已知函数f ( x ) = ∫ 0 x e t 2 sin t d t f(x)=\int_{0}^{x} e^{t^2} \sin t dt f ( x ) = ∫ 0 x e t 2 sin t d t g ( x ) = ∫ 0 x e t 2 d t ⋅ sin 2 x g(x)=\int_{0}^{x} e^{t^2} dt \cdot \sin^2 x g ( x ) = ∫ 0 x e t 2 d t ⋅ sin 2 x x = 0 x=0 x = 0 f ( x ) f(x) f ( x ) g ( x ) g(x) g ( x ) x = 0 x=0 x = 0 f ( x ) f(x) f ( x ) ( 0 , 0 ) (0,0) ( 0 , 0 ) y = g ( x ) y=g(x) y = g ( x ) x = 0 x=0 x = 0 f ( x ) f(x) f ( x ) ( 0 , 0 ) (0,0) ( 0 , 0 ) y = f ( x ) y=f(x) y = f ( x ) ( 0 , 0 ) (0,0) ( 0 , 0 ) y = f ( x ) y=f(x) y = f ( x ) y = g ( x ) y=g(x) y = g ( x )
已知级数：( 1 ) ∑ n = 1 ∞ sin n 3 π n 2 + 1 (1) \sum_{n=1}^{\infty} \sin \frac{n^3 \pi}{n^2+1} ( 1 ) ∑ n = 1 ∞ sin n 2 + 1 n 3 π ( 2 ) ∑ n = 1 ∞ ( − 1 ) n ( 1 n 2 3 − tan 1 n 2 3 ) (2) \sum_{n=1}^{\infty} (-1)^n \left( \frac{1}{\sqrt[3]{n^2}} - \tan \frac{1}{\sqrt[3]{n^2}} \right) ( 2 ) ∑ n = 1 ∞ ( − 1 ) n ( 3 n 2 1 − tan 3 n 2 1 )
设函数f ( x ) f(x) f ( x ) [ 0 , + ∞ ) [0,+\infty) [ 0 , + ∞ ) lim x → + ∞ f ( x ) \lim_{x \to +\infty} f(x) lim x → + ∞ f ( x ) lim x → + ∞ f ′ ( x ) \lim_{x \to +\infty} f'(x) lim x → + ∞ f ′ ( x ) lim x → + ∞ f ′ ( x ) \lim_{x \to +\infty} f'(x) lim x → + ∞ f ′ ( x ) lim x → + ∞ f ( x ) \lim_{x \to +\infty} f(x) lim x → + ∞ f ( x ) lim x → + ∞ ∫ 0 x f ( t ) d t x \lim_{x \to +\infty} \frac{\int_{0}^{x} f(t) dt}{x} lim x → + ∞ x ∫ 0 x f ( t ) d t lim x → + ∞ f ( x ) \lim_{x \to +\infty} f(x) lim x → + ∞ f ( x ) lim x → + ∞ f ( x ) \lim_{x \to +\infty} f(x) lim x → + ∞ f ( x ) lim x → + ∞ ∫ 0 x f ( t ) d t x \lim_{x \to +\infty} \frac{\int_{0}^{x} f(t) dt}{x} lim x → + ∞ x ∫ 0 x f ( t ) d t
设函数f ( x , y ) f(x,y) f ( x , y ) ∫ − 2 2 d x ∫ 4 − x 2 4 f ( x , y ) d y = \int_{-2}^{2} dx \int_{4-x^2}^{4} f(x,y) dy= ∫ − 2 2 d x ∫ 4 − x 2 4 f ( x , y ) d y = ∫ 0 4 [ ∫ − 2 − 4 − y f ( x , y ) d x + ∫ 4 − y 2 f ( x , y ) d x ] d y \int_{0}^{4} \left[ \int_{-2}^{-\sqrt{4-y}} f(x,y) dx + \int_{\sqrt{4-y}}^{2} f(x,y) dx \right] dy ∫ 0 4 [ ∫ − 2 − 4 − y f ( x , y ) d x + ∫ 4 − y 2 f ( x , y ) d x ] d y ∫ 0 4 [ ∫ − 2 4 − y f ( x , y ) d x + ∫ 4 − y 2 f ( x , y ) d x ] d y \int_{0}^{4} \left[ \int_{-2}^{\sqrt{4-y}} f(x,y) dx + \int_{\sqrt{4-y}}^{2} f(x,y) dx \right] dy ∫ 0 4 [ ∫ − 2 4 − y f ( x , y ) d x + ∫ 4 − y 2 f ( x , y ) d x ] d y ∫ 0 4 [ ∫ − 2 − 4 − y f ( x , y ) d x + ∫ 2 4 − y f ( x , y ) d x ] d y \int_{0}^{4} \left[ \int_{-2}^{-\sqrt{4-y}} f(x,y) dx + \int_{2}^{\sqrt{4-y}} f(x,y) dx \right] dy ∫ 0 4 [ ∫ − 2 − 4 − y f ( x , y ) d x + ∫ 2 4 − y f ( x , y ) d x ] d y 2 ∫ 0 4 d y [ ∫ 4 − y 2 f ( x , y ) d x 2 \int_{0}^{4} dy \left[ \int_{\sqrt{4-y}}^{2} f(x,y) dx \right. 2 ∫ 0 4 d y [ ∫ 4 − y 2 f ( x , y ) d x
二次型f ( x 1 , x 2 , x 3 ) = x 1 2 + 2 x 1 x 2 + 2 x 1 x 3 f(x_1,x_2,x_3)=x_1^2+2x_1x_2+2x_1x_3 f ( x 1 , x 2 , x 3 ) = x 1 2 + 2 x 1 x 2 + 2 x 1 x 3
设α 1 , α 2 , α 3 , α 4 \alpha_1,\alpha_2,\alpha_3,\alpha_4 α 1 , α 2 , α 3 , α 4 n n n α 1 , α 2 \alpha_1,\alpha_2 α 1 , α 2 α 1 , α 2 , α 3 \alpha_1,\alpha_2,\alpha_3 α 1 , α 2 , α 3 α 1 + α 2 + α 4 = 0 \alpha_1+\alpha_2+\alpha_4=0 α 1 + α 2 + α 4 = 0 O − x y z O-xyz O − x yz x , y , z x,y,z x , y , z x α 1 + y α 2 + z α 3 = α 4 x\alpha_1+y\alpha_2+z\alpha_3=\alpha_4 x α 1 + y α 2 + z α 3 = α 4
设n n n A , B , C A,B,C A , B , C r ( A ) + r ( B ) + r ( C ) = r ( A B C ) + 2 n r(A)+r(B)+r(C)=r(ABC)+2n r ( A ) + r ( B ) + r ( C ) = r ( A BC ) + 2 n r ( A B C ) + n = r ( A B ) + r ( C ) r(ABC)+n=r(AB)+r(C) r ( A BC ) + n = r ( A B ) + r ( C ) r ( A B ) + n = r ( A ) + r ( B ) r(AB)+n=r(A)+r(B) r ( A B ) + n = r ( A ) + r ( B ) r ( A ) = r ( B ) = r ( C ) = n r(A)=r(B)=r(C)=n r ( A ) = r ( B ) = r ( C ) = n r ( A B ) = r ( B C ) = n r(AB)=r(BC)=n r ( A B ) = r ( BC ) = n
设二维随机变量( X , Y ) (X,Y) ( X , Y ) N ( 0 , 0 ; 1 , 1 ; ρ ) N(0,0;1,1;\rho) N ( 0 , 0 ; 1 , 1 ; ρ ) ρ ∈ ( − 1 , 1 ) \rho \in (-1,1) ρ ∈ ( − 1 , 1 ) a , b a,b a , b a 2 + b 2 = 1 a^2+b^2=1 a 2 + b 2 = 1 D ( a X + b Y ) D(aX+bY) D ( a X + bY ) 1 + ∣ ρ ∣ 1+|\rho| 1 + ∣ ρ ∣ 1 + ρ 2 1+\rho^2 1 + ρ 2
设X 1 , X 2 , … , X 20 X_1,X_2,\dots,X_{20} X 1 , X 2 , … , X 20 B ( 1 , 0.1 ) B(1,0.1) B ( 1 , 0.1 ) T = ∑ i = 1 20 X i T=\sum_{i=1}^{20} X_i T = ∑ i = 1 20 X i P { T ≤ 1 } ≈ P\{T \leq 1\} \approx P { T ≤ 1 } ≈ 1 e 2 \frac{1}{e^2} e 2 1 2 e 2 \frac{2}{e^2} e 2 2 3 e 2 \frac{3}{e^2} e 2 3 4 e 2 \frac{4}{e^2} e 2 4
