b 2 + 5 a b 4 ≡ 0 ( mod 5 ) , 分别使用连续三整数乘积被 3 3 3 a 5 ≡ a ( m o d 5 ) a^5\equiv a\pmod5 a 5 ≡ a ( mod 5 ) z = 1 + i z=1+\mathrm i z = 1 + i Re ( z 2 − z ) = − 1 \operatorname{Re}(z^2-z)=-1 Re ( z 2 − z ) = − 1 Re ( z 4 − z ) = − 5 \operatorname{Re}(z^4-z)=-5 Re ( z 4 − z ) = − 5
第 9 题
题目
椭圆 x 2 a 2 + y 2 b 2 = 1 \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 a 2 x 2 + b 2 y 2 = 1 a > b > 0 a>b>0 a > b > 0 l 1 : y = − 1 2 x l_1:y=-\dfrac12x l 1 : y = − 2 1 x l 2 : y = 1 2 x l_2:y=\dfrac12x l 2 : y = 2 1 x P P P P P P P M ∥ l 1 PM\parallel l_1 P M ∥ l 1 l 2 l_2 l 2 M M M P N ∥ l 2 PN\parallel l_2 P N ∥ l 2 l 1 l_1 l 1 N N N ∣ P M ∣ 2 + ∣ P N ∣ 2 |PM|^2+|PN|^2 ∣ P M ∣ 2 + ∣ P N ∣ 2
A. a b = 2 ab=2 ab = 2
B. a b = 3 ab=3 ab = 3
C. a b = 2 \dfrac ab=2 b a = 2
D. a b = 3 \dfrac ab=3 b a = 3
解答
设 P = ( x , y ) P=(x,y) P = ( x , y ) ( 1 , − 1 / 2 ) (1,-1/2) ( 1 , − 1/2 ) l 2 l_2 l 2 y − x / 2 y-x/2 y − x /2 ( 1 , 1 / 2 ) (1,1/2) ( 1 , 1/2 ) l 1 l_1 l 1 − y − x / 2 -y-x/2 − y − x /2
P M 2 + P N 2 = 5 4 [ ( y − x / 2 ) 2 + ( y + x / 2 ) 2 ] = 5 8 x 2 + 5 2 y 2 . PM^2+PN^2=\frac54\left[(y-x/2)^2+(y+x/2)^2\right]
=\frac58x^2+\frac52y^2. P M 2 + P N 2 = 4 5 [ ( y − x /2 ) 2 + ( y + x /2 ) 2 ] = 8 5 x 2 + 2 5 y 2 . 在椭圆上恒定，当且仅当两个轴端点的值相同，即 ( 5 / 8 ) a 2 = ( 5 / 2 ) b 2 (5/8)a^2=(5/2)b^2 ( 5/8 ) a 2 = ( 5/2 ) b 2 a / b = 2 a/b=2 a / b = 2 ( 5 / 8 ) a 2 ( x 2 / a 2 + y 2 / b 2 ) (5/8)a^2(x^2/a^2+y^2/b^2) ( 5/8 ) a 2 ( x 2 / a 2 + y 2 / b 2 )
第 10 题
题目
已知 z 1 , z 2 z_1,z_2 z 1 , z 2 ∣ z 1 + z 2 ∣ ∣ z 1 z 2 ∣ \dfrac{|z_1+z_2|}{\sqrt{|z_1z_2|}} ∣ z 1 z 2 ∣ ∣ z 1 + z 2 ∣
A. 有最大值 2 2 2
B. 无最大值
C. 有最小值 2 \sqrt2 2
D. 无最小值
解答
记模长为 r 1 , r 2 r_1,r_2 r 1 , r 2 θ \theta θ ∣ θ ∣ < π / 2 |\theta|<\pi/2 ∣ θ ∣ < π /2
( ∣ z 1 + z 2 ∣ ∣ z 1 z 2 ∣ ) 2 = r 1 r 2 + r 2 r 1 + 2 cos θ > 2. \left(\frac{|z_1+z_2|}{\sqrt{|z_1z_2|}}\right)^2
=\frac{r_1}{r_2}+\frac{r_2}{r_1}+2\cos\theta>2. ( ∣ z 1 z 2 ∣ ∣ z 1 + z 2 ∣ ) 2 = r 2 r 1 + r 1 r 2 + 2 cos θ > 2. 取 z 1 = N + i , z 2 = 1 + N i z_1=N+\mathrm i,z_2=1+N\mathrm i z 1 = N + i , z 2 = 1 + N i cos θ = 2 N / ( N 2 + 1 ) → 0 \cos\theta=2N/(N^2+1)\to0 cos θ = 2 N / ( N 2 + 1 ) → 0 2 \sqrt2 2 z 2 = N z 1 z_2=Nz_1 z 2 = N z 1 ( N + 1 ) / N → ∞ (N+1)/\sqrt N\to\infty ( N + 1 ) / N → ∞
第 11 题
题目
已知函数 f ( x ) = sin x sin 2 x f(x)=\sin x\sin2x f ( x ) = sin x sin 2 x
A. f ( x ) f(x) f ( x )
B. f ( x ) f(x) f ( x )
C. f ( x ) = a f(x)=a f ( x ) = a ( 0 , 2 π ) (0,2\pi) ( 0 , 2 π )
D. f ( x ) = 7 9 f(x)=\dfrac79 f ( x ) = 9 7
解答
