函数f(x)=x+2cosx在区间[ 0，π/2 ]上的最大值为()。
A: pi/6+sqrt(3)
B: pi/3+sqrt(3)
C: pi/6+sqrt(2)
D: pi/3+sqrt(2)
A: pi/6+sqrt(3)
B: pi/3+sqrt(3)
C: pi/6+sqrt(2)
D: pi/3+sqrt(2)
举一反三
- 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: $\frac{\pi}{4}$.
- \(\int_{-\sqrt{2}}^{\sqrt{2}}{\sqrt{8-2 { { y}^{2}}}dy}\)=( )。 A: \(\sqrt{2}(\pi -2)\) B: \(\sqrt{2}(\pi +2)\) C: \(2\sqrt{2}(\pi +2)\) D: \(2\sqrt{2}(\pi -2)\)
- 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: $$\left\{ \matrix{ x = {{\sqrt 3 } \over 3}\cos t\cr y = {{\sqrt 3 } \over 3}\cos t \cr z = {{\sqrt 3 } \over 3}\sin t\cr} \right.(0 \le t \le 2\pi )$$
- 求函数$y = \root 3 \of {x + \sqrt x } $的导数$y' = $( ) A: ${{1 + 2\sqrt x } \over {\root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$ B: $ {{1 + 2\sqrt x } \over {6\root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$ C: $ {{1 + 2\sqrt x } \over {6\sqrt x \cdot \root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$ D: $ {{1 + 2\sqrt x } \over {\sqrt x \cdot \root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$
- $\arctan (-\sqrt{3})=$ A: $-\frac{\pi}{3}$ B: $\frac{\pi}{6}$ C: $\frac{\pi}{3}$ D: $-\frac{\pi}{6}$