答案速览
题号
1
2
3
4
5
6
7
8
9
答案
A
2 3 \frac23 3 2 B
B
ABC
A
D
B
C
题号
10
11
12
13
14
15
16
17
18
答案
ABC
AC
BC
BCD
D
D
AD
AC
C
题号
19
20
21
22
23
24
25
26
27
答案
CD
ACD
C
AB
B
6 3 \frac{\sqrt6}{3} 3 6 ABD
3 + 3 2 \frac{3+\sqrt3}{2} 2 3 + 3 C
题号
28
29
30
31
32
33
34
35
答案
C
B
C
BD
ABD
BCD
B
A
逐题解析
第 1 题
题目
一个四面体的棱长分别为 6 , 6 , 6 , 6 , 6 , 9 6,6,6,6,6,9 6 , 6 , 6 , 6 , 6 , 9
A. 21 \sqrt{21} 21
B. 21 2 \frac{\sqrt{21}}2 2 21
C. 210 3 \frac{\sqrt{210}}3 3 210
D. 210 6 \frac{\sqrt{210}}6 6 210
解答
令长度为 9 9 9 U = ( − 9 / 2 , 0 , 0 ) , V = ( 9 / 2 , 0 , 0 ) U=(-9/2,0,0),V=(9/2,0,0) U = ( − 9/2 , 0 , 0 ) , V = ( 9/2 , 0 , 0 ) W = ( 0 , 3 , 3 3 / 2 ) , T = ( 0 , − 3 , 3 3 / 2 ) W=(0,3,3\sqrt3/2),T=(0,-3,3\sqrt3/2) W = ( 0 , 3 , 3 3 /2 ) , T = ( 0 , − 3 , 3 3 /2 ) U V , W T UV,WT U V , W T ( 0 , 0 , h ) (0,0,h) ( 0 , 0 , h )
81 4 + h 2 = 9 + ( 3 3 2 − h ) 2 , \frac{81}{4}+h^2=9+\left(\frac{3\sqrt3}{2}-h\right)^2, 4 81 + h 2 = 9 + ( 2 3 3 − h ) 2 , 故 h = − 3 / 2 h=-\sqrt3/2 h = − 3 /2 81 / 4 + 3 / 4 = 21 81/4+3/4=21 81/4 + 3/4 = 21
第 2 题
题目
求 ∫ − 1 1 ( 1 − sin x ) x 2 d x \displaystyle\int_{-1}^{1}(1-\sin x)x^2\,dx ∫ − 1 1 ( 1 − sin x ) x 2 d x
解答
x 2 sin x x^2\sin x x 2 sin x ∫ − 1 1 x 2 d x = 2 / 3 \int_{-1}^1x^2\,dx=2/3 ∫ − 1 1 x 2 d x = 2/3
第 3 题
题目
P P P A ( 0 , 2 ) , B ( 0 , − 1 ) A(0,2),B(0,-1) A ( 0 , 2 ) , B ( 0 , − 1 ) ∣ A P ∣ ∣ B P ∣ 2 |AP|\,|BP|^2 ∣ A P ∣ ∣ B P ∣ 2
A. 3 2 3\sqrt2 3 2
B. 3 3 3\sqrt3 3 3
C. 6 6 6
D. 8 8 8
解答
写 P = ( u , v ) P=(u,v) P = ( u , v ) u 2 + v 2 = 1 u^2+v^2=1 u 2 + v 2 = 1 A P 2 = 5 − 4 v AP^2=5-4v A P 2 = 5 − 4 v B P 2 = 2 + 2 v BP^2=2+2v B P 2 = 2 + 2 v 4 ( 5 − 4 v ) ( 1 + v ) 2 4(5-4v)(1+v)^2 4 ( 5 − 4 v ) ( 1 + v ) 2 24 ( 1 + v ) ( 1 − 2 v ) 24(1+v)(1-2v) 24 ( 1 + v ) ( 1 − 2 v ) [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] v = 1 / 2 v=1/2 v = 1/2 27 27 27 3 3 3\sqrt3 3 3
第 4 题
题目
A = { 1 , 2 , … , 15 } A=\{1,2,\ldots,15\} A = { 1 , 2 , … , 15 } B = { 1 , 2 , 3 , 4 , 5 } B=\{1,2,3,4,5\} B = { 1 , 2 , 3 , 4 , 5 } f f f A A A B B B f ( x ) = f ( y ) f(x)=f(y) f ( x ) = f ( y ) ( x , y ) (x,y) ( x , y )
A. 30 30 30
B. 45 45 45
C. 60 60 60
D. 75 75 75
解答
设五个纤维的大小为 n 1 , … , n 5 n_1,\ldots,n_5 n 1 , … , n 5 x = y x=y x = y ∑ n i 2 \sum n_i^2 ∑ n i 2 ∑ n i = 15 \sum n_i=15 ∑ n i = 15 15 2 / 5 = 45 15^2/5=45 1 5 2 /5 = 45
第 5 题
题目
已知 a n + 1 = a n 2 − 3 a n + 4 a_{n+1}=a_n^2-3a_n+4 a n + 1 = a n 2 − 3 a n + 4 a 1 = 3 a_1=3 a 1 = 3 { a n } \{a_n\} { a n } { a n } \{a_n\} { a n } lim n → ∞ ( 1 a 1 − 1 + ⋯ + 1 a n − 1 ) = 1 \displaystyle\lim_{n\to\infty}\left(\frac1{a_1-1}+\cdots+\frac1{a_n-1}\right)=1 n → ∞ lim ( a 1 − 1 1 + ⋯ + a n − 1 1 ) = 1 a 100 = 1 a_{100}=1 a 100 = 1
解答
a n + 1 − a n = ( a n − 2 ) 2 a_{n+1}-a_n=(a_n-2)^2 a n + 1 − a n = ( a n − 2 ) 2 a 1 = 3 a_1=3 a 1 = 3 a n ≥ 3 a_n\ge3 a n ≥ 3 1 1 1 b n = a n − 2 b_n=a_n-2 b n = a n − 2 b 1 = 1 b_1=1 b 1 = 1 b n + 1 = b n ( b n + 1 ) b_{n+1}=b_n(b_n+1) b n + 1 = b n ( b n + 1 )
1 a n − 1 = 1 b n + 1 = 1 b n − 1 b n + 1 . \frac1{a_n-1}=\frac1{b_n+1}=\frac1{b_n}-\frac1{b_{n+1}}. a n − 1 1 = b n + 1 1 = b n 1 − b n + 1 1 . 前 n n n 1 − 1 / b n + 1 → 1 1-1/b_{n+1}\to1 1 − 1/ b n + 1 → 1
第 6 题
题目
棱长为 1 1 1 A B C D − A 1 B 1 C 1 D 1 ABCD-A_1B_1C_1D_1 A B C D − A 1 B 1 C 1 D 1 E E E A A 1 AA_1 A A 1 A C AC A C B D BD B D O O O α \alpha α E E E O C 1 OC_1 O C 1 α \alpha α
