问题
解答题
已知函数f(x)＝x2＋xsin x＋cos x.
(1)若曲线y＝f(x)在点(a，f(a))处与直线y＝b相切，求a与b的值；
(2)若曲线y＝f(x)与直线y＝b有两个不同交点，求b的取值范围．
答案
(1) a＝0，b＝1.(2) b>1
(1)由f(x)＝x2＋xsin x＋cos x，
得f′(x)＝x(2＋cos x)，
∵y＝f(x)在点(a，f(a))处与直线y＝b相切．
∴f′(a)＝a(2＋cos a)＝0且b＝f(a)，
则a＝0，b＝f(0)＝1.
(2)令f′(x)＝0，得x＝0.
∴当x>0时，f′(x)>0，f(x)在(0，＋∞)递增．
当x<0时，f′(x)<0，f(x)在(－∞，0)上递减．
∴f(x)的最小值为f(0)＝1.
由于函数f(x)在区间(－∞，0)和(0，＋∞)上均单调，
所以当b>1时曲线y＝f(x)与直线y＝b有且仅有两个不同交点．
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