按广东版源 PDF 的九题编号。第 2 题源答案的分母写错，正确参数为 27 / 256 27/256 27/256
答案速览
题号
答案
题号
答案
1
sin ( π / 2021 ) 1 − cos ( π / 2021 ) \dfrac{\sin(\pi/2021)}{1-\cos(\pi/2021)} 1 − cos ( π /2021 ) sin ( π /2021 ) 6
[ − 1 / 4 , 7 ] [-1/4,7] [ − 1/4 , 7 ]
2
27 / 256 27/256 27/256 7
命题成立
3
f ( x ) = 1 − x 2 f(x)=1-x^2 f ( x ) = 1 − x 2 8
a n = n ( n + 2 ) a_n=n(n+2) a n = n ( n + 2 )
4
6 / 2 \sqrt6/2 6 /2 9
命题成立
5
1 k + ( − 1 ) n k ( k − 1 ) n − 1 \dfrac1k+\dfrac{(-1)^n}{k(k-1)^{n-1}} k 1 + k ( k − 1 ) n − 1 ( − 1 ) n
逐题解析
第 1 题
题目
计算 ∑ k = 1 2020 sin k π 2021 \displaystyle\sum_{k=1}^{2020}\sin\frac{k\pi}{2021} k = 1 ∑ 2020 sin 2021 k π
解答
令 t = π / 2021 t=\pi/2021 t = π /2021 2 sin ( t / 2 ) sin ( j t ) = cos ( ( j − 1 / 2 ) t ) − cos ( ( j + 1 / 2 ) t ) 2\sin(t/2)\sin(jt)=\cos((j-1/2)t)-\cos((j+1/2)t) 2 sin ( t /2 ) sin ( j t ) = cos (( j − 1/2 ) t ) − cos (( j + 1/2 ) t )
∑ j = 1 2020 sin ( j t ) = cos ( t / 2 ) − cos ( ( 2020 + 1 / 2 ) t ) 2 sin ( t / 2 ) = sin ( 1010 t ) sin ( 2021 t / 2 ) sin ( t / 2 ) . \sum_{j=1}^{2020}\sin(jt)
=\frac{\cos(t/2)-\cos((2020+1/2)t)}{2\sin(t/2)}
=\frac{\sin(1010t)\sin(2021t/2)}{\sin(t/2)}. j = 1 ∑ 2020 sin ( j t ) = 2 sin ( t /2 ) cos ( t /2 ) − cos (( 2020 + 1/2 ) t ) = sin ( t /2 ) sin ( 1010 t ) sin ( 2021 t /2 ) . 因 2021 t / 2 = π / 2 2021t/2=\pi/2 2021 t /2 = π /2 sin ( 1010 t ) = cos ( t / 2 ) \sin(1010t)=\cos(t/2) sin ( 1010 t ) = cos ( t /2 ) cot ( t / 2 ) = sin t / ( 1 − cos t ) \cot(t/2)=\sin t/(1-\cos t) cot ( t /2 ) = sin t / ( 1 − cos t )
第 2 题
题目
抛物线 y = x 2 y=x^2 y = x 2 x = a y 2 + 1 x=ay^2+1 x = a y 2 + 1 a a a
解答
设切点为 ( u , u 2 ) (u,u^2) ( u , u 2 ) u = a u 4 + 1 u=au^4+1 u = a u 4 + 1 2 u 2u 2 u 1 / ( 2 a u 2 ) 1/(2au^2) 1/ ( 2 a u 2 ) 4 a u 3 = 1 4au^3=1 4 a u 3 = 1 u ≠ 0 u\ne0 u = 0 a = 1 / ( 4 u 3 ) a=1/(4u^3) a = 1/ ( 4 u 3 ) u = u / 4 + 1 u=u/4+1 u = u /4 + 1 u = 4 / 3 u=4/3 u = 4/3
a = 1 4 ( 4 / 3 ) 3 = 27 256 . a=\frac1{4(4/3)^3}=\frac{27}{256}. a = 4 ( 4/3 ) 3 1 = 256 27 . 代回两条曲线和斜率均一致，故确有切点。答案校注： 源答案表写成 27 / 265 27/265 27/265
第 3 题
题目