A. 6 4 \frac{\sqrt6}{4} 4 6
B. 2 2 \frac{\sqrt2}{2} 2 2
C. 3 2 \frac{\sqrt3}{2} 2 3
D. 1 1 1
解答
以 A A A A B , A D , A A 1 AB,AD,AA_1 A B , A D , A A 1 O = ( 1 / 2 , 1 / 2 , 0 ) O=(1/2,1/2,0) O = ( 1/2 , 1/2 , 0 ) C 1 = ( 1 , 1 , 1 ) C_1=(1,1,1) C 1 = ( 1 , 1 , 1 ) ( 1 , 1 , 2 ) (1,1,2) ( 1 , 1 , 2 ) E = ( 0 , 0 , 1 / 2 ) E=(0,0,1/2) E = ( 0 , 0 , 1/2 ) x + y + 2 z = 1 x+y+2z=1 x + y + 2 z = 1 B = ( 1 , 0 , 0 ) , D = ( 0 , 1 , 0 ) , E B=(1,0,0),D=(0,1,0),E B = ( 1 , 0 , 0 ) , D = ( 0 , 1 , 0 ) , E D − B = ( − 1 , 1 , 0 ) , E − B = ( − 1 , 0 , 1 / 2 ) D-B=(-1,1,0),E-B=(-1,0,1/2) D − B = ( − 1 , 1 , 0 ) , E − B = ( − 1 , 0 , 1/2 ) 6 / 2 \sqrt6/2 6 /2 6 / 4 \sqrt6/4 6 /4
第 7 题
题目
a n = ( n − 1 2 ) ( 9 10 ) n a_n=(n-\frac12)(\frac9{10})^n a n = ( n − 2 1 ) ( 10 9 ) n b n = a n + 1 − a n b_n=a_{n+1}-a_n b n = a n + 1 − a n { a n } \{a_n\} { a n } a 9 a_9 a 9 { b n + 1 / b n } \{b_{n+1}/b_n\} { b n + 1 / b n } 0.27 0.27 0.27 { a n } \{a_n\} { a n } a 1 a_1 a 1 { b n + 1 / b n } \{b_{n+1}/b_n\} { b n + 1 / b n } − 0.9 -0.9 − 0.9
解答
有
b n = 19 − 2 n 20 ( 9 10 ) n , b n + 1 b n = 9 10 − 9 / 5 19 − 2 n . b_n=\frac{19-2n}{20}\left(\frac9{10}\right)^n,\qquad
\frac{b_{n+1}}{b_n}=\frac9{10}-\frac{9/5}{19-2n}. b n = 20 19 − 2 n ( 10 9 ) n , b n b n + 1 = 10 9 − 19 − 2 n 9/5 . b n > 0 b_n>0 b n > 0 n ≤ 9 n\le9 n ≤ 9 a 10 a_{10} a 10 a n > 0 a_n>0 a n > 0 n ≤ 9 n\le9 n ≤ 9 n = 9 n=9 n = 9 − 0.9 -0.9 − 0.9 n ≥ 10 n\ge10 n ≥ 10 n = 10 n=10 n = 10 2.7 2.7 2.7
第 8 题
题目
A , B , C , D , E A,B,C,D,E A , B , C , D , E A A A B B B C C C D D D E E E
A. 1 1 1
B. 2 2 2
C. 3 3 3
D. 4 4 4
解答
D D D A A A D D D C C C B , E B,E B , E B B B C , D C,D C , D E E E E E E C , D C,D C , D
第 9 题
题目
椭圆 x 2 6 + y 2 2 = 1 \frac{x^2}{6}+\frac{y^2}{2}=1 6 x 2 + 2 y 2 = 1 F ( 2 , 0 ) F(2,0) F ( 2 , 0 ) l l l F F F A , B A,B A , B P P P x = 3 x=3 x = 3 △ A B P \triangle ABP △ A B P S △ P A B S_{\triangle PAB} S △ P A B
A. 3 2 \frac{\sqrt3}{2} 2 3
B. 3 \sqrt3 3
C. 3 3 2 \frac{3\sqrt3}{2} 2 3 3
D. 2 3 2\sqrt3 2 3
解答
取弦方向单位向量 ( u , v ) (u,v) ( u , v ) u 2 + v 2 = 1 u^2+v^2=1 u 2 + v 2 = 1 ( 2 + t u , t v ) (2+tu,tv) ( 2 + t u , t v )
( 3 − 2 u 2 ) t 2 + 4 u t − 2 = 0. (3-2u^2)t^2+4ut-2=0. ( 3 − 2 u 2 ) t 2 + 4 u t − 2 = 0. 记 q = 3 − 2 u 2 q=3-2u^2 q = 3 − 2 u 2 L = 2 6 / q L=2\sqrt6/q L = 2 6 / q H = ( 2 − 2 u 2 / q , − 2 u v / q ) H=(2-2u^2/q,-2uv/q) H = ( 2 − 2 u 2 / q , − 2 uv / q )
H ± 3 L 2 ( − v , u ) . H\pm\frac{\sqrt3 L}{2}(-v,u). H ± 2 3 L ( − v , u ) . 其横坐标为 3 3 3 3 q = 2 q − 2 u 2 ∓ 3 2 v 3q=2q-2u^2\mp3\sqrt2v 3 q = 2 q − 2 u 2 ∓ 3 2 v 3 = ∓ 3 2 v 3=\mp3\sqrt2v 3 = ∓ 3 2 v v 2 = 1 / 2 , q = 2 , L = 6 v^2=1/2,q=2,L=\sqrt6 v 2 = 1/2 , q = 2 , L = 6 3 3 / 2 3\sqrt3/2 3 3 /2
第 10 题
题目
已知 ( 2 + 3 ) n = x n + y n 3 (2+\sqrt3)^n=x_n+y_n\sqrt3 ( 2 + 3 ) n = x n + y n 3 x n + 1 = 2 x n + 3 y n x_{n+1}=2x_n+3y_n x n + 1 = 2 x n + 3 y n y n + 1 = x n + 2 y n y_{n+1}=x_n+2y_n y n + 1 = x n + 2 y n lim n → ∞ ( 2 − 3 ) n = 0 \lim_{n\to\infty}(2-\sqrt3)^n=0 lim n → ∞ ( 2 − 3 ) n = 0 x n = 2019 x_n=2019 x n = 2019
解答
乘以 2 + 3 2+\sqrt3 2 + 3 0 < 2 − 3 < 1 0<2-\sqrt3<1 0 < 2 − 3 < 1 x n 2 − 3 y n 2 = 1 x_n^2-3y_n^2=1 x n 2 − 3 y n 2 = 1 x n = 2019 x_n=2019 x n = 2019 3 3 3 0 0 0 1 1 1
第 11 题
题目
