答案速览
题号
答案
1
{ 2 , 5 , 7 } \{2,5,7\} { 2 , 5 , 7 }
2
必要不充分
3
非零偶函数；中心为( 3 π / 2 , 0 ) (3\pi/2,0) ( 3 π /2 , 0 )
4
1 / 2 1/2 1/2
5
1 / 2 1/2 1/2
6
3 / 2 \sqrt3/2 3 /2
7
C
8
20 2 20\sqrt2 20 2
9
y = ± ( 1 + 3 ) x y=\pm(1+\sqrt3)x y = ± ( 1 + 3 ) x
10
9 / 8 9/8 9/8
11
k 1 + ⋯ + k n = 2 n + 1 + n − 2 k_1+\cdots+k_n=2^{n+1}+n-2 k 1 + ⋯ + k n = 2 n + 1 + n − 2
12
平行；正弦值15 / 10 \sqrt{15}/10 15 /10
13
y = x − 1 y=x-1 y = x − 1 8 8 8
14
见证明
逐题解析
第 1 题
题目
U = { 1 , 2 , 3 , 4 , 5 , 6 , 7 } U=\{1,2,3,4,5,6,7\} U = { 1 , 2 , 3 , 4 , 5 , 6 , 7 } A ∩ ∁ U B = { 1 , 3 , 4 , 6 } A\cap\complement_UB=\{1,3,4,6\} A ∩ ∁ U B = { 1 , 3 , 4 , 6 } ∁ U A ∪ B \complement_UA\cup B ∁ U A ∪ B { 2 , 5 , 7 } \{2,5,7\} { 2 , 5 , 7 } { 1 , 3 , 4 , 6 } \{1,3,4,6\} { 1 , 3 , 4 , 6 } U U U ∅ \varnothing ∅
解答
由德摩根律，∁ U A ∪ B = ∁ U ( A ∩ ∁ U B ) \complement_UA\cup B=\complement_U(A\cap\complement_UB) ∁ U A ∪ B = ∁ U ( A ∩ ∁ U B ) U ∖ { 1 , 3 , 4 , 6 } = { 2 , 5 , 7 } U\setminus\{1,3,4,6\}=\{2,5,7\} U ∖ { 1 , 3 , 4 , 6 } = { 2 , 5 , 7 }
第 2 题
题目
m , n m,n m , n α , β \alpha,\beta α , β m ⊥ α , n ⊂ β m\perp\alpha,n\subset\beta m ⊥ α , n ⊂ β m ⊥ n m\perp n m ⊥ n α ∥ β \alpha\parallel\beta α ∥ β
解答
若α ∥ β \alpha\parallel\beta α ∥ β m m m β \beta β β \beta β n n n m ⊥ n m\perp n m ⊥ n m m m m m m α : z = 0 , m \alpha:z=0,m α : z = 0 , m z z z β : y = 0 , n \beta:y=0,n β : y = 0 , n x x x
第 3 题
题目
f ( x ) = a sin x − b cos x f(x)=a\sin x-b\cos x f ( x ) = a sin x − b cos x a 2 + b 2 > 0 a^2+b^2>0 a 2 + b 2 > 0 x = π / 4 x=\pi/4 x = π /4 f ( x + π / 4 ) f(x+\pi/4) f ( x + π /4 ) ( π , 0 ) (\pi,0) ( π , 0 ) ( 3 π / 2 , 0 ) (3\pi/2,0) ( 3 π /2 , 0 )
解答
在π / 4 \pi/4 π /4 a + b = 0 a+b=0 a + b = 0 a > 0 a>0 a > 0 f ( x ) = a ( sin x + cos x ) f(x)=a(\sin x+\cos x) f ( x ) = a ( sin x + cos x )
