题目 #17461

【讲解】  二次函数y = − x +1的图象与轴有交点,\(x \frac{ax^{2}}{1} \frac{4}{0}\)\(\Delta =\)(4)^{2}− − 4  且a  0,解得a  4且a \(0. \frac{a}{4}\)故答案为:a  且a  0. 抛物线开口向下,a  0,  对称轴为x =.即−\(\frac{b}{2} =\).\(\frac{1}{bc} \frac{a}{a} \frac{1}{b}\)b  0,c    0,2 + = 0; 0. 故①错误,②正确; 设抛物线与轴的交点为(x,0),(\(x_{2}\),0),\(\frac{x_{1}}{x_{2}}\)且(\(x_{2}\),0)在正半轴上, 0   3, x =,\(\frac{对称轴是直线}{x_{2}} \frac{1}{x_{2}}x_{1}\)+  = 1,解得 = 2 −\(x_{1}\),\(\frac{2}{2}\) 0  −\(x_{1}\) 3, −1  0,\(\frac{x_{1}}{1}\) = − 时,a − b + c  0,\(\frac{x}{2}\) + = 0,\(\frac{a}{a} \frac{b}{c}\)3 +  0, 故③错误,  a + b = 0,\(\frac{2}{b}\) = −2a,  + − 4ac = 4a − a − 4ac\(\frac{4}{2}\frac{a}{a} \frac{b}{4} \frac{2}{c} =\)− ac = 2a(1− 2),  c  1, −1 2c  0,   0,\(\frac{a}{4}\) a + b − 4ac  0,即4a + b  4ac, 故④正确;根据题意, n =\(am^{2}\)+ mb + c  c,  − 2  0,\(fracam^2\)amm fracam2) ( −  0,  a  0, m(m −  0,frac2)0解得m  或m  2, B 故⑤错误;综上可知:②④正确,故选: