)
A: $\frac{\pi }{3},\,\frac{\pi }{4},\,\frac{2\pi }{3}$
B: $-\frac{\pi }{3}\,,\frac{\pi }{4}\,,\frac{\pi }{3}$
C: $\frac{\pi }{6},\,\pi ,\,\frac{\pi }{6}$
D: $\frac{2\pi }{3},\,\frac{\pi }{3},\,\frac{\pi }{3}$
举一反三
- $\arctan (-\sqrt{3})=$ A: $-\frac{\pi}{3}$ B: $\frac{\pi}{6}$ C: $\frac{\pi}{3}$ D: $-\frac{\pi}{6}$
- Solve $\int_{-\frac{1}{2}}^1{1-x^2}dx=$? A: $\frac{\pi}{3}+\frac{\sqrt{3}}{8}$. B: $\frac{\pi}{2}$. C: $\frac{\pi}{6}+\frac{\sqrt{3}}{4}$. D: $\frac{\pi}{4}$.
- 函数$f(x)=\sin x + \cos x,x \in [0,2 \pi]$的上凸区间为 A: $[0,\frac{\pi}{4}] \cup [\frac{5}{4} \pi,2 \pi] $ B: $[\frac{\pi}{4},\frac{5}{4} \pi]$ C: $[0,\frac{3}{4}\pi] \cup [\frac{7}{4} \pi,2 \pi] $ D: $[\frac{3}{4} \pi,\frac{7}{4} \pi] $
- 这时线圈平面法线方向与该处磁感强度的方向的夹<br/>角为____________________. A: `\frac{1}{3}\pi` B: `\frac{1}{6}\pi` C: `\frac{1}{2}\pi` D: `\frac{2}{3}\pi`
- 函数$f(x) =sin^3 x, x \in [0,2 \pi]$的单调递减区间为 A: $[\frac{\pi}{2},\frac{3}{2} \pi]$ B: $[\frac{3}{2} \pi,2 \pi]$ C: $[0,\frac{\pi}{2}]$ D: $[0,2 \pi]$
内容
- 0
函数\(f(x) = x^2,\; x \in [-\pi,\pi]\)的Fourier级数为 A: \(\frac{\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \sin nx ,\; x \in [-\pi,\pi]\) B: \(\frac{\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos nx ,\; x \in [-\pi,\pi]\) C: \(\frac{2\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \sin nx ,\; x \in [-\pi,\pi]\) D: \(\frac{2\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos nx ,\; x \in [-\pi,\pi]\)
- 1
积分$\int_0^1 x \arctan xdx=$()。 A: $\frac{\pi}{4}+\frac{1}{2}$ B: $\frac{\pi}{4}$ C: $\frac{\pi}{4}-\frac{1}{2}$ D: $\frac{1}{2}$
- 2
半径为$R$, 密度为$1$的均匀平面薄板关于其切线的转动惯量为 A: $\frac{3\pi R^4}{4}$ B: $\frac{5\pi R^4}{4}$ C: $\frac{5\pi R^3}{4}$ D: $\frac{4\pi R^3}{3}$
- 3
\(已知曲面\Sigma:x^2+y^2+z^2=a^2被平面z=h(0 A: \[2\pi a \ln\frac{a}{h}\] B: \[3\pi a \ln\frac{a}{h}\] C: \[4\pi a \ln\frac{a}{h}\] D: \[\pi a \ln\frac{a}{h}\]
- 4
由曲线 \(y= { { x}^{2}},x= { { y}^{2}}\)所围成的图形绕 \(y\)轴旋转所得旋转体的体积为=( )。 A: \(\frac{3}{5}\pi \) B: \(\frac{3}{8}\pi \) C: \(\frac{3}{10}\pi \) D: \(\frac{3}{20}\pi \)