f ( x + π / 4 ) = 2 a cos x . f(x+\pi/4)=\sqrt2a\cos x. f ( x + π /4 ) = 2 a cos x . 这是偶函数而非奇函数，对称中心为( π / 2 + k π , 0 ) (\pi/2+k\pi,0) ( π /2 + k π , 0 ) ( 3 π / 2 , 0 ) (3\pi/2,0) ( 3 π /2 , 0 ) a 2 + b 2 > 0 a^2+b^2>0 a 2 + b 2 > 0
第 4 题
题目
固定非零向量a ⃗ , b ⃗ \vec a,\vec b a , b ∣ a ⃗ − b ⃗ ∣ = 4 |\vec a-\vec b|=4 ∣ a − b ∣ = 4 a ⃗ − b ⃗ \vec a-\vec b a − b a ⃗ \vec a a π / 6 \pi/6 π /6 c ⃗ \vec c c ( c ⃗ − a ⃗ ) ⋅ ( c ⃗ − b ⃗ ) = − 15 / 4 (\vec c-\vec a)\cdot(\vec c-\vec b)=-15/4 ( c − a ) ⋅ ( c − b ) = − 15/4 c ⃗ \vec c c λ \lambda λ ∣ c ⃗ − λ a ⃗ ∣ |\vec c-\lambda\vec a| ∣ c − λ a ∣ 1 , 2 3 , 1 2 , 3 4 1,\frac23,\frac12,\frac34 1 , 3 2 , 2 1 , 4 3
解答
令M = ( a ⃗ + b ⃗ ) / 2 M=(\vec a+\vec b)/2 M = ( a + b ) /2
∣ c ⃗ − M ∣ 2 = ∣ a ⃗ − b ⃗ ∣ 2 4 − 15 4 = 1 4 . |\vec c-M|^2=\frac{|\vec a-\vec b|^2}{4}-\frac{15}4=\frac14. ∣ c − M ∣ 2 = 4 ∣ a − b ∣ 2 − 4 15 = 4 1 . 所以c ⃗ \vec c c M M M 1 / 2 1/2 1/2 λ a ⃗ \lambda\vec a λ a L = R a ⃗ L=\mathbb R\vec a L = R a
d ( M , L ) = 1 2 ∣ a ⃗ − b ⃗ ∣ sin ( π / 6 ) = 1. d(M,L)=\frac12|\vec a-\vec b|\sin(\pi/6)=1. d ( M , L ) = 2 1 ∣ a − b ∣ sin ( π /6 ) = 1. 故两集合最短距离为1 − 1 / 2 = 1 / 2 1-1/2=1/2 1 − 1/2 = 1/2 1 / 2 1/2 1/2
第 5 题
题目
随机变量 X X X 0 , 1 , 2 0,1,2 0 , 1 , 2 a , b , c a,b,c a , b , c 4 a , b , c 4a,b,c 4 a , b , c D ( X ) D(X) D ( X ) 1 6 , 1 3 , 1 2 , 1 \frac16,\frac13,\frac12,1 6 1 , 3 1 , 2 1 , 1
解答
概率非负且a + b + c = 1 a+b+c=1 a + b + c = 1 b 2 = 4 a c b^2=4ac b 2 = 4 a c b = 2 a c b=2\sqrt{ac} b = 2 a c a + c = 1 \sqrt a+\sqrt c=1 a + c = 1 p = c , q = a p=\sqrt c,q=\sqrt a p = c , q = a ( q 2 , 2 p q , p 2 ) (q^2,2pq,p^2) ( q 2 , 2 pq , p 2 ) p + q = 1 p+q=1 p + q = 1
E ( X ) = 2 p , E ( X 2 ) = 2 p q + 4 p 2 , D ( X ) = 2 p q ≤ 1 / 2. E(X)=2p,\quad E(X^2)=2pq+4p^2,\quad D(X)=2pq\le1/2. E ( X ) = 2 p , E ( X 2 ) = 2 pq + 4 p 2 , D ( X ) = 2 pq ≤ 1/2. p = q = 1 / 2 p=q=1/2 p = q = 1/2
第 6 题
题目
