依題可得
c=-2,∴原函數(shù)y=ax²+bx-2,
∵其有與x軸兩交點(diǎn),∴Δ＞0解得b²+8a＞0,
①當(dāng)a＞0時(shí),則∵在0＜x＜1中有零點(diǎn)且拋物線(xiàn)開(kāi)口朝上,
∵當(dāng)x=0時(shí),y＜0,可推得當(dāng)x2＜x＜x1時(shí),y＜0
∴當(dāng)x=1時(shí),y＞0,解得a+b-2＞0即a+b＞2
而且易得知拋物線(xiàn)對(duì)稱(chēng)軸x=-b/2a＜1又∵a＞0,∴有2a+b＞0,又∵a＞0,∴2a+b＞a+b＞2,
依照上面推理得3,4錯(cuò)誤,又∵a＞0,∴3a+b＞a﹢b＞2,∴1,2正確；
②當(dāng)a＜0時(shí),則∵在0＜x＜1中有零點(diǎn)且拋物線(xiàn)開(kāi)口朝下,則
a,x1＜x2時(shí),則有當(dāng)x=1時(shí),y＞0即a+b＞2.
b,x1＞x2時(shí),則有當(dāng)x=1時(shí),y＜0即a+b＜2且0＜-b/2a＜1則有b＞0
綜上所述,3,4錯(cuò)誤,1,2正確,選B