A: \( \left\{ {\matrix{ { { x^2} + { { \left( {y - {1 \over 2}} \right)}^2} = {1 \over 4}} \cr {z = 0} \cr } } \right. \)
B: \( \left\{ {\matrix{ { { z^2} = y} \cr {x = 0} \cr } } \right. \)
C: \( \left\{ {\matrix{ {z = {y^2}} \cr {x = 0} \cr } } \right. \)
D: \( \left\{ {\matrix{ { { y^2} + { { \left( {x - {1 \over 2}} \right)}^2} = {1 \over 4}} \cr {z = 0} \cr } } \right. \)
举一反三
- 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: $$\left\{ \matrix{ x = {{\sqrt 3 } \over 3}\cos t\cr y = {{\sqrt 3 } \over 3}\cos t \cr z = {{\sqrt 3 } \over 3}\sin t\cr} \right.(0 \le t \le 2\pi )$$
- 向量组\(\left( {\matrix{ { - 1} \cr 3 \cr 1 \cr } } \right),\left( {\matrix{ 2 \cr 1 \cr 0 \cr } } \right),\left( {\matrix{ 1 \cr 4 \cr 1 \cr } } \right) \)线性相关.
- 设矩阵\(A = \left( {\matrix{ \matrix{ x \cr 0 \cr y \cr} & \matrix{ 0 \cr 2 \cr 0 \cr} & \matrix{ y \cr 0 \cr - 2 \cr} \cr } } \right)\)的一个特征值为\(-3\),且\(A\)的三个特征值之积为\(-12\),则\(x =\)______
- 下列函数中,( )是初等函数. A: \(y = \arcsin ({x^2} + 2)\) B: \(f(x) = \left\{ \matrix{ 0,x \notin Q \ \cr 1,x \in Q \ \cr} \right.\) C: \(y = \sqrt { - {x^2} + 1} \) D: \(f(x) = \left\{ \matrix{ {x^2},0 \le x < 1 \ \cr x + 1,x > 1 \ \cr} \right.\)
- 在其定义区间上连续的函数是( )。 A: \(f(x) = \left\{ {\matrix{ {x\quad ,{\rm{0}} \le x \le {\rm{1}}} \cr {1 - x\quad ,1 < x \le 2} \cr } } \right.\) B: \(f(x) = \left\{ {\matrix{ {x\quad ,0 < x \le 1 } \cr {2 - x\quad ,1 < x \le 2} \cr } } \right.\) C: \(f(x) = \left\{ {\matrix{ {x\;\quad ,0 \le x < 1} \cr {0\;\quad \quad ,x = 1} \cr {2 - x\quad ,1 < x \le 2} \cr } } \right.\) D: \(f(x) = \left\{ {\matrix{ { { 1 \over {x - 1}}\quad ,0 \le x \le 1} \cr {0\quad ,1 \le x \le 2} \cr } } \right.\)
内容
- 0
下列函数中,在其定义域内处处连续的是( )。 A: \(f(x) = \left\{ {\matrix{ { { {1 - {x^2}} \over {1 + x}}\quad ,x \ne 1} \cr {0\quad \quad ,x = 1} \cr } } \right.\) B: \(f(x) = \left\{ {\matrix{ {\ln x\quad ,x > 0} \cr { { x^2}\quad ,x \le 0} \cr } } \right.\) C: \(f(x) = \left\{ {\matrix{ { { {\sqrt {x + 1} - 1} \over {\sqrt x }}\quad ,x > 0} \cr {1\quad ,x\le 0} \cr } } \right.\) D: \(f(x) = \left\{ {\matrix{ { { x^2} + 2x\quad ,x \le 0} \cr { { e^x}\quad ,x > 0} \cr } } \right.\)
- 1
已知直线的一般方程\( \left\{ {\matrix{ {x - 2y - z + 4 = 0} \cr {5x + y - 2z + 8 = 0} \cr } } \right. \), 则其点向式方程为( ) A: \( { { x - 2} \over 2} = {y \over { - 3}} = { { z - 4} \over {11}} \) B: \( {x \over 5} = {y \over { - 3}} = { { z - 4} \over {11}} \) C: \( { { x - 2} \over 5} = { { y + 1} \over { - 3}} = { { z - 4} \over {11}} \) D: \( { { x - 2} \over 2} = { { y + 1} \over { - 3}} = { { z - 4} \over {11}} \)
- 2
设3阶实对称矩阵\( A \)的秩为2,且\( {A^2} - A = O \) ,则\( A \)相似于( ) A: \( \left( {\matrix{ 1 & {} & {} \cr {} & { - 1} & {} \cr {} & {} & 0 \cr } } \right) \) B: \( \left( {\matrix{ 1 & {} & {} \cr {} & 1 & {} \cr {} & {} & 0 \cr } } \right) \) C: \( \left( {\matrix{ { - 1} & {} & {} \cr {} & { - 1} & {} \cr {} & {} & 0 \cr } } \right) \) D: \( \left( {\matrix{ 1 & 1 & {} \cr {} & 1 & {} \cr {} & {} & 0 \cr } } \right) \)
- 3
\(a = 2\)时,函数\(f(x) = \left\{ {\matrix{ {a\quad \quad ,x = 0} \cr {2 { { \sin x} \over x},x \ne 0} \cr } } \right.\)在\(x = 0\)处连续。( )
- 4
在平面\( xoz \)内一动点,它与原点的距离等于它与点\( (5, - 3,1) \)的距离,此动点的轨迹方程为( ) A: \( \left\{ {\matrix{ {10x - 2z - 35 = 0} \cr {y = 9} \cr } } \right. \) B: \( \left\{ {\matrix{ {10x + 2z - 35 = 0} \cr {y = 9} \cr } } \right. \) C: \( \left\{ {\matrix{ {10x - 2z + 35 = 0} \cr {y = 9} \cr } } \right. \) D: \( \left\{ {\matrix{ {10x + 2z + 35 = 0} \cr {y = 9} \cr } } \right. \)