f ( x ) = 2 sin 2 x cos x f(x)=2\sin^2x\cos x f ( x ) = 2 sin 2 x cos x x = 0 x=0 x = 0 f ( π − x ) = − f ( x ) f(\pi-x)=-f(x) f ( π − x ) = − f ( x ) ( π / 2 , 0 ) (\pi/2,0) ( π /2 , 0 ) a = 0 a=0 a = 0 ( 0 , 2 π ) (0,2\pi) ( 0 , 2 π ) π / 2 , π , 3 π / 2 \pi/2,\pi,3\pi/2 π /2 , π , 3 π /2 t = cos x t=\cos x t = cos x f = 2 t ( 1 − t 2 ) f=2t(1-t^2) f = 2 t ( 1 − t 2 ) 4 / ( 3 3 ) < 7 / 9 4/(3\sqrt3)<7/9 4/ ( 3 3 ) < 7/9 16 / 27 < 49 / 81 16/27<49/81 16/27 < 49/81
第 12 题
题目
已知实数 x , y x,y x , y 5 x 2 − y 2 − 4 x y = 5 5x^2-y^2-4xy=5 5 x 2 − y 2 − 4 x y = 5 2 x 2 + y 2 2x^2+y^2 2 x 2 + y 2
解答
由
3 ( 2 x 2 + y 2 ) − ( 5 x 2 − y 2 − 4 x y ) = ( x + 2 y ) 2 ≥ 0 3(2x^2+y^2)-(5x^2-y^2-4xy)=(x+2y)^2\ge0 3 ( 2 x 2 + y 2 ) − ( 5 x 2 − y 2 − 4 x y ) = ( x + 2 y ) 2 ≥ 0 得 2 x 2 + y 2 ≥ 5 / 3 2x^2+y^2\ge5/3 2 x 2 + y 2 ≥ 5/3 x = − 2 y , y 2 = 5 / 27 x=-2y,y^2=5/27 x = − 2 y , y 2 = 5/27 5 / 3 5/3 5/3
第 13 题
题目
已知 △ A B C \triangle ABC △ A B C A , B , C A,B,C A , B , C a , b , c a,b,c a , b , c b cos C + ( a + c ) ( b sin C − 1 ) = 0 b\cos C+(a+c)(b\sin C-1)=0 b cos C + ( a + c ) ( b sin C − 1 ) = 0 a + c = 3 a+c=\sqrt3 a + c = 3 △ A B C \triangle ABC △ A B C
A. 面积的最大值为 3 3 16 \dfrac{3\sqrt3}{16} 16 3 3
B. 周长的最大值为 3 3 2 \dfrac{3\sqrt3}{2} 2 3 3
C. B = π 3 B=\dfrac\pi3 B = 3 π
D. B = π 4 B=\dfrac\pi4 B = 4 π
解答
由投影关系 b cos C = a − c cos B b\cos C=a-c\cos B b cos C = a − c cos B b sin C = c sin B b\sin C=c\sin B b sin C = c sin B
a − c cos B + 3 c sin B = 3 = a + c . a-c\cos B+\sqrt3c\sin B=\sqrt3=a+c. a − c cos B + 3 c sin B = 3 = a + c . 除以 c > 0 c>0 c > 0 − cos B + 3 sin B = 1 -\cos B+\sqrt3\sin B=1 − cos B + 3 sin B = 1 2 sin ( B − π / 6 ) = 1 2\sin(B-\pi/6)=1 2 sin ( B − π /6 ) = 1 0 < B < π 0<B<\pi 0 < B < π B = π / 3 B=\pi/3 B = π /3
面积 S = a c sin B / 2 = 3 a c / 4 ≤ 3 3 / 16 S=ac\sin B/2=\sqrt3ac/4\le3\sqrt3/16 S = a c sin B /2 = 3 a c /4 ≤ 3 3 /16 a = c = 3 / 2 a=c=\sqrt3/2 a = c = 3 /2
b = a 2 + c 2 − a c = 3 − 3 a c , b=\sqrt{a^2+c^2-ac}=\sqrt{3-3ac}, b = a 2 + c 2 − a c = 3 − 3 a c , 故周长范围为 [ 3 3 / 2 , 2 3 ) [3\sqrt3/2,2\sqrt3) [ 3 3 /2 , 2 3 )
第 14 题
题目
两个半径为 1 1 1 d d d V V V lim d → + ∞ V d 3 \displaystyle\lim_{d\to+\infty}\dfrac{V}{d^3} d → + ∞ lim d 3 V
解答
包含两球的球半径至少为 ( d + 2 ) / 2 (d+2)/2 ( d + 2 ) /2 d + 2 d+2 d + 2 d / 2 + 1 d/2+1 d /2 + 1
lim d → ∞ V d 3 = lim d → ∞ 4 π 3 ( 1 2 + 1 d ) 3 = π 6 . \lim_{d\to\infty}\frac{V}{d^3}
=\lim_{d\to\infty}\frac{4\pi}{3}\left(\frac12+\frac1d\right)^3
=\frac\pi6. d → ∞ lim d 3 V = d → ∞ lim 3 4 π ( 2 1 + d 1 ) 3 = 6 π . 第 15 题
题目
椭圆 x 2 4 + y 2 9 = 1 \dfrac{x^2}{4}+\dfrac{y^2}{9}=1 4 x 2 + 9 y 2 = 1
A. 16 16 16
B. 12 12 12
C. 10 10 10
D. 18 18 18
解答
设两条正交直线对应的半径长为 r 1 , r 2 r_1,r_2 r 1 , r 2 θ \theta θ S = 2 r 1 r 2 S=2r_1r_2 S = 2 r 1 r 2