l 1 , l 2 l_1,l_2 l 1 , l 2 A 1 ∈ l 1 , A 2 ∈ l 2 A_1\in l_1,A_2\in l_2 A 1 ∈ l 1 , A 2 ∈ l 2 P P P A 1 A 2 A_1A_2 A 1 A 2 l 1 ∥ l 2 l_1\parallel l_2 l 1 ∥ l 2 P P P l 1 ∥ l 2 l_1\parallel l_2 l 1 ∥ l 2 l 1 , l 2 l_1,l_2 l 1 , l 2 P P P
解答
两平行线可写成 y = h 1 , y = h 2 y=h_1,y=h_2 y = h 1 , y = h 2 ( h 1 + h 2 ) / 2 (h_1+h_2)/2 ( h 1 + h 2 ) /2 u , v \mathbf u,\mathbf v u , v ( s u + t v ) / 2 (s\mathbf u+t\mathbf v)/2 ( s u + t v ) /2
第 12 题
题目
f ( x ) = x − 1 e x f(x)=\dfrac{x-1}{e^x} f ( x ) = e x x − 1 f ( x ) ≥ 0 f(x)\ge0 f ( x ) ≥ 0 0 < x < 1 0<x<1 0 < x < 1 x ≥ 1 x\ge1 x ≥ 1 f ( x ) f(x) f ( x ) y = 2 x − 1 y=2x-1 y = 2 x − 1 f ( x ) f(x) f ( x ) y = 1 3 ( x − 1 ) y=\frac13(x-1) y = 3 1 ( x − 1 )
解答
e x > 0 e^x>0 e x > 0 f ( x ) ≥ 0 f(x)\ge0 f ( x ) ≥ 0 x ≥ 1 x\ge1 x ≥ 1 g ( x ) = ( x − 1 ) e − x − 2 x + 1 g(x)=(x-1)e^{-x}-2x+1 g ( x ) = ( x − 1 ) e − x − 2 x + 1
g ′ ( x ) = ( 2 − x ) e − x − 2. g'(x)=(2-x)e^{-x}-2. g ′ ( x ) = ( 2 − x ) e − x − 2. 当 x < 0 x<0 x < 0 g ′ > 0 g'>0 g ′ > 0 0 < x < 2 0<x<2 0 < x < 2 g ′ < 0 g'<0 g ′ < 0 x ≥ 2 x\ge2 x ≥ 2 g ′ < 0 g'<0 g ′ < 0 g ( 0 ) = 0 g(0)=0 g ( 0 ) = 0 g g g 0 0 0 ( x − 1 ) / 3 (x-1)/3 ( x − 1 ) /3 1 , ln 3 1,\ln3 1 , ln 3
第 13 题
题目
A B AB A B O O O C C C C O ⊥ A B CO\perp AB C O ⊥ A B M M M A C AC A C C H ⊥ M B CH\perp MB C H ⊥ M B H H H C C C O H = 1 2 A M OH=\frac12AM O H = 2 1 A M O H = 1 2 A H OH=\frac12AH O H = 2 1 A H △ A H M ∼ △ B A M \triangle AHM\sim\triangle BAM △ A H M ∼ △ B A M △ B H O ∼ △ B A M \triangle BHO\sim\triangle BAM △ B H O ∼ △ B A M
解答
题面校注：需明确 C C C C O ⊥ A B CO\perp AB C O ⊥ A B C C C
按相似缩放，取 A = ( − 1 , 0 ) , B = ( 1 , 0 ) , C = ( 0 , 1 ) A=(-1,0),B=(1,0),C=(0,1) A = ( − 1 , 0 ) , B = ( 1 , 0 ) , C = ( 0 , 1 ) M = ( − 1 / 2 , 1 / 2 ) M=(-1/2,1/2) M = ( − 1/2 , 1/2 ) H = ( − 1 / 5 , 2 / 5 ) H=(-1/5,2/5) H = ( − 1/5 , 2/5 )
O H = 1 5 , A H = 2 5 , A M = 1 2 , B M = 10 2 , B H = 4 10 , H M = 1 10 . OH=\frac1{\sqrt5},\quad AH=\frac2{\sqrt5},\quad
AM=\frac1{\sqrt2},\quad BM=\frac{\sqrt{10}}2,\quad
BH=\frac4{\sqrt{10}},\quad HM=\frac1{\sqrt{10}}. O H = 5 1 , A H = 5 2 , A M = 2 1 , B M = 2 10 , B H = 10 4 , H M = 10 1 . 故 B 正确、A 错。三边比给出
A H B A = H M A M = A M B M = 1 5 , B H B A = H O A M = B O B M = 2 5 , \frac{AH}{BA}=\frac{HM}{AM}=\frac{AM}{BM}=\frac1{\sqrt5},
\qquad
\frac{BH}{BA}=\frac{HO}{AM}=\frac{BO}{BM}=\sqrt{\frac25}, B A A H = A M H M = B M A M = 5 1 , B A B H = A M H O = B M B O = 5 2 , 从而 C、D 均正确。
第 14 题
题目
圆 x 2 + y 2 = 4 x^2+y^2=4 x 2 + y 2 = 4 ( x 0 , y 0 ) (x_0,y_0) ( x 0 , y 0 ) y 2 = 8 x y^2=8x y 2 = 8 x A , B A,B A , B ∠ A O B = 90 ∘ \angle AOB=90^\circ ∠ A O B = 9 0 ∘ O O O x 0 x_0 x 0
A. 2 2 2
B. 1 1 1
C. 2 3 \frac23 3 2
D. 1 2 \frac12 2 1
解答
切线为 x 0 x + y 0 y = 4 x_0x+y_0y=4 x 0 x + y 0 y = 4 x 0 = 0 x_0=0 x 0 = 0 x = y 2 / 8 x=y^2/8 x = y 2 /8 y A y B = − 32 / x 0 y_Ay_B=-32/x_0 y A y B = − 32/ x 0 ∠ A O B = 90 ∘ \angle AOB=90^\circ ∠ A O B = 9 0 ∘
0 = x A x B + y A y B = ( y A y B ) 2 64 + y A y B . 0=x_Ax_B+y_Ay_B=\frac{(y_Ay_B)^2}{64}+y_Ay_B. 0 = x A x B + y A y B = 64 ( y A y B ) 2 + y A y B . 交点不为原点，故 y A y B = − 64 y_Ay_B=-64 y A y B = − 64 x 0 = 1 / 2 x_0=1/2 x 0 = 1/2 y 0 = ± 15 / 2 y_0=\pm\sqrt{15}/2 y 0 = ± 15 /2
第 15 题
题目
求 f ( x ) = ∣ x − 1 ∣ + ∣ x − 3 ∣ + 2 e x f(x)=|x-1|+|x-3|+2e^x f ( x ) = ∣ x − 1∣ + ∣ x − 3∣ + 2 e x
