• 2021-04-14 问题

    【单选题】对任意实数x 1 , y 1 , x 2 , y 2 , x 1 < x 2 , y 1 < y 2 , 分布函数P{x 1 <X≤x 2 , y 1 <Y≤y 2 }=? A. F(x 2 , y 2 )+ F(x 1 , y 1 )+ F(x 1 , y 2 )+ F(x 2 , y 1 ) B. F(x 2 , y 2 )- F(x 1 , y 1 )+ F(x 1 , y 2 )- F(x 2 , y 1 ) C. F(x 2 , y 2 )+ F(x 1 , y 1 )- F(x 1 , y 2 )- F(x 2 , y 1 ) D. F(x 2 , y 2 )- F(x 1 , y 1 )- F(x 1 , y 2 )+ F(x 2 , y 1 )

    【单选题】对任意实数x 1 , y 1 , x 2 , y 2 , x 1 < x 2 , y 1 < y 2 , 分布函数P{x 1 <X≤x 2 , y 1 <Y≤y 2 }=? A. F(x 2 , y 2 )+ F(x 1 , y 1 )+ F(x 1 , y 2 )+ F(x 2 , y 1 ) B. F(x 2 , y 2 )- F(x 1 , y 1 )+ F(x 1 , y 2 )- F(x 2 , y 1 ) C. F(x 2 , y 2 )+ F(x 1 , y 1 )- F(x 1 , y 2 )- F(x 2 , y 1 ) D. F(x 2 , y 2 )- F(x 1 , y 1 )- F(x 1 , y 2 )+ F(x 2 , y 1 )

  • 2022-07-23 问题

    函数$f(x,y)=\sin x\cdot \ln (1+y)$在点$(0,0)$处带有Peano型余项的3阶Taylor公式为$f(x,y)=$ A: $xy+\frac{1}{2}x{{y}^{2}}+o({{(\sqrt{{{x}^{2}}+{{y}^{2}}})}^{3}})$ B: $xy-\frac{1}{2}x{{y}^{2}}+o({{(\sqrt{{{x}^{2}}+{{y}^{2}}})}^{3}})$ C: $xy-x{{y}^{2}}+o({{(\sqrt{{{x}^{2}}+{{y}^{2}}})}^{3}})$ D: $xy+x{{y}^{2}}+o({{(\sqrt{{{x}^{2}}+{{y}^{2}}})}^{3}})$

    函数$f(x,y)=\sin x\cdot \ln (1+y)$在点$(0,0)$处带有Peano型余项的3阶Taylor公式为$f(x,y)=$ A: $xy+\frac{1}{2}x{{y}^{2}}+o({{(\sqrt{{{x}^{2}}+{{y}^{2}}})}^{3}})$ B: $xy-\frac{1}{2}x{{y}^{2}}+o({{(\sqrt{{{x}^{2}}+{{y}^{2}}})}^{3}})$ C: $xy-x{{y}^{2}}+o({{(\sqrt{{{x}^{2}}+{{y}^{2}}})}^{3}})$ D: $xy+x{{y}^{2}}+o({{(\sqrt{{{x}^{2}}+{{y}^{2}}})}^{3}})$

  • 2022-07-26 问题

    17e0b849d3a4a3b.jpg,计算[img=19x34]17e0ab14a855463.jpg[/img]的实验命令为( ). A: syms x; f=diff((1+sin(x)^2)/cos(x),1)f=2*sin(x) + (sin(x)*(sin(x)^2 + 1))/cos(x)^2 B: f=diff((1+sinx^2)/cosx,1)f=1/2/x^(1/2)/(1-x)^(1/2) C: syms x;f=diff((1+sinx^2)/cosx,1)f=2*sin(x) + (sin(x)*(sin(x)^2 + 1))/cos(x)^2

    17e0b849d3a4a3b.jpg,计算[img=19x34]17e0ab14a855463.jpg[/img]的实验命令为( ). A: syms x; f=diff((1+sin(x)^2)/cos(x),1)f=2*sin(x) + (sin(x)*(sin(x)^2 + 1))/cos(x)^2 B: f=diff((1+sinx^2)/cosx,1)f=1/2/x^(1/2)/(1-x)^(1/2) C: syms x;f=diff((1+sinx^2)/cosx,1)f=2*sin(x) + (sin(x)*(sin(x)^2 + 1))/cos(x)^2

