Question

設(shè)a∈R,函數(shù)f(x)=x·|x-a|+2x.

(1) 若a=2,求函數(shù)f(x)在區(qū)間[0,3]上的最大值;

(2) 若a>2,寫出函數(shù)f(x)的單調(diào)區(qū)間(不必證明);

(3) 若存在a∈[-2,4],使得關(guān)于x的方程f(x)=t·f(a)有3個(gè)不相等的實(shí)數(shù)解,求實(shí)數(shù)t的取值范圍.

Answer 1

 (1) 當(dāng)a=2,x∈[0,3]時(shí),f(x)=x·|x-2|+2x=

作函數(shù)圖象(圖象略),可知函數(shù)f(x)在區(qū)間[0,3]上是單調(diào)增函數(shù),所以f(x)的最大值為f(3)=9.

(2) f(x)=

①當(dāng)x≥a時(shí),f(x)=-,

因?yàn)閍>2,所以<a,

所以函數(shù)f(x)在[a,+∞)上單調(diào)遞增.

②當(dāng)x<a時(shí),f(x)=-+,

因?yàn)閍>2,所以<a,所以函數(shù)f(x)在-∞,上單調(diào)遞增,在上單調(diào)遞減.

綜上,函數(shù)f(x)的單調(diào)增區(qū)間是和[a,+∞),單調(diào)減區(qū)間是.

(第11題)

(3) ①當(dāng)-2≤a≤2時(shí),≤0,≥0,所以函數(shù)f(x)在(-∞,+∞)上是增函數(shù),

關(guān)于x的方程f(x)=t·f(a)不可能有三個(gè)不相等的實(shí)數(shù)解.

②當(dāng)2<a≤4時(shí),由(2)知函數(shù)f(x)在區(qū)間和[a,+∞)上分別是增函數(shù),在區(qū)間上是減函數(shù),所以當(dāng)且僅當(dāng)2a<t·f(a)<時(shí),方程f(x)=t·f(a)有三個(gè)不相等的實(shí)數(shù)解,

即1<t<=.

令g(a)=a+,g (a)在a∈(2,4]時(shí)是增函數(shù),故g(a)max=5,

所以,實(shí)數(shù)t的取值范圍是.