A: 0
B: 2
C: 3
D: 4
举一反三
- 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}} }}$
- 函数\( f\left( x \right) = 2x + 3\root 3 \of { { x^2}} \)的极大值为 ______ ______
- 下列方程中，不是全微分方程的为（ ）。 A: \(\left( {3{x^2} + 6x{y^2}} \right)dx + \left( {6{x^2}y + 4{y^2}} \right)dy = 0\) B: \({e^y}dx + \left( {x \cdot {e^y} - 2y} \right)dy = 0\) C: \(y\left( {x - 2y} \right)dx - {x^2}dy = 0\) D: \(\left( { { x^2} - y} \right)dx - xdy = 0\)
- 函数\( y = \left( {2x - 5} \right)\root 3 \of { { x^2}} \)的驻点为x= _______ ______
- 函数\(y = {\left( { - 2x + 1} \right)^4}\)的导数为（ ）. A: \( - 8{\left( { - 2x + 1} \right)^3}\) B: \(8{\left( { - 2x + 1} \right)^3}\) C: \(4{\left( { - 2x + 1} \right)^3}\) D: \(- 4{\left( { - 2x + 1} \right)^3}\)
内容
- 0
以点\( (2, - 1,2) \) 为球心，3为半径的球面方程为（ ） A: \( {\left( {x + 2} \right)^2} + {(y - 1)^2} + {(z + 2)^2} = 9 \) B: \( {\left( {x + 2} \right)^2} + {(y - 1)^2} + {(z + 2)^2} = 3 \) C: \( {\left( {x - 2} \right)^2} + {(y + 1)^2} + {(z - 2)^2} = 9 \) D: \( {\left( {x - 2} \right)^2} + {(y + 1)^2} + {(z - 2)^2} = 3 \)
- 1
\(\left\{ {\left( {x,y} \right)\left| {2 \le {x^2} + {y^2} \le 4} \right.} \right\}\)是闭区域.
- 2
函数$z=\arcsin\dfrac{1}{~\sqrt{x+y}~}$的定义域为( ) A: $\left\{(x,y)\left|~x+y\geq<br/>0\right.\right\}$; B: $\left\{(x,y)\left|~x+y\geq<br/>1~\text{或}~x+y\leq<br/>-1 \right.\right\}$; C: $\left\{(x,y)\left|~x+y\geq<br/>1\right.\right\}$; D: $\left\{(x,y)\left|~x+y\geq<br/>\dfrac{4}{~\pi^2~}\right.\right\}$．
- 3
由\( y = {x^2} - 1,\;y = 0 \)围成的平面图形面积可表示为( )。 A: \( \int_{ - 1}^1 {\left( { - {x^2} + 1} \right)} dx \) B: \( \int_{ - 1}^1 {\left( { { x^2} - 1} \right)} dx \) C: \( \int_0^1 {\left( { - {x^2} + 1} \right)} dx \) D: \( \int_0^1 {\left( { { x^2} - 1} \right)} dx \)
- 4
\( \sin x \)的麦克劳林公式为（ ）. A: \( \sin x = x - { { {x^3}} \over {3!}} + { { {x^5}} \over {5!}} - \cdots + {( - 1)^n} { { {x^{2n + 1}}} \over {\left( {2n + 1} \right)!}} + o\left( { { x^{2n + 2}}} \right) \) B: \( \sin x = 1 - { { {x^2}} \over {2!}} + { { {x^4}} \over {4!}} - { { {x^6}} \over {6!}} + \cdots + {( - 1)^n} { { {x^{2n}}} \over {\left( {2n} \right)!}} + o\left( { { x^{2n + 1}}} \right) \) C: \( \sin x = 1 + x + { { {x^2}} \over 2} + \cdots + { { {x^n}} \over {n!}} + o\left( { { x^n}} \right) \)