A. 3 ln 2 3\ln2 3 ln 2
B. 4 ln 2 4\ln2 4 ln 2
C. 5 5 5
D. 6 6 6
解答
当 x ≤ 1 x\le1 x ≤ 1 f ( x ) = 4 − 2 x + 2 e x ≥ 6 f(x)=4-2x+2e^x\ge6 f ( x ) = 4 − 2 x + 2 e x ≥ 6 x = 0 x=0 x = 0 1 ≤ x ≤ 3 1\le x\le3 1 ≤ x ≤ 3 f ( x ) = 2 + 2 e x > 6 f(x)=2+2e^x>6 f ( x ) = 2 + 2 e x > 6 x ≥ 3 x\ge3 x ≥ 3 f ( x ) = 2 x − 4 + 2 e x > 6 f(x)=2x-4+2e^x>6 f ( x ) = 2 x − 4 + 2 e x > 6 6 6 6
第 16 题
题目
已知 x 3 + 27 y 3 + 9 x y = 1 x^3+27y^3+9xy=1 x 3 + 27 y 3 + 9 x y = 1 x 3 y x^3y x 3 y 1 3 \frac13 3 1 − 27 64 -\frac{27}{64} − 64 27 − 3 3 -\frac{\sqrt3}{3} − 3 3
解答
利用三立方恒等式，题设等价于
( x + 3 y − 1 ) ( x − 3 y ) 2 + ( 3 y + 1 ) 2 + ( x + 1 ) 2 2 = 0. (x+3y-1)\frac{(x-3y)^2+(3y+1)^2+(x+1)^2}{2}=0. ( x + 3 y − 1 ) 2 ( x − 3 y ) 2 + ( 3 y + 1 ) 2 + ( x + 1 ) 2 = 0. 因此解集为直线 y = ( 1 − x ) / 3 y=(1-x)/3 y = ( 1 − x ) /3 ( − 1 , − 1 / 3 ) (-1,-1/3) ( − 1 , − 1/3 ) x 3 y = x 3 ( 1 − x ) / 3 x^3y=x^3(1-x)/3 x 3 y = x 3 ( 1 − x ) /3 x 2 ( 3 − 4 x ) / 3 x^2(3-4x)/3 x 2 ( 3 − 4 x ) /3 9 / 256 9/256 9/256 1 / 3 > 9 / 256 1/3>9/256 1/3 > 9/256 1 / 3 1/3 1/3
第 17 题
题目
从 { 1 , 2 , 3 , 4 , 5 , 6 } \{1,2,3,4,5,6\} { 1 , 2 , 3 , 4 , 5 , 6 } A A A B B B A A A x 1 , x 2 x_1,x_2 x 1 , x 2 B B B y 1 , y 2 y_1,y_2 y 1 , y 2 X = x 1 − x 2 , Y = y 1 − y 2 X=x_1-x_2,Y=y_1-y_2 X = x 1 − x 2 , Y = y 1 − y 2 E ( X ) = 7 2 E(X)=\frac72 E ( X ) = 2 7 E ( X ) = 7 4 E(X)=\frac74 E ( X ) = 4 7 P ( X = Y ) = 2 5 P(X=Y)=\frac25 P ( X = Y ) = 5 2 P ( X = Y ) = 3 10 P(X=Y)=\frac3{10} P ( X = Y ) = 10 3
解答
题面校注：现明确在全部 ( 6 3 ) = 20 \binom63=20 ( 3 6 ) = 20 A A A
当 X = k X=k X = k 6 − k 6-k 6 − k k − 1 k-1 k − 1 k = 2 , 3 , 4 , 5 k=2,3,4,5 k = 2 , 3 , 4 , 5 4 , 6 , 6 , 4 4,6,6,4 4 , 6 , 6 , 4
E ( X ) = 2 ⋅ 4 + 3 ⋅ 6 + 4 ⋅ 6 + 5 ⋅ 4 20 = 7 2 . E(X)=\frac{2\cdot4+3\cdot6+4\cdot6+5\cdot4}{20}=\frac72. E ( X ) = 20 2 ⋅ 4 + 3 ⋅ 6 + 4 ⋅ 6 + 5 ⋅ 4 = 2 7 . X = Y X=Y X = Y A A A 123 , 124 , 135 , 145 , 236 , 246 , 356 , 456 123,124,135,145,236,246,356,456 123 , 124 , 135 , 145 , 236 , 246 , 356 , 456 2 / 5 2/5 2/5
第 18 题
题目
若 x 1 4 + x 2 4 + ⋯ + x n 4 = 799 x_1^4+x_2^4+\cdots+x_n^4=799 x 1 4 + x 2 4 + ⋯ + x n 4 = 799 n n n
A. 11 11 11
B. 13 13 13
C. 15 15 15
D. 17 17 17
解答
偶数四次方模 16 16 16 0 0 0 16 16 16 1 1 1 799 ≡ 15 ( m o d 16 ) 799\equiv15\pmod {16} 799 ≡ 15 ( mod 16 ) 15 15 15 n ≥ 15 n\ge15 n ≥ 15
799 = 5 4 + 3 4 + 3 4 + 1 4 + ⋯ + 1 4 ⏟ 12 项 , 799=5^4+3^4+3^4+\underbrace{1^4+\cdots+1^4}_{12\text{ 项}}, 799 = 5 4 + 3 4 + 3 4 + 12 项 1 4 + ⋯ + 1 4 , 恰有十五项，选 C。
第 19 题
题目
判断正确选项：
A. sin 2 π 7 + sin 4 π 7 + sin 6 π 7 = 1 \sin\frac{2\pi}7+\sin\frac{4\pi}7+\sin\frac{6\pi}7=1 sin 7 2 π + sin 7 4 π + sin 7 6 π = 1
B. cos 2 π 7 + cos 4 π 7 + cos 6 π 7 = 1 2 \cos\frac{2\pi}7+\cos\frac{4\pi}7+\cos\frac{6\pi}7=\frac12 cos 7 2 π + cos 7 4 π + cos 7 6 π = 2 1
C. sin 6 ∘ sin 42 ∘ sin 66 ∘ sin 78 ∘ = 1 16 \sin6^\circ\sin42^\circ\sin66^\circ\sin78^\circ=\frac1{16} sin 6 ∘ sin 4 2 ∘ sin 6 6 ∘ sin 7 8 ∘ = 16 1
D. 将 C 中正弦全换成余弦后等式成立。
解答
A 的前两个正弦均大于 2 / 2 \sqrt2/2 2 /2 1 1 1 − 1 / 2 -1/2 − 1/2
对于 C，连续使用倍角公式，
8 cos 12 ∘ cos 24 ∘ cos 48 ∘ = sin 96 ∘ sin 12 ∘ = 1 2 sin 6 ∘ . 8\cos12^\circ\cos24^\circ\cos48^\circ
=\frac{\sin96^\circ}{\sin12^\circ}
=\frac1{2\sin6^\circ}. 8 cos 1 2 ∘ cos 2 4 ∘ cos 4 8 ∘ = sin 1 2 ∘ sin 9 6 ∘ = 2 sin 6 ∘ 1 . 将 sin 42 ∘ , sin 66 ∘ , sin 78 ∘ \sin42^\circ,\sin66^\circ,\sin78^\circ sin 4 2 ∘ , sin 6 6 ∘ , sin 7 8 ∘