P , Q , M P,Q,M P , Q , M C : x 2 / a 2 + y 2 / b 2 = 1 C:x^2/a^2+y^2/b^2=1 C : x 2 / a 2 + y 2 / b 2 = 1 a > b > 0 a>b>0 a > b > 0 O O O △ P Q M \triangle PQM △ P QM M ( 2 2 a , 2 2 b ) M(\frac{\sqrt2}2a,\frac{\sqrt2}2b) M ( 2 2 a , 2 2 b ) P Q PQ P Q − 1 2 -\frac12 − 2 1 2 3 , 3 3 , 2 2 , 3 2 \frac{\sqrt2}3,\frac{\sqrt3}3,\frac{\sqrt2}2,\frac{\sqrt3}2 3 2 , 3 3 , 2 2 , 2 3
解答
设H H H P Q PQ P Q H = − M / 2 H=-M/2 H = − M /2 H = ( − a / ( 2 2 ) , − b / ( 2 2 ) ) H=(-a/(2\sqrt2),-b/(2\sqrt2)) H = ( − a / ( 2 2 ) , − b / ( 2 2 ))
k = − b 2 x H a 2 y H = − b a . k=-\frac{b^2x_H}{a^2y_H}=-\frac ba. k = − a 2 y H b 2 x H = − a b . 由k = − 1 / 2 k=-1/2 k = − 1/2 b / a = 1 / 2 b/a=1/2 b / a = 1/2 e = 1 − b 2 / a 2 = 3 / 2 e=\sqrt{1-b^2/a^2}=\sqrt3/2 e = 1 − b 2 / a 2 = 3 /2
第 7 题
题目
a 1 = 1 2 a_1=\frac12 a 1 = 2 1 a n + 1 = e a n − 1 a_{n+1}=e^{a_n-1} a n + 1 = e a n − 1 T n T_n T n n n n a 100 < 1 / 2 a_{100}<1/2 a 100 < 1/2 a 100 > 1 a_{100}>1 a 100 > 1 T 99 ∈ ( 0 , 1 / 100 ) T_{99}\in(0,1/100) T 99 ∈ ( 0 , 1/100 ) T 99 ∈ ( 1 / 100 , 1 / 10 ) T_{99}\in(1/100,1/10) T 99 ∈ ( 1/100 , 1/10 )
解答
在0 < x < 1 0<x<1 0 < x < 1 e x − 1 > x e^{x-1}>x e x − 1 > x 1 1 1 a n a_n a n ( 0 , 1 ) (0,1) ( 0 , 1 ) a n ≤ n / ( n + 1 ) a_n\le n/(n+1) a n ≤ n / ( n + 1 ) n n n
a n + 1 ≤ e − 1 / ( n + 1 ) < n + 1 n + 2 , a_{n+1}\le e^{-1/(n+1)}<\frac{n+1}{n+2}, a n + 1 ≤ e − 1/ ( n + 1 ) < n + 2 n + 1 , 最后用e u > 1 + u e^u>1+u e u > 1 + u u > 0 u>0 u > 0
0 < T 99 < ∏ n = 1 99 n n + 1 = 1 100 . 0<T_{99}<\prod_{n=1}^{99}\frac n{n+1}=\frac1{100}. 0 < T 99 < n = 1 ∏ 99 n + 1 n = 100 1 . 故C正确，D错误。
第 8 题
题目
在 △ A B C \triangle ABC △ A B C D D D B C BC B C E E E B C BC B C A E = A C = B E AE=AC=BE A E = A C = B E D E = 1 DE=1 D E = 1 cos C = 1 / 3 \cos C=1/3 cos C = 1/3 △ A B C \triangle ABC △ A B C
解答
