设f(x)可导,F(x)=f(x)(1+|sinx|),则f(0)=0是F(x)在x=0处可导的( )A.充分必要条件B.
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解答:解;∵f(x)可导∴f′(0)存在,f(x)在x=0连续
又∵F′(0)=
lim
x→0
F(x)?F(0)
x
=
lim
x→0
f(x)(1+|sinx|)?f(0)
x
∴F′?(0)=
lim
x→0?
f(x)(1?sinx)?f(0)
x
=
lim
x→0?
f(x)?f(0)
x
?
lim
x→0?
f(x)
sinx
x
=f′(0)-f(0)
F′+(0)=
lim
x→0+
f(x)(1+sinx)?f(0)
x
=
lim
x→0+
f(x)?f(0)
x
+
lim
x→0+
f(x)
sinx
x
=f′(0)+f(0)
∴F′(0)??F′-(0)=F′+(0)?f′(0)-f(0)=f′(0)+f(0)?f(0)=0
故选:A.
又∵F′(0)=
lim
x→0
F(x)?F(0)
x
=
lim
x→0
f(x)(1+|sinx|)?f(0)
x
∴F′?(0)=
lim
x→0?
f(x)(1?sinx)?f(0)
x
=
lim
x→0?
f(x)?f(0)
x
?
lim
x→0?
f(x)
sinx
x
=f′(0)-f(0)
F′+(0)=
lim
x→0+
f(x)(1+sinx)?f(0)
x
=
lim
x→0+
f(x)?f(0)
x
+
lim
x→0+
f(x)
sinx
x
=f′(0)+f(0)
∴F′(0)??F′-(0)=F′+(0)?f′(0)-f(0)=f′(0)+f(0)?f(0)=0
故选:A.
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