对于 D，由积化和差得 4 cos 18 ∘ cos 42 ∘ cos 78 ∘ = cos 54 ∘ = sin 36 ∘ 4\cos18^\circ\cos42^\circ\cos78^\circ=\cos54^\circ=\sin36^\circ 4 cos 1 8 ∘ cos 4 2 ∘ cos 7 8 ∘ = cos 5 4 ∘ = sin 3 6 ∘ cos 36 ∘ − cos 72 ∘ = 1 / 2 \cos36^\circ-\cos72^\circ=1/2 cos 3 6 ∘ − cos 7 2 ∘ = 1/2 2 cos 6 ∘ cos 66 ∘ = cos 36 ∘ 2\cos6^\circ\cos66^\circ=\cos36^\circ 2 cos 6 ∘ cos 6 6 ∘ = cos 3 6 ∘
cos 36 ∘ sin 36 ∘ 8 cos 18 ∘ = sin 72 ∘ 16 cos 18 ∘ = 1 16 . \frac{\cos36^\circ\sin36^\circ}{8\cos18^\circ}
=\frac{\sin72^\circ}{16\cos18^\circ}=\frac1{16}. 8 cos 1 8 ∘ cos 3 6 ∘ sin 3 6 ∘ = 16 cos 1 8 ∘ sin 7 2 ∘ = 16 1 . 所以 CD 正确。
第 20 题
题目
若 A ∩ B = ∅ , A ∪ B = N ∗ A\cap B=\varnothing,A\cup B=\mathbb N^* A ∩ B = ∅ , A ∪ B = N ∗ ( A , B ) (A,B) ( A , B ) A A A 3 3 3 B B B 3 k ± 1 3k\pm1 3 k ± 1 A A A B B B A A A B B B A A A B B B
解答
题面校注：D 的条件已完整补写；所有被排除的三项数列均要求元素互异，无穷数列要求递增。
A 确为按模 3 3 3 1 1 1
将正整数中的所有递增无穷等差数列枚举为 P 1 , P 2 , … P_1,P_2,\ldots P 1 , P 2 , … P n P_n P n a n a_n a n P n P_n P n A = { a n : n ≥ 1 } A=\{a_n:n\ge1\} A = { a n : n ≥ 1 } B = N ∗ ∖ A B=\mathbb N^*\setminus A B = N ∗ ∖ A A A A P n P_n P n A A A B B B
等比情形同样构造。正整数无穷递增等比数列的公比必为整数：若既约公比为 u / v u/v u / v v k v^k v k v = 1 v=1 v = 1 x , y x,y x , y x 2 / y , y 2 / x , x y x^2/y,y^2/x,\sqrt{xy} x 2 / y , y 2 / x , x y
第 21 题
题目
四面体 P A B C PABC P A B C A B = P A = 2 A C = 2 AB=PA=2AC=2 A B = P A = 2 A C = 2 P A ⊥ PA\perp P A ⊥ A B C ABC A B C ∠ B A C = 120 ∘ \angle BAC=120^\circ ∠ B A C = 12 0 ∘
A. 20 π 3 \frac{20\pi}3 3 20 π
B. 20 3 π 3 \frac{20\sqrt3\pi}3 3 20 3 π
C. 40 π 3 \frac{40\pi}3 3 40 π
D. 40 3 π 3 \frac{40\sqrt3\pi}3 3 40 3 π
解答
底面 A B = 2 , A C = 1 AB=2,AC=1 A B = 2 , A C = 1 120 ∘ 120^\circ 12 0 ∘ B C = 7 BC=\sqrt7 B C = 7 R 0 = 7 / ( 2 sin 120 ∘ ) = 7 / 3 R_0=\sqrt7/(2\sin120^\circ)=\sqrt{7/3} R 0 = 7 / ( 2 sin 12 0 ∘ ) = 7/3 P A PA P A 2 2 2 A , P A,P A , P 1 1 1 R 2 = 7 / 3 + 1 = 10 / 3 R^2=7/3+1=10/3 R 2 = 7/3 + 1 = 10/3 4 π R 2 = 40 π / 3 4\pi R^2=40\pi/3 4 π R 2 = 40 π /3
第 22 题
题目
u = x + 3 y x 2 + y 2 u=\dfrac{x+\sqrt3y}{\sqrt{x^2+y^2}} u = x 2 + y 2 x + 3 y D = { ( x , y ) ∣ x 2 + ( y − 2 ) 2 ≤ 1 } D=\{(x,y)\mid x^2+(y-2)^2\le1\} D = {( x , y ) ∣ x 2 + ( y − 2 ) 2 ≤ 1 } ( x , y ) ∈ D (x,y)\in D ( x , y ) ∈ D u min = 1 u_{\min}=1 u m i n = 1 u max = 2 u_{\max}=2 u m a x = 2 u min = 0 u_{\min}=0 u m i n = 0 u max = 3 u_{\max}=3 u m a x = 3
解答
原点到圆心距离为 2 2 2 1 1 1 θ ∈ [ π / 3 , 2 π / 3 ] \theta\in[\pi/3,2\pi/3] θ ∈ [ π /3 , 2 π /3 ] u = cos θ + 3 sin θ = 2 cos ( θ − π / 3 ) u=\cos\theta+\sqrt3\sin\theta=2\cos(\theta-\pi/3) u = cos θ + 3 sin θ = 2 cos ( θ − π /3 ) [ 1 , 2 ] [1,2] [ 1 , 2 ]
第 23 题
题目
a , b ≥ 0 a,b\ge0 a , b ≥ 0 a b ( a + 8 b ) = 20 ab(a+8b)=20 ab ( a + 8 b ) = 20 a + 3 b a+3b a + 3 b
A. 4 4 4
B. 5 5 5
C. 60 3 \sqrt[3]{60} 3 60
D. 4 60 3 3 \frac{4\sqrt[3]{60}}3 3 4 3 60
解答
约束迫使 a , b > 0 a,b>0 a , b > 0 t = a + 3 b , u = b / t t=a+3b,u=b/t t = a + 3 b , u = b / t 0 < u < 1 / 3 0<u<1/3 0 < u < 1/3
20 = t 3 u ( 1 − 3 u ) ( 1 + 5 u ) = t 3 ( u + 2 u 2 − 15 u 3 ) . 20=t^3u(1-3u)(1+5u)=t^3(u+2u^2-15u^3). 20 = t 3 u ( 1 − 3 u ) ( 1 + 5 u ) = t 3 ( u + 2 u 2 − 15 u 3 ) . 括号内函数的导数为 ( 1 − 5 u ) ( 1 + 9 u ) (1-5u)(1+9u) ( 1 − 5 u ) ( 1 + 9 u ) u = 1 / 5 u=1/5 u = 1/5 4 / 25 4/25 4/25 t 3 ≥ 125 t^3\ge125 t 3 ≥ 125 t ≥ 5 t\ge5 t ≥ 5 a = 2 , b = 1 a=2,b=1 a = 2 , b = 1