设A C = A E = B E = L AC=AE=BE=L A C = A E = B E = L C = ( 0 , 0 ) , B = ( d , 0 ) C=(0,0),B=(d,0) C = ( 0 , 0 ) , B = ( d , 0 ) cos C = 1 / 3 \cos C=1/3 cos C = 1/3 A = ( L / 3 , 2 2 L / 3 ) A=(L/3,2\sqrt2L/3) A = ( L /3 , 2 2 L /3 ) E = ( e , 0 ) E=(e,0) E = ( e , 0 ) A E = A C AE=AC A E = A C e ( e − 2 L / 3 ) = 0 e(e-2L/3)=0 e ( e − 2 L /3 ) = 0 E = C E=C E = C B E = A C = B C BE=AC=BC B E = A C = B C D D D B E = B C = L BE=BC=L B E = B C = L D E = L / 2 = 1 DE=L/2=1 D E = L /2 = 1 4 2 / 3 4\sqrt2/3 4 2 /3 E = C E=C E = C
按通常图形中的内部点约定补足E E E B C BC B C e = 2 L / 3 e=2L/3 e = 2 L /3 B E = L BE=L B E = L d = 5 L / 3 d=5L/3 d = 5 L /3 D D D 5 L / 6 5L/6 5 L /6 D E = L / 6 = 1 DE=L/6=1 D E = L /6 = 1 L = 6 , d = 10 L=6,d=10 L = 6 , d = 10 1 2 ⋅ 10 ⋅ 4 2 = 20 2 \frac12\cdot10\cdot4\sqrt2=20\sqrt2 2 1 ⋅ 10 ⋅ 4 2 = 20 2
第 9 题
题目
双曲线 x 2 / a 2 − y 2 / b 2 = 1 x^2/a^2-y^2/b^2=1 x 2 / a 2 − y 2 / b 2 = 1 F 1 , F 2 F_1,F_2 F 1 , F 2 F 1 F_1 F 1 l l l M , N M,N M , N ∣ M N ∣ = ∣ N F 2 ∣ |MN|=|NF_2| ∣ M N ∣ = ∣ N F 2 ∣
解答
设a = 1 , b = t > 0 , c = 1 + t 2 a=1,b=t>0,c=\sqrt{1+t^2} a = 1 , b = t > 0 , c = 1 + t 2 ( x , y ) = ( − c + t q , − q ) (x,y)=(-c+tq,-q) ( x , y ) = ( − c + tq , − q )
( t 2 − t − 2 ) q 2 − 2 c t q + t 2 = 0. (t^2-t^{-2})q^2-2ctq+t^2=0. ( t 2 − t − 2 ) q 2 − 2 c tq + t 2 = 0. 要交两支，斜率绝对值1 / t 1/t 1/ t t t t t > 1 t>1 t > 1 0 < q 1 < q 2 0<q_1<q_2 0 < q 1 < q 2 M M M N N N M F 1 = q 1 c , N F 1 = q 2 c MF_1=q_1c,NF_1=q_2c M F 1 = q 1 c , N F 1 = q 2 c N F 2 = N F 1 − 2 NF_2=NF_1-2 N F 2 = N F 1 − 2 M N = N F 2 MN=NF_2 M N = N F 2 q 1 c = 2 q_1c=2 q 1 c = 2
代入q = 2 / c q=2/c q = 2/ c
t 2 ( t − 2 ) 2 = 4. t^2(t-2)^2=4. t 2 ( t − 2 ) 2 = 4. 在1 < t < 2 1<t<2 1 < t < 2 t ( 2 − t ) ≤ 1 t(2-t)\le1 t ( 2 − t ) ≤ 1 t ≥ 2 t\ge2 t ≥ 2 t = 1 + 3 t=1+\sqrt3 t = 1 + 3 2 / c 2/c 2/ c t 4 / [ 4 ( t 2 − 1 ) ] > 1 t^4/[4(t^2-1)]>1 t 4 / [ 4 ( t 2 − 1 )] > 1 y = ± ( 1 + 3 ) x y=\pm(1+\sqrt3)x y = ± ( 1 + 3 ) x a a a
第 10 题