r 1 − 2 r 2 − 2 = ( cos 2 θ 4 + sin 2 θ 9 ) ( sin 2 θ 4 + cos 2 θ 9 ) = 1 36 + 25 1296 sin 2 θ cos 2 θ . r_1^{-2}r_2^{-2}
=\left(\frac{\cos^2\theta}{4}+\frac{\sin^2\theta}{9}\right)
\left(\frac{\sin^2\theta}{4}+\frac{\cos^2\theta}{9}\right)
=\frac1{36}+\frac{25}{1296}\sin^2\theta\cos^2\theta. r 1 − 2 r 2 − 2 = ( 4 cos 2 θ + 9 sin 2 θ ) ( 4 sin 2 θ + 9 cos 2 θ ) = 36 1 + 1296 25 sin 2 θ cos 2 θ . 因此 r 1 r 2 ∈ [ 72 / 13 , 6 ] r_1r_2\in[72/13,6] r 1 r 2 ∈ [ 72/13 , 6 ] S ∈ [ 144 / 13 , 12 ] S\in[144/13,12] S ∈ [ 144/13 , 12 ] 12 12 12
第 16 题
题目
a 1 , a 2 , … , a 8 a_1,a_2,\ldots,a_8 a 1 , a 2 , … , a 8 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 1,2,3,4,5,6,7,8 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 a 1 + a 3 + a 5 + a 7 = a 2 + a 4 + a 6 + a 8 a_1+a_3+a_5+a_7=a_2+a_4+a_6+a_8 a 1 + a 3 + a 5 + a 7 = a 2 + a 4 + a 6 + a 8
解答
奇数位置四个数之和必须为 18 18 18
1278 , 1368 , 1458 , 1467 , 2358 , 2367 , 2457 , 3456 , 1278,\ 1368,\ 1458,\ 1467,\ 2358,\ 2367,\ 2457,\ 3456, 1278 , 1368 , 1458 , 1467 , 2358 , 2367 , 2457 , 3456 , 每串数字表示一个集合。这个清单可按最小元素 1 , 2 , 3 1,2,3 1 , 2 , 3 4 4 4 18 18 18 4 ! 4! 4 ! 4 ! 4! 4 ! 8 ( 4 ! ) 2 = 4608 8(4!)^2=4608 8 ( 4 ! ) 2 = 4608
第 17 题
题目
甲乙丙丁四个人背后的号码为 1 , 2 , 3 , 4 1,2,3,4 1 , 2 , 3 , 4 2 2 2 3 3 3 2 2 2 4 4 4 2 2 2 3 3 3 1 1 1 3 3 3
解答
已明确号码为 1 , 2 , 3 , 4 1,2,3,4 1 , 2 , 3 , 4 3 3 3 2 2 2 1 1 1 4 4 4 2 2 2 1 , 4 1,4 1 , 4 2 2 2 3 3 3
所以乙不为 3 3 3 2 2 2 1 1 1 2 2 2 3 3 3 4 4 4 3 3 3
第 18 题
题目
已知函数 f ( x ) = sin 3 x + 2 cos 3 x 2 sin 2 x + cos 2 x f(x)=\dfrac{\sin^3x+2\cos^3x}{2\sin^2x+\cos^2x} f ( x ) = 2 sin 2 x + cos 2 x sin 3 x + 2 cos 3 x n ∈ N ∗ n\in\mathbb N^* n ∈ N ∗ ∫ 0 2 n π f ( x ) d x \displaystyle\int_0^{2n\pi}f(x)\,dx ∫ 0 2 nπ f ( x ) d x
A. 与 n n n
B. 0 0 0
C. 1 1 1
D. 2 2 2
解答
分母 2 sin 2 x + cos 2 x 2\sin^2x+\cos^2x 2 sin 2 x + cos 2 x f ( x + π ) = − f ( x ) f(x+\pi)=-f(x) f ( x + π ) = − f ( x ) 2 π 2\pi 2 π ∫ 0 2 n π f ( x ) d x = 0 \int_0^{2n\pi}f(x)\,dx=0 ∫ 0 2 nπ f ( x ) d x = 0
第 19 题
题目
求函数 f ( x ) = [ 2 x ] − 2 [ 1 x ] f(x)=\left[\dfrac2x\right]-2\left[\dfrac1x\right] f ( x ) = [ x 2 ] − 2 [ x 1 ]
解答
定义域为 x ≠ 0 x\ne0 x = 0 t = 1 / x = k + r t=1/x=k+r t = 1/ x = k + r k ∈ Z , 0 ≤ r < 1 k\in\mathbb Z,0\le r<1 k ∈ Z , 0 ≤ r < 1
[ 2 t ] − 2 [ t ] = [ 2 r ] ∈ { 0 , 1 } . [2t]-2[t]=[2r]\in\{0,1\}. [ 2 t ] − 2 [ t ] = [ 2 r ] ∈ { 0 , 1 } . x = 1 x=1 x = 1 0 0 0 x = 4 / 3 x=4/3 x = 4/3 1 1 1 { 0 , 1 } \{0,1\} { 0 , 1 }
第 20 题
题目
求满足 m ∣ 2016 m\mid2016 m ∣ 2016 n ∣ 2016 n\mid2016 n ∣ 2016 m n ∤ 2016 mn\nmid2016 mn ∤ 2016 ( m , n ) (m,n) ( m , n )
解答