设x 1 , x 2 , … , x n x_1,x_2,\dots,x_n x 1 , x 2 , … , x n N ( μ , 2 ) N(\mu,2) N ( μ , 2 ) X ‾ = 1 n ∑ i = 1 n x i \overline{X}=\frac{1}{n}\sum_{i=1}^{n}x_i X = n 1 ∑ i = 1 n x i Z α Z_{\alpha} Z α α \alpha α H 0 : μ ≤ 1 H_0:\mu \leq 1 H 0 : μ ≤ 1 H 1 : μ > 1 H_1:\mu>1 H 1 : μ > 1 α \alpha α { ( x 1 , x 2 , … , x n ) ∣ X ‾ > 1 + 2 n Z α } \left\{ (x_1,x_2,\dots,x_n) \mid \overline{X}>1+\frac{2}{n}Z_{\alpha} \right\} { ( x 1 , x 2 , … , x n ) ∣ X > 1 + n 2 Z α } { ( x 1 , x 2 , … , x n ) ∣ X ‾ > 1 + 2 n Z α } \left\{ (x_1,x_2,\dots,x_n) \mid \overline{X}>1+\sqrt{\frac{2}{n}}Z_{\alpha} \right\} { ( x 1 , x 2 , … , x n ) ∣ X > 1 + n 2 Z α }
lim x → 0 + x x − 1 ln x ⋅ ln ( 1 − x ) = \lim_{x \to 0^+} \frac{x^x - 1}{\ln x \cdot \ln (1-x)}= lim x → 0 + l n x ⋅ l n ( 1 − x ) x x − 1 =
已知函数f ( x ) = { 0 , 0 ≤ x l t ; 1 2 x 2 , 1 2 ≤ x ≤ 1 f(x)=\begin{cases}0, & 0 \leq x<\frac{1}{2} \\ x^2, & \frac{1}{2} \leq x \leq 1\end{cases} f ( x ) = { 0 , x 2 , 0 ≤ x 2 1 ≤ x ≤ 1 lt ; 2 1 ∑ n = 1 ∞ b n sin n π x \sum_{n=1}^{\infty} b_n \sin n\pi x ∑ n = 1 ∞ b n sin nπ x S ( x ) S(x) S ( x ) S ( − 7 2 ) = S\left(-\frac{7}{2}\right)= S ( − 2 7 ) =
已知函数U ( x , y , z ) = x y 2 z 3 U(x,y,z)=xy^2z^3 U ( x , y , z ) = x y 2 z 3 n = ( 2 , 2 , − 1 ) \boldsymbol{n}=(2,2,-1) n = ( 2 , 2 , − 1 ) ∂ U ∂ n ∣ ( 1 , 1 , 1 ) = \frac{\partial U}{\partial \boldsymbol{n}} \big|_{(1,1,1)}= ∂ n ∂ U ( 1 , 1 , 1 ) =
已知有向曲线L L L y = 1 − x 2 y=1-x^2 y = 1 − x 2 ( 1 , 0 ) (1,0) ( 1 , 0 ) ( − 1 , 0 ) (-1,0) ( − 1 , 0 ) ∫ L ( y + cos x ) d x + ( 2 x + cos y ) d y = \int_{L} (y+\cos x)dx + (2x+\cos y)dy= ∫ L ( y + cos x ) d x + ( 2 x + cos y ) d y =
设矩阵A = ( 4 2 − 3 a 3 − 4 b 5 − 7 ) A=\begin{pmatrix} 4 & 2 & -3 \\ a & 3 & -4 \\ b & 5 & -7 \end{pmatrix} A = 4 a b 2 3 5 − 3 − 4 − 7 A 2 X = 0 A^2X=0 A 2 X = 0 A X = 0 AX=0 A X = 0
设A , B A,B A , B P ( A ) = 2 P ( B ) P(A)=2P(B) P ( A ) = 2 P ( B ) P ( A ∪ B ) = 5 8 P(A \cup B)=\frac{5}{8} P ( A ∪ B ) = 8 5 A , B A,B A , B A , B A,B A , B
计算∫ 0 1 1 ( x + 1 ) ( x 2 − 2 x + 2 ) d x \int_{0}^{1} \frac{1}{(x+1)(x^2-2x+2)} dx ∫ 0 1 ( x + 1 ) ( x 2 − 2 x + 2 ) 1 d x
已知函数f ( u ) f(u) f ( u ) ( 0 , + ∞ ) (0,+\infty) ( 0 , + ∞ ) g ( x , y ) = f ( x y ) g(x,y)=f\left( \frac{x}{y} \right) g ( x , y ) = f ( y x ) g ( x , y ) g(x,y) g ( x , y ) x 2 ∂ 2 g ∂ x 2 + x y ∂ 2 g ∂ x ∂ y + y 2 ∂ 2 g ∂ y 2 = 1 x^2 \frac{\partial^2 g}{\partial x^2} + xy \frac{\partial^2 g}{\partial x \partial y} + y^2 \frac{\partial^2 g}{\partial y^2}=1 x 2 ∂ x 2 ∂ 2 g + x y ∂ x ∂ y ∂ 2 g + y 2 ∂ y 2 ∂ 2 g = 1 g ( x , x ) = 1 g(x,x)=1 g ( x , x ) = 1 ∂ g ∂ x ∣ ( x , x ) = 2 x \left. \frac{\partial g}{\partial x} \right|_{(x,x)}=\frac{2}{x} ∂ x ∂ g ( x , x ) = x 2 f ( u ) f(u) f ( u )
设函数f ( x ) f(x) f ( x ) ( a , b ) (a,b) ( a , b ) f ′ ( x ) f'(x) f ′ ( x ) ( a , b ) (a,b) ( a , b ) ( a , b ) (a,b) ( a , b ) x 1 , x 2 , x 3 x_1,x_2,x_3 x 1 , x 2 , x 3 x 1 < x 2 < x 3 x_1<x_2<x_3 x 1 < x 2 < x 3 \frac{f(x_2)-f(x_1)}{x_2-x_1} < \frac{f(x_3)-f(x_2)}{x_3-x_2}
设Σ \Sigma Σ { x = 0 y = 0 \begin{cases} x=0 \\ y=0 \end{cases} { x = 0 y = 0 { x = t y = t z = t \begin{cases} x=t \\ y=t \\ z=t \end{cases} ⎩ ⎨ ⎧ x = t y = t z = t t t t Σ 1 \Sigma_1 Σ 1 Σ \Sigma Σ x + y + z = 0 x+y+z=0 x + y + z = 0 x + y + z = 1 x+y+z=1 x + y + z = 1 ∬ Σ 1 x d y d z + ( y + 1 ) d z d x + ( z + 2 ) d x d y \iint_{\Sigma_1} xdydz + (y+1)dzdx + (z+2)dxdy ∬ Σ 1 x d y d z + ( y + 1 ) d z d x + ( z + 2 ) d x d y
设矩阵A = ( 0 − 1 2 − 1 0 2 − 1 − 1 a ) A=\begin{pmatrix} 0 & -1 & 2 \\ -1 & 0 & 2 \\ -1 & -1 & a \end{pmatrix} A = 0 − 1 − 1 − 1 0 − 1 2 2 a 1 1 1 A A A a a a A α = α + β A\alpha=\alpha+\beta A α = α + β A 2 α = α + 2 β A^2\alpha=\alpha+2\beta A 2 α = α + 2 β α , β \alpha,\beta α , β
投保人的损失事件发生时，保险公司的赔付额Y Y Y X X X Y = { 0 , X ≤ 100 X − 100 , X > 100 Y=\begin{cases}0, & X \leq 100 \\ X-100, & X>100\end{cases} Y = { 0 , X − 100 , X ≤ 100 X > 100 X X X f ( x ) = { 2 × 100 2 ( 100 + x ) 3 , x > 0 0 , x ≤ 0 f(x)=\begin{cases}\frac{2 \times 100^2}{(100+x)^3}, & x>0 \\ 0, & x \leq 0\end{cases} f ( x ) = { ( 100 + x ) 3 2 × 10 0 2 , 0 , x > 0 x ≤ 0 P { Y > 0 } P\{Y>0\} P { Y > 0 } E ( Y ) E(Y) E ( Y ) N N N M M M N N N N = n ( n ≥ 1 ) N=n(n \geq 1) N = n ( n ≥ 1 ) M M M B ( n , P ) B(n,P) B ( n , P ) P = P { Y > 0 } P=P\{Y>0\} P = P { Y > 0 } M M M
