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: $\int_{0}^{\frac{\pi }{2}}{\cos (\sin x)dx}\lt 0\lt \int_{0}^{\frac{\pi }{2}}{\sin (\sin x)dx}$
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);
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);
∫(上限л/2下限0)sin^2(x/2)dx
∫(上限л/2下限0)sin^2(x/2)dx
\( \lim \limits_{x \to 0} { { \sqrt {1 + x\sin x} - \cos x} \over { { {\sin }^2}{x \over 2}}} = \)______ 。
\( \lim \limits_{x \to 0} { { \sqrt {1 + x\sin x} - \cos x} \over { { {\sin }^2}{x \over 2}}} = \)______ 。
MATLAB中,A=[0:0.5:2]*pi,那么sin(A)= [0 1 0 -1 0]。
MATLAB中,A=[0:0.5:2]*pi,那么sin(A)= [0 1 0 -1 0]。
\(\lim \limits_{x \to 0} 2 { { \tan x - \sin x} \over { { {\sin }^3}x}}{\rm{ = }}\)______ 。
\(\lim \limits_{x \to 0} 2 { { \tan x - \sin x} \over { { {\sin }^3}x}}{\rm{ = }}\)______ 。
\(\mathop {\lim }\limits_{x \to 0} 2 { { \tan x - \sin x} \over { { {\sin }^3}x}}{\rm{ = }}\)______。______
\(\mathop {\lim }\limits_{x \to 0} 2 { { \tan x - \sin x} \over { { {\sin }^3}x}}{\rm{ = }}\)______。______
sin(π/2)值为() A: 0 B: 1 C: -1
sin(π/2)值为() A: 0 B: 1 C: -1
函数[img=79x27]180355ae2690a03.png[/img]在x=2处的二阶泰勒展开式为 A: exp(sin(2))+cos(2)*exp(sin(2))*(x-2)+exp(sin(2))*(sin(2)/2-cos(2)^2/2)*(x-2)^2 B: exp(sin(2))+cos(2)*exp(sin(2))*(x-2)-exp(sin(2))*(sin(2)/2-cos(2)^2/2)*(x-2)^2 C: exp(sin(2))+cos(2)*exp(sin(2))*(x-2)-exp(sin(2))*(sin(2)/2+cos(2)^2/2)*(x-2)^2 D: exp(sin(2))+cos(2)*exp(sin(2))*(x-2)+exp(sin(2))*(sin(2)/2+cos(2)^2/2)*(x-2)^2
函数[img=79x27]180355ae2690a03.png[/img]在x=2处的二阶泰勒展开式为 A: exp(sin(2))+cos(2)*exp(sin(2))*(x-2)+exp(sin(2))*(sin(2)/2-cos(2)^2/2)*(x-2)^2 B: exp(sin(2))+cos(2)*exp(sin(2))*(x-2)-exp(sin(2))*(sin(2)/2-cos(2)^2/2)*(x-2)^2 C: exp(sin(2))+cos(2)*exp(sin(2))*(x-2)-exp(sin(2))*(sin(2)/2+cos(2)^2/2)*(x-2)^2 D: exp(sin(2))+cos(2)*exp(sin(2))*(x-2)+exp(sin(2))*(sin(2)/2+cos(2)^2/2)*(x-2)^2
求极限\( \lim \limits_{x \to 0} { { {x^2}\sin {1 \over x}} \over {\sin x}}{\rm{ = }}\)______
求极限\( \lim \limits_{x \to 0} { { {x^2}\sin {1 \over x}} \over {\sin x}}{\rm{ = }}\)______