已明确 m , n m,n m , n 2016 = 2 5 ⋅ 3 2 ⋅ 7 2016=2^5\cdot3^2\cdot7 2016 = 2 5 ⋅ 3 2 ⋅ 7 6 ⋅ 3 ⋅ 2 = 36 6\cdot3\cdot2=36 6 ⋅ 3 ⋅ 2 = 36 36 2 = 1296 36^2=1296 3 6 2 = 1296 2016 2016 2016 e e e ( e + 1 ) ( e + 2 ) / 2 (e+1)(e+2)/2 ( e + 1 ) ( e + 2 ) /2 21 ⋅ 6 ⋅ 3 = 378 21\cdot6\cdot3=378 21 ⋅ 6 ⋅ 3 = 378 1296 − 378 = 918 1296-378=918 1296 − 378 = 918
第 21 题
题目
正方形 A B C D ABCD A B C D O O O △ O A B , △ O B C , △ O C D , △ O D A \triangle OAB,\triangle OBC,\triangle OCD,\triangle ODA △ O A B , △ O B C , △ O C D , △ O D A O O O
解答
将正方形取为 A = ( 0 , 0 ) , B = ( 1 , 0 ) , C = ( 1 , 1 ) , D = ( 0 , 1 ) A=(0,0),B=(1,0),C=(1,1),D=(0,1) A = ( 0 , 0 ) , B = ( 1 , 0 ) , C = ( 1 , 1 ) , D = ( 0 , 1 ) O = ( x , y ) O=(x,y) O = ( x , y )
若 x = 1 / 2 x=1/2 x = 1/2 O A = O B , O C = O D OA=OB,OC=OD O A = O B , O C = O D
y = 1 2 , 或 y 2 + 1 4 = 1 , 或 ( y − 1 ) 2 + 1 4 = 1. y=\frac12,\qquad\text{或}\qquad y^2+\frac14=1,\qquad\text{或}\qquad (y-1)^2+\frac14=1. y = 2 1 , 或 y 2 + 4 1 = 1 , 或 ( y − 1 ) 2 + 4 1 = 1. 得到五点 y = 1 / 2 , ± 3 / 2 , 1 ± 3 / 2 y=1/2,\pm\sqrt3/2,1\pm\sqrt3/2 y = 1/2 , ± 3 /2 , 1 ± 3 /2 y = 1 / 2 y=1/2 y = 1/2
若 x ≠ 1 / 2 , y ≠ 1 / 2 x\ne1/2,y\ne1/2 x = 1/2 , y = 1/2 O O O O O O 1 1 1 1 1 1 1 / 2 1/\sqrt2 1/ 2 O O O 1 1 1 9 9 9
第 22 题
题目
已知所有元素均为非负实数的集合 A A A a i , a j ∈ A a_i,a_j\in A a i , a j ∈ A a i ⩾ a j a_i\geqslant a_j a i ⩾ a j a i + a j ∈ A a_i+a_j\in A a i + a j ∈ A a i − a j ∈ A a_i-a_j\in A a i − a j ∈ A A A A A A A
A. 3 3 3
B. 4 4 4
C. 5 5 5
D. 6 6 6
解答
三元素条件要求互异，否则任取同一元素三次即已构成等差数列；现已明确。若 A A A M M M ( M , M ) (M,M) ( M , M ) 0 ∈ A 0\in A 0 ∈ A
对任意 x ∈ A ∖ { 0 , M } x\in A\setminus\{0,M\} x ∈ A ∖ { 0 , M } M + x > M M+x>M M + x > M M − x ∈ A M-x\in A M − x ∈ A x = M / 2 x=M/2 x = M /2 0 , x , M 0,x,M 0 , x , M 3 , 5 3,5 3 , 5
若有六个元素，设为 0 < a < b < M − b < M − a < M 0<a<b<M-b<M-a<M 0 < a < b < M − b < M − a < M a , b a,b a , b b − a b-a b − a A A A a a a 0 , a , b 0,a,b 0 , a , b a + b ∈ A a+b\in A a + b ∈ A a + b < M a+b<M a + b < M M − b M-b M − b M − a M-a M − a M − a = 2 b M-a=2b M − a = 2 b 0 , b , 2 b 0,b,2b 0 , b , 2 b M − b = 2 a M-b=2a M − b = 2 a 0 , a , 2 a 0,a,2a 0 , a , 2 a
令 φ = ( 1 + 5 ) / 2 \varphi=(1+\sqrt5)/2 φ = ( 1 + 5 ) /2 A = { 0 , 1 , φ , 1 + φ } A=\{0,1,\varphi,1+\varphi\} A = { 0 , 1 , φ , 1 + φ } 1 + φ 1+\varphi 1 + φ ( 1 + φ ) − 1 = φ (1+\varphi)-1=\varphi ( 1 + φ ) − 1 = φ ( 1 + φ ) − φ = 1 (1+\varphi)-\varphi=1 ( 1 + φ ) − φ = 1 1 , φ 1,\varphi 1 , φ 2 2 2 2 φ 2\varphi 2 φ
第 23 题
题目
已知关于 z z z z 2017 − 1 = 0 z^{2017}-1=0 z 2017 − 1 = 0 z i z_i z i i = 1 , 2 , … , 2017 i=1,2,\ldots,2017 i = 1 , 2 , … , 2017 ∑ i = 1 2017 1 2 − z i \displaystyle\sum_{i=1}^{2017}\dfrac1{2-z_i} i = 1 ∑ 2017 2 − z i 1
A. 是比 2017 2 \dfrac{2017}{2} 2 2017
