(1)求导函数,可得f′(x)= ∵f(x)在[1,+∞)上递增, ∴在[1,+∞)上,f′(x)=≥0恒成立 ∴在[1,+∞)上,a≥ ∴a≥2 ∴a的取值范围为[2,+∞); (2)由f′(x)=,x∈[1,4] ①当a≥2时,在x∈[1,4]上,f"(x)≥0,∴fmin(x)=f(1)=a(8分) ②当0≤a≤1时,在x∈[1,4]上,f"(x)≤0,∴fmin(x)=f(4)=2a-2ln2(10分) ③当1<a<2时,在x∈[1,]上f"(x)≤0,在x∈[,4]上f"(x)≥0 此时fmin(x)=f()=2-2ln2+2lna 综上所述:fmin(x)= | 2a-2ln20≤a≤1 | 2-2ln2+2lna1<a≤2 | a2<a |
| | (13分) |