第 24 题
题目
长方体底面 A B C D ABCD A B C D 1 1 1 B B 1 = 2 BB_1=2 B B 1 = 2 M , N , P M,N,P M , N , P C C 1 , C D , D D 1 CC_1,CD,DD_1 C C 1 , C D , D D 1 P P P B M N BMN B M N
解答
取 A = ( 0 , 0 , 0 ) , B = ( 1 , 0 , 0 ) , D = ( 0 , 1 , 0 ) A=(0,0,0),B=(1,0,0),D=(0,1,0) A = ( 0 , 0 , 0 ) , B = ( 1 , 0 , 0 ) , D = ( 0 , 1 , 0 ) M = ( 1 , 1 , 1 ) , N = ( 1 / 2 , 1 , 0 ) , P = ( 0 , 1 , 1 ) M=(1,1,1),N=(1/2,1,0),P=(0,1,1) M = ( 1 , 1 , 1 ) , N = ( 1/2 , 1 , 0 ) , P = ( 0 , 1 , 1 ) B M N BMN B M N − 2 x − y + z + 2 = 0 -2x-y+z+2=0 − 2 x − y + z + 2 = 0 2 / 6 = 6 / 3 2/\sqrt6=\sqrt6/3 2/ 6 = 6 /3
第 25 题
题目
若对任意 c ∈ R c\in\mathbb R c ∈ R a , b a,b a , b f ( a ) − f ( b ) a − b = f ′ ( c ) \dfrac{f(a)-f(b)}{a-b}=f'(c) a − b f ( a ) − f ( b ) = f ′ ( c ) f f f x 3 − 3 x 2 + 3 x x^3-3x^2+3x x 3 − 3 x 2 + 3 x 1 / ( x 2 + 1 ) 1/(x^2+1) 1/ ( x 2 + 1 ) e x + 1 e^{x+1} e x + 1 sin ( 2 x + 1 ) \sin(2x+1) sin ( 2 x + 1 )
解答
按本卷定义，右边是 f ′ ( c ) f'(c) f ′ ( c ) a ≠ b a\ne b a = b
A 的导数 3 ( x − 1 ) 2 3(x-1)^2 3 ( x − 1 ) 2 1 1 1 f ′ ( 1 ) = 0 f'(1)=0 f ′ ( 1 ) = 0
B 的导数 − 2 x / ( 1 + x 2 ) 2 -2x/(1+x^2)^2 − 2 x / ( 1 + x 2 ) 2 x = − 1 / 3 x=-1/\sqrt3 x = − 1/ 3
C 固定 b = a + 1 b=a+1 b = a + 1 ( e − 1 ) e a + 1 (e-1)e^{a+1} ( e − 1 ) e a + 1 f ′ ( c ) = e c + 1 f'(c)=e^{c+1} f ′ ( c ) = e c + 1
D 的导数为 2 cos ( 2 x + 1 ) 2\cos(2x+1) 2 cos ( 2 x + 1 ) 2 2 2
第 26 题
题目
已知 ∣ a ⃗ ∣ = ∣ b ⃗ ∣ = 1 |\vec a|=|\vec b|=1 ∣ a ∣ = ∣ b ∣ = 1 a ⃗ ⋅ b ⃗ = 1 2 \vec a\cdot\vec b=\frac12 a ⋅ b = 2 1 ( c ⃗ − a ⃗ ) ⋅ ( c ⃗ − b ⃗ ) = 0 (\vec c-\vec a)\cdot(\vec c-\vec b)=0 ( c − a ) ⋅ ( c − b ) = 0 ∣ d ⃗ − c ⃗ ∣ = 1 |\vec d-\vec c|=1 ∣ d − c ∣ = 1 ∣ d ⃗ ∣ |\vec d| ∣ d ∣
解答
令 m = ( a + b ) / 2 \mathbf m=(\mathbf a+\mathbf b)/2 m = ( a + b ) /2
∣ c − m ∣ 2 = ∣ a − b ∣ 2 4 = 1 4 , ∣ m ∣ = 3 2 . |\mathbf c-\mathbf m|^2=\frac{|\mathbf a-\mathbf b|^2}{4}=\frac14,
\qquad |\mathbf m|=\frac{\sqrt3}{2}. ∣ c − m ∣ 2 = 4 ∣ a − b ∣ 2 = 4 1 , ∣ m ∣ = 2 3 . 故 ∣ d ∣ ≤ ∣ m ∣ + ∣ c − m ∣ + ∣ d − c ∣ = ( 3 + 3 ) / 2 |\mathbf d|\le|\mathbf m|+|\mathbf c-\mathbf m|+|\mathbf d-\mathbf c|=(3+\sqrt3)/2 ∣ d ∣ ≤ ∣ m ∣ + ∣ c − m ∣ + ∣ d − c ∣ = ( 3 + 3 ) /2 m \mathbf m m
第 27 题
题目
a a a 4444 4444 4444^{4444} 444 4 4444 b b b a a a c c c b b b c c c
A. 5 5 5
B. 6 6 6
C. 7 7 7
D. 16 16 16
解答
题面校注：对照另一份同年试题记录，恢复为对 4444 4444 4444^{4444} 444 4 4444 16 16 16 9 9 9 7 7 7
记 N = 4444 4444 N=4444^{4444} N = 444 4 4444 N < 10 17776 N<10^{17776} N < 1 0 17776 1 ≤ a ≤ 9 ⋅ 17776 < 160000 1\le a\le9\cdot17776<160000 1 ≤ a ≤ 9 ⋅ 17776 < 160000 a a a b ≤ 54 b\le54 b ≤ 54 c ≤ 13 c\le13 c ≤ 13 9 9 9
N ≡ 7 4444 ≡ 7 ( m o d 9 ) N\equiv7^{4444}\equiv7\pmod9 N ≡ 7 4444 ≡ 7 ( mod 9 ) （7 3 ≡ 1 7^3\equiv1 7 3 ≡ 1 1 ≤ c ≤ 13 1\le c\le13 1 ≤ c ≤ 13 7 7 7 7 7 7
第 28 题
题目
正整数 x , y x,y x , y 1 x + 1 y = 3 100 \frac1x+\frac1y=\frac3{100} x 1 + y 1 = 100 3 ( x , y ) (x,y) ( x , y )
A. 24 24 24
B. 25 25 25
C. 12 12 12
D. 13 13 13
解答
原式等价于 ( 3 x − 100 ) ( 3 y − 100 ) = 10000 (3x-100)(3y-100)=10000 ( 3 x − 100 ) ( 3 y − 100 ) = 10000 1 / x < 3 / 100 1/x<3/100 1/ x < 3/100 x > 100 / 3 x>100/3 x > 100/3 3 3 3 2 2 2 2 i 5 j 2^i5^j 2 i 5 j 0 ≤ i , j ≤ 4 0\le i,j\le4 0 ≤ i , j ≤ 4 ( − 1 ) i + j (-1)^{i+j} ( − 1 ) i + j i + j i+j i + j 3 ⋅ 2 + 2 ⋅ 3 = 12 3\cdot2+2\cdot3=12 3 ⋅ 2 + 2 ⋅ 3 = 12