B. 是比 2017 2 \dfrac{2017}{2} 2 2017
C. 是有理数
D. 不是有理数
解答
对 P ( z ) = z 2017 − 1 = ∏ i ( z − z i ) P(z)=z^{2017}-1=\prod_i(z-z_i) P ( z ) = z 2017 − 1 = ∏ i ( z − z i ) z = 2 z=2 z = 2
∑ i = 1 2017 1 2 − z i = P ′ ( 2 ) P ( 2 ) = 2017 ⋅ 2 2016 2 2017 − 1 = 2017 2 ( 1 + 1 2 2017 − 1 ) . \sum_{i=1}^{2017}\frac1{2-z_i}
=\frac{P'(2)}{P(2)}
=\frac{2017\cdot2^{2016}}{2^{2017}-1}
=\frac{2017}{2}\left(1+\frac1{2^{2017}-1}\right). i = 1 ∑ 2017 2 − z i 1 = P ( 2 ) P ′ ( 2 ) = 2 2017 − 1 2017 ⋅ 2 2016 = 2 2017 ( 1 + 2 2017 − 1 1 ) . 这是大于 2017 / 2 2017/2 2017/2
第 24 题
题目
已知复数 x , y x,y x , y x + y = x 4 + y 4 = 1 x+y=x^4+y^4=1 x + y = x 4 + y 4 = 1 x y xy x y
解答
记 s = x y s=xy s = x y x + y = 1 x+y=1 x + y = 1
x 4 + y 4 = ( 1 − 2 s ) 2 − 2 s 2 = 1 − 4 s + 2 s 2 . x^4+y^4=(1-2s)^2-2s^2=1-4s+2s^2. x 4 + y 4 = ( 1 − 2 s ) 2 − 2 s 2 = 1 − 4 s + 2 s 2 . 它等于 1 1 1 s = 0 s=0 s = 0 2 2 2 s s s t 2 − t + s = 0 t^2-t+s=0 t 2 − t + s = 0 x , y x,y x , y
第 25 题
题目
已知函数 f f f f ( m + 1 , n + 1 ) = f ( m , n ) + f ( m + 1 , n ) + n f(m+1,n+1)=f(m,n)+f(m+1,n)+n f ( m + 1 , n + 1 ) = f ( m , n ) + f ( m + 1 , n ) + n f ( m , 1 ) = 1 f(m,1)=1 f ( m , 1 ) = 1 f ( 1 , n ) = n f(1,n)=n f ( 1 , n ) = n m , n ∈ N ∗ m,n\in\mathbb N^* m , n ∈ N ∗
A. 使 f ( 2 , n ) ⩾ 100 f(2,n)\geqslant100 f ( 2 , n ) ⩾ 100 n n n 11 11 11
B. 使 f ( 2 , n ) ⩾ 100 f(2,n)\geqslant100 f ( 2 , n ) ⩾ 100 n n n 13 13 13
C. 使 f ( 3 , n ) ⩾ 100 f(3,n)\geqslant100 f ( 3 , n ) ⩾ 100 n n n 19 19 19
D. 使 f ( 3 , n ) ⩾ 100 f(3,n)\geqslant100 f ( 3 , n ) ⩾ 100 n n n 20 20 20
解答
取 m = 1 m=1 m = 1 f ( 2 , n + 1 ) − f ( 2 , n ) = 2 n f(2,n+1)-f(2,n)=2n f ( 2 , n + 1 ) − f ( 2 , n ) = 2 n f ( 2 , 1 ) = 1 f(2,1)=1 f ( 2 , 1 ) = 1 f ( 2 , n ) = n ( n − 1 ) + 1 f(2,n)=n(n-1)+1 f ( 2 , n ) = n ( n − 1 ) + 1 n = 10 n=10 n = 10 91 91 91 n = 11 n=11 n = 11 111 111 111 n n n
再取 m = 2 m=2 m = 2 f ( 3 , n + 1 ) − f ( 3 , n ) = n 2 + 1 f(3,n+1)-f(3,n)=n^2+1 f ( 3 , n + 1 ) − f ( 3 , n ) = n 2 + 1
f ( 3 , n ) = n + ( n − 1 ) n ( 2 n − 1 ) 6 . f(3,n)=n+\frac{(n-1)n(2n-1)}6. f ( 3 , n ) = n + 6 ( n − 1 ) n ( 2 n − 1 ) . n = 7 n=7 n = 7 98 98 98 n = 8 n=8 n = 8 148 148 148 n n n 8 8 8
第 26 题
题目
已知 f ( x ) f(x) f ( x ) ( 0 , + ∞ ) (0,+\infty) ( 0 , + ∞ ) g ( x ) = sup 0 < t ⩽ x f ( t ) g(x)=\sup_{0<t\leqslant x}f(t) g ( x ) = sup 0 < t ⩽ x f ( t ) t t t
A. g ( x ) g(x) g ( x )
B. g ( x ) g(x) g ( x )
C. g ( x ) g(x) g ( x )
D. g ( x ) g(x) g ( x )
解答
题面校注：原稿写 max 0 < n ≤ x f ( n ) \max_{0<n\le x}f(n) max 0 < n ≤ x f ( n ) f ( t ) = e − t f(t)=e^{-t} f ( t ) = e − t 0 < x < 1 0<x<1 0 < x < 1
因为 f f f g g g x 1 < x 2 x_1<x_2 x 1 < x 2 g ( x 1 ) ≤ g ( x 2 ) g(x_1)\le g(x_2) g ( x 1 ) ≤ g ( x 2 )
