解答:解:(1)∵f(x)=a
x+x
2-xlna,
∴f′(x)=a
xlna+2x-lna,
∴f′(0)=0,f(0)=1
即函數(shù)f(x)圖象在點(diǎn)(0,1)處的切線斜率為0,
∴圖象在點(diǎn)(0,f(0))處的切線方程為y=1;(3分)
(2)由于f'(x)=a
xlna+2x-lna=2x+(a
x-1)lna>0
①當(dāng)a>1,y=2x單調(diào)遞增,lna>0,所以y=(a
x-1)lna單調(diào)遞增,故y=2x+(a
x-1)lna單調(diào)遞增,
∴2x+(a
x-1)lna>2×0+(a
0-1)lna=0,即f'(x)>f'(0),所以x>0
故函數(shù)f(x)在(0,+∞)上單調(diào)遞增;
②當(dāng)0<a<1,y=2x單調(diào)遞增,lna<0,所以y=(a
x-1)lna單調(diào)遞增,故y=2x+(a
x-1)lna單調(diào)遞增,
∴2x+(a
x-1)lna>2×0+(a
0-1)lna=0,即f'(x)>f'(0),所以x>0
故函數(shù)f(x)在(0,+∞)上單調(diào)遞增;
綜上,函數(shù)f(x)單調(diào)增區(qū)間(0,+∞);(8分)
(3)因?yàn)榇嬖趚
1,x
2∈[-1,1],使得|f(x
1)-f(x
2)|≥e-1,
所以當(dāng)x∈[-1,1]時(shí),|(f(x))
max-(f(x))
min|
=(f(x))
max-(f(x))
min≥e-1,(12分)
由(2)知,f(x)在[-1,0]上遞減,在[0,1]上遞增,
所以當(dāng)x∈[-1,1]時(shí),(f(x))
min=f(0)=1,
(f(x))
max=max{f(-1),f(1)},
而f(1)-f(-1)=(a+1-lna)-(
+1+lna)=a-
-2lna,
記g(t)=t-
-2lnt(t>0),
因?yàn)間′(t)=1+
-
=(
-1)
2≥0(當(dāng)t=1時(shí)取等號),
所以g(t)=t-
-2lnt在t∈(0,+∞)上單調(diào)遞增,而g(1)=0,
所以當(dāng)t>1時(shí),g(t)>0;當(dāng)0<t<1時(shí),g(t)<0,
也就是當(dāng)a>1時(shí),f(1)>f(-1);
當(dāng)0<a<1時(shí),f(1)<f(-1)(14分)
①當(dāng)a>1時(shí),由f(1)-f(0)≥e-1⇒a-lna≥e-1⇒a≥e,
②當(dāng)0<a<1時(shí),由f(-1)-f(0)≥e-1⇒
+lna≥e-1⇒0<a≤
,
綜上知,所求a的取值范圍為a∈(0,
]∪[e,+∞).(16分)