2.12 f(x) 即为分段函数：f(x) = x^2 x > 1；f(x) = (1/2)(1+a+b) x = 1 ;f(x) = ax+b x < 1 .函数要连续，则 (1/2)(1+a+b) = 1， (1/2)(1+a+b) = a+b得 a+b = 1
