题目

设函数f(x)=2|x-1|+x-1，g(x)=16x2-8x+1，记f(x)≤1的解集为M,g(x)≤4的解集为N. (1)求M； (2)当x∈M∩N时，证明：. 答案： 解析 (1)f(x)= 当x≥1时,由f(x)=3x-3≤1得x≤,故1≤x≤; 当x<1时,由f(x)=1-x≤1得x≥0,故0≤x<1. 所以f(x)≤1的解集为M=. (2)证明:由g(x)=16x2-8x+1≤4得, 解得. 因此,故. 当x∈M∩N时, f(x)=1-x,于是x2f(x)+x·[f(x)]2 =xf(x)[x+f(x)]=x·f(x)=x(1-x)=.Your book is under the desk. Please________.A.pick it B.pick up it C.pick it up