下列信号中，（）信号的频谱是连续的。 A: $x(t) = A\sin (\omega t + {\varphi _1}) + B\sin (3\omega t + {\varphi _2})$ B: $x(t) = 5\sin 30t + 3\sin \sqrt {50} t$ C: $x(t) = {e^{ - at}}\sin {\omega _0}t$

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2022-06-07 问题

下列信号中，（）信号的频谱是连续的。 A: $x(t) = A\sin (\omega t + {\varphi _1}) + B\sin (3\omega t + {\varphi _2})$ B: $x(t) = 5\sin 30t + 3\sin \sqrt {50} t$ C: $x(t) = {e^{ - at}}\sin {\omega _0}t$

2022-06-12 问题

8. 下列不等式正确的是 A: $0\lt \int_{0}^{\frac{\pi }{2}}{\sin (\sin x)dx}\lt \int_{0}^{\frac{\pi }{2}}{\cos (\sin x)dx}$ B: $0\lt \int_{0}^{\frac{\pi }{2}}{\cos (\sin x)dx}\lt \int_{0}^{\frac{\pi }{2}}{\sin (\sin x)dx}$ C: $\int_{0}^{\frac{\pi }{2}}{\sin (\sin x)dx}\lt 0\lt \int_{0}^{\frac{\pi }{2}}{\cos (\sin x)dx}$ D: $\int_{0}^{\frac{\pi }{2}}{\cos (\sin x)dx}\lt 0\lt \int_{0}^{\frac{\pi }{2}}{\sin (\sin x)dx}$

8. 下列不等式正确的是 A: $0\lt \int_{0}^{\frac{\pi }{2}}{\sin (\sin x)dx}\lt \int_{0}^{\frac{\pi }{2}}{\cos (\sin x)dx}$ B: $0\lt \int_{0}^{\frac{\pi }{2}}{\cos (\sin x)dx}\lt \int_{0}^{\frac{\pi }{2}}{\sin (\sin x)dx}$ C: $\int_{0}^{\frac{\pi }{2}}{\sin (\sin x)dx}\lt 0\lt \int_{0}^{\frac{\pi }{2}}{\cos (\sin x)dx}$ D: $\int_{0}^{\frac{\pi }{2}}{\cos (\sin x)dx}\lt 0\lt \int_{0}^{\frac{\pi }{2}}{\sin (\sin x)dx}$

2022-06-19 问题

求方程的根的程序( )。 A: A.solve(sin(x)-2x+0.5=0,x); B: B.solve(sin(x)-2x+0.5=0,'x'); C: C.solve('sin(x)-2x+0.5=0','x'); D: D.solve('sin(x)-2x+0.5=0',x);

求方程的根的程序( )。 A: A.solve(sin(x)-2*x+0.5=0,x); B: B.solve(sin(x)-2*x+0.5=0,'x'); C: C.solve('sin(x)-2*x+0.5=0','x'); D: D.solve('sin(x)-2*x+0.5=0',x);

2022-06-09 问题

设$z = {e^{x - 2y}}$，而$x = \sin t,\;y = {t^3},$则$ { { dz} \over {dt}} = $( ) A: ${e^{\sin t - 2{t^3}}}$ B: ${e^{\sin t - 2{t^3}}}\left( {\cos t - 6{t^2}} \right)$ C: ${e^{\sin t - 2{t^3}}}\ {\sin t } $ D: ${e^{\sin t - 2{t^3}}}\,{t^3}$

设$z = {e^{x - 2y}}$，而$x = \sin t,\;y = {t^3},$则$ { { dz} \over {dt}} = $( ) A: ${e^{\sin t - 2{t^3}}}$ B: ${e^{\sin t - 2{t^3}}}\left( {\cos t - 6{t^2}} \right)$ C: ${e^{\sin t - 2{t^3}}}\ {\sin t } $ D: ${e^{\sin t - 2{t^3}}}\,{t^3}$

2022-06-07 问题

$ \lim \limits_{x \to 0} { { \sqrt {1 + x\sin x} - \cos x} \over { { {\sin }^2}{x \over 2}}} = $______ 。