写出一个函数 f ( x ) f(x) f ( x ) x , y ∈ R x,y\in\mathbb R x , y ∈ R f ( x − f ( y ) ) = f ( f ( y ) ) + 2 x f ( y ) + f ( x ) − 1 f(x-f(y))=f(f(y))+2xf(y)+f(x)-1 f ( x − f ( y )) = f ( f ( y )) + 2 x f ( y ) + f ( x ) − 1
解答
取 f ( x ) = 1 − x 2 f(x)=1-x^2 f ( x ) = 1 − x 2 h = f ( y ) = 1 − y 2 h=f(y)=1-y^2 h = f ( y ) = 1 − y 2
f ( x − h ) = 1 − x 2 + 2 x h − h 2 , f(x-h)=1-x^2+2xh-h^2, f ( x − h ) = 1 − x 2 + 2 x h − h 2 , 而 f ( h ) + 2 x h + f ( x ) − 1 = ( 1 − h 2 ) + 2 x h + ( 1 − x 2 ) − 1 f(h)+2xh+f(x)-1=(1-h^2)+2xh+(1-x^2)-1 f ( h ) + 2 x h + f ( x ) − 1 = ( 1 − h 2 ) + 2 x h + ( 1 − x 2 ) − 1 x , y x,y x , y
第 4 题
题目
空间区域 { ( x , y , z ) ∣ x 2 + y 2 + z 2 ≤ 1 , z ≥ 0 } \{(x,y,z)\mid x^2+y^2+z^2\le1, z\ge0\} {( x , y , z ) ∣ x 2 + y 2 + z 2 ≤ 1 , z ≥ 0 } d d d d d d
解答
四点两两等距 d d d O O O O O O h = 2 / 3 d h=\sqrt{2/3}\,d h = 2/3 d
证明引理。正四面体的外心 G G G O O O G G G O O O Q Q Q R ( P ) = max i ∣ P − V i ∣ R(P)=\max_i|P-V_i| R ( P ) = max i ∣ P − V i ∣ V i V_i V i G G G G G G R ( O ) ≥ R ( Q ) R(O)\ge R(Q) R ( O ) ≥ R ( Q ) Q Q Q R R R Q Q Q R R R h h h R ( O ) ≥ h R(O)\ge h R ( O ) ≥ h
本题球半径为 1 1 1 d ≤ 1 / 2 / 3 = 6 / 2 d\le1/\sqrt{2/3}=\sqrt6/2 d ≤ 1/ 2/3 = 6 /2 ( 0 , 0 , 1 ) (0,0,1) ( 0 , 0 , 1 ) z = 0 z=0 z = 0 1 / 2 1/\sqrt2 1/ 2 120 ∘ 120^\circ 12 0 ∘ 3 / 2 \sqrt{3/2} 3/2 1 + 1 / 2 = 3 / 2 \sqrt{1+1/2}=\sqrt{3/2} 1 + 1/2 = 3/2
第 5 题
题目
k k k k ≥ 3 k\ge3 k ≥ 3 n n n
解答
记第 n n n p n p_n p n p 0 = 1 p_0=1 p 0 = 1 1 / ( k − 1 ) 1/(k-1) 1/ ( k − 1 )
p n + 1 = 1 − p n k − 1 . p_{n+1}=\frac{1-p_n}{k-1}. p n + 1 = k − 1 1 − p n . 定点为 1 / k 1/k 1/ k q n = p n − 1 / k q_n=p_n-1/k q n = p n − 1/ k q n + 1 = − q n / ( k − 1 ) q_{n+1}=-q_n/(k-1) q n + 1 = − q n / ( k − 1 ) q 0 = ( k − 1 ) / k q_0=(k-1)/k q 0 = ( k − 1 ) / k
p n = 1 k + ( − 1 ) n k ( k − 1 ) n − 1 . p_n=\frac1k+\frac{(-1)^n}{k(k-1)^{n-1}}. p n = k 1 + k ( k − 1 ) n − 1 ( − 1 ) n . n = 0 n=0 n = 0 p 0 = 1 p_0=1 p 0 = 1 n ≥ 1 n\ge1 n ≥ 1
第 6 题
题目
