a与b满足关系：b-2a＜0．（4分）下面给出证明：任取-2＜x1＜x2．∵f（x）=ax+bx+2=a+b−2ax+2，∴f（x1）-f（x2）=（a+b−2ax1+2）-（a+b−2ax2+2）=（b-2a）（1x1+2−1x2+2）=（b-2a）x2−x1(x1+2)(x2+2)．（8分）...