已知x+y=2sin(θ+π4)，x-y=2sin(θ-π4)，则x2+y2=．（ ）
A: 1
B: 2
C: 2
D: 4
A: 1
B: 2
C: 2
D: 4
举一反三
- y=arcsin(4x+1)的反函数为 A: y=(sinx-1)/4, x∈R B: y=sin[(x-1)/4], x∈R C: y=sin[(x-1)/4], x∈[-π/2,π/2] D: y=(sinx-1)/4, x∈[-π/2,π/2]
- 曲线积分$$\int_{(0,0}^{(x,y)}(2x\cos y-y^2\sin x)dx+(2y\cos x-x^2\sin y)dy=$$ A: $y^2\cos x+x^2\cos y$ B: $x^2\cos x+y^2\cos y$ C: $x^2\sin y+y^2\sin x$ D: $x^2\sin x+y^2\sin y$
- 求微分方程[img=143x21]17da5f14490e50e.png[/img]的通解,实验命令为(). A: dsolve(D2y-2*Dy+5*y=sin(2*x),x)ans =exp(x)*sin(2*x)*C2+exp(x)*cos(2*x)*C1+1/17*sin(2*x)+4/17*cos(2*x) B: dsolve('D2y-2*Dy+5*y=sin(2*x)','x')ans =cos(2*x)*(sin(4*x)/17 - cos(4*x)/68 + 1/4) - sin(2*x)*(cos(4*x)/17 + sin(4*x)/68) + C1*cos(2*x)*exp(x) - C2*sin(2*x)*exp(x) C: dsolve(D2y-2*Dy+5*y=sin(2*x),'x','y')ans =exp(x)*sin(2*x)*C2+exp(x)*cos(2*x)*C1+1/17*sin(2*x)+4/17*cos(2*x)
- 已知 \( y = \sin x + \ln 2 \)，则 \( y' = \cos x + {1 \over 2} \)（ ）.
- 下列函数在点$(0,0)$的重极限存在的是 A: $f(x,y)=\frac{y^2}{x^2+y^2}$ B: $f(x,y)=(x+y)\sin\frac{1}{x}\sin\frac{1}{y}$ C: $f(x,y)=\frac{x^2y^2}{x^2y^2+(x-y)^2}$ D: $f(x,y)=\frac{x^2y^2}{x^3+y^3}$