设f(x)在区间(一∞,+∞)上连续,且满足f(x)=∫0xf(x—t)sin tdt+x,则在(一∞,+∞)上,当x≠0时,f(x) ( )

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问题 设f(x)在区间(一∞,+∞)上连续,且满足f(x)=∫0xf(x—t)sin tdt+x,则在(一∞,+∞)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 作积分变量代换,令x—t=u,得
  f(x)=∫0xf(u)sin(x—u)d(—u)+x=∫0xf(u)sin(x—u)du+x
    =sin x.∫0xf(u)cos udu—cos x.∫0xf(u)sin udu+x,
f’(x)=cos x.∫0xf(u)cos udu+sin x.cos x.f(x)+sin x.∫0xf(u)sin udu—cos x.sin x.f(x)+1
    =cos x.∫0xf(u)cos udu+sin x.∫0xf(u)sin udu+1,
  f"(x)=一sin x.∫0xf(u)cos udu+cos2x.f(x)+cos x.∫0xf(u)sin udu+sin2x.f(x)
    =f(x)一f(x)+x=x.
  所以f(x)=+C1x+C2.又因f(0)=0,f(0)=1,所以C1=1,C2=0.
  从而f(x)=+1),故应选C.
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