已知函数f ( x ) = ∫ 0 x e cos t d t f(x)=\int_{0}^{x} e^{\cos t} dt f ( x ) = ∫ 0 x e c o s t d t g ( x ) = ∫ 0 sin x e t 2 d t g(x)=\int_{0}^{\sin x} e^{t^2} dt g ( x ) = ∫ 0 s i n x e t 2 d t f ( x ) f(x) f ( x ) g ( x ) g(x) g ( x ) f ( x ) f(x) f ( x ) g ( x ) g(x) g ( x ) f ( x ) f(x) f ( x ) g ( x ) g(x) g ( x ) f ( x ) f(x) f ( x ) g ( x ) g(x) g ( x )
设P = P ( x , y , z ) P=P(x,y,z) P = P ( x , y , z ) Q = Q ( x , y , z ) Q=Q(x,y,z) Q = Q ( x , y , z ) Σ \Sigma Σ z = 1 − x 2 − y 2 ( x ≥ 0 , y ≥ 0 ) z=\sqrt{1-x^2-y^2}(x \geq 0,y \geq 0) z = 1 − x 2 − y 2 ( x ≥ 0 , y ≥ 0 ) ∬ Σ P d y d z + Q d z d x = \iint_{\Sigma} Pdydz + Qdzdx= ∬ Σ P d y d z + Q d z d x = ∬ Σ ( x z P + y z Q ) d x d y \iint_{\Sigma} \left( \frac{x}{z}P + \frac{y}{z}Q \right) dxdy ∬ Σ ( z x P + z y Q ) d x d y ∬ Σ ( − x z P + y z Q ) d x d y \iint_{\Sigma} \left( -\frac{x}{z}P + \frac{y}{z}Q \right) dxdy ∬ Σ ( − z x P + z y Q ) d x d y ∬ Σ ( x z P − y z Q ) d x d y \iint_{\Sigma} \left( \frac{x}{z}P - \frac{y}{z}Q \right) dxdy ∬ Σ ( z x P − z y Q ) d x d y ∬ Σ ( − x z P − y z Q ) d x d y \iint_{\Sigma} \left( -\frac{x}{z}P - \frac{y}{z}Q \right) dxdy ∬ Σ ( − z x P − z y Q ) d x d y
已知幂级数∑ n = 0 ∞ a n x n \sum_{n=0}^{\infty} a_n x^n ∑ n = 0 ∞ a n x n ln ( 2 + x ) \ln(2+x) ln ( 2 + x ) ∑ n = 0 ∞ n a 2 n = \sum_{n=0}^{\infty} n a_{2n}= ∑ n = 0 ∞ n a 2 n = − 1 6 -\frac{1}{6} − 6 1 − 1 3 -\frac{1}{3} − 3 1 1 6 \frac{1}{6} 6 1 1 3 \frac{1}{3} 3 1
设函数f ( x ) f(x) f ( x ) ( − 1 , 1 ) (-1,1) ( − 1 , 1 ) lim x → 0 f ( x ) = 0 \lim_{x \to 0} f(x)=0 lim x → 0 f ( x ) = 0 lim x → 0 f ( x ) x = m \lim_{x \to 0} \frac{f(x)}{x}=m lim x → 0 x f ( x ) = m f ′ ( 0 ) = m f'(0)=m f ′ ( 0 ) = m f ′ ( 0 ) = m f'(0)=m f ′ ( 0 ) = m lim x → 0 f ( x ) x = m \lim_{x \to 0} \frac{f(x)}{x}=m lim x → 0 x f ( x ) = m lim x → 0 f ′ ( x ) = m \lim_{x \to 0} f'(x)=m lim x → 0 f ′ ( x ) = m f ′ ( 0 ) = m f'(0)=m f ′ ( 0 ) = m f ′ ( 0 ) = m f'(0)=m f ′ ( 0 ) = m lim x → 0 f ′ ( x ) = m \lim_{x \to 0} f'(x)=m lim x → 0 f ′ ( x ) = m
在空间直角坐标系O − x y z O-xyz O − x yz π i : a i x + b i y + c i z = d i ( i = 1 , 2 , 3 ) \pi_i:a_ix+b_iy+c_iz=d_i(i=1,2,3) π i : a i x + b i y + c i z = d i ( i = 1 , 2 , 3 ) α i = ( a i , b i , c i ) \alpha_i=(a_i,b_i,c_i) α i = ( a i , b i , c i ) β i = ( a i , b i , c i , d i ) \beta_i=(a_i,b_i,c_i,d_i) β i = ( a i , b i , c i , d i ) r ( α 1 α 2 α 3 ) = m r\begin{pmatrix} \alpha_1 \\ \alpha_2 \\ \alpha_3 \end{pmatrix}=m r α 1 α 2 α 3 = m r ( β 1 β 2 β 3 ) = n r\begin{pmatrix} \beta_1 \\ \beta_2 \\ \beta_3 \end{pmatrix}=n r β 1 β 2 β 3 = n m = 1 , n = 2 m=1,n=2 m = 1 , n = 2 m = n = 2 m=n=2 m = n = 2 m = 2 , n = 3 m=2,n=3 m = 2 , n = 3 m = n = 3 m=n=3 m = n = 3
设向量α 1 = ( a 1 − 1 1 ) \alpha_1=\begin{pmatrix} a \\ 1 \\ -1 \\ 1 \end{pmatrix} α 1 = a 1 − 1 1 α 2 = ( 1 1 b ) \alpha_2=\begin{pmatrix} 1 \\ 1 \\ b \end{pmatrix} α 2 = 1 1 b α 3 = ( 1 a − 1 1 ) \alpha_3=\begin{pmatrix} 1 \\ a \\ -1 \\ 1 \end{pmatrix} α 3 = 1 a − 1 1 α 1 , α 2 , α 3 \alpha_1,\alpha_2,\alpha_3 α 1 , α 2 , α 3 a = 1 , b ≠ − 1 a=1,b \neq -1 a = 1 , b = − 1 a = 1 , b = − 1 a=1,b=-1 a = 1 , b = − 1 a ≠ − 2 , b = 2 a \neq -2,b=2 a = − 2 , b = 2 − 2 + 6 -2+\sqrt{6} − 2 + 6
设A A A α \alpha α A α = 0 A\alpha=0 A α = 0 β T α = 0 \beta^T\alpha=0 β T α = 0 β \beta β A β = β A\beta=\beta A β = β A 3 A^3 A 3 A 3 A^3 A 3 A 5 A^5 A 5 A 5 A^5 A 5
设随机变量X X X Y Y Y X ∼ N ( 0 , 2 ) X \sim N(0,2) X ∼ N ( 0 , 2 ) Y ∼ N ( − 2 , 2 ) Y \sim N(-2,2) Y ∼ N ( − 2 , 2 ) P { 2 X + Y < a } = P { X > Y } P\{2X+Y<a\}=P\{X>Y\} P { 2 X + Y < a } = P { X > Y } a = a= a = − 2 − 10 -2-\sqrt{10} − 2 − 10 − 2 + 10 -2+\sqrt{10} − 2 + 10 − 2 − 6 -2-\sqrt{6} − 2 − 6 − 2 + 6 -2+\sqrt{6} − 2 + 6
设随机变量X X X f ( x ) = { 2 ( 1 − x ) , 0 < x l t ; 1 0 , 其他 f(x)=\begin{cases}2(1-x), & 0<x<1 \\ 0, & \text{其他}\end{cases} f ( x ) = { 2 ( 1 − x ) , 0 , 0 < x 其他 lt ; 1 X = x X=x X = x Y Y Y ( x , 1 ) (x,1) ( x , 1 ) cov ( X , Y ) = \text{cov}(X,Y)= cov ( X , Y ) = − 1 36 -\frac{1}{36} − 36 1 − 1 72 -\frac{1}{72} − 72 1 1 72 \frac{1}{72} 72 1 1 36 \frac{1}{36} 36 1
设随机变量X X X Y Y Y λ \lambda λ Z = ∣ X − Y ∣ Z=|X-Y| Z = ∣ X − Y ∣ Z Z Z X X X X + Y 2 \frac{X+Y}{2} 2 X + Y X + Y X+Y X + Y X Y XY X Y
