已知函數(shù)f(x)=ax2-2ax+2+b(a≠0),在區(qū)間[2,3]上有最大值5,最小值2.
(1)求a,b的值;
(2)若b<1,g(x)=f(x)-(2m)•x在[2,4]上單調(diào),求m的取值范圍.
分析:(1)函數(shù)對(duì)稱軸為x=2,當(dāng)a>0時(shí),函數(shù)開(kāi)口向上,在區(qū)間[2,3]單增,則可知在2處去最小值,在處去最大值,分類討論即可求出a,b的值;
(2)若b<1,則根據(jù)(1)中求得值,即可確定a,b的值,從而求出函數(shù)g(x)解析式,根據(jù)函數(shù)的單調(diào)性,可求出m的取值范圍.
解答:解(1)f(x)=a(x-1)
2+2+b-a
①當(dāng)a>0時(shí),f(x)在[2,3]上為增函數(shù)
故
? ?②當(dāng)a<0時(shí),f(x)在[2,3]上為減函數(shù)
故
??(2)∵b<1
∴a=1b=0即f(x)=x
2-2x+2g(x)=x
2-2x+2-(2
m)x=x
2-(2+2
m)x+2
≤2或
≥4,
∴2
m≤2或2
m≥6,即m≤1或m≥log
26 點(diǎn)評(píng):此題主要考查函數(shù)的單調(diào)性及最值的計(jì)算.