固定 x 0 > 0 x_0>0 x 0 > 0 ε > 0 \varepsilon>0 ε > 0 f f f x 0 x_0 x 0 δ < x 0 / 2 \delta<x_0/2 δ < x 0 /2 ∣ t − x 0 ∣ < 2 δ |t-x_0|<2\delta ∣ t − x 0 ∣ < 2 δ ∣ f ( t ) − f ( x 0 ) ∣ < ε |f(t)-f(x_0)|<\varepsilon ∣ f ( t ) − f ( x 0 ) ∣ < ε H = sup 0 < t ≤ x 0 − δ f ( t ) H=\sup_{0<t\le x_0-\delta}f(t) H = sup 0 < t ≤ x 0 − δ f ( t ) ∣ x − x 0 ∣ < δ / 2 |x-x_0|<\delta/2 ∣ x − x 0 ∣ < δ /2 ( x 0 − δ , x ] (x_0-\delta,x] ( x 0 − δ , x ] f f f ( f ( x 0 ) − ε , f ( x 0 ) + ε ) (f(x_0)-\varepsilon,f(x_0)+\varepsilon) ( f ( x 0 ) − ε , f ( x 0 ) + ε )
max { H , f ( x 0 ) − ε } ≤ g ( x ) ≤ max { H , f ( x 0 ) + ε } . \max\{H,f(x_0)-\varepsilon\}\le g(x)\le
\max\{H,f(x_0)+\varepsilon\}. max { H , f ( x 0 ) − ε } ≤ g ( x ) ≤ max { H , f ( x 0 ) + ε } . g ( x 0 ) g(x_0) g ( x 0 ) ∣ g ( x ) − g ( x 0 ) ∣ ≤ 2 ε |g(x)-g(x_0)|\le2\varepsilon ∣ g ( x ) − g ( x 0 ) ∣ ≤ 2 ε g g g
第 27 题
题目
已知对任意实数 x x x a cos x + b cos 3 x ⩽ 1 a\cos x+b\cos3x\leqslant1 a cos x + b cos 3 x ⩽ 1
A. ∣ a − 2 b ∣ ⩽ 2 |a-2b|\leqslant2 ∣ a − 2 b ∣ ⩽ 2
B. ∣ a + b ∣ ⩽ 1 |a+b|\leqslant1 ∣ a + b ∣ ⩽ 1
A. ∣ a − b ∣ ⩽ 2 |a-b|\leqslant\sqrt2 ∣ a − b ∣ ⩽ 2
B. ∣ a + 2 b ∣ ⩽ 2 |a+2b|\leqslant2 ∣ a + 2 b ∣ ⩽ 2
解答
将 x x x x + π x+\pi x + π ∣ a cos x + b cos 3 x ∣ ≤ 1 |a\cos x+b\cos3x|\le1 ∣ a cos x + b cos 3 x ∣ ≤ 1 x = π / 3 , 0 , π / 4 x=\pi/3,0,\pi/4 x = π /3 , 0 , π /4
∣ a − 2 b ∣ ≤ 2 , ∣ a + b ∣ ≤ 1 , ∣ a − b ∣ ≤ 2 . |a-2b|\le2,\qquad |a+b|\le1,\qquad |a-b|\le\sqrt2. ∣ a − 2 b ∣ ≤ 2 , ∣ a + b ∣ ≤ 1 , ∣ a − b ∣ ≤ 2 . 最后
a + 2 b = 4 3 ( a + b ) − 1 3 ( a − 2 b ) , a+2b=\frac43(a+b)-\frac13(a-2b), a + 2 b = 3 4 ( a + b ) − 3 1 ( a − 2 b ) , 三角不等式给出 ∣ a + 2 b ∣ ≤ 4 / 3 + 2 / 3 = 2 |a+2b|\le4/3+2/3=2 ∣ a + 2 b ∣ ≤ 4/3 + 2/3 = 2
第 28 题
题目
5 5 5 i i i x i x_i x i i = 1 , 2 , 3 , 4 , 5 i=1,2,3,4,5 i = 1 , 2 , 3 , 4 , 5 y i y_i y i i = 1 , 2 , 3 , 4 , 5 i=1,2,3,4,5 i = 1 , 2 , 3 , 4 , 5
A. x 1 + x 2 + x 3 + x 4 + x 5 x_1+x_2+x_3+x_4+x_5 x 1 + x 2 + x 3 + x 4 + x 5
B. y 1 + y 2 + y 3 + y 4 + y 5 y_1+y_2+y_3+y_4+y_5 y 1 + y 2 + y 3 + y 4 + y 5
C. x 1 2 + x 2 2 + x 3 2 + x 4 2 + x 5 2 x_1^2+x_2^2+x_3^2+x_4^2+x_5^2 x 1 2 + x 2 2 + x 3 2 + x 4 2 + x 5 2
D. y 1 2 + y 2 2 + y 3 2 + y 4 2 + y 5 2 y_1^2+y_2^2+y_3^2+y_4^2+y_5^2 y 1 2 + y 2 2 + y 3 2 + y 4 2 + y 5 2
解答
每场必分胜负，共有 ( 5 2 ) = 10 \binom52=10 ( 2 5 ) = 10 1 1 1 10 10 10
若五人胜场数为 ( 0 , 1 , 2 , 3 , 4 ) (0,1,2,3,4) ( 0 , 1 , 2 , 3 , 4 ) 30 30 30 20 20 20 ( 4 , 3 , 2 , 1 , 0 ) (4,3,2,1,0) ( 4 , 3 , 2 , 1 , 0 ) ( 2 , 2 , 2 , 2 , 2 ) (2,2,2,2,2) ( 2 , 2 , 2 , 2 , 2 )
第 29 题
题目
若存在满足下列三个条件的集合 A , B , C A,B,C A , B , C n n n