第 29 题
题目
方程 x 4 + 2 x 3 + 2 = 0 x^4+2x^3+2=0 x 4 + 2 x 3 + 2 = 0 z 1 , z 2 , z 3 , z 4 z_1,z_2,z_3,z_4 z 1 , z 2 , z 3 , z 4 ( z 1 z 2 + z 3 z 4 ) ( z 1 z 4 + z 2 z 3 ) ( z 1 z 3 + z 2 z 4 ) (z_1z_2+z_3z_4)(z_1z_4+z_2z_3)(z_1z_3+z_2z_4) ( z 1 z 2 + z 3 z 4 ) ( z 1 z 4 + z 2 z 3 ) ( z 1 z 3 + z 2 z 4 )
A. − 6 -6 − 6
B. 8 8 8
C. 6 6 6
D. − 8 -8 − 8
解答
记四根的基本对称式为 e 1 , e 2 , e 3 , e 4 e_1,e_2,e_3,e_4 e 1 , e 2 , e 3 , e 4
( z 1 z 2 + z 3 z 4 ) ( z 1 z 3 + z 2 z 4 ) ( z 1 z 4 + z 2 z 3 ) = e 3 2 + e 1 2 e 4 − 4 e 2 e 4 . (z_1z_2+z_3z_4)(z_1z_3+z_2z_4)(z_1z_4+z_2z_3)
=e_3^2+e_1^2e_4-4e_2e_4. ( z 1 z 2 + z 3 z 4 ) ( z 1 z 3 + z 2 z 4 ) ( z 1 z 4 + z 2 z 3 ) = e 3 2 + e 1 2 e 4 − 4 e 2 e 4 . 其中无重复指标的项贡献 e 4 ∑ z i 2 e_4\sum z_i^2 e 4 ∑ z i 2 ∑ z i 2 = e 1 2 − 2 e 2 \sum z_i^2=e_1^2-2e_2 ∑ z i 2 = e 1 2 − 2 e 2 e 2 e 4 e_2e_4 e 2 e 4 e 1 = − 2 , e 2 = e 3 = 0 , e 4 = 2 e_1=-2,e_2=e_3=0,e_4=2 e 1 = − 2 , e 2 = e 3 = 0 , e 4 = 2 8 8 8
第 30 题
题目
x ∈ R , z ∈ C x\in\mathbb R,z\in\mathbb C x ∈ R , z ∈ C z z z π 4 \frac\pi4 4 π ∣ x − z ∣ + ∣ 2 + i − x ∣ + ∣ 2 + i − z ∣ |x-z|+|2+i-x|+|2+i-z| ∣ x − z ∣ + ∣2 + i − x ∣ + ∣2 + i − z ∣
A. 2 2 2\sqrt2 2 2
B. 3 3 3
C. 10 \sqrt{10} 10
D. 11 \sqrt{11} 11
解答
将复数视为点，记 W = ( 2 , 1 ) W=(2,1) W = ( 2 , 1 ) Q = z Q=z Q = z y = x > 0 y=x>0 y = x > 0 P = ( x , 0 ) P=(x,0) P = ( x , 0 ) W W W y = x y=x y = x W ′ = ( 1 , 2 ) W'=(1,2) W ′ = ( 1 , 2 ) W ′ ′ = ( 2 , − 1 ) W''=(2,-1) W ′′ = ( 2 , − 1 )
W Q + Q P + P W = W ′ Q + Q P + P W ′ ′ ≥ W ′ W ′ ′ = 10 . WQ+QP+PW=W'Q+QP+PW''\ge W'W''=\sqrt{10}. W Q + QP + P W = W ′ Q + QP + P W ′′ ≥ W ′ W ′′ = 10 . 线段 W ′ W ′ ′ W'W'' W ′ W ′′ Q = ( 5 / 4 , 5 / 4 ) Q=(5/4,5/4) Q = ( 5/4 , 5/4 ) P = ( 5 / 3 , 0 ) P=(5/3,0) P = ( 5/3 , 0 )
第 31 题
题目
直三棱柱 A B C − A 1 B 1 C 1 ABC-A_1B_1C_1 A B C − A 1 B 1 C 1 ∠ A B C = 90 ∘ \angle ABC=90^\circ ∠ A B C = 9 0 ∘ A B = 6 AB=6 A B = 6 B C = B B 1 = 3 2 BC=BB_1=3\sqrt2 B C = B B 1 = 3 2 P ∈ B 1 C P\in B_1C P ∈ B 1 C A 1 P + B P A_1P+BP A 1 P + B P 3 + 10 \sqrt{3+\sqrt{10}} 3 + 10 3 10 3\sqrt{10} 3 10 6 + 3 2 6+3\sqrt2 6 + 3 2 9 2 9\sqrt2 9 2
解答
题面校注：必须明确为直三棱柱；斜三棱柱只有所列棱长不足以决定答案。
令 B = ( 0 , 0 , 0 ) , A = ( 6 , 0 , 0 ) , C = ( 0 , 3 2 , 0 ) , B 1 = ( 0 , 0 , 3 2 ) B=(0,0,0),A=(6,0,0),C=(0,3\sqrt2,0),B_1=(0,0,3\sqrt2) B = ( 0 , 0 , 0 ) , A = ( 6 , 0 , 0 ) , C = ( 0 , 3 2 , 0 ) , B 1 = ( 0 , 0 , 3 2 ) P = ( 0 , 3 2 t , 3 2 ( 1 − t ) ) P=(0,3\sqrt2t,3\sqrt2(1-t)) P = ( 0 , 3 2 t , 3 2 ( 1 − t )) 0 ≤ t ≤ 1 0\le t\le1 0 ≤ t ≤ 1
F ( t ) = 36 + 36 t 2 + 9 + ( 6 t − 3 ) 2 . F(t)=\sqrt{36+36t^2}+\sqrt{9+(6t-3)^2}. F ( t ) = 36 + 36 t 2 + 9 + ( 6 t − 3 ) 2 . 平面向量 ( 6 , 6 t ) (6,6t) ( 6 , 6 t ) ( 3 , 3 − 6 t ) (3,3-6t) ( 3 , 3 − 6 t ) ( 9 , 3 ) (9,3) ( 9 , 3 ) F ( t ) ≥ 3 10 F(t)\ge3\sqrt{10} F ( t ) ≥ 3 10 t = 1 / 3 t=1/3 t = 1/3 F F F F ( 0 ) = 6 + 3 2 , F ( 1 ) = 9 2 F(0)=6+3\sqrt2,F(1)=9\sqrt2 F ( 0 ) = 6 + 3 2 , F ( 1 ) = 9 2
第 32 题
题目
令 z k = cos 2 k − 2 5 π + i sin 2 k − 2 5 π z_k=\cos\frac{2k-2}{5}\pi+i\sin\frac{2k-2}{5}\pi z k = cos 5 2 k − 2 π + i sin 5 2 k − 2 π k = 1 , … , 5 k=1,\ldots,5 k = 1 , … , 5 a i = ∏ i ≠ j ( z i − z j ) a_i=\prod_{i\ne j}(z_i-z_j) a i = ∏ i = j ( z i − z j ) a 1 a 3 a 4 = 125 a_1a_3a_4=125 a 1 a 3 a 4 = 125 a 1 a 2 a 3 a 4 a 5 = 5 5 a_1a_2a_3a_4a_5=5^5 a 1 a 2 a 3 a 4 a 5 = 5 5 a 2 a 4 2 = 125 a_2a_4^2=125 a 2 a 4 2 = 125 a 1 = 5 a_1=5 a 1 = 5
