1
首先使f(x)在[0,2]內(nèi)有定義
即0≤x≤2時,u(x)=x^2+tx+1>0恒成立
x=0時,u=1
0-x-1/x
∵x+1/x≥2,(x=1時取等號）
∴-(x+1/x)≤-2
∴t>-2
u(x) =(x+t/2)²+1-t²/4
當(dāng)-20
{g(2)=2t+3>0
{0為空集嗎？為空集嗎？｛Δ=（t-1)²-4>0 ==>t>3{g(0)=1>0==>t∈R{g(2)=2t+3>0 ==>t>-3/2 {0<-(t-1)/2<2 ==>-30 ==>t>3?（t-1)²-4>0 ==>(t-1)²>4==>t-1>2[在t>-2的前提下】==>t>3