A , B , C A,B,C A , B , C M = { 1 , 2 , … , n } M=\{1,2,\ldots,n\} M = { 1 , 2 , … , n } 3 3 3 A ∪ B ∪ C = M A\cup B\cup C=M A ∪ B ∪ C = M A A A B B B 3 3 3 C C C 集合 A , B , C A,B,C A , B , C S 1 , S 2 , S 3 S_1,S_2,S_3 S 1 , S 2 , S 3 S 1 = S 2 = S 3 S_1=S_2=S_3 S 1 = S 2 = S 3
下列说法正确的是（ ）
A. 8 8 8
B. 60 60 60
C. 68 68 68
D. 80 80 80
解答
每组的和必须为 T = n ( n + 1 ) / 6 T=n(n+1)/6 T = n ( n + 1 ) /6 3 3 3 C C C
n = 8 n=8 n = 8 A = { 5 , 7 } , B = { 4 , 8 } , C = { 1 , 2 , 3 , 6 } A=\{5,7\},B=\{4,8\},C=\{1,2,3,6\} A = { 5 , 7 } , B = { 4 , 8 } , C = { 1 , 2 , 3 , 6 } 12 12 12
n = 60 n=60 n = 60 T = 610 T=610 T = 610 3 3 3 3 ( 1 + ⋯ + 20 ) = 630 > T 3(1+\cdots+20)=630>T 3 ( 1 + ⋯ + 20 ) = 630 > T
n = 68 n=68 n = 68 3 3 3 759 759 759 C C C { 2 , 10 , 11 } \{2,10,11\} { 2 , 10 , 11 } 782 = T 782=T 782 = T A A A B B B 3 3 3 34 2 − 3 ⋅ 11 2 = 793 34^2-3\cdot11^2=793 3 4 2 − 3 ⋅ 1 1 2 = 793 11 11 11 782 782 782 B B B 782 782 782
n = 80 n=80 n = 80 3 3 3 1053 1053 1053 C C C { 13 , 14 } \{13,14\} { 13 , 14 } 1080 = T 1080=T 1080 = T 3 3 3 40 2 − 3 ⋅ 13 2 = 1093 40^2-3\cdot13^2=1093 4 0 2 − 3 ⋅ 1 3 2 = 1093 13 13 13 1080 1080 1080 1080 1080 1080
第 30 题
题目
已知非零实数 a , b , c , A , B , C a,b,c,A,B,C a , b , c , A , B , C a x 2 + b x + c ⩾ 0 ax^2+bx+c\geqslant0 a x 2 + b x + c ⩾ 0 A x 2 + B x + C ⩾ 0 Ax^2+Bx+C\geqslant0 A x 2 + B x + C ⩾ 0 a A = b B = c C \dfrac aA=\dfrac bB=\dfrac cC A a = B b = C c
A. 充分非必要条件
B. 必要非充分条件
C. 充要条件
D. 既非充分也非必要条件
解答
解集相同不推出系数成比例：x 2 + x + 1 x^2+x+1 x 2 + x + 1 x 2 + 2 x + 2 x^2+2x+2 x 2 + 2 x + 2 R \mathbb R R
系数成比例也不推出解集相同，因为比例可以为负。例如 x 2 + x + 1 ≥ 0 x^2+x+1\ge0 x 2 + x + 1 ≥ 0 R \mathbb R R − x 2 − x − 1 ≥ 0 -x^2-x-1\ge0 − x 2 − x − 1 ≥ 0 − 1 -1 − 1
第 31 题
题目
一个人每次投篮独立，命中率为 2 3 \dfrac23 3 2 4 4 4
解答
已补明各次投篮独立。恰在第四次停止，需要第三、四次命中，第二次未中，第一次任意；这同时保证此前没有连续两次命中。因此概率为 ( 1 / 3 ) ( 2 / 3 ) 2 = 4 / 27 (1/3)(2/3)^2=4/27 ( 1/3 ) ( 2/3 ) 2 = 4/27
第 32 题
题目
已知 0 < P ( A ) < 1 0<P(A)<1 0 < P ( A ) < 1 0 < P ( B ) < 1 0<P(B)<1 0 < P ( B ) < 1 P ( A ∣ B ) = 1 P(A\mid B)=1 P ( A ∣ B ) = 1
A. P ( A ‾ ∣ B ‾ ) = 0 P(\overline A\mid\overline B)=0 P ( A ∣ B ) = 0
B. P ( B ‾ ∣ A ‾ ) = 1 P(\overline B\mid\overline A)=1 P ( B ∣ A ) = 1
C. P ( A ∪ B ) = P ( A ) P(A\cup B)=P(A) P ( A ∪ B ) = P ( A )
D. P ( B ‾ ∣ A ) = 1 P(\overline B\mid A)=1 P ( B ∣ A ) = 1
解答
P ( A ∣ B ) = 1 P(A\mid B)=1 P ( A ∣ B ) = 1 B ∖ A B\setminus A B ∖ A
P ( B ‾ ∣ A ‾ ) = 1 , P ( A ∪ B ) = P ( A ) . P(\overline B\mid\overline A)=1,\qquad P(A\cup B)=P(A). P ( B ∣ A ) = 1 , P ( A ∪ B ) = P ( A ) . 故 B、C 成立。又
P ( A ‾ ∣ B ‾ ) = 1 − P ( A ) 1 − P ( B ) > 0 , P ( B ‾ ∣ A ) = 1 − P ( B ) P ( A ) < 1 , P(\overline A\mid\overline B)=\frac{1-P(A)}{1-P(B)}>0,\qquad