解答
五个 z i z_i z i z 5 − 1 z^5-1 z 5 − 1 a i = 5 z i 4 a_i=5z_i^4 a i = 5 z i 4 ζ = e 2 π i / 5 \zeta=e^{2\pi i/5} ζ = e 2 π i /5 a i = 5 ζ 4 ( i − 1 ) a_i=5\zeta^{4(i-1)} a i = 5 ζ 4 ( i − 1 ) 4 ( 0 + 2 + 3 ) = 20 4(0+2+3)=20 4 ( 0 + 2 + 3 ) = 20 4 ( 0 + 1 + 2 + 3 + 4 ) = 40 4(0+1+2+3+4)=40 4 ( 0 + 1 + 2 + 3 + 4 ) = 40 4 + 2 ⋅ 12 = 28 4+2\cdot12=28 4 + 2 ⋅ 12 = 28 5 5 5 5 5 5
第 33 题
题目
f ( x ) = 9 x − 3 x + 1 + a f(x)=9^x-3^{x+1}+a f ( x ) = 9 x − 3 x + 1 + a f ( x ) < 0 f(x)<0 f ( x ) < 0 x ∈ ( 0 , 1 ) x\in(0,1) x ∈ ( 0 , 1 ) a ∈ ( − ∞ , − 3 ) a\in(-\infty,-3) a ∈ ( − ∞ , − 3 ) x ∈ ( 0 , 1 ) x\in(0,1) x ∈ ( 0 , 1 ) f ( x ) < 0 f(x)<0 f ( x ) < 0 a ∈ ( − ∞ , 9 / 4 ) a\in(-\infty,9/4) a ∈ ( − ∞ , 9/4 ) f ( x ) = 3 x a f(x)=3^xa f ( x ) = 3 x a [ 0 , 1 ] [0,1] [ 0 , 1 ] a ∈ ( − ∞ , 0 ] a\in(-\infty,0] a ∈ ( − ∞ , 0 ] [ 0 , 1 ] [0,1] [ 0 , 1 ] a ∈ ( − ∞ , 0 ] a\in(-\infty,0] a ∈ ( − ∞ , 0 ]
解答
令 t = 3 x t=3^x t = 3 x x ∈ ( 0 , 1 ) x\in(0,1) x ∈ ( 0 , 1 ) t ∈ ( 1 , 3 ) t\in(1,3) t ∈ ( 1 , 3 ) f = t 2 − 3 t + a f=t^2-3t+a f = t 2 − 3 t + a a ≤ 0 a\le0 a ≤ 0 t = 3 t=3 t = 3 a = 0 a=0 a = 0 a − 9 / 4 < 0 a-9/4<0 a − 9/4 < 0
对于后两个选项，方程为 t ( t − 3 ) = a ( t − 1 ) t(t-3)=a(t-1) t ( t − 3 ) = a ( t − 1 ) t ∈ [ 1 , 3 ] t\in[1,3] t ∈ [ 1 , 3 ] t = 1 t=1 t = 1 ( 1 , 3 ] (1,3] ( 1 , 3 ]
a = h ( t ) = t ( t − 3 ) t − 1 = t − 2 − 2 t − 1 , h ′ ( t ) = 1 + 2 ( t − 1 ) 2 > 0. a=h(t)=\frac{t(t-3)}{t-1}=t-2-\frac2{t-1},\qquad
h'(t)=1+\frac2{(t-1)^2}>0. a = h ( t ) = t − 1 t ( t − 3 ) = t − 2 − t − 1 2 , h ′ ( t ) = 1 + ( t − 1 ) 2 2 > 0. 其值域为 ( − ∞ , 0 ] (-\infty,0] ( − ∞ , 0 ]
第 34 题
题目
求 min a , b ∈ R max x ∈ [ 0 , 4 ] ∣ ( x − 2 ) 3 + a x + b ∣ \displaystyle\min_{a,b\in\mathbb R}\max_{x\in[0,4]}|(x-2)^3+ax+b| a , b ∈ R min x ∈ [ 0 , 4 ] max ∣ ( x − 2 ) 3 + a x + b ∣
A. 1 1 1
B. 2 2 2
C. 2 3 27 \frac{2\sqrt3}{27} 27 2 3
D. 4 3 27 \frac{4\sqrt3}{27} 27 4 3
解答
令 t = x − 2 t=x-2 t = x − 2 p ( t ) = t 3 + a t + c p(t)=t^3+at+c p ( t ) = t 3 + a t + c c = 2 a + b c=2a+b c = 2 a + b t ∈ [ − 2 , 2 ] t\in[-2,2] t ∈ [ − 2 , 2 ] ∣ p ( t ) ∣ ≤ M |p(t)|\le M ∣ p ( t ) ∣ ≤ M
[ p ( 2 ) − p ( − 2 ) ] − 2 [ p ( 1 ) − p ( − 1 ) ] = 12 [p(2)-p(-2)]-2[p(1)-p(-1)]=12 [ p ( 2 ) − p ( − 2 )] − 2 [ p ( 1 ) − p ( − 1 )] = 12 给出 12 ≤ 6 M 12\le6M 12 ≤ 6 M M ≥ 2 M\ge2 M ≥ 2 p ( t ) = t 3 − 3 t p(t)=t^3-3t p ( t ) = t 3 − 3 t [ − 2 , 2 ] [-2,2] [ − 2 , 2 ] t = ± 1 t=\pm1 t = ± 1 2 2 2 2 2 2 a = − 3 , b = 6 a=-3,b=6 a = − 3 , b = 6
第 35 题
题目
在 △ A B C \triangle ABC △ A B C sin 2 A − sin 2 C = sin A sin B − sin 2 B \sin^2A-\sin^2C=\sin A\sin B-\sin^2B sin 2 A − sin 2 C = sin A sin B − sin 2 B c = 2 c=2 c = 2 sin 2 C \sin2C sin 2 C sin 2 C \sin^2C sin 2 C
A. 3 \sqrt3 3
B. 2 2 2
C. 5 \sqrt5 5
D. 2 2 2\sqrt2 2 2
解答
题面校注：现将来源不明的 sin 2 C \sin2C sin 2 C sin 2 C \sin^2C sin 2 C
由正弦定理，条件等价于 a 2 + b 2 − c 2 = a b a^2+b^2-c^2=ab a 2 + b 2 − c 2 = ab cos C = 1 / 2 \cos C=1/2 cos C = 1/2 C = 60 ∘ C=60^\circ C = 6 0 ∘ c 2 = a 2 + b 2 − a b = 4 c^2=a^2+b^2-ab=4 c 2 = a 2 + b 2 − ab = 4 4 = ( a − b ) 2 + a b ≥ a b 4=(a-b)^2+ab\ge ab 4 = ( a − b ) 2 + ab ≥ ab
S = 1 2 a b sin 60 ∘ ≤ 3 . S=\frac12ab\sin60^\circ\le\sqrt3. S = 2 1 ab sin 6 0 ∘ ≤ 3 . a = b = c = 2 a=b=c=2 a = b = c = 2 3 \sqrt3 3