题目
f ( x ) = ∣ x 2 + a ∣ + ∣ x ∣ f(x)=|x^2+a|+|x| f ( x ) = ∣ x 2 + a ∣ + ∣ x ∣ x ∈ [ − 1 , 1 ] x\in[-1,1] x ∈ [ − 1 , 1 ] M ( a ) M(a) M ( a ) M ( a ) M(a) M ( a )
解答
令t = ∣ x ∣ ∈ [ 0 , 1 ] t=|x|\in[0,1] t = ∣ x ∣ ∈ [ 0 , 1 ] t = 1 , 1 / 2 t=1,1/2 t = 1 , 1/2
M ( a ) ≥ max { a + 2 , 1 / 4 − a } ≥ 9 / 8. M(a)\ge\max\{a+2,1/4-a\}\ge9/8. M ( a ) ≥ max { a + 2 , 1/4 − a } ≥ 9/8. 取a = − 7 / 8 a=-7/8 a = − 7/8 t 2 ≤ 7 / 8 t^2\le7/8 t 2 ≤ 7/8 7 / 8 − t 2 + t ≤ 9 / 8 7/8-t^2+t\le9/8 7/8 − t 2 + t ≤ 9/8 t 2 ≥ 7 / 8 t^2\ge7/8 t 2 ≥ 7/8 t 2 + t − 7 / 8 t^2+t-7/8 t 2 + t − 7/8 t t t t = 1 t=1 t = 1 9 / 8 9/8 9/8 M M M 9 / 8 9/8 9/8
第 11 题
题目
{ a n } \{a_n\} { a n } d ≠ 0 d\ne0 d = 0 { a k n } \{a_{k_n}\} { a k n } k 1 = 3 , k 2 = 5 , k 3 = 9 k_1=3,k_2=5,k_3=9 k 1 = 3 , k 2 = 5 , k 3 = 9 k 1 + k 2 + ⋯ + k n k_1+k_2+\cdots+k_n k 1 + k 2 + ⋯ + k n b n = a n + 1 a n + 2 + a n a n + 2 b_n=\dfrac{a_{n+1}}{a_{n+2}}+\sqrt{\dfrac{a_n}{a_{n+2}}} b n = a n + 2 a n + 1 + a n + 2 a n
∑ j = 1 n 1 j ( j + 1 ) 2 b j < n n + 1 . \sum_{j=1}^n\frac1{j(j+1)\sqrt{2b_j}}<\sqrt{\frac n{n+1}}. j = 1 ∑ n j ( j + 1 ) 2 b j 1 < n + 1 n . 解答
设等差数列首项为A A A a 5 2 = a 3 a 9 a_5^2=a_3a_9 a 5 2 = a 3 a 9 ( A + 4 d ) 2 = ( A + 2 d ) ( A + 8 d ) (A+4d)^2=(A+2d)(A+8d) ( A + 4 d ) 2 = ( A + 2 d ) ( A + 8 d ) A d = 0 Ad=0 A d = 0 d ≠ 0 d\ne0 d = 0 A = 0 A=0 A = 0 a n = ( n − 1 ) d a_n=(n-1)d a n = ( n − 1 ) d 2 d , 4 d 2d,4d 2 d , 4 d 2 2 2 k n = 2 n + 1 k_n=2^n+1 k n = 2 n + 1 2 n + 1 + n − 2 2^{n+1}+n-2 2 n + 1 + n − 2
又
2 b j = 1 + j − 1 j + 1 > j j + 1 + j − 1 j . \sqrt{2b_j}=1+\sqrt{\frac{j-1}{j+1}}
>\sqrt{\frac j{j+1}}+\sqrt{\frac{j-1}j}. 2 b j = 1 + j + 1 j − 1 > j + 1 j + j j − 1 . 不等号成立，因为两边之差为
j + 1 − j j + 1 ( 1 − j − 1 j ) > 0. \frac{\sqrt{j+1}-\sqrt j}{\sqrt{j+1}}
\left(1-\sqrt{\frac{j-1}j}\right)>0. j + 1 j + 1 − j ( 1 − j j − 1 ) > 0. 而有理化给