P(\overline B\mid A)=1-\frac{P(B)}{P(A)}<1, P ( A ∣ B ) = 1 − P ( B ) 1 − P ( A ) > 0 , P ( B ∣ A ) = 1 − P ( A ) P ( B ) < 1 , A、D 不成立。选 BC。
第 33 题
题目
已知实数 x , y x,y x , y
{ ( x − 1 ) ( y 2 + 6 ) = x ( y 2 + 1 ) , ( y − 1 ) ( x 2 + 6 ) = y ( x 2 + 1 ) , \begin{cases}
(x-1)(y^2+6)=x(y^2+1),\\
(y-1)(x^2+6)=y(x^2+1),
\end{cases} { ( x − 1 ) ( y 2 + 6 ) = x ( y 2 + 1 ) , ( y − 1 ) ( x 2 + 6 ) = y ( x 2 + 1 ) , 则（ ）
A. ( x − 5 2 ) 2 + ( y − 5 2 ) 2 = 1 2 \left(x-\dfrac52\right)^2+\left(y-\dfrac52\right)^2=\dfrac12 ( x − 2 5 ) 2 + ( y − 2 5 ) 2 = 2 1
B. x = y x=y x = y
C. 有 3 3 3 ( x , y ) (x,y) ( x , y )
D. 有 4 4 4 ( x , y ) (x,y) ( x , y )
解答
两式整理为 y 2 = 5 x − 6 , x 2 = 5 y − 6 y^2=5x-6,x^2=5y-6 y 2 = 5 x − 6 , x 2 = 5 y − 6
( x − 5 / 2 ) 2 + ( y − 5 / 2 ) 2 = 1 / 2 , (x-5/2)^2+(y-5/2)^2=1/2, ( x − 5/2 ) 2 + ( y − 5/2 ) 2 = 1/2 , A 正确；相减得 ( x − y ) ( x + y + 5 ) = 0 (x-y)(x+y+5)=0 ( x − y ) ( x + y + 5 ) = 0 x + y ≥ 4 x+y\ge4 x + y ≥ 4 x = y x=y x = y x 2 − 5 x + 6 = 0 x^2-5x+6=0 x 2 − 5 x + 6 = 0 ( 2 , 2 ) , ( 3 , 3 ) (2,2),(3,3) ( 2 , 2 ) , ( 3 , 3 )
第 34 题
题目
在 △ A B C \triangle ABC △ A B C sin 2 A = sin 2 B + sin B sin C \sin^2A=\sin^2B+\sin B\sin C sin 2 A = sin 2 B + sin B sin C A − 2 B A-2B A − 2 B
解答
利用平方差恒等式和 A + B = π − C A+B=\pi-C A + B = π − C
sin 2 A − sin 2 B = sin ( A + B ) sin ( A − B ) = sin C sin ( A − B ) . \sin^2A-\sin^2B=\sin(A+B)\sin(A-B)=\sin C\sin(A-B). sin 2 A − sin 2 B = sin ( A + B ) sin ( A − B ) = sin C sin ( A − B ) . 因 sin C > 0 \sin C>0 sin C > 0 sin ( A − B ) = sin B \sin(A-B)=\sin B sin ( A − B ) = sin B 0 < A − B < π 0<A-B<\pi 0 < A − B < π A − B = B A-B=B A − B = B A − B = π − B A-B=\pi-B A − B = π − B A = π A=\pi A = π A = 2 B A=2B A = 2 B 0 0 0
第 35 题
题目
已知 Q ( x ) = a 2017 x 2017 + a 2016 x 2016 + ⋯ + a 1 x + a 0 Q(x)=a_{2017}x^{2017}+a_{2016}x^{2016}+\cdots+a_1x+a_0 Q ( x ) = a 2017 x 2017 + a 2016 x 2016 + ⋯ + a 1 x + a 0 x x x Q ( x ) > 0 Q(x)>0 Q ( x ) > 0 a i ∈ { − 1 , 1 } a_i\in\{-1,1\} a i ∈ { − 1 , 1 } i = 0 , 1 , 2 , … , 2017 i=0,1,2,\ldots,2017 i = 0 , 1 , 2 , … , 2017 a 0 , a 1 , a 2 , … , a 2017 a_0,a_1,a_2,\ldots,a_{2017} a 0 , a 1 , a 2 , … , a 2017 − 1 -1 − 1
解答
设负系数个数为 m m m x = 1 x=1 x = 1 2018 − 2 m > 0 2018-2m>0 2018 − 2 m > 0 m ≤ 1008 m\le1008 m ≤ 1008
此界可达到。取
Q ( x ) = 1 − x + x 2 − x 3 + ⋯ − x 2015 + x 2016 + x 2017 . Q(x)=1-x+x^2-x^3+\cdots-x^{2015}+x^{2016}+x^{2017}. Q ( x ) = 1 − x + x 2 − x 3 + ⋯ − x 2015 + x 2016 + x 2017 . 其中恰有 1008 1008 1008 x > 0 x>0 x > 0
Q ( x ) = 1 + x 2017 1 + x + x 2017 > 0. Q(x)=\frac{1+x^{2017}}{1+x}+x^{2017}>0. Q ( x ) = 1 + x 1 + x 2017 + x 2017 > 0. 因此负系数最多为 1008 1008 1008
