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【简答题】
已知函数 ． （1）若曲线 的一条切线的斜率是2，求切点坐标； （2）求 在点 处的切线方程．
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反馈参考答案：
举一反三
【单选题】氢原子轨道的能级高低是( )
A.
E ls 2s 2p 3s 3p 3d 4s
B.
E ls 2s 2p 3s 3p < E 4s < E 3d
C.
.E ls 2s = E 2p < E 3s=E 3p < E 4s < E 3d
D.
E ls 2s = E 2p < E 3s=E 3p=E 3d < E 4s
【简答题】What makes them take different positions on the question of creativity? A: This is mainly due to the difference in their way of thinking.
【单选题】氢原子轨道的能级高低是( )
A.
E 1s 2s2p3s3p3d4s
B.
E 1s 2s2p3s3p< E 4s < E 3d
C.
E 1s 2s= E 2p < E 3s=E 3p < E 4s < E 3d
D.
E 1s 2s= E 2p < E 3s=E 3p=E 3d < E 4s
【单选题】氢原子轨道的能级高低是
A.
E( 1 s )< E (2 s )=E (2 p )< E (3 s )=E (3 p )=E (3 d )< E (4 s )
B.
E (1 s )< E (2 s )< E (2 p )< E (3 s )< E (3 p )< E (3 d )< E (4 s )
C.
E (1 s )< E (2 s )< E (2 p )< E (3 s )< E (3 p )< E (4 s )< E (3 d )
D.
E (1 s )< E (2 s )=E (2 p )< E (3 s )=E (3 p )< E (4 s )< E (3 d )
【简答题】What makes them take different positions on the questions of creativity?
【单选题】氢原子轨道的能级高低是:
A.
E ls < E 2s < E 2p < E 3s < E 3p < E 3d < E 4s
B.
E ls < E 2s < E 2p < E 3s < E 3p E 4s E 3d
C.
E ls< E 2s=E 2pE3s=E 3pE4s E3d
D.
E ls< E 2s=E 2pE3s=E 3p=E 3d E4s
【单选题】氢原子轨道的能级高低是
A.
E (1 s )< E (2 s )< E (2 p )< E (3 s )< E (3 p )< E (3 d )< E (4 s )
B.
E (1 s )< E (2 s )< E (2 p )< E (3 s )< E (3 p )< E (4 s )< E (3 d )
C.
E (1 s )< E (2 s )=E (2 p )< E (3 s )=E (3 p )< E (4 s )< E (3 d )
D.
E( 1 s )< E (2 s )=E (2 p )< E (3 s )=E (3 p )=E (3 d )< E (4 s )
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