2022-06-04 问题

设$z = {e^{x - 2y}}$，而$x = \sin t$，$y = {t^3}$，则全导数$ { { dz} \over {dt}} = $ A: ${e^{\sin t - {t^3}}}(\cos t - 6{t^2})$ B: ${e^{\sin t - 2{t^3}}}(\sin t - 6{t^2})$ C: ${e^{\cos t - 2{t^3}}}(\cos t - 6{t^2})$ D: ${e^{\sin t - 2{t^3}}}(\cos t - 6{t^2})$

设$z = {e^{x - 2y}}$，而$x = \sin t$，$y = {t^3}$，则全导数$ { { dz} \over {dt}} = $ A: ${e^{\sin t - {t^3}}}(\cos t - 6{t^2})$ B: ${e^{\sin t - 2{t^3}}}(\sin t - 6{t^2})$ C: ${e^{\cos t - 2{t^3}}}(\cos t - 6{t^2})$ D: ${e^{\sin t - 2{t^3}}}(\cos t - 6{t^2})$

2022-06-27 问题

曲线$x={{\sin }^{2}}t, y=\sin t\cos t, z={{\cos }^{2}}t$在$t=\frac{\text{ }\!\!\pi\!\!\text{ }}{2}$所对应的点处的切向向量为 A: $(0,-1,1)$ B: $(1,-1,0)$ C: $(0,1,1)$ D: $(0,-1,0)$

2022-07-01 问题

$\lim \limits_{x \to 0} 2 { { \tan x - \sin x} \over { { {\sin }^3}x}}{\rm{ = }}$______ 。

2022-06-14 问题

曲线$\left\{ \matrix{ {x^2} + {y^2} + {z^2} = 9 \cr y = x \cr} \right.$的参数方程为( ). A: $$\left\{ \matrix{ x = \sqrt 3 \cos t \cr y = \sqrt 3 \cos t \cr z = \sqrt 3 \sin t \cr} \right.(0 \le t \le 2\pi )$$ B: $$\left\{ \matrix{ x = {3 \over {\sqrt 2 }}\cos t\cr y = {3 \over {\sqrt 2 }}\cos t \cr z = 3\sin t \cr} \right.(0 \le t \le 2\pi )$$ C: $$\left\{ \matrix{ x = \cos t\cr y = \cos t\cr z = \sin t \cr} \right.(0 \le t \le 2\pi )$$ D: $$\left\{ \matrix{ x = {{\sqrt 3 } \over 3}\cos t\cr y = {{\sqrt 3 } \over 3}\cos t \cr z = {{\sqrt 3 } \over 3}\sin t\cr} \right.(0 \le t \le 2\pi )$$

曲线$\left\{ \matrix{ {x^2} + {y^2} + {z^2} = 9 \cr y = x \cr} \right.$的参数方程为( ). A: $$\left\{ \matrix{ x = \sqrt 3 \cos t \cr y = \sqrt 3 \cos t \cr z = \sqrt 3 \sin t \cr} \right.(0 \le t \le 2\pi )$$ B: $$\left\{ \matrix{ x = {3 \over {\sqrt 2 }}\cos t\cr y = {3 \over {\sqrt 2 }}\cos t \cr z = 3\sin t \cr} \right.(0 \le t \le 2\pi )$$ C: $$\left\{ \matrix{ x = \cos t\cr y = \cos t\cr z = \sin t \cr} \right.(0 \le t \le 2\pi )$$ D: $$\left\{ \matrix{ x = {{\sqrt 3 } \over 3}\cos t\cr y = {{\sqrt 3 } \over 3}\cos t \cr z = {{\sqrt 3 } \over 3}\sin t\cr} \right.(0 \le t \le 2\pi )$$

2022-06-07 问题

$\mathop {\lim }\limits_{x \to 0} 2 { { \tan x - \sin x} \over { { {\sin }^3}x}}{\rm{ = }}$。

$\mathop {\lim }\limits_{x \to 0} 2 { { \tan x - \sin x} \over { { {\sin }^3}x}}{\rm{ = }}$______。______

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