j j + 1 − j − 1 j = 1 j ( j + 1 ) ( j / ( j + 1 ) + ( j − 1 ) / j ) . \sqrt{\frac j{j+1}}-\sqrt{\frac{j-1}j}
=\frac1{j(j+1)\left(\sqrt{j/(j+1)}+\sqrt{(j-1)/j}\right)}. j + 1 j − j j − 1 = j ( j + 1 ) ( j / ( j + 1 ) + ( j − 1 ) / j ) 1 . 故题目左端每一项严格小于上述差，相加望远镜消去即得严格小于n / ( n + 1 ) \sqrt{n/(n+1)} n / ( n + 1 )
第 12 题
题目
四棱柱 A B C D − A ′ B ′ C ′ D ′ ABCD-A'B'C'D' A B C D − A ′ B ′ C ′ D ′ D C C ′ D ′ ⊥ DCC'D'\perp D C C ′ D ′ ⊥ A B C D ABCD A B C D A B C D ABCD A B C D D A = A B = B C = 1 DA=AB=BC=1 D A = A B = B C = 1 D D ′ = D C DD'=DC D D ′ = D C ∠ A D C = π / 3 \angle ADC=\pi/3 ∠ A D C = π /3 M M M C D CD C D A ′ M ∥ A'M\parallel A ′ M ∥ B ′ B C C ′ B'BCC' B ′ B C C ′ ∠ D ′ D C = π / 3 \angle D'DC=\pi/3 ∠ D ′ D C = π /3 A ′ M A'M A ′ M D A A ′ D ′ DAA'D' D A A ′ D ′
解答
设D = ( 0 , 0 , 0 ) , C = ( 2 , 0 , 0 ) , A = ( 1 / 2 , 3 / 2 , 0 ) , B = ( 3 / 2 , 3 / 2 , 0 ) D=(0,0,0),C=(2,0,0),A=(1/2,\sqrt3/2,0),B=(3/2,\sqrt3/2,0) D = ( 0 , 0 , 0 ) , C = ( 2 , 0 , 0 ) , A = ( 1/2 , 3 /2 , 0 ) , B = ( 3/2 , 3 /2 , 0 ) v = ( u , 0 , w ) v=(u,0,w) v = ( u , 0 , w ) w ≠ 0 w\ne0 w = 0 u 2 + w 2 = 4 u^2+w^2=4 u 2 + w 2 = 4 M = ( 1 , 0 , 0 ) M=(1,0,0) M = ( 1 , 0 , 0 )
A ′ M → = B C → − v . \overrightarrow{A'M}=\overrightarrow{BC}-v. A ′ M = B C − v . 右端是平面B ′ B C C ′ B'BCC' B ′ B C C ′ A ′ A' A ′ A ′ M A'M A ′ M
第二问取上方棱柱，∠ D ′ D C = π / 3 \angle D'DC=\pi/3 ∠ D ′ D C = π /3 v = ( 1 , 0 , 3 ) v=(1,0,\sqrt3) v = ( 1 , 0 , 3 ) A ′ M → = ( − 1 / 2 , − 3 / 2 , − 3 ) \overrightarrow{A'M}=(-1/2,-\sqrt3/2,-\sqrt3) A ′ M = ( − 1/2 , − 3 /2 , − 3 ) 2 2 2 D A A ′ D ′ DAA'D' D A A ′ D ′
n = ( 3 / 2 , − 3 / 2 , − 3 / 2 ) , ∣ n ∣ = 15 / 2. n=(3/2,-\sqrt3/2,-\sqrt3/2),\qquad |n|=\sqrt{15}/2. n = ( 3/2 , − 3 /2 , − 3 /2 ) , ∣ n ∣ = 15 /2. 其与A ′ M → \overrightarrow{A'M} A ′ M 3 / 2 3/2 3/2 ( 3 / 2 ) / ( 2 ⋅ 15 / 2 ) = 15 / 10 (3/2)/(2\cdot\sqrt{15}/2)=\sqrt{15}/10 ( 3/2 ) / ( 2 ⋅ 15 /2 ) = 15 /10
