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∵函數y=ax1+x的圖像關於直線y=x對稱∴利用反函數的性質,依題知(1,a2)與(a2,1)皆在原函數圖像上,(1,a2)與(a2,1)為不同的點,即a≠2;∴a×a21+a2 ;=1∴a=-1或a=2(舍去)故可得a= -1
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