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設f(x)=x-asinx-b,下麵即證f(x)至少存在一個不超過a+b的正零點,顯然f(x)連續 f(0)=-b=0 若f(a+b)=0,則原命題成立; 若f(a+b)>0,則f(x)在[0,a+b]的兩個端點函數值异號,且f(x)連續,由零點定理存在x0屬於(0,a+b)使得f(x0)=0,證畢.
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