已知平面区域D = { ( x , y ) ∣ 1 − y 2 ≤ x ≤ 1 , − 1 ≤ y ≤ 1 } D=\{(x,y) \mid \sqrt{1-y^2} \leq x \leq 1,-1 \leq y \leq 1\} D = {( x , y ) ∣ 1 − y 2 ≤ x ≤ 1 , − 1 ≤ y ≤ 1 } ∬ D x x 2 + y 2 d σ \iint_{D} \frac{x}{\sqrt{x^2+y^2}} d\sigma ∬ D x 2 + y 2 x d σ
设f ( x , y ) = x 3 + y 3 − ( x + y ) 2 + 3 f(x,y)=x^3+y^3-(x+y)^2+3 f ( x , y ) = x 3 + y 3 − ( x + y ) 2 + 3 z = f ( x , y ) z=f(x,y) z = f ( x , y ) ( 1 , 1 , 1 ) (1,1,1) ( 1 , 1 , 1 ) T T T T T T x o y xoy x oy D D D T T T f ( x , y ) f(x,y) f ( x , y ) D D D
设f ( x ) f(x) f ( x ) f ′ ( 0 ) = f ′ ( 1 ) f'(0)=f'(1) f ′ ( 0 ) = f ′ ( 1 ) ∣ f ′ ′ ( x ) ∣ ≤ 1 |f''(x)| \leq 1 ∣ f ′′ ( x ) ∣ ≤ 1 ∣ f ( x ) − f ( 0 ) ( 1 − x ) − f ( 1 ) x ∣ ≤ x ( 1 − x ) 2 |f(x)-f(0)(1-x)-f(1)x| \leq \frac{x(1-x)}{2} ∣ f ( x ) − f ( 0 ) ( 1 − x ) − f ( 1 ) x ∣ ≤ 2 x ( 1 − x ) ∣ ∫ 0 1 f ( x ) d x − f ( 0 ) + f ( 1 ) 2 ∣ ≤ 1 12 \left| \int_{0}^{1} f(x)dx - \frac{f(0)+f(1)}{2} \right| \leq \frac{1}{12} ∫ 0 1 f ( x ) d x − 2 f ( 0 ) + f ( 1 ) ≤ 12 1
已知有向曲线L L L x 2 + y 2 + z 2 = 2 x x^2+y^2+z^2=2x x 2 + y 2 + z 2 = 2 x 2 x − z − 1 = 0 2x-z-1=0 2 x − z − 1 = 0 z z z z z z ∫ L ( 6 x y z − y z 2 ) d x + 2 x 2 z d y + x y z d z \int_{L} (6xyz-yz^2)dx + 2x^2zdy + xyzdz ∫ L ( 6 x yz − y z 2 ) d x + 2 x 2 z d y + x yz d z
已知数列{ x n } , { y n } , { z n } \{x_n\},\{y_n\},\{z_n\} { x n } , { y n } , { z n } x 0 = − 1 , y 0 = 0 , z 0 = 2 x_0=-1,y_0=0,z_0=2 x 0 = − 1 , y 0 = 0 , z 0 = 2 { x n = − 2 x n − 1 + 2 z n − 1 y n = − 2 y n − 1 − 2 z n − 1 z n = − 6 x n − 1 − 3 y n − 1 + 3 z n − 1 \begin{cases} x_n=-2x_{n-1}+2z_{n-1} \\ y_n=-2y_{n-1}-2z_{n-1} \\ z_n=-6x_{n-1}-3y_{n-1}+3z_{n-1} \end{cases} ⎩ ⎨ ⎧ x n = − 2 x n − 1 + 2 z n − 1 y n = − 2 y n − 1 − 2 z n − 1 z n = − 6 x n − 1 − 3 y n − 1 + 3 z n − 1 α n = ( x n y n z n ) \alpha_n=\begin{pmatrix} x_n \\ y_n \\ z_n \end{pmatrix} α n = x n y n z n α n = A α n − 1 \alpha_n=A\alpha_{n-1} α n = A α n − 1 A n A^n A n x n , y n , z n ( n = 1 , 2 , … ) x_n,y_n,z_n(n=1,2,\dots) x n , y n , z n ( n = 1 , 2 , … )
设总体X ∼ U ( 0 , θ ) X \sim U(0,\theta) X ∼ U ( 0 , θ ) θ \theta θ X 1 , X 2 , … , X n X_1,X_2,\dots,X_n X 1 , X 2 , … , X n X ( n ) = max ( X 1 , X 2 , … , X n ) X_{(n)}=\max(X_1,X_2,\dots,X_n) X ( n ) = max ( X 1 , X 2 , … , X n ) T c = c X ( n ) T_c=cX_{(n)} T c = c X ( n ) c c c T c T_c T c θ \theta θ h ( c ) = E ( T c − θ ) 2 h(c)=E(T_c-\theta)^2 h ( c ) = E ( T c − θ ) 2 c c c h ( c ) h(c) h ( c )
曲线y = x ln ( e + 1 x − 1 ) y=x\ln\left(e+\frac{1}{x-1}\right) y = x ln ( e + x − 1 1 ) y = x + e y=x+e y = x + e y = x + 1 e y=x+\frac{1}{e} y = x + e 1 y = x y=x y = x y = x − 1 e y=x-\frac{1}{e} y = x − e 1
若微分方程y ′ ′ + a y ′ + b y = 0 y''+ay'+by=0 y ′′ + a y ′ + b y = 0 ( − ∞ , + ∞ ) (-\infty,+\infty) ( − ∞ , + ∞ ) a<0,b>0
B. a > 0 , b > 0 a>0,b>0 a > 0 , b > 0 a = 0 , b > 0 a=0,b>0 a = 0 , b > 0 a=0,b<0
设函数y = f ( x ) y=f(x) y = f ( x ) { x = 2 t + ∣ t ∣ y = ∣ t ∣ sin t \begin{cases}x=2t+|t| \\ y=|t|\sin t\end{cases} { x = 2 t + ∣ t ∣ y = ∣ t ∣ sin t f ( x ) f(x) f ( x ) f ′ ( 0 ) f'(0) f ′ ( 0 ) f ′ ( 0 ) f'(0) f ′ ( 0 ) f ′ ( x ) f'(x) f ′ ( x ) x = 0 x=0 x = 0 f ′ ( x ) f'(x) f ′ ( x ) f ′ ′ ( 0 ) f''(0) f ′′ ( 0 ) f ′ ′ ( 0 ) f''(0) f ′′ ( 0 ) f ′ ( x ) f'(x) f ′ ( x ) x = 0 x=0 x = 0
已知a n < b n ( n = 1 , 2 , … ) a_n<b_n(n=1,2,\dots) a n < b n ( n = 1 , 2 , … ) ∑ n = 1 ∞ a n \sum_{n=1}^{\infty}a_n ∑ n = 1 ∞ a n ∑ n = 1 ∞ b n \sum_{n=1}^{\infty}b_n ∑ n = 1 ∞ b n ∑ n = 1 ∞ a n \sum_{n=1}^{\infty}a_n ∑ n = 1 ∞ a n ∑ n = 1 ∞ b n \sum_{n=1}^{\infty}b_n ∑ n = 1 ∞ b n
已知A , B , C A,B,C A , B , C A B C = O ABC=O A BC = O r ( O A B C E E ) = r 1 r\begin{pmatrix} O & ABC \\ E & E \end{pmatrix}=r_1 r ( O E A BC E ) = r 1 r ( A B C O E ) = r 2 r\begin{pmatrix} AB & C \\ O & E \end{pmatrix}=r_2 r ( A B O C E ) = r 2 r ( E A B A B O ) = r 3 r\begin{pmatrix} E & AB \\ AB & O \end{pmatrix}=r_3 r ( E A B A B O ) = r 3 r 1 ≤ r 2 ≤ r 3 r_1 \leq r_2 \leq r_3 r 1 ≤ r 2 ≤ r 3 r 1 ≤ r 3 ≤ r 2 r_1 \leq r_3 \leq r_2 r 1 ≤ r 3 ≤ r 2 r 3 ≤ r 2 ≤ r 1 r_3 \leq r_2 \leq r_1 r 3 ≤ r 2 ≤ r 1 r 2 ≤ r 1 ≤ r 3 r_2 \leq r_1 \leq r_3 r 2 ≤ r 1 ≤ r 3
下列矩阵中不能相似于对角矩阵的是（）
A. ( 1 1 a 0 2 2 0 0 3 ) \begin{pmatrix}1&1&a\\0&2&2\\0&0&3\end{pmatrix} 1 0 0 1 2 0 a 2 3 ( 1 1 a 1 2 0 a 0 3 ) \begin{pmatrix}1&1&a\\1&2&0\\a&0&3\end{pmatrix} 1 1 a 1 2 0 a 0 3 ( 1 1 a 0 2 0 0 0 2 ) \begin{pmatrix}1&1&a\\0&2&0\\0&0&2\end{pmatrix} 1 0 0 1 2 0 a 0 2 ( 1 1 a 0 2 2 0 0 2 ) \begin{pmatrix}1&1&a\\0&2&2\\0&0&2\end{pmatrix} 1 0 0 1 2 0 a 2 2
已知向量γ \gamma γ α 1 = ( 1 2 3 ) , α 2 = ( 2 1 1 ) \alpha_1=\begin{pmatrix}1\\2\\3\end{pmatrix},\alpha_2=\begin{pmatrix}2\\1\\1\end{pmatrix} α 1 = 1 2 3 , α 2 = 2 1 1 β 1 = ( 2 5 9 ) , β 2 = ( 1 0 1 ) \beta_1=\begin{pmatrix}2\\5\\9\end{pmatrix},\beta_2=\begin{pmatrix}1\\0\\1\end{pmatrix} β 1 = 2 5 9 , β 2 = 1 0 1 γ = \gamma= γ = k ( 3 3 4 ) , k ∈ R k\begin{pmatrix}3\\3\\4\end{pmatrix},k \in R k 3 3 4 , k ∈ R k ( 3 5 10 ) , k ∈ R k\begin{pmatrix}3\\5\\10\end{pmatrix},k \in R k 3 5 10 , k ∈ R k ( − 1 1 2 ) , k ∈ R k\begin{pmatrix}-1\\1\\2\end{pmatrix},k \in R k − 1 1 2 , k ∈ R k ( 1 5 8 ) , k ∈ R k\begin{pmatrix}1\\5\\8\end{pmatrix},k \in R k 1 5 8 , k ∈ R
设随机变量X X X E ( ∣ X − E X ∣ ) = E(|X-EX|)= E ( ∣ X − EX ∣ ) = 1 e \frac{1}{e} e 1 1 2 \frac{1}{2} 2 1 2 e \frac{2}{e} e 2 2 π \sqrt{2\pi} 2 π
X 1 , X 2 , … , X n X_1,X_2,\dots,X_n X 1 , X 2 , … , X n N ( μ 1 , σ 2 ) N(\mu_1,\sigma^2) N ( μ 1 , σ 2 ) Y 1 , Y 2 , … , Y m Y_1,Y_2,\dots,Y_m Y 1 , Y 2 , … , Y m N ( μ 2 , 2 σ 2 ) N(\mu_2,2\sigma^2) N ( μ 2 , 2 σ 2 ) X ‾ = 1 n ∑ i = 1 n X i \overline{X}=\frac{1}{n}\sum_{i=1}^nX_i X = n 1 ∑ i = 1 n X i Y ‾ = 1 m ∑ i = 1 m Y i \overline{Y}=\frac{1}{m}\sum_{i=1}^mY_i Y = m 1 ∑ i = 1 m Y i S 1 2 = 1 n − 1 ∑ i = 1 n ( X i − X ‾ ) 2 S_1^2=\frac{1}{n-1}\sum_{i=1}^n(X_i-\overline{X})^2 S 1 2 = n − 1 1 ∑ i = 1 n ( X i − X ) 2 S 2 2 = 1 m − 1 ∑ i = 1 m ( Y i − Y ‾ ) 2 S_2^2=\frac{1}{m-1}\sum_{i=1}^m(Y_i-\overline{Y})^2 S 2 2 = m − 1 1 ∑ i = 1 m ( Y i − Y ) 2 S 1 2 S 2 2 ∼ F ( n , m ) \frac{S_1^2}{S_2^2} \sim F(n,m) S 2 2 S 1 2 ∼ F ( n , m ) S 1 2 S 2 2 ∼ F ( n − 1 , m − 1 ) \frac{S_1^2}{S_2^2} \sim F(n-1,m-1) S 2 2 S 1 2 ∼ F ( n − 1 , m − 1 ) 2 S 1 2 S 2 2 ∼ F ( n , m ) \frac{2S_1^2}{S_2^2} \sim F(n,m) S 2 2 2 S 1 2 ∼ F ( n , m ) 2 S 1 2 S 2 2 ∼ F ( n − 1 , m − 1 ) \frac{2S_1^2}{S_2^2} \sim F(n-1,m-1) S 2 2 2 S 1 2 ∼ F ( n − 1 , m − 1 )
设X 1 , X 2 X_1,X_2 X 1 , X 2 N ( μ , σ 2 ) N(\mu,\sigma^2) N ( μ , σ 2 ) σ ( σ > 0 ) \sigma(\sigma>0) σ ( σ > 0 ) σ ^ = a ∣ X 1 − X 2 ∣ \hat{\sigma}=a|X_1-X_2| σ ^ = a ∣ X 1 − X 2 ∣ σ \sigma σ a = a= a = π 2 \frac{\sqrt{\pi}}{2} 2 π 2 π 2 \frac{\sqrt{2\pi}}{2} 2 2 π π \sqrt{\pi} π 2 π \sqrt{2\pi} 2 π
当x → 0 x \to 0 x → 0 f ( x ) = a x + b x 2 + ln ( 1 + x ) f(x)=ax+bx^2+\ln(1+x) f ( x ) = a x + b x 2 + ln ( 1 + x ) g ( x ) = e x 2 − cos x g(x)=e^{x^2}-\cos x g ( x ) = e x 2 − cos x a b = ab= ab =
曲面z = x + 2 y + ln ( 1 + x 2 + y 2 ) z=x+2y+\ln(1+x^2+y^2) z = x + 2 y + ln ( 1 + x 2 + y 2 ) ( 0 , 0 , 0 ) (0,0,0) ( 0 , 0 , 0 )
设f ( x ) f(x) f ( x ) f ( x ) = 1 − x f(x)=1-x f ( x ) = 1 − x x ∈ [ 0 , 1 ] x \in [0,1] x ∈ [ 0 , 1 ] f ( x ) = a 0 2 + ∑ n = 1 ∞ a n cos n π x f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos n\pi x f ( x ) = 2 a 0 + ∑ n = 1 ∞ a n cos nπ x ∑ n = 1 ∞ a 2 n = \sum_{n=1}^{\infty}a_{2n}= ∑ n = 1 ∞ a 2 n =
设连续函数f ( x ) f(x) f ( x ) f ( x + 2 ) − f ( x ) = x f(x+2)-f(x)=x f ( x + 2 ) − f ( x ) = x ∫ 0 2 f ( x ) d x = 0 \int_{0}^{2}f(x)dx=0 ∫ 0 2 f ( x ) d x = 0 ∫ 1 3 f ( x ) d x = \int_{1}^{3}f(x)dx= ∫ 1 3 f ( x ) d x =
已知向量α 1 = ( 1 0 1 1 ) \alpha_1=\begin{pmatrix}1\\0\\1\\1\end{pmatrix} α 1 = 1 0 1 1 α 2 = ( − 1 − 1 0 1 ) \alpha_2=\begin{pmatrix}-1\\-1\\0\\1\end{pmatrix} α 2 = − 1 − 1 0 1 α 3 = ( 0 1 − 1 1 ) \alpha_3=\begin{pmatrix}0\\1\\-1\\1\end{pmatrix} α 3 = 0 1 − 1 1 β = ( 1 1 1 − 1 ) \beta=\begin{pmatrix}1\\1\\1\\-1\end{pmatrix} β = 1 1 1 − 1 γ = k 1 α 1 + k 2 α 2 + k 3 α 3 \gamma=k_1\alpha_1+k_2\alpha_2+k_3\alpha_3 γ = k 1 α 1 + k 2 α 2 + k 3 α 3 γ T α i = β T α i ( i = 1 , 2 , 3 ) \gamma^T\alpha_i=\beta^T\alpha_i(i=1,2,3) γ T α i = β T α i ( i = 1 , 2 , 3 ) k 1 2 + k 2 2 + k 3 2 = k_1^2+k_2^2+k_3^2= k 1 2 + k 2 2 + k 3 2 =
设随机变量X X X Y Y Y X ∼ B ( 1 , 1 3 ) X \sim B(1,\frac{1}{3}) X ∼ B ( 1 , 3 1 ) Y ∼ B ( 2 , 1 2 ) Y \sim B(2,\frac{1}{2}) Y ∼ B ( 2 , 2 1 ) P { X = Y } = P\{X=Y\}= P { X = Y } =
设函数f ( x ) f(x) f ( x ) [ − a , a ] [-a,a] [ − a , a ] f ( 0 ) = 0 f(0)=0 f ( 0 ) = 0 ξ ∈ ( − a , a ) \xi \in (-a,a) ξ ∈ ( − a , a ) ∫ − a a f ( x ) d x = f ′ ′ ( ξ ) a 3 3 \int_{-a}^{a}f(x)dx=f''(\xi)\frac{a^3}{3} ∫ − a a f ( x ) d x = f ′′ ( ξ ) 3 a 3 f ( x ) f(x) f ( x ) ( − a , a ) (-a,a) ( − a , a ) η ∈ ( − a , a ) \eta \in (-a,a) η ∈ ( − a , a ) ∣ f ′ ′ ( η ) ∣ ≥ 1 2 a 2 ∣ f ( a ) − f ( − a ) ∣ |f''(\eta)| \geq \frac{1}{2a^2}|f(a)-f(-a)| ∣ f ′′ ( η ) ∣ ≥ 2 a 2 1 ∣ f ( a ) − f ( − a ) ∣
已知二次型g ( y 1 , y 2 , y 3 ) = y 1 2 + y 2 2 + y 3 2 + 2 y 2 y 3 g(y_1,y_2,y_3)=y_1^2+y_2^2+y_3^2+2y_2y_3 g ( y 1 , y 2 , y 3 ) = y 1 2 + y 2 2 + y 3 2 + 2 y 2 y 3 x = P y x=Py x = P y f ( x 1 , x 2 , x 3 ) f(x_1,x_2,x_3) f ( x 1 , x 2 , x 3 ) g ( y 1 , y 2 , y 3 ) g(y_1,y_2,y_3) g ( y 1 , y 2 , y 3 ) x = Q y x=Qy x = Q y f ( x 1 , x 2 , x 3 ) f(x_1,x_2,x_3) f ( x 1 , x 2 , x 3 ) g ( y 1 , y 2 , y 3 ) g(y_1,y_2,y_3) g ( y 1 , y 2 , y 3 )
设二维随机变量( X , Y ) (X,Y) ( X , Y ) f ( x , y ) = { 2 π ( x 2 + y 2 ) , x 2 + y 2 ≤ 1 0 , 其他 f(x,y)=\begin{cases}\frac{2}{\pi}(x^2+y^2), & x^2+y^2 \leq 1 \\ 0, & \text{其他}\end{cases} f ( x , y ) = { π 2 ( x 2 + y 2 ) , 0 , x 2 + y 2 ≤ 1 其他 X X X Y Y Y X X X Y Y Y Z = X 2 + Y 2 Z=X^2+Y^2 Z = X 2 + Y 2
设lim x → 1 f ( x ) ln x = 1 \lim_{x \to 1} \frac{f(x)}{\ln x}=1 lim x → 1 l n x f ( x ) = 1 f ( 1 ) = 0 f(1)=0 f ( 1 ) = 0 lim x → 1 f ( x ) = 0 \lim_{x \to 1}f(x)=0 lim x → 1 f ( x ) = 0 f ( 1 ) = 0 , f ′ ( 1 ) = 1 f(1)=0,f'(1)=1 f ( 1 ) = 0 , f ′ ( 1 ) = 1 lim x → 1 f ′ ( x ) = 1 \lim_{x \to 1}f'(x)=1 lim x → 1 f ′ ( x ) = 1
设f ( u ) f(u) f ( u ) z = x y f ( y x ) z=xyf\left( \frac{y}{x} \right) z = x y f ( x y ) x ∂ z ∂ x + y ∂ z ∂ y = y 2 ( ln y − ln x ) x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=y^2(\ln y-\ln x) x ∂ x ∂ z + y ∂ y ∂ z = y 2 ( ln y − ln x ) f ( 1 ) = 1 2 , f ′ ( 1 ) = 0 f(1)=\frac{1}{2},f'(1)=0 f ( 1 ) = 2 1 , f ′ ( 1 ) = 0 f ( 1 ) = 0 , f ′ ( 1 ) = 1 2 f(1)=0,f'(1)=\frac{1}{2} f ( 1 ) = 0 , f ′ ( 1 ) = 2 1 f ( 1 ) = 1 2 , f ′ ( 1 ) = 1 f(1)=\frac{1}{2},f'(1)=1 f ( 1 ) = 2 1 , f ′ ( 1 ) = 1 f ( 1 ) = 0 , f ′ ( 1 ) = 1 f(1)=0,f'(1)=1 f ( 1 ) = 0 , f ′ ( 1 ) = 1
设数列{ x n } \{x_n\} { x n } − π 2 ≤ x n ≤ π 2 -\frac{\pi}{2} \leq x_n \leq \frac{\pi}{2} − 2 π ≤ x n ≤ 2 π lim n → ∞ cos ( sin x n ) \lim_{n \to \infty}\cos(\sin x_n) lim n → ∞ cos ( sin x n ) lim n → ∞ x n \lim_{n \to \infty}x_n lim n → ∞ x n lim n → ∞ sin ( cos x n ) \lim_{n \to \infty}\sin(\cos x_n) lim n → ∞ sin ( cos x n ) lim n → ∞ x n \lim_{n \to \infty}x_n lim n → ∞ x n lim n → ∞ cos ( sin x n ) \lim_{n \to \infty}\cos(\sin x_n) lim n → ∞ cos ( sin x n ) lim n → ∞ sin x n \lim_{n \to \infty}\sin x_n lim n → ∞ sin x n lim n → ∞ x n \lim_{n \to \infty}x_n lim n → ∞ x n lim n → ∞ sin ( cos x n ) \lim_{n \to \infty}\sin(\cos x_n) lim n → ∞ sin ( cos x n ) lim n → ∞ cos x n \lim_{n \to \infty}\cos x_n lim n → ∞ cos x n lim n → ∞ x n \lim_{n \to \infty}x_n lim n → ∞ x n
若I 1 = ∫ 0 1 x 2 ( 1 + cos x ) d x I_1=\int_{0}^{1}\frac{x}{2(1+\cos x)}dx I 1 = ∫ 0 1 2 ( 1 + c o s x ) x d x I 2 = ∫ 0 1 ln ( 1 + x ) 1 + cos x d x I_2=\int_{0}^{1}\frac{\ln(1+x)}{1+\cos x}dx I 2 = ∫ 0 1 1 + c o s x l n ( 1 + x ) d x I 3 = ∫ 0 1 2 x 1 + sin x d x I_3=\int_{0}^{1}\frac{2x}{1+\sin x}dx I 3 = ∫ 0 1 1 + s i n x 2 x d x I 1 < I 2 < I 3 I_1<I_2<I_3 I 1 < I 2 < I 3 I 2 < I 1 < I 3 I_2<I_1<I_3 I 2 < I 1 < I 3 I 1 < I 3 < I 2 I_1<I_3<I_2 I 1 < I 3 < I 2 I 3 < I 2 < I 1 I_3<I_2<I_1 I 3 < I 2 < I 1
下列4个条件中，3阶矩阵A A A A A A A A A A A A A A A
设A , B A,B A , B n n n E E E n n n A x = 0 Ax=0 A x = 0 B x = 0 Bx=0 B x = 0 ( A O E B ) y = 0 \begin{pmatrix}A&O\\E&B\end{pmatrix}y=0 ( A E O B ) y = 0 ( E A O A B ) y = 0 \begin{pmatrix}E&A\\O&AB\end{pmatrix}y=0 ( E O A A B ) y = 0 ( A B O B ) y = 0 \begin{pmatrix}A&B\\O&B\end{pmatrix}y=0 ( A O B B ) y = 0 ( B A O A ) y = 0 \begin{pmatrix}B&A\\O&A\end{pmatrix}y=0 ( B O A A ) y = 0 ( A B B O A ) y = 0 \begin{pmatrix}AB&B\\O&A\end{pmatrix}y=0 ( A B O B A ) y = 0 ( B A A O B ) y = 0 \begin{pmatrix}BA&A\\O&B\end{pmatrix}y=0 ( B A O A B ) y = 0
设α 1 = ( λ , 1 , 1 ) T \alpha_1=(\lambda,1,1)^T α 1 = ( λ , 1 , 1 ) T α 2 = ( 1 , λ , 1 ) T \alpha_2=(1,\lambda,1)^T α 2 = ( 1 , λ , 1 ) T α 3 = ( 1 , 1 , λ ) T \alpha_3=(1,1,\lambda)^T α 3 = ( 1 , 1 , λ ) T α 4 = ( 1 , λ , λ 2 ) T \alpha_4=(1,\lambda,\lambda^2)^T α 4 = ( 1 , λ , λ 2 ) T { α 1 , α 2 , α 3 } \{\alpha_1,\alpha_2,\alpha_3\} { α 1 , α 2 , α 3 } { α 1 , α 2 , α 4 } \{\alpha_1,\alpha_2,\alpha_4\} { α 1 , α 2 , α 4 } λ \lambda λ { 0 , 1 } \{0,1\} { 0 , 1 } { λ ∣ λ ∈ R , λ ≠ − 2 } \{\lambda \mid \lambda \in R,\lambda \neq -2\} { λ ∣ λ ∈ R , λ = − 2 } { λ ∣ λ ∈ R , λ ≠ − 1 , λ ≠ − 2 } \{\lambda \mid \lambda \in R,\lambda \neq -1,\lambda \neq -2\} { λ ∣ λ ∈ R , λ = − 1 , λ = − 2 } { λ ∣ λ ∈ R , λ ≠ − 1 } \{\lambda \mid \lambda \in R,\lambda \neq -1\} { λ ∣ λ ∈ R , λ = − 1 }
设随机变量X X X ( 0 , 3 ) (0,3) ( 0 , 3 ) Y Y Y X X X Y Y Y − 1 -1 − 1 D ( 2 X − Y + 1 ) = D(2X-Y+1)= D ( 2 X − Y + 1 ) =
设随机变量X 1 , X 2 , … , X n X_1,X_2,\dots,X_n X 1 , X 2 , … , X n X 1 X_1 X 1 E ( X 1 k ) = μ k ( k = 1 , 2 , 3 , 4 ) E(X_1^k)=\mu_k(k=1,2,3,4) E ( X 1 k ) = μ k ( k = 1 , 2 , 3 , 4 ) ε > 0 \varepsilon>0 ε > 0 P { ∣ 1 n ∑ i = 1 n X i 2 − μ 2 ∣ ≥ ε } ≤ P\left\{ \left| \frac{1}{n}\sum_{i=1}^nX_i^2 - \mu_2 \right| \geq \varepsilon \right\} \leq P { n 1 ∑ i = 1 n X i 2 − μ 2 ≥ ε } ≤ μ 4 − μ 2 2 n ε 2 \frac{\mu_4-\mu_2^2}{n\varepsilon^2} n ε 2 μ 4 − μ 2 2 μ 4 − μ 2 2 n ε 2 \frac{\mu_4-\mu_2^2}{\sqrt{n}\varepsilon^2} n ε 2 μ 4 − μ 2 2 μ 2 − μ 1 2 n ε 2 \frac{\mu_2-\mu_1^2}{n\varepsilon^2} n ε 2 μ 2 − μ 1 2 μ 2 − μ 1 2 n ε 2 \frac{\mu_2-\mu_1^2}{\sqrt{n}\varepsilon^2} n ε 2 μ 2 − μ 1 2
设随机变量X ∼ N ( 0 , 1 ) X \sim N(0,1) X ∼ N ( 0 , 1 ) X = x X=x X = x Y ∼ N ( x , 1 ) Y \sim N(x,1) Y ∼ N ( x , 1 ) X X X Y Y Y 1 4 \frac{1}{4} 4 1 1 2 \frac{1}{2} 2 1 3 3 \frac{\sqrt{3}}{3} 3 3 2 2 \frac{\sqrt{2}}{2} 2 2
设函数y ( x ) y(x) y ( x ) y ′ + 1 2 x y = 2 + x y'+\frac{1}{2\sqrt{x}}y=2+\sqrt{x} y ′ + 2 x 1 y = 2 + x y ( 1 ) = 3 y(1)=3 y ( 1 ) = 3 y = y ( x ) y=y(x) y = y ( x )
已知平面区域D = { ( x , y ) ∣ y − 2 ≤ x ≤ 4 − y 2 , 0 ≤ y ≤ 2 } D=\{(x,y) \mid y-2 \leq x \leq \sqrt{4-y^2},0 \leq y \leq 2\} D = {( x , y ) ∣ y − 2 ≤ x ≤ 4 − y 2 , 0 ≤ y ≤ 2 } I = ∬ D ( x − y ) 2 x 2 + y 2 d x d y I=\iint_{D} \frac{(x-y)^2}{x^2+y^2} dxdy I = ∬ D x 2 + y 2 ( x − y ) 2 d x d y
已知曲线L L L Σ : 4 x 2 + y 2 + z 2 = 1 \Sigma:4x^2+y^2+z^2=1 Σ : 4 x 2 + y 2 + z 2 = 1 x ≥ 0 , y ≥ 0 , z ≥ 0 x \geq 0,y \geq 0,z \geq 0 x ≥ 0 , y ≥ 0 , z ≥ 0 Σ \Sigma Σ L L L Σ \Sigma Σ I = ∮ L ( y z 2 − cos z ) d x + 2 x z 2 d y + ( 2 x y z + x sin z ) d z I=\oint_{L} (yz^2-\cos z)dx + 2xz^2dy + (2xyz+x\sin z)dz I = ∮ L ( y z 2 − cos z ) d x + 2 x z 2 d y + ( 2 x yz + x sin z ) d z
设函数f ( x ) f(x) f ( x ) ( − ∞ , + ∞ ) (-\infty,+\infty) ( − ∞ , + ∞ ) f ′ ′ ( x ) ≥ 0 f''(x) \geq 0 f ′′ ( x ) ≥ 0 a , b a,b a , b f ( a + b 2 ) ≤ 1 b − a ∫ a b f ( x ) d x f\left( \frac{a+b}{2} \right) \leq \frac{1}{b-a}\int_{a}^{b}f(x)dx f ( 2 a + b ) ≤ b − a 1 ∫ a b f ( x ) d x
设二次型f ( x 1 , x 2 , x 3 ) = ∑ i = 1 3 ∑ j = 1 3 i j x i x j f(x_1,x_2,x_3)=\sum_{i=1}^3\sum_{j=1}^3 ij x_i x_j f ( x 1 , x 2 , x 3 ) = ∑ i = 1 3 ∑ j = 1 3 ij x i x j f ( x 1 , x 2 , x 3 ) f(x_1,x_2,x_3) f ( x 1 , x 2 , x 3 ) x = Q y x=Qy x = Q y f ( x 1 , x 2 , x 3 ) f(x_1,x_2,x_3) f ( x 1 , x 2 , x 3 ) f ( x 1 , x 2 , x 3 ) = 0 f(x_1,x_2,x_3)=0 f ( x 1 , x 2 , x 3 ) = 0
设X 1 , X 2 , … , X n X_1,X_2,\dots,X_n X 1 , X 2 , … , X n θ \theta θ Y 1 , Y 2 , … , Y m Y_1,Y_2,\dots,Y_m Y 1 , Y 2 , … , Y m 2 θ 2\theta 2 θ θ ( θ > 0 ) \theta(\theta>0) θ ( θ > 0 ) θ \theta θ θ ^ \hat{\theta} θ ^ D ( θ ^ ) D(\hat{\theta}) D ( θ ^ )
f ( x ) = { e x − 1 , x ≤ 0 ln ( 1 + 2 x ) sin x , x > 0 f(x)=\begin{cases}e^x-1, & x \leq 0 \\ \frac{\ln(1+2x)}{\sin x}, & x>0\end{cases} f ( x ) = { e x − 1 , s i n x l n ( 1 + 2 x ) , x ≤ 0 x > 0 x = 0 x=0 x = 0 f ( x ) f(x) f ( x )
设函数f ( x , y ) f(x,y) f ( x , y ) f ( x + 1 , e x ) = x ( x + 1 ) 2 f(x+1,e^x)=x(x+1)^2 f ( x + 1 , e x ) = x ( x + 1 ) 2 f ( x , x 2 ) = 2 x 2 ln x f(x,x^2)=2x^2\ln x f ( x , x 2 ) = 2 x 2 ln x d f ( 1 , 1 ) = df(1,1)= df ( 1 , 1 ) = 5 2 d x + 1 2 d y \frac{5}{2}dx+\frac{1}{2}dy 2 5 d x + 2 1 d y − 5 2 d x + 1 2 d y -\frac{5}{2}dx+\frac{1}{2}dy − 2 5 d x + 2 1 d y 5 2 d x − 1 2 d y \frac{5}{2}dx-\frac{1}{2}dy 2 5 d x − 2 1 d y − 5 2 d x − 1 2 d y -\frac{5}{2}dx-\frac{1}{2}dy − 2 5 d x − 2 1 d y
设函数f ( x ) = sin x 1 + x 2 f(x)=\frac{\sin x}{1+x^2} f ( x ) = 1 + x 2 s i n x x = 0 x=0 x = 0 a x + b x 2 + c x 3 ax+bx^2+cx^3 a x + b x 2 + c x 3 a = 1 , b = 0 , c = − 7 6 a=1,b=0,c=-\frac{7}{6} a = 1 , b = 0 , c = − 6 7 a = 1 , b = 0 , c = 7 6 a=1,b=0,c=\frac{7}{6} a = 1 , b = 0 , c = 6 7 a = − 1 , b = − 1 , c = − 7 6 a=-1,b=-1,c=-\frac{7}{6} a = − 1 , b = − 1 , c = − 6 7 a = − 1 , b = − 1 , c = 7 6 a=-1,b=-1,c=\frac{7}{6} a = − 1 , b = − 1 , c = 6 7
