1由于 f(1+)= lim =1x-i+ xf(1-) = lim (ax2 + bx) = a + bx-1f(-1t) = lim,(ax2 + bx) = a- bX-1f(-1-) = lim,1x=-i+ x且f(αx)连续，即 f(1+) =f(1-) =f(1),f(-1+) = f(-1-) = f(-1)则有 α+b= 1,α一b=-1. 解得α=0,b=110𝑓 1 + = lim 𝑥→1+ 1 𝑥 = 1 𝑓 1 − = lim 𝑥→1+ (𝑎𝑥 2 + 𝑏𝑥) = 𝑎 + 𝑏 𝑓 −1 + = lim 𝑥→−1+ (𝑎𝑥 2 + 𝑏𝑥) = 𝑎 − 𝑏 𝑓 −1 − = lim 𝑥→−1+ 1 𝑥 = −1 𝑎 + 𝑏 = 1, 𝑎 − 𝑏 = −1. 𝑎 = 0, 𝑏 = 1. 由于 且𝑓(𝑥)连续，即 则有 解得 𝑓 1 + = 𝑓 1 − = 𝑓(1), 𝑓 −1 + = 𝑓 −1 − = 𝑓(−1) 10