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解:(1)由于:g(x)=λx+sinx则:g'(x)=λ+cosx又g(x)=λx+sinx在区间[-1,1]上为减函数,则:g'(x)<=0 则:λ+cosx<=0λ<=-cosx则:λ<=(-cosx)min由于:X=0时,-cosx最小=-1则:λ<=-1A={λ|λ<=-1}(2)由于:g(x)<=t^2+λt+1在x∈[-1,1]上恒成立,且λ∈A则:[g(x)]max<=t^2+λt+1在[-1,1]上恒成立又:在[-1,1]上,[g(x)]max=g(-1)则:t^2+λt+1≥-λ-sin1恒成立λ(t+1)≥-t^2-1-sin1①若... |