设函数f ( x ) f(x) f ( x ) [ 0 , 1 ] [0,1] [ 0 , 1 ] ∫ 0 1 f ( x ) d x = \int_{0}^{1}f(x)dx= ∫ 0 1 f ( x ) d x = lim n → ∞ ∑ k = 1 n f ( 2 k − 1 2 n ) 1 2 n \lim_{n \to \infty}\sum_{k=1}^n f\left( \frac{2k-1}{2n} \right)\frac{1}{2n} lim n → ∞ ∑ k = 1 n f ( 2 n 2 k − 1 ) 2 n 1 lim n → ∞ ∑ k = 1 n f ( 2 k − 1 2 n ) 1 n \lim_{n \to \infty}\sum_{k=1}^n f\left( \frac{2k-1}{2n} \right)\frac{1}{n} lim n → ∞ ∑ k = 1 n f ( 2 n 2 k − 1 ) n 1 lim n → ∞ ∑ k = 1 2 n f ( k − 1 2 n ) 1 n \lim_{n \to \infty}\sum_{k=1}^{2n} f\left( \frac{k-1}{2n} \right)\frac{1}{n} lim n → ∞ ∑ k = 1 2 n f ( 2 n k − 1 ) n 1 lim n → ∞ ∑ k = 1 2 n f ( k 2 n ) 2 n \lim_{n \to \infty}\sum_{k=1}^{2n} f\left( \frac{k}{2n} \right)\frac{2}{n} lim n → ∞ ∑ k = 1 2 n f ( 2 n k ) n 2
设二次型f ( x 1 , x 2 , x 3 ) = ( x 1 + x 2 ) 2 + ( x 2 + x 3 ) 2 − ( x 3 − x 1 ) 2 f(x_1,x_2,x_3)=(x_1+x_2)^2+(x_2+x_3)^2-(x_3-x_1)^2 f ( x 1 , x 2 , x 3 ) = ( x 1 + x 2 ) 2 + ( x 2 + x 3 ) 2 − ( x 3 − x 1 ) 2
设α 1 = ( 1 0 1 ) \alpha_1=\begin{pmatrix}1\\0\\1\end{pmatrix} α 1 = 1 0 1 α 2 = ( 1 2 1 ) \alpha_2=\begin{pmatrix}1\\2\\1\end{pmatrix} α 2 = 1 2 1 α 3 = ( 3 1 2 ) \alpha_3=\begin{pmatrix}3\\1\\2\end{pmatrix} α 3 = 3 1 2 l 1 , l 2 l_1,l_2 l 1 , l 2 β 1 , β 2 , β 3 \beta_1,\beta_2,\beta_3 β 1 , β 2 , β 3 l 1 , l 2 l_1,l_2 l 1 , l 2
设A , B A,B A , B n n n r ( A O O A T A ) = 2 r ( A ) r\begin{pmatrix}A&O\\O&A^TA\end{pmatrix}=2r(A) r ( A O O A T A ) = 2 r ( A ) r ( A A B O A T ) = 2 r ( A ) r\begin{pmatrix}A&AB\\O&A^T\end{pmatrix}=2r(A) r ( A O A B A T ) = 2 r ( A ) r ( A B A O A A T ) = 2 r ( A ) r\begin{pmatrix}A&BA\\O&AA^T\end{pmatrix}=2r(A) r ( A O B A A A T ) = 2 r ( A ) r ( A O B A A T ) = 2 r ( A ) r\begin{pmatrix}A&O\\BA&A^T\end{pmatrix}=2r(A) r ( A B A O A T ) = 2 r ( A )
设A , B A,B A , B 0<P(B)<1 ，下列命题中为假命题的是（）
(A) 若P ( A ∣ B ) = P ( A ) P(A|B)=P(A) P ( A ∣ B ) = P ( A ) P ( A ∣ B ‾ ) = P ( A ) P(A|\overline{B})=P(A) P ( A ∣ B ) = P ( A ) P ( A ∣ B ) > P ( A ) P(A|B)>P(A) P ( A ∣ B ) > P ( A ) P ( A ‾ ∣ B ‾ ) > P ( A ‾ ) P(\overline{A}|\overline{B})>P(\overline{A}) P ( A ∣ B ) > P ( A ) P ( A ∣ B ) > P ( A ∣ B ‾ ) P(A|B)>P(A|\overline{B}) P ( A ∣ B ) > P ( A ∣ B ) P ( A ∣ B ) > P ( A ) P(A|B)>P(A) P ( A ∣ B ) > P ( A ) P ( A ∣ A ∪ B ) > P ( A ‾ ∣ A ∪ B ) P(A|A \cup B)>P(\overline{A}|A \cup B) P ( A ∣ A ∪ B ) > P ( A ∣ A ∪ B ) P ( A ) > P ( B ) P(A)>P(B) P ( A ) > P ( B )
设( X 1 , Y 1 ) , ( X 2 , Y 2 ) , … , ( X n , Y n ) (X_1,Y_1),(X_2,Y_2),\dots,(X_n,Y_n) ( X 1 , Y 1 ) , ( X 2 , Y 2 ) , … , ( X n , Y n ) N ( μ 1 , μ 2 ; σ 1 2 , σ 2 2 ; ρ ) N(\mu_1,\mu_2;\sigma_1^2,\sigma_2^2;\rho) N ( μ 1 , μ 2 ; σ 1 2 , σ 2 2 ; ρ ) θ = μ 1 − μ 2 \theta=\mu_1-\mu_2 θ = μ 1 − μ 2 X ‾ = 1 n ∑ i = 1 n X i \overline{X}=\frac{1}{n}\sum_{i=1}^nX_i X = n 1 ∑ i = 1 n X i Y ‾ = 1 n ∑ i = 1 n Y i \overline{Y}=\frac{1}{n}\sum_{i=1}^nY_i Y = n 1 ∑ i = 1 n Y i θ ^ = X ‾ − Y ‾ \hat{\theta}=\overline{X}-\overline{Y} θ ^ = X − Y θ ^ \hat{\theta} θ ^ θ \theta θ D ( θ ^ ) = σ 1 2 + σ 2 2 n D(\hat{\theta})=\frac{\sigma_1^2+\sigma_2^2}{n} D ( θ ^ ) = n σ 1 2 + σ 2 2 θ ^ \hat{\theta} θ ^ θ \theta θ D ( θ ^ ) = σ 1 2 + σ 2 2 n D(\hat{\theta})=\frac{\sigma_1^2+\sigma_2^2}{n} D ( θ ^ ) = n σ 1 2 + σ 2 2 θ ^ \hat{\theta} θ ^ θ \theta θ D ( θ ^ ) = σ 1 2 + σ 2 2 − 2 ρ σ 1 σ 2 n D(\hat{\theta})=\frac{\sigma_1^2+\sigma_2^2-2\rho\sigma_1\sigma_2}{n} D ( θ ^ ) = n σ 1 2 + σ 2 2 − 2 ρ σ 1 σ 2 θ ^ \hat{\theta} θ ^ θ \theta θ D ( θ ^ ) = σ 1 2 + σ 2 2 − 2 ρ σ 1 σ 2 n D(\hat{\theta})=\frac{\sigma_1^2+\sigma_2^2-2\rho\sigma_1\sigma_2}{n} D ( θ ^ ) = n σ 1 2 + σ 2 2 − 2 ρ σ 1 σ 2
设X 1 , X 2 , … , X 16 X_1,X_2,\dots,X_{16} X 1 , X 2 , … , X 16 N ( μ , 4 ) N(\mu,4) N ( μ , 4 ) H 0 : μ ≤ 10 H_0:\mu \leq 10 H 0 : μ ≤ 10 H 1 : μ > 10 H_1:\mu>10 H 1 : μ > 10 Φ ( x ) \Phi(x) Φ ( x ) W = { X ‾ ≥ 11 } W=\{\overline{X} \geq 11\} W = { X ≥ 11 } X ‾ = 1 16 ∑ i = 1 16 X i \overline{X}=\frac{1}{16}\sum_{i=1}^{16}X_i X = 16 1 ∑ i = 1 16 X i μ = 11.5 \mu=11.5 μ = 11.5 1 − Φ ( 0.5 ) 1-\Phi(0.5) 1 − Φ ( 0.5 ) 1 − Φ ( 1 ) 1-\Phi(1) 1 − Φ ( 1 ) 1 − Φ ( 1.5 ) 1-\Phi(1.5) 1 − Φ ( 1.5 ) 1 − Φ ( 2 ) 1-\Phi(2) 1 − Φ ( 2 )
