设f(x)在区间(-∞,+∞)上连续,且满足f(x)=∫0xf(x-t)sintdt+x.则在(-∞,+∞)上,当x≠0时,f(x) ( )

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问题 设f(x)在区间(-∞,+∞)上连续,且满足f(x)=∫0xf(x-t)sintdt+x.则在(-∞,+∞)上,当x≠0时,f(x)    (    )

选项 A、恒为正.
B、恒为负.
C、与x同号.
D、与x异号.

答案C

解析 令x-t=u,作积分变量代换,得
f(x)=∫x0sin(x-u)d(-u)+x=∫0xf(u) sin(x-u)d(-u)+x
=sinx∫0xf(u)cosudu-cosx∫0xf(u)sinudu+x,
fˊ(x)=cosx∫0xf(u)cosudu+sinx·cosx·f(x)+sinx∫0xf(u)sinudu-cosx·sinx·f(x)+1
= cosx∫0xf(u)cosudu+sinx∫0xf(u)sinudu+1,
f″(x)=-sinx∫0xf(u)cosudu+cos2·f(x)+ cosx∫0xf(u)sinudu+sin2·f(x)
=f(x)-f(x)+x,
所以.又因f(0)=0,fˊ(0)=1,所以C1=1,所以C2=0.从而[img][/img]
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