求函数 f ( x ) = 5 + 6 cos x − 3 cos 2 x − 4 cos 3 x + 1 4 sin 3 x 2 f(x)=5+6\cos x-3\cos^2x-4\cos^3x+\frac14\sin\frac{3x}{2} f ( x ) = 5 + 6 cos x − 3 cos 2 x − 4 cos 3 x + 4 1 sin 2 3 x
解答
设 c = cos x ∈ [ − 1 , 1 ] c=\cos x\in[-1,1] c = cos x ∈ [ − 1 , 1 ]
5 + 6 c − 3 c 2 − 4 c 3 = ( 1 + c ) 2 ( 5 − 4 c ) . 5+6c-3c^2-4c^3=(1+c)^2(5-4c). 5 + 6 c − 3 c 2 − 4 c 3 = ( 1 + c ) 2 ( 5 − 4 c ) . 它非负，且由求导 6 − 6 c − 12 c 2 = 6 ( 1 + c ) ( 1 − 2 c ) 6-6c-12c^2=6(1+c)(1-2c) 6 − 6 c − 12 c 2 = 6 ( 1 + c ) ( 1 − 2 c ) 27 / 4 27/4 27/4 c = 1 / 2 c=1/2 c = 1/2 − 1 ≤ sin ( 3 x / 2 ) ≤ 1 -1\le\sin(3x/2)\le1 − 1 ≤ sin ( 3 x /2 ) ≤ 1 [ − 1 / 4 , 7 ] [-1/4,7] [ − 1/4 , 7 ] x = π / 3 x=\pi/3 x = π /3 c = 1 / 2 c=1/2 c = 1/2 sin ( 3 x / 2 ) = 1 \sin(3x/2)=1 sin ( 3 x /2 ) = 1 x = π x=\pi x = π c = − 1 c=-1 c = − 1 sin ( 3 x / 2 ) = − 1 \sin(3x/2)=-1 sin ( 3 x /2 ) = − 1 [ − 1 / 4 , 7 ] [-1/4,7] [ − 1/4 , 7 ]
第 7 题
题目
设 a , b , c a,b,c a , b , c p ≥ 5 p\ge5 p ≥ 5 p ∣ a ( p − 1 ) / 2 + b ( p − 1 ) / 2 + c ( p − 1 ) / 2 p\mid a^{(p-1)/2}+b^{(p-1)/2}+c^{(p-1)/2} p ∣ a ( p − 1 ) /2 + b ( p − 1 ) /2 + c ( p − 1 ) /2 p ∣ a b c p\mid abc p ∣ ab c
解答
反设 p ∤ a b c p\nmid abc p ∤ ab c a ( p − 1 ) / 2 , b ( p − 1 ) / 2 , c ( p − 1 ) / 2 a^{(p-1)/2},b^{(p-1)/2},c^{(p-1)/2} a ( p − 1 ) /2 , b ( p − 1 ) /2 , c ( p − 1 ) /2 p p p 1 1 1 p p p 1 1 1 − 1 -1 − 1 − 3 , − 1 , 1 , 3 -3,-1,1,3 − 3 , − 1 , 1 , 3 p ≥ 5 p\ge5 p ≥ 5 p p p p p p a , b , c a,b,c a , b , c p ∣ a b c p\mid abc p ∣ ab c
第 8 题
题目
数列 { a n } \{a_n\} { a n } a 1 = 3 a_1=3 a 1 = 3 m , n m,n m , n a 2 m + n = 2 a m + a n + 2 m 2 + 4 m n a_{2m+n}=2a_m+a_n+2m^2+4mn a 2 m + n = 2 a m + a n + 2 m 2 + 4 mn a n a_n a n
解答
在给定关系中取 m = n = 1 m=n=1 m = n = 1 a 3 = 2 a 1 + a 1 + 2 + 4 = 15 a_3=2a_1+a_1+2+4=15 a 3 = 2 a 1 + a 1 + 2 + 4 = 15 ( m , n ) = ( 2 , 1 ) (m,n)=(2,1) ( m , n ) = ( 2 , 1 ) ( 1 , 3 ) (1,3) ( 1 , 3 ) a 5 a_5 a 5