第 13 题
题目
抛物线 x 2 = 4 y x^2=4y x 2 = 4 y A A A l l l x x x B B B B B B l l l l ′ l' l ′ C , D C,D C , D C C C l ′ l' l ′ y y y K K K A A A 2 2 2 l l l O C , O D , A K , A C OC,OD,AK,AC O C , O D , A K , A C △ O K D , △ O K C , △ A K C \triangle OKD,\triangle OKC,\triangle AKC △ O K D , △ O K C , △ A K C S 1 , S 2 , S 3 S_1,S_2,S_3 S 1 , S 2 , S 3 S 3 S 2 ( S 1 S 2 − 1 ) \dfrac{S_3}{S_2}\left(\dfrac{S_1}{S_2}-1\right) S 2 S 3 ( S 2 S 1 − 1 )
解答
设A = ( 2 t , t 2 ) A=(2t,t^2) A = ( 2 t , t 2 ) t > 0 t>0 t > 0 y = t x − t 2 y=tx-t^2 y = t x − t 2 2 2 2 t = 1 t=1 t = 1 l : y = x − 1 l:y=x-1 l : y = x − 1
一般情形B = ( t , 0 ) B=(t,0) B = ( t , 0 ) l ′ : y = − x / t + 1 l':y=-x/t+1 l ′ : y = − x / t + 1 K = ( 0 , 1 ) K=(0,1) K = ( 0 , 1 ) C = ( 2 u , u 2 ) , D = ( 2 v , v 2 ) C=(2u,u^2),D=(2v,v^2) C = ( 2 u , u 2 ) , D = ( 2 v , v 2 ) u > 0 > v u>0>v u > 0 > v u v = − 1 , u + v = − 2 / t uv=-1,u+v=-2/t uv = − 1 , u + v = − 2/ t 0 < u < 1 0<u<1 0 < u < 1 t = 2 u / ( 1 − u 2 ) t=2u/(1-u^2) t = 2 u / ( 1 − u 2 )
S 2 = u , S 1 = − v = 1 / u , S 3 = u ( t 2 + 1 ) . S_2=u,\quad S_1=-v=1/u,\quad S_3=u(t^2+1). S 2 = u , S 1 = − v = 1/ u , S 3 = u ( t 2 + 1 ) . 目标为( t 2 + 1 ) ( u − 2 − 1 ) (t^2+1)(u^{-2}-1) ( t 2 + 1 ) ( u − 2 − 1 ) r = u 2 ∈ ( 0 , 1 ) r=u^2\in(0,1) r = u 2 ∈ ( 0 , 1 )
( 1 + r ) 2 r ( 1 − r ) ≥ 8 , \frac{(1+r)^2}{r(1-r)}\ge8, r ( 1 − r ) ( 1 + r ) 2 ≥ 8 , 因为( 1 + r ) 2 − 8 r ( 1 − r ) = ( 3 r − 1 ) 2 ≥ 0 (1+r)^2-8r(1-r)=(3r-1)^2\ge0 ( 1 + r ) 2 − 8 r ( 1 − r ) = ( 3 r − 1 ) 2 ≥ 0 r = 1 / 3 r=1/3 r = 1/3 t = 3 t=\sqrt3 t = 3
第 14 题
题目
f ( x ) = ln ( x + 1 ) − a x − 1 f(x)=\ln(x+1)-ax-1 f ( x ) = ln ( x + 1 ) − a x − 1 a ≤ − 3 / 10 a\le-3/10 a ≤ − 3/10 f f f x 0 x_0 x 0
1 1 − a ≤ x 0 ≤ 1 − a , \frac1{1-a}\le x_0\le1-a, 1 − a 1 ≤ x 0 ≤ 1 − a , 以及
