已知函数f(x)是二次函数,不等式f(x)≥0的解集为{x|-2≤x≤3},且f(x)在区间[-1,1]上的最小值是4.(Ⅰ)求f(x)的解析式;(Ⅱ)设g(x)=x+5-f(x),若对任意的x∈(-

题目简介

已知函数f(x)是二次函数,不等式f(x)≥0的解集为{x|-2≤x≤3},且f(x)在区间[-1,1]上的最小值是4.(Ⅰ)求f(x)的解析式;(Ⅱ)设g(x)=x+5-f(x),若对任意的x∈(-

题目详情

已知函数f(x)是二次函数,不等式f(x)≥0的解集为{x|-2≤x≤3},且f(x)在区间[-1,1]上的最小值是4.
(Ⅰ)求f(x)的解析式;
(Ⅱ)设g(x)=x+5-f(x),若对任意的x∈(-∞,-
3
4
]
g(
x
m
)-g(x-1)≤4[m2g(x)+g(m)]
均成立,求实数m的取值范围.
题型:解答题难度:中档来源:不详

答案

(Ⅰ)由f(x)≥0解集为{x|-2≤x≤3},可设f(x)=a(x+2)(x-3)=a(x2-x-6),且a<0
对称轴x=class="stub"1
2
,开口向下,f(x)min=f(-1)=-4a=4,解得a=-1,f(x)=-x2+x+6;…(5分)
(Ⅱ)g(x)=x+5+x2-x-6=x2-1,g(class="stub"x
m
)-g(x-1)≤4[m2g(x)+g(m)]
恒成立
x2
m2
-1-(x-1)2+1≤4[m2(x2-1)+m2-1]
x∈(-∞,-class="stub"3
4
]
恒成立
化简(class="stub"1
m2
-4m2)x2x2-2x-3
,即class="stub"1
m2
-4m2
-class="stub"3
x2
-class="stub"2
x
+1
x∈(-∞,-class="stub"3
4
]
恒成立…(8分)
y=-class="stub"3
x2
-class="stub"2
x
+1
,记t=class="stub"1
x
∈[-class="stub"4
3
,0)
,则y=-3t2-2t+1,
二次函数开口向下,对称轴为t0=-class="stub"1
3
,当t=-class="stub"4
3
ymin=-class="stub"5
3

class="stub"1
m2
-4m2≤-class="stub"5
3
…(10分)
所以(3m2+1)(4m2-3)≥0,解得m≤-
3
2
m≥
3
2
…(12分)

更多内容推荐