2 a 2 + 19 = a 5 = 2 a 1 + a 3 + 2 + 12 = 35 , 2a_2+19=a_5=2a_1+a_3+2+12=35, 2 a 2 + 19 = a 5 = 2 a 1 + a 3 + 2 + 12 = 35 , 故 a 2 = 8 a_2=8 a 2 = 8 m = 1 , n = r − 2 m=1,n=r-2 m = 1 , n = r − 2 r ≥ 3 r\ge3 r ≥ 3 a r = a r − 2 + 4 r a_r=a_{r-2}+4r a r = a r − 2 + 4 r a 1 = 3 , a 2 = 8 a_1=3,a_2=8 a 1 = 3 , a 2 = 8 a r = r ( r + 2 ) a_r=r(r+2) a r = r ( r + 2 ) r ( r + 2 ) − ( r − 2 ) r = 4 r r(r+2)-(r-2)r=4r r ( r + 2 ) − ( r − 2 ) r = 4 r
最后对任意正整数 m , n m,n m , n
2 m ( m + 2 ) + n ( n + 2 ) + 2 m 2 + 4 m n = ( 2 m + n ) ( 2 m + n + 2 ) . 2m(m+2)+n(n+2)+2m^2+4mn=(2m+n)(2m+n+2). 2 m ( m + 2 ) + n ( n + 2 ) + 2 m 2 + 4 mn = ( 2 m + n ) ( 2 m + n + 2 ) . 所以该通项满足全部原条件，而不只是上述特例。
第 9 题
题目
f ( x ) f(x) f ( x ) n n n n ≥ 1 n\ge1 n ≥ 1 g ( x ) = f ( x ) − f ′ ( x ) g(x)=f(x)-f'(x) g ( x ) = f ( x ) − f ′ ( x ) f f f n n n g g g n n n
解答
设 f f f r 1 < ⋯ < r k r_1<\cdots<r_k r 1 < ⋯ < r k m 1 , … , m k m_1,\ldots,m_k m 1 , … , m k ∑ i = 1 k m i = n \sum_{i=1}^k m_i=n ∑ i = 1 k m i = n f ( x ) = ( x − r i ) m i u i ( x ) f(x)=(x-r_i)^{m_i}u_i(x) f ( x ) = ( x − r i ) m i u i ( x ) u i ( r i ) ≠ 0 u_i(r_i)\ne0 u i ( r i ) = 0
g ( x ) = ( x − r i ) m i − 1 [ ( x − r i ) ( u i ( x ) − u i ′ ( x ) ) − m i u i ( x ) ] . g(x)=(x-r_i)^{m_i-1}\bigl[(x-r_i)(u_i(x)-u_i'(x))-m_i u_i(x)\bigr]. g ( x ) = ( x − r i ) m i − 1 [ ( x − r i ) ( u i ( x ) − u i ′ ( x )) − m i u i ( x ) ] . 方括号在 r i r_i r i − m i u i ( r i ) ≠ 0 -m_i u_i(r_i)\ne0 − m i u i ( r i ) = 0 r i r_i r i g g g m i − 1 m_i-1 m i − 1 n − k n-k n − k
当 x ≠ r 1 , … , r k x\ne r_1,\ldots,r_k x = r 1 , … , r k g ( x ) = 0 g(x)=0 g ( x ) = 0
H ( x ) : = f ′ ( x ) f ( x ) = ∑ i = 1 k m i x − r i = 1. H(x):=\frac{f'(x)}{f(x)}=\sum_{i=1}^k\frac{m_i}{x-r_i}=1. H ( x ) := f ( x ) f ′ ( x ) = i = 1 ∑ k x − r i m i = 1. 在每个相邻根之间的区间 ( r i , r i + 1 ) (r_i,r_{i+1}) ( r i , r i + 1 ) H ′ ( x ) = − ∑ j m j / ( x − r j ) 2 < 0 H'(x)=-\sum_j m_j/(x-r_j)^2<0 H ′ ( x ) = − ∑ j m j / ( x − r j ) 2 < 0 + ∞ +\infty + ∞ − ∞ -\infty − ∞ H ( x ) = 1 H(x)=1 H ( x ) = 1 ( r k , + ∞ ) (r_k,+\infty) ( r k , + ∞ ) H H H + ∞ +\infty + ∞ 0 0 0 k k k k = 1 k=1 k = 1
连同原根保留的重数，g g g ( n − k ) + k = n (n-k)+k=n ( n − k ) + k = n f ′ f' f ′ n − 1 n-1 n − 1 g = f − f ′ g=f-f' g = f − f ′ n n n