  • 2022-07-26 问题

    17e0b849b7d64bd.jpg,计算[img=19x34]17e0ab14a855463.jpg[/img]实验命令为(). A: syms x;f=diff(asinsqrt(x))f=1/2/x^(1/2)/(1-x)^(1/2) B: f=diff(asin(sqrt(x)))f=1/2/x^(1/2)/(1-x)^(1/2) C: syms x;diff(asin(sqrt(x)))f=1/2/x^(1/2)/(1-x)^(1/2)

    17e0b849b7d64bd.jpg,计算[img=19x34]17e0ab14a855463.jpg[/img]实验命令为(). A: syms x;f=diff(asinsqrt(x))f=1/2/x^(1/2)/(1-x)^(1/2) B: f=diff(asin(sqrt(x)))f=1/2/x^(1/2)/(1-x)^(1/2) C: syms x;diff(asin(sqrt(x)))f=1/2/x^(1/2)/(1-x)^(1/2)

  • 2022-10-27 问题

    17da42840675a6d.jpg,计算[img=19x34]17da4275482315f.jpg[/img]实验命令为(). A: syms x;f=diff(asinsqrt(x))f=1/2/x^(1/2)/(1-x)^(1/2) B: f=diff(asin(sqrt(x)))f=1/2/x^(1/2)/(1-x)^(1/2) C: syms x;diff(asin(sqrt(x)))f=1/2/x^(1/2)/(1-x)^(1/2)

    17da42840675a6d.jpg,计算[img=19x34]17da4275482315f.jpg[/img]实验命令为(). A: syms x;f=diff(asinsqrt(x))f=1/2/x^(1/2)/(1-x)^(1/2) B: f=diff(asin(sqrt(x)))f=1/2/x^(1/2)/(1-x)^(1/2) C: syms x;diff(asin(sqrt(x)))f=1/2/x^(1/2)/(1-x)^(1/2)

  • 2022-07-23 问题

    若\( \int {f(x)dx = {x^2} + C} \),则\( \int {xf(1 - {x^2})dx = } \)( ) A: \( 2{(1 - {x^2})^2} + C \) B: \( - {1 \over 2}{(1 - {x^2})^2} + C \) C: \( {1 \over 2}{(1 - {x^2})^2} + C \) D: \( - 2{(1 - {x^2})^2} + C \)

    若\( \int {f(x)dx = {x^2} + C} \),则\( \int {xf(1 - {x^2})dx = } \)( ) A: \( 2{(1 - {x^2})^2} + C \) B: \( - {1 \over 2}{(1 - {x^2})^2} + C \) C: \( {1 \over 2}{(1 - {x^2})^2} + C \) D: \( - 2{(1 - {x^2})^2} + C \)

  • 2021-04-14 问题

    青书学堂: 二次型 f( x 1 , x 2 , x 3 )=2 x 1 2 +5 x 2 2 +5 x 3 2 +4 x 1 x 2 −8 x 2 x 3 ,则 f的矩阵为 。

    青书学堂: 二次型 f( x 1 , x 2 , x 3 )=2 x 1 2 +5 x 2 2 +5 x 3 2 +4 x 1 x 2 −8 x 2 x 3 ,则 f的矩阵为 。

  • 2022-06-07 问题

    设函数f(x)=a|x|(a>0),且f(2)=4,则(  ) A: f(-1)>f(-2) B: f(1)>f(2) C: f(2)<f(-2) D: f(-3)>f(-2)

    设函数f(x)=a|x|(a>0),且f(2)=4,则(  ) A: f(-1)>f(-2) B: f(1)>f(2) C: f(2)<f(-2) D: f(-3)>f(-2)

  • 2021-04-14 问题

    【单选题】若f(x- 1)= x 2 (x-1),则f(x) =(). A. x(x+ 1) 2 B. x(x- 1) 2 C. x 2 (x+1) D. x 2 (x- 1)

    【单选题】若f(x- 1)= x 2 (x-1),则f(x) =(). A. x(x+ 1) 2 B. x(x- 1) 2 C. x 2 (x+1) D. x 2 (x- 1)

  • 2022-07-24 问题

    满足方程f(x)+2f(x)dx=x2的解f(x)是:() A: -(1/2)e+x+1/2 B: (1/2)e+x-1/2 C: ce+x-1/2 D: ce+x+1/2

    满足方程f(x)+2f(x)dx=x2的解f(x)是:() A: -(1/2)e+x+1/2 B: (1/2)e+x-1/2 C: ce+x-1/2 D: ce+x+1/2

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