设a为实数，函数f（x）=x3+ax2+(a-2)x的才，则曲线y=f（x）在原点处的切线方程为
A．y=-3x
B．y=-2x
C．y=3x
D．y=2x
答案：B
f'（x）="3" x2+2ax+（a-2） 因为f'（x）是偶函数，所以f'（x）="-" f'（x）所以
3 x2+2ax+（a-2）="3" （-x2）-2ax+（a-2） 得到a=0。
由于 f（x）的切线斜率k="3" x2+2ax+（a-2）代入（0，0）则切线的斜率k=3*0+0-2=-2。设y=ax+b，代入k=-2，点（0；0）所以y=-2x
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