解：f（x）=e^x/a+a/ e^x在R上為偶函式 f（x）= f（-x） e^x/a+a/ e^x= e^-x/a+a/ e^-x e^x/a+a/ e^x=1/a*e^x +a*e^x (1/a-a) *e^x=(1/a-a) *e^-x 當1/a-a不等於0，則不成立上式 故1/a-a=0 a^2=1 a=1或-1 證明: 當a=1時f（x）= e^x+1/e^x f（x+1）= e^[x+1]+1/e^[x+1] f（x+1）-f（x）= e^[x+1]+1/e^[x+1]- e^x-1/e^x =(e-1) e^x+[(1-e)/e] e^[-x]= [(e-1) e^[2x+1]+(1-e) ]/e^[x+1] =(e-1) [e^[2x+1]-1]/ e^[x+1] 因x>0,所以e-1>0 ,e^[x+1]>0, [e^[2x+1]-1]>0 所以f（x+1）-f（x）>0 得f（x+1）> f（x）所以得證 當為-1時依舊成立同理可得
解：f（x）=e^x/a+a/ e^x在R上為偶函式 f（x）= f（-x） e^x/a+a/ e^x= e^-x/a+a/ e^-x e^x/a+a/ e^x=1/a*e^x +a*e^x (1/a-a) *e^x=(1/a-a) *e^-x 當1/a-a不等於0，則不成立上式 故1/a-a=0 a^2=1 a=1或-1 證明: 當a=1時f（x）= e^x+1/e^x f（x+1）= e^[x+1]+1/e^[x+1] f（x+1）-f（x）= e^[x+1]+1/e^[x+1]- e^x-1/e^x =(e-1) e^x+[(1-e)/e] e^[-x]= [(e-1) e^[2x+1]+(1-e) ]/e^[x+1] =(e-1) [e^[2x+1]-1]/ e^[x+1] 因x>0,所以e-1>0 ,e^[x+1]>0, [e^[2x+1]-1]>0 所以f（x+1）-f（x）>0 得f（x+1）> f（x）所以得證 當為-1時依舊成立同理可得