• 2022-07-26
    曲线\( \left\{ {\matrix{ { { x^2} + {y^2} = {z^2}} \cr { { z^2} = y} \cr } } \right. \)在坐标面\( yoz \) 上的投影曲线方程为( )
    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: \( \left\{ {\matrix{ { { y^2} + { { \left( {x - {1 \over 2}} \right)}^2} = {1 \over 4}} \cr {z = 0} \cr } } \right. \)
  • B

    举一反三

    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: \( \left\{ {\matrix{ {10x + 2z + 35 = 0} \cr {y = 9} \cr } } \right. \)