3 − 10 a + 8 a 2 − 2 a 3 2 ( 2 − a ) 2 ≤ f ( x 0 + 1 ) ≤ a 2 − 3 a + 1 2 − a . \frac{3-10a+8a^2-2a^3}{2(2-a)^2}\le f(x_0+1)\le\frac{a^2-3a+1}{2-a}. 2 ( 2 − a ) 2 3 − 10 a + 8 a 2 − 2 a 3 ≤ f ( x 0 + 1 ) ≤ 2 − a a 2 − 3 a + 1 . （ln 2 ≈ 0.693 , ln 3 ≈ 1.098 \ln2\approx0.693,\ln3\approx1.098 ln 2 ≈ 0.693 , ln 3 ≈ 1.098
解答
令b = − a ≥ 3 / 10 b=-a\ge3/10 b = − a ≥ 3/10 f ( x ) = ln ( 1 + x ) + b x − 1 f(x)=\ln(1+x)+bx-1 f ( x ) = ln ( 1 + x ) + b x − 1 x > − 1 x>-1 x > − 1 1 / ( 1 + x ) + b > 0 1/(1+x)+b>0 1/ ( 1 + x ) + b > 0 − ∞ , + ∞ -\infty,+\infty − ∞ , + ∞ x 0 x_0 x 0 f ( 0 ) = − 1 f(0)=-1 f ( 0 ) = − 1 x 0 > 0 x_0>0 x 0 > 0
由ln ( 1 + x ) ≤ x \ln(1+x)\le x ln ( 1 + x ) ≤ x 0 = f ( x 0 ) ≤ ( 1 + b ) x 0 − 1 0=f(x_0)\le(1+b)x_0-1 0 = f ( x 0 ) ≤ ( 1 + b ) x 0 − 1 x 0 ≥ 1 / ( 1 + b ) x_0\ge1/(1+b) x 0 ≥ 1/ ( 1 + b )
f ( 1 + b ) = ln ( 2 + b ) + b ( 1 + b ) − 1 > ln 2 + 3 10 13 10 − 1 > 0 , f(1+b)=\ln(2+b)+b(1+b)-1
>\ln2+\frac3{10}\frac{13}{10}-1>0, f ( 1 + b ) = ln ( 2 + b ) + b ( 1 + b ) − 1 > ln 2 + 10 3 10 13 − 1 > 0 , 故x 0 < 1 + b x_0<1+b x 0 < 1 + b
利用f ( x 0 ) = 0 f(x_0)=0 f ( x 0 ) = 0
f ( x 0 + 1 ) = b + ln ( 1 + 1 x 0 + 1 ) . f(x_0+1)=b+\ln\left(1+\frac1{x_0+1}\right). f ( x 0 + 1 ) = b + ln ( 1 + x 0 + 1 1 ) . 右端随x 0 x_0 x 0 t > 0 t>0 t > 0 t − t 2 / 2 ≤ ln ( 1 + t ) ≤ t t-t^2/2\le\ln(1+t)\le t t − t 2 /2 ≤ ln ( 1 + t ) ≤ t x 0 ≤ 1 + b x_0\le1+b x 0 ≤ 1 + b
f ( x 0 + 1 ) ≥ b + 1 2 + b − 1 2 ( 2 + b ) 2 = 3 + 10 b + 8 b 2 + 2 b 3 2 ( 2 + b ) 2 . f(x_0+1)\ge b+\frac1{2+b}-\frac1{2(2+b)^2}
=\frac{3+10b+8b^2+2b^3}{2(2+b)^2}. f ( x 0 + 1 ) ≥ b + 2 + b 1 − 2 ( 2 + b ) 2 1 = 2 ( 2 + b ) 2 3 + 10 b + 8 b 2 + 2 b 3 . 由x 0 ≥ 1 / ( 1 + b ) x_0\ge1/(1+b) x 0 ≥ 1/ ( 1 + b )
f ( x 0 + 1 ) ≤ b + 1 + b 2 + b = b 2 + 3 b + 1 2 + b . f(x_0+1)\le b+\frac{1+b}{2+b}
=\frac{b^2+3b+1}{2+b}. f ( x 0 + 1 ) ≤ b + 2 + b 1 + b = 2 + b b 2 + 3 b + 1 . 代回b = − a b=-a b = − a