记满足下列条件的函数f(x)的集合为M:当绝对值x1≤1,绝对值x2≤1时,∣f(x1)-f(x2)∣≤4∣x1-x2∣,若有函数g(x)=x2(这是平方)+2x-1,则g(x)与M的关系是( )A.g(x)∈M B.g(x)不属于M A.
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记满足下列条件的函数f(x)的集合为M:|x1|≤1,|x2|≤1时,∣f(x1)-f(x2)∣≤4∣x1-x2∣,若有函数g(x)=x^+2x-1,则g(x)与M的关系是( )解::|x1|≤1,|x2|≤1时,|g(x1)-g(x2)|=|(x1^+2x1-1)-(x2^+2x2-1)|=|(x1^-x2^)+2(x1-x2)|=|(x1-x2)(x1+x2)+2(x1-x2)|=|(x1-x2)(x1+x2+2)|=|x1-x2|·|x1+x2+2|≤|x1-x2|·(|x1|+|x2|+|2|)≤|x1-x2|·(|x1|+|x2|+|2|)∵|x1|≤1,|x2|≤1∴|x1-x2|·(|x1|+|x2|+|2|)≤|x1-x2|·(1+1+2)=4∣x1-x2∣∴当|x1|≤1,|x2|≤1时,∣g(x1)-g(x2)∣≤4∣x1-x2∣,g(x)≤4∣x1-x2∣成立∴g(x)属于M 。